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Welcome to Chemistry

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Welcome to Chemistry

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12-5: Reaction Mechanism

12-5-1: Learning Objectives

Learning Objectives

One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.

In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:

2C8H18(l)+25O2( g)⟶16CO2( g)+18H2O(g)2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l})+25 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{~g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})

For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction, involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.

The overall sequence of elementary reactions is the mechanism of the reaction.

12-5-2: Molecularity and the Rate-Determining Step

To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.

2CO+2NO2→2CO2+N2O4\mathrm{2CO + 2NO_2 \rightarrow 2CO_2 + N_2O_4}

From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO2\mathrm{NO}_{2} with a molecule of CO\mathrm{CO} that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:

 rate =k[NO2]2\text { rate }=k\left[\mathrm{NO}_{2}\right]^{2}

The fact that the reaction is second order in [NO2]\left[\mathrm{NO}_{2}\right] and independent of [CO][\mathrm{CO}] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be

 rate =k[NO2][CO].\text { rate }=k\left[\mathrm{NO}_{2}\right][\mathrm{CO}] .

The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:

15-2-1: Two-Step Mechanism

StepsReactionReaction Type
step 1NO2+NO2⟶ slow NO3+NO\mathrm{NO}_{2}+\mathrm{NO}_{2} \stackrel{\text { slow }}{\longrightarrow} \mathrm{NO}_{3}+\mathrm{NO}elementary reaction
step 2NO3+CO→NO2+CO2‟\underline{\mathrm{NO}_{3}+\mathrm{CO} \rightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}}elementary reaction
sumNO2+CO→NO+CO2\mathrm{NO}_{2}+\mathrm{CO} \rightarrow \mathrm{NO}+\mathrm{CO}_{2}overall reaction

According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The NO3\mathrm{NO_3} molecule is intermediate in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.

The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.

12-5-3: Using Molecularity to Describe a Rate Law

The molecularity of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)

About the image

The Basis for Writing Rate Laws of Elementary Reactions. This diagram illustrates how the number of possible collisions per unit time between two reactant species, A and B, depends on the number of A and B particles present. The number of collisions between A and B particles increases as the product of the number of particles, not as the sum. This is why the rate law for an elementary reaction depends on the product of the concentrations of the species that collide in that step. (CC BY-NC-SA; anonymous)

Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (Table 14.6.1). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction ( A→\mathrm{A} \rightarrow products) is

 rate =k[A].\text { rate }=k[A] .

For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure 14.6.1 For a bimolecular elementary reaction of the form A+B→\mathrm{A}+\mathrm{B} \rightarrow products, the general rate law is

 rate =k[A][B].\text { rate }=k[A][B] .

Elementary ReactionMolecularityRateReaction Order
A→ products \mathrm{A} \rightarrow \text { products }Unimolecular rate =k[ A]\text { rate }=k[\mathrm{~A}]first
2 A→ products 2\mathrm{~A} \rightarrow \text { products }Bimolecular rate =k[ A]2\text { rate }=k[\mathrm{~A}]^2second
A+B→ products \mathrm{A}+\mathrm{B} \rightarrow \text { products }Bimolecular rate =k[ A][B]\text { rate }=k[\mathrm{~A}][\mathrm{B}]second
2 A+B→ products 2 \mathrm{~A}+\mathrm{B} \rightarrow \text { products }Termolecular rate =k[ A]2[ B]\text { rate }=k[\mathrm{~A}]^2[\mathrm{~B}]third
A+B+C→ products \mathrm{A}+\mathrm{B}+\mathrm{C} \rightarrow \text { products }Termolecular rate =k[ A][B][C]\text { rate }=k[\mathrm{~A}][\mathrm{B}][\mathrm{C}]third

For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law cannot be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step).

12-5-4: Identifying the Rate-Determining Step

Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the ratedetermining step, that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.

Look at the rate laws for each elementary reaction in the example as well as for the overall reaction.

Rate laws for each elementary reaction in our example as well as for the overall reaction

StepsReactionRate
step 1NO2+NO2→k1NO3+NO\mathrm{NO}_2+\mathrm{NO}_2 \stackrel{\mathrm{k}_1}{\rightarrow} \mathrm{NO}_3+\mathrm{NO} rate =k1[NO2]2( predicted) \text { rate }=k_1\left[\mathrm{NO}_2\right]^2(\text { predicted) }
step 2NO3+CO→k2NO2+CO2‟\underline{\mathrm{NO}_3+\mathrm{CO} \stackrel{k_2}{\rightarrow} \mathrm{NO}_2+\mathrm{CO}_2} rate =k2[NO3][CO]( predicted )\text { rate }=k_2\left[\mathrm{NO}_3\right][\mathrm{CO}](\text { predicted })
step 3NO2+CO→kNO+CO2\mathrm{NO}_2+\mathrm{CO} \stackrel{k}{\rightarrow} \mathrm{NO}+\mathrm{CO}_2 rate =k[NO2]2( observed) \text { rate }=k\left[\mathrm{NO}_2\right]^2(\text { observed) }

The experimentally determined rate law for the reaction of NO2\mathrm{NO}_{2} with COC O is the same as the predicted rate law for step 1 . This tells us that the first elementary reaction is the rate-determining step, so kk for the overall reaction must equal k1k_{1}. That is, NO3\mathrm{NO}_{3} is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.

Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.

12-5-4-1: Example-A Reaction with an Intermediate

In an alternative mechanism for the reaction of NO2\mathrm{NO}_{2} with CO\mathrm{CO} with N2O4\mathrm{N}_{2} \mathrm{O}_{4} appearing as an intermediate.

alternative mechanism for the reaction of NO2\mathrm{NO}_{2} with CO\mathrm{CO} with N2O4\mathrm{N}_{2} \mathrm{O}_{4} appearing as an intermediate.

Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate =k[NO2]2t=\mathrm{k[{NO}_{2}]^{2}t})

Given: elementary reactions Asked for: rate law for each elementary reaction and overall rate law

Strategy

  • Determine the rate law for each elementary reaction in the reaction.

  • Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.

Solution

View solution

A The rate law for step 1 is rate =k1[NO2]2=k_{1}\left[\mathrm{NO}_{2}\right]^{2}; for step 2 , it is rate =k2[ N2O4][CO]=k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right][\mathrm{CO}].

B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate == k1[NO2]2k_{1}\left[\mathrm{NO}_{2}\right]^{2}. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4\mathrm{N}_{2} \mathrm{O}_{4} as an intermediate, and the one described previously, with NO3\mathrm{NO}_{3} as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3\mathrm{NO}_{3} and N2O4\mathrm{N}_{2} \mathrm{O}_{4}, directly.

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12-5: Reaction Mechanism

12-5-1: Learning Objectives

Learning Objectives

One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.

In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:

2C8H18(l)+25O2( g)⟶16CO2( g)+18H2O(g)2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l})+25 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{~g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})

For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction, involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.

The overall sequence of elementary reactions is the mechanism of the reaction.

12-5-2: Molecularity and the Rate-Determining Step

To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.

2CO+2NO2→2CO2+N2O4\mathrm{2CO + 2NO_2 \rightarrow 2CO_2 + N_2O_4}

From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO2\mathrm{NO}_{2} with a molecule of CO\mathrm{CO} that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:

 rate =k[NO2]2\text { rate }=k\left[\mathrm{NO}_{2}\right]^{2}

The fact that the reaction is second order in [NO2]\left[\mathrm{NO}_{2}\right] and independent of [CO][\mathrm{CO}] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be

 rate =k[NO2][CO].\text { rate }=k\left[\mathrm{NO}_{2}\right][\mathrm{CO}] .

The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:

15-2-1: Two-Step Mechanism

StepsReactionReaction Type
step 1NO2+NO2⟶ slow NO3+NO\mathrm{NO}_{2}+\mathrm{NO}_{2} \stackrel{\text { slow }}{\longrightarrow} \mathrm{NO}_{3}+\mathrm{NO}elementary reaction
step 2NO3+CO→NO2+CO2‟\underline{\mathrm{NO}_{3}+\mathrm{CO} \rightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}}elementary reaction
sumNO2+CO→NO+CO2\mathrm{NO}_{2}+\mathrm{CO} \rightarrow \mathrm{NO}+\mathrm{CO}_{2}overall reaction

According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The NO3\mathrm{NO_3} molecule is intermediate in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.

The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.

