# Presentation problem: 02-20 | Toshiki's Note
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# Presentation problem: 02-20 ​

Author:Anda Toshiki
Updated:a minute ago
Words:412
Reading:2 min

## Question ​

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

C 2H 5 OH(g)⟶CH 3 CHO(g)+H 2(g)\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{g}) \longrightarrow \mathrm{CH}_3 \mathrm{CHO}(g)+\mathrm{H}_2(g) C2​H5​OH(g)⟶CH3​CHO(g)+H2​(g)

The pressure of C 2H 5 OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}C2​H5​OH was measured as a function of time and the following data were obtained:

Time (s) | P C 2H 5 OH (torr) P_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \text { (torr) }PC2​H5​OH​ (torr)
--- | ---
0 | 250.
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 C 2H 5 OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}C2​H5​OH after 900.s 900 . \mathrm{s}900.s from the start of the reaction. (Hint: To determine the order of the reaction with respect to C 2H 5 OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}C2​H5​OH, compare how the pressure of C 2H 5 OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}C2​H5​OH decreases with each time listing.)

## Solution ​

◎

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

k=p 0[C 2H 5 OH]−p[C 2H 5 OH]t=250 torr −185 torr 500 s=0.13 torr s −1 k=0.13 torr s s−1×1 atm760 torr =1.71×1 0−4 atm s−1 rate =k p(C 2H 5 OH)=−kt+p 0(C 2H 5 OH)p(C 2H 5 OH)=−(1.71×1 0−4 atm s−1)×900 s+0.33 atm p(C 2H 5 OH)=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} ​k=tp0​[C2​H5​OH]−p[C2​H5​OH]​=500 s250 torr −185 torr ​=0.13 torr s −1k=0.13 torr s s−1×760 torr 1 atm​=1.71×10−4 atm s−1 rate =kp(C2​H5​OH)=−kt+p0​(C2​H5​OH)p(C2​H5​OH)=−(1.71×10−4 atm s−1)×900 s+0.33 atmp(C2​H5​OH)=0.176 atm s−1​