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52 lines
2.3 KiB
Markdown
52 lines
2.3 KiB
Markdown
# Problem: 03-02-1
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## Question
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$$
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2 \mathrm{~N}_2 \mathrm{O}_5(g) \rightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)
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$$
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The decomposition of $\mathrm{N}_2 \mathrm{O}_5(g)$ is represented by the equation above. A sample of $\mathrm{N}_2 \mathrm{O}_5(g)$ is monitored as it decomposes, and the concentration of $\mathrm{N}_2 \mathrm{O}_5$ as a function of time is recorded. The results are shown in the table below.
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| Time (s) | $\mathrm{[N_2O_5]}$ |
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| -------- | ------------------- |
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| 0 | 1.000 |
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| 25.0 | 0.801 |
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| 50.0 | 0.642 |
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| 75.0 | 0.515 |
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Calculate the average rate of the reaction between $50.0$ and $75.0$ seconds.
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## Solution
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$$
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a \mathrm{~A}+b \mathrm{~B} \rightarrow c \mathrm{C}+d \mathrm{D}
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$$
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the rate of reaction is defined as
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$$
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\color{cyan}\text { rate }=-\frac{1}{a} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{b} \frac{\Delta[\mathrm{B}]}{\Delta t}=+\frac{1}{c} \frac{\Delta[\mathrm{C}]}{\Delta t}=+\frac{1}{d} \frac{\Delta[\mathrm{D}]}{\Delta t}
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$$
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Notice that the rate of change in concentration of each species is divided by its coefficient from the balanced chemical equation ( $a$, $b, c$, or $d$ ). This ensures that the calculated reaction rate is the same no matter which reactant or product is monitored for changes in concentration.
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In this case, the monitored species was $\mathrm{N}_2 \mathrm{O}_5$. With that in mind, let's write the reaction rate in terms of the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ :
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$$
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\text { rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta t}
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$$
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Since the coefficient for $\mathrm{N}_2 \mathrm{O}_5$ in the balanced equation is 2 , we divided the rate of change in concentration of $\mathrm{N}_2 \mathrm{O}_5$ by 2 .
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Additionally, since $\mathrm{N}_2 \mathrm{O}_5$ is being consumed in the reaction, we included a negative sign in front of the expression.
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Now, let's plug in the information from the table to calculate the average reaction rate between $50.0$ and $75.0$ seconds:
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$$
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\begin{aligned}
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\text { rate } & =-\frac{1}{2} \frac{(0.515 M-0.642 M)}{(75.0 \mathrm{~s}-50.0 \mathrm{~s})} \\
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& =2.54 \times 10^{-3} M \mathrm{~s}^{-1}
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\end{aligned}
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$$
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So, the average rate of the reaction between $50.0$ and $75.0$ seconds is $\bold{2.54 \times 10^{-3} \mathrm{M} \mathrm{s}^{-1}}$.
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