12-5-3: Using Molecularity to Describe a Rate Law

The molecularity of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)

About the image

The Basis for Writing Rate Laws of Elementary Reactions. This diagram illustrates how the number of possible collisions per unit time between two reactant species, A and B, depends on the number of A and B particles present. The number of collisions between A and B particles increases as the product of the number of particles, not as the sum. This is why the rate law for an elementary reaction depends on the product of the concentrations of the species that collide in that step. (CC BY-NC-SA; anonymous)

Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (Table 14.6.1). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction ( A→\mathrm{A} \rightarrow products) is

 rate =k[A].\text { rate }=k[A] .

For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure 14.6.1 For a bimolecular elementary reaction of the form A+B→\mathrm{A}+\mathrm{B} \rightarrow products, the general rate law is

 rate =k[A][B].\text { rate }=k[A][B] .

Elementary ReactionMolecularityRateReaction Order
A→ products \mathrm{A} \rightarrow \text { products }Unimolecular rate =k[ A]\text { rate }=k[\mathrm{~A}]first
2 A→ products 2\mathrm{~A} \rightarrow \text { products }Bimolecular rate =k[ A]2\text { rate }=k[\mathrm{~A}]^2second
A+B→ products \mathrm{A}+\mathrm{B} \rightarrow \text { products }Bimolecular rate =k[ A][B]\text { rate }=k[\mathrm{~A}][\mathrm{B}]second
2 A+B→ products 2 \mathrm{~A}+\mathrm{B} \rightarrow \text { products }Termolecular rate =k[ A]2[ B]\text { rate }=k[\mathrm{~A}]^2[\mathrm{~B}]third
A+B+C→ products \mathrm{A}+\mathrm{B}+\mathrm{C} \rightarrow \text { products }Termolecular rate =k[ A][B][C]\text { rate }=k[\mathrm{~A}][\mathrm{B}][\mathrm{C}]third

For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law cannot be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step).

12-5-4: Identifying the Rate-Determining Step

Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the ratedetermining step, that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.

Look at the rate laws for each elementary reaction in the example as well as for the overall reaction.

Rate laws for each elementary reaction in our example as well as for the overall reaction

StepsReactionRate
step 1NO2+NO2→k1NO3+NO\mathrm{NO}_2+\mathrm{NO}_2 \stackrel{\mathrm{k}_1}{\rightarrow} \mathrm{NO}_3+\mathrm{NO} rate =k1[NO2]2( predicted) \text { rate }=k_1\left[\mathrm{NO}_2\right]^2(\text { predicted) }
step 2NO3+CO→k2NO2+CO2‟\underline{\mathrm{NO}_3+\mathrm{CO} \stackrel{k_2}{\rightarrow} \mathrm{NO}_2+\mathrm{CO}_2} rate =k2[NO3][CO]( predicted )\text { rate }=k_2\left[\mathrm{NO}_3\right][\mathrm{CO}](\text { predicted })
step 3NO2+CO→kNO+CO2\mathrm{NO}_2+\mathrm{CO} \stackrel{k}{\rightarrow} \mathrm{NO}+\mathrm{CO}_2 rate =k[NO2]2( observed) \text { rate }=k\left[\mathrm{NO}_2\right]^2(\text { observed) }

The experimentally determined rate law for the reaction of NO2\mathrm{NO}_{2} with COC O is the same as the predicted rate law for step 1 . This tells us that the first elementary reaction is the rate-determining step, so kk for the overall reaction must equal k1k_{1}. That is, NO3\mathrm{NO}_{3} is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.

Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.

12-5-4-1: Example-A Reaction with an Intermediate

In an alternative mechanism for the reaction of NO2\mathrm{NO}_{2} with CO\mathrm{CO} with N2O4\mathrm{N}_{2} \mathrm{O}_{4} appearing as an intermediate.

alternative mechanism for the reaction of NO2\mathrm{NO}_{2} with CO\mathrm{CO} with N2O4\mathrm{N}_{2} \mathrm{O}_{4} appearing as an intermediate.

Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate =k[NO2]2t=\mathrm{k[{NO}_{2}]^{2}t})

Given: elementary reactions Asked for: rate law for each elementary reaction and overall rate law

Strategy

  • Determine the rate law for each elementary reaction in the reaction.

  • Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.

Solution

View solution

A The rate law for step 1 is rate =k1[NO2]2=k_{1}\left[\mathrm{NO}_{2}\right]^{2}; for step 2 , it is rate =k2[ N2O4][CO]=k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right][\mathrm{CO}].

B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate == k1[NO2]2k_{1}\left[\mathrm{NO}_{2}\right]^{2}. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4\mathrm{N}_{2} \mathrm{O}_{4} as an intermediate, and the one described previously, with NO3\mathrm{NO}_{3} as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3\mathrm{NO}_{3} and N2O4\mathrm{N}_{2} \mathrm{O}_{4}, directly.

- - + + \ No newline at end of file diff --git a/academic/chemistry/problems/02-20.html b/academic/chemistry/problems/02-20.html index 420bddc1..fa4c9312 100644 --- a/academic/chemistry/problems/02-20.html +++ b/academic/chemistry/problems/02-20.html @@ -6,9 +6,9 @@ Presentation problem: 02-20 | Toshiki's Note - - - + + + @@ -30,10 +30,10 @@ -
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Presentation problem: 02-20

Question

At 500 K500 \mathrm{~K} in the presence of a copper surface, ethanol decomposes according to the equation

C2H5OH(g)⟶CH3CHO(g)+H2(g)\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{g}) \longrightarrow \mathrm{CH}_3 \mathrm{CHO}(g)+\mathrm{H}_2(g)

The pressure of C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} was measured as a function of time and the following data were obtained:

Time (s)PC2H5OH (torr) P_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \text { (torr) }
0250.
100.237
200.224
300.211
400.198
500.185

Since the pressure of a gas is directly proportional to the concentration of gas, we can express the rate law for a gaseous reaction in terms of partial pressures. Using the above data, deduce the rate law, the integrated rate law, and the value of the rate constant, all in terms of pressure units in atm and time in seconds. Predict the pressure of C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} after 900.s900 . \mathrm{s} from the start of the reaction. (Hint: To determine the order of the reaction with respect to C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, compare how the pressure of C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} decreases with each time listing.)

Solution

solution graph

Due to the fact that the graph of p[O2]\mathrm{p}\left[\mathrm{O}_2\right] over time is showing R2\mathrm{R}^2 value of 1 we know we have a zero-order reaction. Therefore:

k=p0[C2H5OH]−p[C2H5OH]t=250 torr −185 torr 500 s=0.13 torr s −1k=0.13 torr s s−1×1 atm760 torr =1.71×10−4 atm s−1 rate =kp(C2H5OH)=−kt+p0(C2H5OH)p(C2H5OH)=−(1.71×10−4 atm s−1)×900 s+0.33 atmp(C2H5OH)=0.176 atm s−1\begin{aligned} & \mathrm{k}=\frac{\mathrm{p}_0\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]-\mathrm{p}\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}{\mathrm{t}}=\frac{250 \text { torr }-185 \text { torr }}{500 \mathrm{~s}}=0.13 \text { torr s }^{-1} \\ & \mathrm{k}=0.13 \text { torr s } \mathrm{s}^{-1} \times \frac{1 \mathrm{~atm}}{760 \text { torr }}=1.71 \times 10^{-4} \mathrm{~atm} \mathrm{~s}^{-1} \\ & \text { rate }=\mathbf{k} \\ & \mathrm{p}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)=-\mathrm{kt}+\mathrm{p}_0\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right) \\ & \mathrm{p}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)=-\left(1.71 \times 10^{-4} \mathrm{~atm} \mathrm{~s}^{-1}\right) \times 900 \mathrm{~s}+0.33 \mathrm{~atm} \\ & \mathrm{p}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)=0.176 \mathrm{~atm} \mathrm{~s}^{-1} \\ & \end{aligned}

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Presentation problem: 02-20

Question

At 500 K500 \mathrm{~K} in the presence of a copper surface, ethanol decomposes according to the equation

C2H5OH(g)⟶CH3CHO(g)+H2(g)\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{g}) \longrightarrow \mathrm{CH}_3 \mathrm{CHO}(g)+\mathrm{H}_2(g)

The pressure of C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} was measured as a function of time and the following data were obtained:

Time (s)PC2H5OH (torr) P_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \text { (torr) }
0250.
100.237
200.224
300.211
400.198
500.185

Since the pressure of a gas is directly proportional to the concentration of gas, we can express the rate law for a gaseous reaction in terms of partial pressures. Using the above data, deduce the rate law, the integrated rate law, and the value of the rate constant, all in terms of pressure units in atm and time in seconds. Predict the pressure of C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} after 900.s900 . \mathrm{s} from the start of the reaction. (Hint: To determine the order of the reaction with respect to C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, compare how the pressure of C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} decreases with each time listing.)

