The reaction $CH_{3}COF + H_{2}O \rightleftharpoons CH_{3}COOH + HF$ is studied under two conditions:
Condition $I$: $[H_{2}O]_{0} = 1.00 \ M$,$[CH_{3}COF]_{0} = 0.01 \ M$
Condition $II$: $[H_{2}O]_{0} = 0.02 \ M$,$[CH_{3}COF]_{0} = 0.80 \ M$
Time $(t)$ min (Condition $I$) / $[CH_{3}COF]$ $M$ Time $(t)$ min (Condition $II$) / $[H_{2}O]$ $M$ $0$ / $0.01000$ $0$ / $0.0200$ $10$ / $0.00867$ $10$ / $0.0176$ $20$ / $0.00735$ $20$ / $0.0156$ $40$ / $0.00540$ $40$ / $0.0122$
Determine the overall order of the reaction and calculate the rate constant.
Condition $I$: $[H_{2}O]_{0} = 1.00 \ M$,$[CH_{3}COF]_{0} = 0.01 \ M$
Condition $II$: $[H_{2}O]_{0} = 0.02 \ M$,$[CH_{3}COF]_{0} = 0.80 \ M$
| Time $(t)$ min (Condition $I$) / $[CH_{3}COF]$ $M$ | Time $(t)$ min (Condition $II$) / $[H_{2}O]$ $M$ |
|---|---|
| $0$ / $0.01000$ | $0$ / $0.0200$ |
| $10$ / $0.00867$ | $10$ / $0.0176$ |
| $20$ / $0.00735$ | $20$ / $0.0156$ |
| $40$ / $0.00540$ | $40$ / $0.0122$ |
Determine the overall order of the reaction and calculate the rate constant.
(N/A) The rate law is given by $Rate = k[CH_{3}COF]^{x}[H_{2}O]^{y}$.
In Condition $I$,$[H_{2}O] \gg [CH_{3}COF]$,so $[H_{2}O]$ is effectively constant. The reaction follows pseudo-first-order kinetics with respect to $CH_{3}COF$. Using $k_{obs, I} = \frac{2.303}{t} \log(\frac{[A]_{0}}{[A]_{t}})$,for $t=10 \ min$,$k_{obs, I} = \frac{2.303}{10} \log(\frac{0.01}{0.00867}) \approx 0.0142 \ min^{-1}$.
In Condition $II$,$[CH_{3}COF] \gg [H_{2}O]$,so $[CH_{3}COF]$ is effectively constant. The reaction follows pseudo-first-order kinetics with respect to $H_{2}O$. Using $k_{obs, II} = \frac{2.303}{t} \log(\frac{[B]_{0}}{[B]_{t}})$,for $t=10 \ min$,$k_{obs, II} = \frac{2.303}{10} \log(\frac{0.02}{0.0176}) \approx 0.0128 \ min^{-1}$.
Comparing the rates,the reaction is $1^{st}$ order with respect to both reactants $(x=1, y=1)$.
Thus,the overall order of the reaction is $1+1 = 2$.
The rate constant $k$ can be calculated as $k = \frac{k_{obs, I}}{[H_{2}O]_{0}} = \frac{0.0142}{1.00} = 0.0142 \ M^{-1} min^{-1}$.
In Condition $I$,$[H_{2}O] \gg [CH_{3}COF]$,so $[H_{2}O]$ is effectively constant. The reaction follows pseudo-first-order kinetics with respect to $CH_{3}COF$. Using $k_{obs, I} = \frac{2.303}{t} \log(\frac{[A]_{0}}{[A]_{t}})$,for $t=10 \ min$,$k_{obs, I} = \frac{2.303}{10} \log(\frac{0.01}{0.00867}) \approx 0.0142 \ min^{-1}$.
In Condition $II$,$[CH_{3}COF] \gg [H_{2}O]$,so $[CH_{3}COF]$ is effectively constant. The reaction follows pseudo-first-order kinetics with respect to $H_{2}O$. Using $k_{obs, II} = \frac{2.303}{t} \log(\frac{[B]_{0}}{[B]_{t}})$,for $t=10 \ min$,$k_{obs, II} = \frac{2.303}{10} \log(\frac{0.02}{0.0176}) \approx 0.0128 \ min^{-1}$.
Comparing the rates,the reaction is $1^{st}$ order with respect to both reactants $(x=1, y=1)$.
Thus,the overall order of the reaction is $1+1 = 2$.
The rate constant $k$ can be calculated as $k = \frac{k_{obs, I}}{[H_{2}O]_{0}} = \frac{0.0142}{1.00} = 0.0142 \ M^{-1} min^{-1}$.
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