Solution

solution graph

Due to the fact that the graph of p[O2]\mathrm{p}\left[\mathrm{O}_2\right] over time is showing R2\mathrm{R}^2 value of 1 we know we have a zero-order reaction. Therefore:

k=p0[C2H5OH]−p[C2H5OH]t=250 torr −185 torr 500 s=0.13 torr s −1k=0.13 torr s s−1×1 atm760 torr =1.71×10−4 atm s−1 rate =kp(C2H5OH)=−kt+p0(C2H5OH)p(C2H5OH)=−(1.71×10−4 atm s−1)×900 s+0.33 atmp(C2H5OH)=0.176 atm s−1\begin{aligned} & \mathrm{k}=\frac{\mathrm{p}_0\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]-\mathrm{p}\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}{\mathrm{t}}=\frac{250 \text { torr }-185 \text { torr }}{500 \mathrm{~s}}=0.13 \text { torr s }^{-1} \\ & \mathrm{k}=0.13 \text { torr s } \mathrm{s}^{-1} \times \frac{1 \mathrm{~atm}}{760 \text { torr }}=1.71 \times 10^{-4} \mathrm{~atm} \mathrm{~s}^{-1} \\ & \text { rate }=\mathbf{k} \\ & \mathrm{p}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)=-\mathrm{kt}+\mathrm{p}_0\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right) \\ & \mathrm{p}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)=-\left(1.71 \times 10^{-4} \mathrm{~atm} \mathrm{~s}^{-1}\right) \times 900 \mathrm{~s}+0.33 \mathrm{~atm} \\ & \mathrm{p}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)=0.176 \mathrm{~atm} \mathrm{~s}^{-1} \\ & \end{aligned}

- - + + \ No newline at end of file diff --git a/academic/chemistry/problems/03-02-1.html b/academic/chemistry/problems/03-02-1.html index 43bf8d6d..2d8325fa 100644 --- a/academic/chemistry/problems/03-02-1.html +++ b/academic/chemistry/problems/03-02-1.html @@ -6,9 +6,9 @@ Problem: 03-02-1 | Toshiki's Note - - - + + + @@ -30,10 +30,10 @@ -
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Problem: 03-02-1

Question

2 N2O5(g)→4NO2(g)+O2(g)2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)

The decomposition of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(g) is represented by the equation above. A sample of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(g) is monitored as it decomposes, and the concentration of N2O5\mathrm{N}_2 \mathrm{O}_5 as a function of time is recorded. The results are shown in the table below.

Time (s)[N2O5]\mathrm{[N_2O_5]}
01.000
25.00.801
50.00.642
75.00.515

Calculate the average rate of the reaction between 50.050.0 and 75.075.0 seconds.

Solution

a A+b B→cC+dDa \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D}

the rate of reaction is defined as

 rate =−1aΔ[A]Δt=−1bΔ[B]Δt=+1cΔ[C]Δt=+1dΔ[D]Δt\color{cyan}\text { rate }=-\frac{1}{a} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{b} \frac{\Delta[\mathrm{B}]}{\Delta t}=+\frac{1}{c} \frac{\Delta[\mathrm{C}]}{\Delta t}=+\frac{1}{d} \frac{\Delta[\mathrm{D}]}{\Delta t}

Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( aa, b,cb, c, or dd ). This ensures that the calculated reaction rate is the same no matter which reactant or product is monitored for changes in concentration. In this case, the monitored species was N2O5\mathrm{N}_2 \mathrm{O}_5. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of N2O5\mathrm{N}_2 \mathrm{O}_5 :

 rate =−12Δ[N2O5]Δt\text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t}

Since the coefficient for N2O5\mathrm{N}_2 \mathrm{O}_5 in the balanced equation is 2 , we divided the rate of change in concentration of N2O5\mathrm{N}_2 \mathrm{O}_5 by 2 . Additionally, since N2O5\mathrm{N}_2 \mathrm{O}_5 is being consumed in the reaction, we included a negative sign in front of the expression.

Now, let's plug in the information from the table to calculate the average reaction rate between 50.050.0 and 75.075.0 seconds:

 rate =−12(0.515M−0.642M)(75.0 s−50.0 s)=2.54×10−3M s−1\begin{aligned} \text { rate } & =-\frac{1}{2} \frac{(0.515 M-0.642 M)}{(75.0 \mathrm{~s}-50.0 \mathrm{~s})} \\ & =2.54 \times 10^{-3} M \mathrm{~s}^{-1} \end{aligned}

So, the average rate of the reaction between 50.050.0 and 75.075.0 seconds is 2.54×10−3Ms−1\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}.

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Problem: 03-02-1

Question

2 N2O5(g)→4NO2(g)+O2(g)2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)

The decomposition of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(g) is represented by the equation above. A sample of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(g) is monitored as it decomposes, and the concentration of N2O5\mathrm{N}_2 \mathrm{O}_5 as a function of time is recorded. The results are shown in the table below.

Time (s)[N2O5]\mathrm{[N_2O_5]}
01.000
25.00.801
50.00.642
75.00.515

Calculate the average rate of the reaction between 50.050.0 and 75.075.0 seconds.

Solution

a A+b B→cC+dDa \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D}

the rate of reaction is defined as

 rate =−1aΔ[A]Δt=−1bΔ[B]Δt=+1cΔ[C]Δt=+1dΔ[D]Δt\color{cyan}\text { rate }=-\frac{1}{a} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{b} \frac{\Delta[\mathrm{B}]}{\Delta t}=+\frac{1}{c} \frac{\Delta[\mathrm{C}]}{\Delta t}=+\frac{1}{d} \frac{\Delta[\mathrm{D}]}{\Delta t}

Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( aa, b,cb, c, or dd ). This ensures that the calculated reaction rate is the same no matter which reactant or product is monitored for changes in concentration. In this case, the monitored species was N2O5\mathrm{N}_2 \mathrm{O}_5. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of N2O5\mathrm{N}_2 \mathrm{O}_5 :

 rate =−12Δ[N2O5]Δt\text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t}

Since the coefficient for N2O5\mathrm{N}_2 \mathrm{O}_5 in the balanced equation is 2 , we divided the rate of change in concentration of N2O5\mathrm{N}_2 \mathrm{O}_5 by 2 . Additionally, since N2O5\mathrm{N}_2 \mathrm{O}_5 is being consumed in the reaction, we included a negative sign in front of the expression.

Now, let's plug in the information from the table to calculate the average reaction rate between 50.050.0 and 75.075.0 seconds:

 rate =−12(0.515M−0.642M)(75.0 s−50.0 s)=2.54×10−3M s−1\begin{aligned} \text { rate } & =-\frac{1}{2} \frac{(0.515 M-0.642 M)}{(75.0 \mathrm{~s}-50.0 \mathrm{~s})} \\ & =2.54 \times 10^{-3} M \mathrm{~s}^{-1} \end{aligned}

So, the average rate of the reaction between 50.050.0 and 75.075.0 seconds is 2.54×10−3Ms−1\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}.

- - + + \ No newline at end of file diff --git a/academic/chemistry/problems/03-02-2.html b/academic/chemistry/problems/03-02-2.html index c6271c4b..ca55e179 100644 --- a/academic/chemistry/problems/03-02-2.html +++ b/academic/chemistry/problems/03-02-2.html @@ -6,9 +6,9 @@ Problem: 03-02-2 | Toshiki's Note - - - + + + @@ -30,10 +30,10 @@ -
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Problem: 03-02-2

Question

The rate law for a particular reaction is rate =k[XY]2=k[\mathrm{XY}]^2. In an experiment, the initial rate of the reaction is determined to be 0.16 mol/(L⋅s)0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s}) when the initial concentration of XY\mathrm{XY} is 0.40 mol/L0.40 \mathrm{~mol} / \mathrm{L}.

  • What is the value of the rate constant, kk, for the reaction?
    • [ ] 0.10 L/(mol⋅s)0.10 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
    • [ ] 0.40 L/(mol⋅s)0.40 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
    • [x] 1.0 L/(mol⋅s)1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
    • [ ] 2.5 L/(mol⋅s)2.5 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})

Solution

lo tind the value of the rate constant for the reaction, let's first solve the rate law for kk :

k= rate [XY]2k=\frac{\text { rate }}{[\mathrm{XY}]^2}

Next, let's plug in the initial rate and concentration given in the text:

k=0.16 mol/(L⋅s)(0.40 mol/L)2=0.16 mol/(L⋅s)0.16 mol2/L2=1.0 L/(mol⋅s)\begin{aligned} k & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{(0.40 \mathrm{~mol} / \mathrm{L})^2} \\ & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{0.16 \mathrm{~mol}^2 / \mathrm{L}^2} \\ & =1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s}) \end{aligned}

So, the value of the rate constant for the reaction is 1.0 L/(mol⋅s)1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})

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Problem: 03-02-2

Question

The rate law for a particular reaction is rate =k[XY]2=k[\mathrm{XY}]^2. In an experiment, the initial rate of the reaction is determined to be 0.16 mol/(L⋅s)0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s}) when the initial concentration of XY\mathrm{XY} is 0.40 mol/L0.40 \mathrm{~mol} / \mathrm{L}.

  • What is the value of the rate constant, kk, for the reaction?
    • [ ] 0.10 L/(mol⋅s)0.10 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
    • [ ] 0.40 L/(mol⋅s)0.40 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
    • [x] 1.0 L/(mol⋅s)1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
    • [ ] 2.5 L/(mol⋅s)2.5 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})

Solution

lo tind the value of the rate constant for the reaction, let's first solve the rate law for kk :

k= rate [XY]2k=\frac{\text { rate }}{[\mathrm{XY}]^2}

Next, let's plug in the initial rate and concentration given in the text:

k=0.16 mol/(L⋅s)(0.40 mol/L)2=0.16 mol/(L⋅s)0.16 mol2/L2=1.0 L/(mol⋅s)\begin{aligned} k & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{(0.40 \mathrm{~mol} / \mathrm{L})^2} \\ & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{0.16 \mathrm{~mol}^2 / \mathrm{L}^2} \\ & =1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s}) \end{aligned}

So, the value of the rate constant for the reaction is 1.0 L/(mol⋅s)1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})

- - + + \ No newline at end of file diff --git a/academic/chemistry/problems/03-02-3.html b/academic/chemistry/problems/03-02-3.html new file mode 100644 index 00000000..0e6c79d9 --- /dev/null +++ b/academic/chemistry/problems/03-02-3.html @@ -0,0 +1,39 @@ + + + + + + Problem 03-02-3 | Toshiki's Note + + + + + + + + + + + + + + + + + + + + + + + + + + +
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Problem 03-02-3

Question

2X(g)+2Y(g)→Q(g)+2R(g)2 \mathrm{X}(g)+2 \mathrm{Y}(g) \rightarrow \mathrm{Q}(g)+2 \mathrm{R}(g)

The reaction represented above is found to be second order with respect to X\mathrm{X} and first order with respect to Y\mathrm{Y}.

What happens to the rate of the reaction when [X][\mathrm{X}] is halved and [Y][\mathrm{Y}] is doubled?

  • [ ] It increases by a factor of 4.
  • [x] It decreases by a factor of 2.
  • [ ] It decreases by a factor of 4.
  • [ ] It does not change.

Solution

To solve this problem, let's first write out the rate law for the reaction. According to the text, the reaction is second order with respect to X\mathrm{X} and first order with respect to Y\mathrm{Y}, so the rate law is rate =k[X]2[Y]=k[\mathrm{X}]^2[\mathrm{Y}]

Now, let's think about how the rate changes when [X][\mathrm{X}] is halved and [Y][\mathrm{Y}] is doubled. Since [X][\mathrm{X}] is raised to the second power in the rate law, halving [X][\mathrm{X}] decreases the reaction rate by a factor of 4 . Similarly, since [Y][\mathrm{Y}] is raised to the first power, doubling [Y][\mathrm{Y}] increases the reaction rate by a factor of 2 . Combined, these changes result in the reaction rate decreasing by a factor of 2 overall. So, when [X][\mathrm{X}] is halved and [Y][\mathrm{Y}] is doubled, the rate of the reaction decreases by a factor of 2 .

+ + + + + \ No newline at end of file diff --git a/academic/literature/index.html b/academic/literature/index.html index 3170c593..0ddcadc6 100644 --- a/academic/literature/index.html +++ b/academic/literature/index.html @@ -6,9 +6,9 @@ Welcome to Literature | Toshiki's Note - - - + + + @@ -30,10 +30,10 @@ -
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Welcome to Literature

Wrote with and ☕ by Anda Toshiki at root@andatoshiki:/~

Copyright © 2023-2023 Anda Toshiki +

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Welcome to Literature

- - + + \ No newline at end of file diff --git a/academic/literature/writing/methods-of-development.html b/academic/literature/writing/methods-of-development.html index 1b39730e..48377864 100644 --- a/academic/literature/writing/methods-of-development.html +++ b/academic/literature/writing/methods-of-development.html @@ -6,9 +6,9 @@ Patterns of Organization and Methods of Development | Toshiki's Note - - - + + + @@ -30,10 +30,10 @@ -
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Patterns of Organization and Methods of Development

Patterns of organization can help your readers follow the ideas within your essay and your paragraphs, but they can also work as methods of development to help you recognize and further develop ideas and relationships in your writing. Here are some strategies that can help you with both organization and development in your essays.

Major Patterns of Organization

A fruit pie.

Read the following sentences:

  • Now take the pie out of the oven and let it cool on the stovetop.
  • Mix the dry ingredients with the liquid ingredients.
  • Set the pie crust aside while you make the filling.

How did it feel to read the above list? A bit confusing, I would guess. That’s because the steps for making a pie were not well organized, and the steps don’t include enough detail for us to know exactly what we should do. (Like what are the dry and liquid ingredients?) We all know that starting instructions from the beginning and giving each detailed step in the order it should happen is vital to having a good outcome, in this case a yummy pie! But it’s not always so simple to know how to organize or develop ideas, and sometimes there’s more than one way, which complicates things even further.

First, let’s take a look at a couple of ways to think about organization.

General to Specific or Specific to General

It might be useful to think about organizing your topic like a triangle:

Two triangles. The first is an inverted pyramid for General to Specific, the second is a pyramid for Specific to General.

The first triangle represents starting with the most general, big picture information first, moving then to more detailed and often more personal information later in the paper. The second triangle represents an organizational structure that starts with the specific, small scale information first and then moves to the more global, big picture stuff.

For example, if your topic is traffic in Vancouver, British Columbia, an essay that uses the general-to-specific organizational structure might begin this way:

Many people consider Vancouver, British Columbia, to be a relaxed place to live. They would be shocked to know how bad the traffic is traveling major arteries into the city and even driving around the city itself.

An essay that uses the specific-to-general structure might start like this:

Transit is crowded, parking is expensive, and vehicles stop and go through the main streets of the city of Vancouver, British Columbia, and that is just once travelers brave the crowded arteries to enter the city; Vancouver’s traffic problem does not lend itself to the relaxed atmosphere many believe the city to have.

What’s the difference between these two introductions? And how might they appeal to the intended audience for this essay in different ways? The first introduction is looking at the big picture of the problem and mentions pollution’s impact on all citizens in Portland, while the second introduction focuses on one specific family. The first helps readers see how vast the problem really is, and the second helps connect readers to a real family, making an emotional appeal from the very beginning. Neither introduction is necessarily better. You’ll choose one over the other based on the kind of tone you’d like to create and how you’d like to affect your audience. It’s completely up to you to make this decision.

Does the Triangle Mean the Essay Keeps Getting More Specific or More Broad until the Very End?

The triangle is kind of a general guide, meaning you’re allowed to move around within it all you want. For example, it’s possible that each of your paragraphs will be its own triangle, starting with the general or specific and moving out or in. However, if you begin very broadly, it might be effective to end your essay in a more specific, personal way. And if you begin with a personal story, consider ending your essay by touching on the global impact and importance of your topic.

Are There Other Ways to Think about Organizing My Ideas?

Yes! Rather than thinking about which of your ideas are most specific or personal or which are more broad or universal, you might consider one of the following ways of organizing your ideas:

  • Most important information first (consider what you want readers to focus on first)
  • Chronological order (the order in time that events take place)
  • Compare and contrast (ideas are organized together because of their relationship to each other)

The section on Methods of Development, below, offers more detail about some of these organizational patterns, along with some others.

Choose one of the following topics, and practice writing a few opening sentences like we did above, once using the general-to-specific format and once using the specific-to-general. Which do you like better? What audience would be attracted to which one? Share with peers to see how others tackled this challenge. How would you rewrite their sentences? Why? Discuss your changes and listen to how your peers have revised your sentences. Taking in other people’s ideas will help you see new ways to approach your own writing and thinking.
Topics:

  • Facing fears
  • Safety in sports
  • Community policing
  • Educating prisoners
  • Sex education

Methods of Development

The methods of development covered here are best used as ways to look at what’s already happening in your draft and to consider how you might emphasize or expand on any existing patterns. You might already be familiar with some of these patterns because teachers will sometimes assign them as the purpose for writing an essay. For example, you might have been asked to write a cause-and-effect essay or a comparison-and-contrast essay.

It’s important to emphasize here that patterns of organization or methods of developing content usually happen naturally as a consequence of the way the writer engages with and organizes information while writing. That is to say, most writers don’t sit down and say, “I think I’ll write a cause-and-effect essay today.” Instead, a writer might be more likely to be interested in a topic, say, the state of drinking water in the local community, and as the writer begins to explore the topic, certain cause-and-effect relationships between environmental pollutants and the community water supply may begin to emerge.

So if these patterns just occur naturally in writing, what’s the use in knowing about them? Well, sometimes you might be revising a draft and notice that some of your paragraphs are a bit underdeveloped. Maybe they lack a clear topic, or maybe they lack support. In either case, you can look to these common methods of development to find ways to sharpen those vague topics or to add support where needed. Do you have a clear cause statement somewhere but you haven’t explored the effects? Are you lacking detail somewhere where a narrative story or historical chronology can help build reader interest and add support? Are you struggling to define an idea that might benefit from some comparison or contrast? Read on to consider some of the ways that these strategies can help you in revision.

Cause and Effect (or Effect and Cause)

Do you see a potential cause-and-effect relationship developing in your draft? The cause-and-effect pattern may be used to identify one or more causes followed by one or more effects or results. Or you may reverse this sequence and describe effects first and then the cause or causes. For example, the causes of water pollution might be followed by its effects on both humans and animals. You may use obvious transitions to clarify cause and effect, such as “What are the results? Here are some of them
” or you might simply use the words cause, effect, and result, to cue the reader about your about the relationships that you’re establishing.

Problem-Solution

At some point does your essay explore a problem or suggest a solution? The problem-solution pattern is commonly used in identifying something that’s wrong and in contemplating what might be done to remedy the situation. There are probably more ways to organize a problem-solution approach, but but here are three possibilities:

  • Describe the problem, followed by the solution.
  • Propose the solution first and then describe the problems that motivated it.
  • Or a problem may be followed by several solutions, one of which is selected as the best.

When the solution is stated at the end of the paper, the pattern is sometimes called the delayed proposal. For a hostile audience, it may be effective to describe the problem, show why other solutions do not work, and finally suggest the favored solution. You can emphasize the words problem and solution to signal these sections of your paper for your reader.

Chronology or Narrative

Do you need to develop support for a topic where telling a story can illustrate some important concept for your readers? Material arranged chronologically is explained as it occurs in time. A chronological or narrative method of development might help you find a way to add both interest and content to your essay. Material arranged chronologically is explained as it occurs in time. This pattern may be used to establish what has happened. Chronology or narrative can be a great way to introduce your essay by providing a background or history behind your topic. Or you may want to tell a story to develop one or more points in the body of your essay. You can use transitional words like then, next, and finally to make the parts of the chronology clear.

Comparison and Contrast

Are you trying to define something? Do you need your readers to understand what something is and what it is not? The comparison-and-contrast method of development is particularly useful in extending a definition, or anywhere you need to show how a subject is like or unlike another subject. For example, the statement is often made that drug abuse is a medical problem instead of a criminal justice issue. An author might attempt to prove this point by comparing drug addiction to AIDS, cancer, or heart disease to redefine the term “addiction” as a medical problem. A statement in opposition to this idea could just as easily establish contrast by explaining all the ways that addiction is different from what we traditionally understand as an illness. In seeking to establish comparison or contrast in your writing, some words or terms that might be useful are by contrast, in comparison, while, some, and others.

Summary

These four methods of development—cause and effect, problem-solution, chronology or narrative, and comparison and contrast—are just a few ways to organize and develop ideas and content in your essays. It’s important to note that they should not be a starting point for writers who want to write something authentic—something that they care deeply about. Instead, they can be a great way to help you look for what’s already happening with your topic or in a draft, to help you to write more, or to help you reorganize some parts of an essay that seem to lack connection or feel disjointed. Look for organizational patterns when you’re reading work by professional writers. Notice where they combine strategies (e.g., a problem-solution pattern that uses cause-and-effect organization, or a comparison-contrast pattern that uses narrative or chronology to develop similarities or differences). Pay attention to how different writers emphasize and develop their main ideas, and use what you find to inspire you in your own writing. Better yet, work on developing completely new patterns of your own.

Reference

Wrote with and ☕ by Anda Toshiki at root@andatoshiki:/~

Copyright © 2023-2023 Anda Toshiki +

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Patterns of Organization and Methods of Development

Patterns of organization can help your readers follow the ideas within your essay and your paragraphs, but they can also work as methods of development to help you recognize and further develop ideas and relationships in your writing. Here are some strategies that can help you with both organization and development in your essays.

Major Patterns of Organization

A fruit pie.

Read the following sentences:

  • Now take the pie out of the oven and let it cool on the stovetop.
  • Mix the dry ingredients with the liquid ingredients.
  • Set the pie crust aside while you make the filling.

How did it feel to read the above list? A bit confusing, I would guess. That’s because the steps for making a pie were not well organized, and the steps don’t include enough detail for us to know exactly what we should do. (Like what are the dry and liquid ingredients?) We all know that starting instructions from the beginning and giving each detailed step in the order it should happen is vital to having a good outcome, in this case a yummy pie! But it’s not always so simple to know how to organize or develop ideas, and sometimes there’s more than one way, which complicates things even further.

First, let’s take a look at a couple of ways to think about organization.

General to Specific or Specific to General

It might be useful to think about organizing your topic like a triangle:

Two triangles. The first is an inverted pyramid for General to Specific, the second is a pyramid for Specific to General.

The first triangle represents starting with the most general, big picture information first, moving then to more detailed and often more personal information later in the paper. The second triangle represents an organizational structure that starts with the specific, small scale information first and then moves to the more global, big picture stuff.

For example, if your topic is traffic in Vancouver, British Columbia, an essay that uses the general-to-specific organizational structure might begin this way:

Many people consider Vancouver, British Columbia, to be a relaxed place to live. They would be shocked to know how bad the traffic is traveling major arteries into the city and even driving around the city itself.

An essay that uses the specific-to-general structure might start like this:

Transit is crowded, parking is expensive, and vehicles stop and go through the main streets of the city of Vancouver, British Columbia, and that is just once travelers brave the crowded arteries to enter the city; Vancouver’s traffic problem does not lend itself to the relaxed atmosphere many believe the city to have.

What’s the difference between these two introductions? And how might they appeal to the intended audience for this essay in different ways? The first introduction is looking at the big picture of the problem and mentions pollution’s impact on all citizens in Portland, while the second introduction focuses on one specific family. The first helps readers see how vast the problem really is, and the second helps connect readers to a real family, making an emotional appeal from the very beginning. Neither introduction is necessarily better. You’ll choose one over the other based on the kind of tone you’d like to create and how you’d like to affect your audience. It’s completely up to you to make this decision.

Does the Triangle Mean the Essay Keeps Getting More Specific or More Broad until the Very End?

The triangle is kind of a general guide, meaning you’re allowed to move around within it all you want. For example, it’s possible that each of your paragraphs will be its own triangle, starting with the general or specific and moving out or in. However, if you begin very broadly, it might be effective to end your essay in a more specific, personal way. And if you begin with a personal story, consider ending your essay by touching on the global impact and importance of your topic.

Are There Other Ways to Think about Organizing My Ideas?

Yes! Rather than thinking about which of your ideas are most specific or personal or which are more broad or universal, you might consider one of the following ways of organizing your ideas:

  • Most important information first (consider what you want readers to focus on first)
  • Chronological order (the order in time that events take place)
  • Compare and contrast (ideas are organized together because of their relationship to each other)

The section on Methods of Development, below, offers more detail about some of these organizational patterns, along with some others.

Choose one of the following topics, and practice writing a few opening sentences like we did above, once using the general-to-specific format and once using the specific-to-general. Which do you like better? What audience would be attracted to which one? Share with peers to see how others tackled this challenge. How would you rewrite their sentences? Why? Discuss your changes and listen to how your peers have revised your sentences. Taking in other people’s ideas will help you see new ways to approach your own writing and thinking.
Topics:

  • Facing fears
  • Safety in sports
  • Community policing
  • Educating prisoners
  • Sex education

Methods of Development

The methods of development covered here are best used as ways to look at what’s already happening in your draft and to consider how you might emphasize or expand on any existing patterns. You might already be familiar with some of these patterns because teachers will sometimes assign them as the purpose for writing an essay. For example, you might have been asked to write a cause-and-effect essay or a comparison-and-contrast essay.

It’s important to emphasize here that patterns of organization or methods of developing content usually happen naturally as a consequence of the way the writer engages with and organizes information while writing. That is to say, most writers don’t sit down and say, “I think I’ll write a cause-and-effect essay today.” Instead, a writer might be more likely to be interested in a topic, say, the state of drinking water in the local community, and as the writer begins to explore the topic, certain cause-and-effect relationships between environmental pollutants and the community water supply may begin to emerge.

So if these patterns just occur naturally in writing, what’s the use in knowing about them? Well, sometimes you might be revising a draft and notice that some of your paragraphs are a bit underdeveloped. Maybe they lack a clear topic, or maybe they lack support. In either case, you can look to these common methods of development to find ways to sharpen those vague topics or to add support where needed. Do you have a clear cause statement somewhere but you haven’t explored the effects? Are you lacking detail somewhere where a narrative story or historical chronology can help build reader interest and add support? Are you struggling to define an idea that might benefit from some comparison or contrast? Read on to consider some of the ways that these strategies can help you in revision.

Cause and Effect (or Effect and Cause)

Do you see a potential cause-and-effect relationship developing in your draft? The cause-and-effect pattern may be used to identify one or more causes followed by one or more effects or results. Or you may reverse this sequence and describe effects first and then the cause or causes. For example, the causes of water pollution might be followed by its effects on both humans and animals. You may use obvious transitions to clarify cause and effect, such as “What are the results? Here are some of them
” or you might simply use the words cause, effect, and result, to cue the reader about your about the relationships that you’re establishing.

Problem-Solution

At some point does your essay explore a problem or suggest a solution? The problem-solution pattern is commonly used in identifying something that’s wrong and in contemplating what might be done to remedy the situation. There are probably more ways to organize a problem-solution approach, but but here are three possibilities:

  • Describe the problem, followed by the solution.
  • Propose the solution first and then describe the problems that motivated it.
  • Or a problem may be followed by several solutions, one of which is selected as the best.

When the solution is stated at the end of the paper, the pattern is sometimes called the delayed proposal. For a hostile audience, it may be effective to describe the problem, show why other solutions do not work, and finally suggest the favored solution. You can emphasize the words problem and solution to signal these sections of your paper for your reader.

Chronology or Narrative

Do you need to develop support for a topic where telling a story can illustrate some important concept for your readers? Material arranged chronologically is explained as it occurs in time. A chronological or narrative method of development might help you find a way to add both interest and content to your essay. Material arranged chronologically is explained as it occurs in time. This pattern may be used to establish what has happened. Chronology or narrative can be a great way to introduce your essay by providing a background or history behind your topic. Or you may want to tell a story to develop one or more points in the body of your essay. You can use transitional words like then, next, and finally to make the parts of the chronology clear.

Comparison and Contrast

Are you trying to define something? Do you need your readers to understand what something is and what it is not? The comparison-and-contrast method of development is particularly useful in extending a definition, or anywhere you need to show how a subject is like or unlike another subject. For example, the statement is often made that drug abuse is a medical problem instead of a criminal justice issue. An author might attempt to prove this point by comparing drug addiction to AIDS, cancer, or heart disease to redefine the term “addiction” as a medical problem. A statement in opposition to this idea could just as easily establish contrast by explaining all the ways that addiction is different from what we traditionally understand as an illness. In seeking to establish comparison or contrast in your writing, some words or terms that might be useful are by contrast, in comparison, while, some, and others.

Summary

These four methods of development—cause and effect, problem-solution, chronology or narrative, and comparison and contrast—are just a few ways to organize and develop ideas and content in your essays. It’s important to note that they should not be a starting point for writers who want to write something authentic—something that they care deeply about. Instead, they can be a great way to help you look for what’s already happening with your topic or in a draft, to help you to write more, or to help you reorganize some parts of an essay that seem to lack connection or feel disjointed. Look for organizational patterns when you’re reading work by professional writers. Notice where they combine strategies (e.g., a problem-solution pattern that uses cause-and-effect organization, or a comparison-contrast pattern that uses narrative or chronology to develop similarities or differences). Pay attention to how different writers emphasize and develop their main ideas, and use what you find to inspire you in your own writing. Better yet, work on developing completely new patterns of your own.

Reference

- - + + \ No newline at end of file diff --git a/academic/physics/index.html b/academic/physics/index.html index 029a2b10..1e0c4e89 100644 --- a/academic/physics/index.html +++ b/academic/physics/index.html @@ -6,9 +6,9 @@ Welcome to Physics | Toshiki's Note - - - + + + @@ -30,10 +30,10 @@ -
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Welcome to Physics

Wrote with and ☕ by Anda Toshiki at root@andatoshiki:/~

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- - + + \ No newline at end of file diff --git a/academic/physics/ipho-formulas-jpn/1.html b/academic/physics/ipho-formulas-jpn/1.html index 243f4469..fc24924f 100644 --- a/academic/physics/ipho-formulas-jpn/1.html +++ b/academic/physics/ipho-formulas-jpn/1.html @@ -6,9 +6,9 @@ Formulas for IPhO æ—„æœŹèȘžç‰ˆ: Section 1 | Toshiki's Note - - - + + + @@ -50,10 +50,10 @@ s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5, c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1 s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26 c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z -M1001 80h400000v40h-400000z">​​

1.13: ç©ćˆ†

  1. ç©ćˆ†ïŒšćŸźćˆ†ăźć…ŹćŒăźć·ŠèŸșべ揳èŸșă‚’ć…„ă‚Œæ›żăˆăŸă‚‚ăźăšćŒ ă˜ïŒˆé€†æŒ”çź—ïŒ‰.äŸ‹ăˆă°,

    ∫xn dx=xn+1/(n+1).\int x^n \mathrm{~d} x=x^{n+1} /(n+1) .

    çœźæ›ç©ćˆ†ăźç‰č戄ăȘ栎搈 :

    ∫f(ax+b)dx=F(ax+b)/a.\int f(a x+b) \mathrm{d} x=F(a x+b) / a .

1.14: 憆錐æ›Č線

  1. 憆錐æ›Č線: a11x2+2a12xy+a22y2+a1x+a2y+a0=a_{11} x^2+2 a_{12} x y+a_{22} y^2+a_1 x+a_2 y+a_0= 0 で, a11=a22a_{11}=a_{22} ăȘらば憆, a11(a11a22−a122)>0a_{11}\left(a_{11} a_{22}-a_{12}^2\right)>0 ăȘă‚‰ă°æ„•ć††, ⋯<0\cdots<0 ăȘらば揌æ›Č線, a11a22−a122=0a_{11} a_{22}-a_{12}^2=0 ăȘă‚‰ă°æ”Ÿç‰©ç·š. æ„•ć†† : l1+l2=2a,α1=α2l_1+l_2=2 a, \alpha_1=\alpha_2 [èšł è€…æłš : 焩ç‚čずæ›Č線䞊たç‚čă‚’ç”ă¶ç›Žç·šăšæŽ„ç·šăšăźăȘă™è§’ ], A=πabA=\pi a b. 揌æ›Č線 : l1−l2=2a,α1+α2=0l_1-l_2=2 a, \alpha_1+\alpha_2=0. 攟物線 :l+h=: l+h= const., α1=α2\alpha_1=\alpha_2.

1.15: æ•°ć€Œèšˆçź— & ć°ćœąèŠć‰‡

  1. æ•°ć€Œèšˆçź—. f(x)=0f(x)=0 ăźè§Łă‚’æ±‚ă‚ă‚‹ăƒ‹ăƒ„ăƒŒăƒˆăƒłæł• :

    xn+1=xn−f(xn)/fâ€Č(xn)x_{n+1}=x_n-f\left(x_n\right) / f^{\prime}\left(x_n\right)

    èż‘äŒŒç©ćˆ†ăźć°ćœąèŠć‰‡ïŒš

    ∫abf(x)dx≈b−a2n[f(x0)+2{f(x1)+⋯ +f(xn−1)}+f(xn)]\begin{array}{r} \int_a^b f(x) \mathrm{d} x \approx \frac{b-a}{2 n}\left[f\left(x_0\right)+2\left\{f\left(x_1\right)+\cdots\right.\right. \left.\left.+f\left(x_{n-1}\right)\right\}+f\left(x_n\right)\right] \end{array}

1.16: ăƒ™ă‚Żăƒˆăƒ«ăźćŸźćˆ† & ç©ćˆ†

  1. ăƒ™ă‚Żăƒˆăƒ«ăźćŸźćˆ†ăšç©ćˆ†ïŒšæˆćˆ†ă”ăšă«ćŸźćˆ†/ç©ćˆ†ă™ă‚‹ïŒŽă‚ă‚‹ă„ăŻç„Ąé™ă«èż‘ă„22ă€ăźăƒ™ă‚Żăƒˆăƒ«ăźć·źă‚’æ±‚ă‚ă‚‹ă“ăšă§ ćŸźćˆ†ă™.

Wrote with and ☕ by Anda Toshiki at root@andatoshiki:/~

Copyright © 2023-2023 Anda Toshiki +M1001 80h400000v40h-400000z">​​

1.13: ç©ćˆ†

  1. ç©ćˆ†ïŒšćŸźćˆ†ăźć…ŹćŒăźć·ŠèŸșべ揳èŸșă‚’ć…„ă‚Œæ›żăˆăŸă‚‚ăźăšćŒ ă˜ïŒˆé€†æŒ”çź—ïŒ‰.äŸ‹ăˆă°,

    ∫xn dx=xn+1/(n+1).\int x^n \mathrm{~d} x=x^{n+1} /(n+1) .

    çœźæ›ç©ćˆ†ăźç‰č戄ăȘ栎搈 :

    ∫f(ax+b)dx=F(ax+b)/a.\int f(a x+b) \mathrm{d} x=F(a x+b) / a .

1.14: 憆錐æ›Č線

  1. 憆錐æ›Č線: a11x2+2a12xy+a22y2+a1x+a2y+a0=a_{11} x^2+2 a_{12} x y+a_{22} y^2+a_1 x+a_2 y+a_0= 0 で, a11=a22a_{11}=a_{22} ăȘらば憆, a11(a11a22−a122)>0a_{11}\left(a_{11} a_{22}-a_{12}^2\right)>0 ăȘă‚‰ă°æ„•ć††, ⋯<0\cdots<0 ăȘらば揌æ›Č線, a11a22−a122=0a_{11} a_{22}-a_{12}^2=0 ăȘă‚‰ă°æ”Ÿç‰©ç·š. æ„•ć†† : l1+l2=2a,α1=α2l_1+l_2=2 a, \alpha_1=\alpha_2 [èšł è€…æłš : 焩ç‚čずæ›Č線䞊たç‚čă‚’ç”ă¶ç›Žç·šăšæŽ„ç·šăšăźăȘă™è§’ ], A=πabA=\pi a b. 揌æ›Č線 : l1−l2=2a,α1+α2=0l_1-l_2=2 a, \alpha_1+\alpha_2=0. 攟物線 :l+h=: l+h= const., α1=α2\alpha_1=\alpha_2.

1.15: æ•°ć€Œèšˆçź— & ć°ćœąèŠć‰‡

  1. æ•°ć€Œèšˆçź—. f(x)=0f(x)=0 ăźè§Łă‚’æ±‚ă‚ă‚‹ăƒ‹ăƒ„ăƒŒăƒˆăƒłæł• :

    xn+1=xn−f(xn)/fâ€Č(xn)x_{n+1}=x_n-f\left(x_n\right) / f^{\prime}\left(x_n\right)

    èż‘äŒŒç©ćˆ†ăźć°ćœąèŠć‰‡ïŒš

    ∫abf(x)dx≈b−a2n[f(x0)+2{f(x1)+⋯ +f(xn−1)}+f(xn)]\begin{array}{r} \int_a^b f(x) \mathrm{d} x \approx \frac{b-a}{2 n}\left[f\left(x_0\right)+2\left\{f\left(x_1\right)+\cdots\right.\right. \left.\left.+f\left(x_{n-1}\right)\right\}+f\left(x_n\right)\right] \end{array}

1.16: ăƒ™ă‚Żăƒˆăƒ«ăźćŸźćˆ† & ç©ćˆ†

  1. ăƒ™ă‚Żăƒˆăƒ«ăźćŸźćˆ†ăšç©ćˆ†ïŒšæˆćˆ†ă”ăšă«ćŸźćˆ†/ç©ćˆ†ă™ă‚‹ïŒŽă‚ă‚‹ă„ăŻç„Ąé™ă«èż‘ă„22ă€ăźăƒ™ă‚Żăƒˆăƒ«ăźć·źă‚’æ±‚ă‚ă‚‹ă“ăšă§ ćŸźćˆ†ă™.
- - + + \ No newline at end of file diff --git a/academic/physics/ipho-formulas-jpn/10.html b/academic/physics/ipho-formulas-jpn/10.html index 3bcd43dd..85b21f23 100644 --- a/academic/physics/ipho-formulas-jpn/10.html +++ b/academic/physics/ipho-formulas-jpn/10.html @@ -6,9 +6,9 @@ Formulas for IPhO æ—„æœŹèȘžç‰ˆ: Section 10 | Toshiki's Note - - - + + + @@ -49,10 +49,10 @@ s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5, c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1 s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26 c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z -M1001 80h400000v40h-400000z">​,Îœ=vnS.

10.10: Carnot ă‚”ă‚€ă‚Żăƒ«

  1. Carnot ă‚”ă‚€ă‚Żăƒ« : 断熱過皋 2 ă€ăšç­‰æž©éŽçš‹ 2 ぀. S−TS-T ćș§æš™ă‚’ç”šă„ă‚‹ă“ăšă«ă‚ˆă‚Š η=(T1−T2)/T1\eta=\left(T_1-T_2\right) / T_1 ă‚’ćŸ—ă‚‹.

10.11: ăƒ’ăƒŒăƒˆăƒăƒłăƒ—

  1. ăƒ’ăƒŒăƒˆăƒăƒłăƒ—: Carnot ă‚”ă‚€ă‚Żăƒ«ăźé€†. η=T1T1−T2\eta=\frac{T_1}{T_1-T_2}.

10.12: ă‚šăƒłăƒˆăƒ­ăƒ”ăƒŒ

  1. ă‚šăƒłăƒˆăƒ­ăƒ”ăƒŒ :dS=dQ/T: \mathrm{d} S=\mathrm{d} Q / T.

10.13: ç†±ćŠ›ć­ŠçŹŹäž€æł•ć‰‡

  1. ç†±ćŠ›ć­ŠçŹŹäž€æł•ć‰‡ : dâ€ČU=dâ€ČA+dâ€ČQ\mathrm{d}^{\prime} U=\mathrm{d}^{\prime} A+\mathrm{d}^{\prime} Q

10.14: 熱抛歊珏äșŒæł•ć‰‡

  1. 熱抛歊珏äșŒæł•ć‰‡ : ΔS≄0\Delta S \geq 0 (ăŸăŸ ηreal ≀ηCarnot )\left.\eta_{\text {real }} \leq \eta_{\text {Carnot }}\right).

10.15: æ°—äœ“ăźă™ă‚‹ä»•äș‹

  1. æ°—äœ“ăźă™ă‚‹ä»•äș‹ïŒˆăƒă‚€ăƒłăƒˆ 10 ă‚‚ć‚ç…§ïŒ‰:

    A=∫p dV, 断熱過皋: A=i2Δ(pV)A=\int p \mathrm{~d} V, \quad \text { 断熱過皋: } A=\frac{i}{2} \Delta(p V)

10.16: Dalton ăźæł•ć‰‡

  1. Dalton ăźæł•ć‰‡: p=p= ∑pi\sum p_i

    èšłè€…æłš

    ç†æƒłæ°—äœ“ăźăżæˆç«‹

10.17: æČžéš°

  1. æČžéš°: éŁœć’Œè’žæ°—ăźćœ§ćŠ› pv=p0.2p_v=p_0 .2 æ¶Čぼ界靱では pv1+pv2=p0p_{v 1}+p_{v 2}=p_0.

10.18: 熱攁

  1. 熱攁: P=kSΔT/lP=k S \Delta T / l ( kk ăŻç†±äŒć°ŽçŽ‡). ç›Žæ”ć›žè·Żă«äŒŒăŠ いる (P↔I,ΔT↔V,k↔1/ρ)(P \leftrightarrow I, \Delta T \leftrightarrow V, k \leftrightarrow 1 / \rho).

10.19: 熱ćźč量

  1. 熱ćźč量 : Q=∫c(T)dTQ=\int c(T) \mathrm{d} T. ć›șäœ“ă§ăŻäœŽæž©ă§ c∝T3c \propto T^3, é«˜æž©ă§ c=3NkBc=3 N k_B Dulong-Petit ăźæł•ć‰‡. ここで NN ăŻç”æ™¶äž­ăźćŽŸć­æ•°ïŒ‰

10.20: èĄšéąćŒ”ćŠ›

  1. èĄšéąćŒ”ćŠ› :

    U=Sσ,F=lσ,p=2σ/RU=S \sigma, F=l \sigma, p=2 \sigma / R

10.21: Stefan-Boltzmann ăźæł•ć‰‡ (灰è‰Č䜓)

  1. Stefan-Boltzmann ăźæł•ć‰‡ (灰è‰Č䜓) : P=ΔσAT4P=\varepsilon \sigma A T^4.

10.22: Wien ăźć€‰äœć‰‡

  1. Wien ăźć€‰äœć‰‡: Îœmax⁥=AkBT/h(A≈\nu_{\max }=A k_B T / h(A \approx 2.8), λmax⁥=hc/Aâ€ČkBT(Aâ€Č≈5)\lambda_{\max }=h c / A^{\prime} k_B T\left(A^{\prime} \approx 5\right).

Wrote with and ☕ by Anda Toshiki at root@andatoshiki:/~

Copyright © 2023-2023 Anda Toshiki +M1001 80h400000v40h-400000z">​,Îœ=vnS.

10.10: Carnot ă‚”ă‚€ă‚Żăƒ«

  1. Carnot ă‚”ă‚€ă‚Żăƒ« : 断熱過皋 2 ă€ăšç­‰æž©éŽçš‹ 2 ぀. S−TS-T ćș§æš™ă‚’ç”šă„ă‚‹ă“ăšă«ă‚ˆă‚Š η=(T1−T2)/T1\eta=\left(T_1-T_2\right) / T_1 ă‚’ćŸ—ă‚‹.

10.11: ăƒ’ăƒŒăƒˆăƒăƒłăƒ—

  1. ăƒ’ăƒŒăƒˆăƒăƒłăƒ—: Carnot ă‚”ă‚€ă‚Żăƒ«ăźé€†. η=T1T1−T2\eta=\frac{T_1}{T_1-T_2}.

10.12: ă‚šăƒłăƒˆăƒ­ăƒ”ăƒŒ

  1. ă‚šăƒłăƒˆăƒ­ăƒ”ăƒŒ :dS=dQ/T: \mathrm{d} S=\mathrm{d} Q / T.

10.13: ç†±ćŠ›ć­ŠçŹŹäž€æł•ć‰‡

  1. ç†±ćŠ›ć­ŠçŹŹäž€æł•ć‰‡ : dâ€ČU=dâ€ČA+dâ€ČQ\mathrm{d}^{\prime} U=\mathrm{d}^{\prime} A+\mathrm{d}^{\prime} Q

10.14: 熱抛歊珏äșŒæł•ć‰‡

  1. 熱抛歊珏äșŒæł•ć‰‡ : ΔS≄0\Delta S \geq 0 (ăŸăŸ ηreal ≀ηCarnot )\left.\eta_{\text {real }} \leq \eta_{\text {Carnot }}\right).

10.15: æ°—äœ“ăźă™ă‚‹ä»•äș‹

  1. æ°—äœ“ăźă™ă‚‹ä»•äș‹ïŒˆăƒă‚€ăƒłăƒˆ 10 ă‚‚ć‚ç…§ïŒ‰:

    A=∫p dV, 断熱過皋: A=i2Δ(pV)A=\int p \mathrm{~d} V, \quad \text { 断熱過皋: } A=\frac{i}{2} \Delta(p V)

10.16: Dalton ăźæł•ć‰‡

  1. Dalton ăźæł•ć‰‡: p=p= ∑pi\sum p_i

    èšłè€…æłš

    ç†æƒłæ°—äœ“ăźăżæˆç«‹

10.17: æČžéš°

  1. æČžéš°: éŁœć’Œè’žæ°—ăźćœ§ćŠ› pv=p0.2p_v=p_0 .2 æ¶Čぼ界靱では pv1+pv2=p0p_{v 1}+p_{v 2}=p_0.

10.18: 熱攁

  1. 熱攁: P=kSΔT/lP=k S \Delta T / l ( kk ăŻç†±äŒć°ŽçŽ‡). ç›Žæ”ć›žè·Żă«äŒŒăŠ いる (P↔I,ΔT↔V,k↔1/ρ)(P \leftrightarrow I, \Delta T \leftrightarrow V, k \leftrightarrow 1 / \rho).

10.19: 熱ćźč量

  1. 熱ćźč量 : Q=∫c(T)dTQ=\int c(T) \mathrm{d} T. ć›șäœ“ă§ăŻäœŽæž©ă§ c∝T3c \propto T^3, é«˜æž©ă§ c=3NkBc=3 N k_B Dulong-Petit ăźæł•ć‰‡. ここで NN ăŻç”æ™¶äž­ăźćŽŸć­æ•°ïŒ‰

10.20: èĄšéąćŒ”ćŠ›

  1. èĄšéąćŒ”ćŠ› :

    U=Sσ,F=lσ,p=2σ/RU=S \sigma, F=l \sigma, p=2 \sigma / R

10.21: Stefan-Boltzmann ăźæł•ć‰‡ (灰è‰Č䜓)

  1. Stefan-Boltzmann ăźæł•ć‰‡ (灰è‰Č䜓) : P=ΔσAT4P=\varepsilon \sigma A T^4.

10.22: Wien ăźć€‰äœć‰‡

  1. Wien ăźć€‰äœć‰‡: Îœmax⁥=AkBT/h(A≈\nu_{\max }=A k_B T / h(A \approx 2.8), λmax⁥=hc/Aâ€ČkBT(Aâ€Č≈5)\lambda_{\max }=h c / A^{\prime} k_B T\left(A^{\prime} \approx 5\right).
- - + + \ No newline at end of file diff --git a/academic/physics/ipho-formulas-jpn/11.html b/academic/physics/ipho-formulas-jpn/11.html index 1c25a48b..6b70011d 100644 --- a/academic/physics/ipho-formulas-jpn/11.html +++ b/academic/physics/ipho-formulas-jpn/11.html @@ -6,9 +6,9 @@ Formulas for IPhO æ—„æœŹèȘžç‰ˆ: Section 11 | Toshiki's Note - - - + + + @@ -40,10 +40,10 @@ s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5, c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1 s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26 c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z -M1001 80h400000v40h-400000z">​) であればćźčæ˜“ă«é€éŽă™ă‚‹.

11.7: Bohr ăƒąăƒ‡ăƒ«

  1. Bohr ăƒąăƒ‡ăƒ« : En∝−1/n2E_n \propto-1 / n^2. (ć€ć…žçš„ă«èšˆçź—ă•ă‚Œă‚‹) ć††è»Œé“ă§ăŻ, è»Œé“ăźé•·ă•ăŒæłąé•· λ=h/mv\lambda=h / m v ăźæ•Žæ•°ć€.

11.8: Compton ćŠč果

  1. Compton ćŠč果: ć…‰ć­ăŒé›»ć­ă‹ă‚‰æ•Łäč±ă•ă‚Œă‚‹ăš, 慉歐ぼ Δλ=λC(1−cos⁥Ξ)\Delta \lambda=\lambda_C(1-\cos \theta)

11.9: 慉電ćŠč果

  1. 慉電ćŠč果: W+mvmax⁥2/2=hÎœW+m v_{\max }^2 / 2=h \nu ( WW は仕äș‹é–ąæ•° )). I−VI-V ă‚°ăƒ©ăƒ•ïŒšć…‰é›»æ”ăŻé˜»æ­ąé›»ćœ§ V=−(hΜ−W)/eV=-(h \nu-W) / e ă§ć§‹ăŸă‚Š, æ­Łæ–čć‘ă«é›»ćœ§ăŒć€§ăăăȘă‚‹ăšç·©ć’Œă™ă‚‹.

11.10: Stefan-Boltzmann ăźæł•ć‰‡

  1. Stefan-Boltzmann ăźæł•ć‰‡ : P=σAT4P=\sigma A T^4