Question

In the reaction,
$CH_{3}COCH_{3\left(g\right)} \to C_{2}H_{4\left(g\right)}+H_{2\left(g\right)}+CO_{\left(g\right)}$
he initial pressure is found to be 0.40 atm and after 10 min, it was 0.50 atm. The rate constant for first order reaction is [$log^{4}$ = 0.6021, log 3.5 = 0.5441]

• A. $0.0133 min^{-1}$
• B. $0.4 s^{-1}$
• C. $10 s^{-1}$
• D. $0.6 min^{-1}$

Answer 1

Answer: (A)

$CH_{3}COOCH_{3\left(l\right)} \to C_{2}H_{4\left(g\right)}+H_{2\left(g\right)}+CO_{\left(g\right)}$
$\begin{matrix}0.4&0&0&0\\ 0.4-x&x&x&x\end{matrix}$
$0.4-x+x+x+x=0.50$
$\Rightarrow \, 2x=0.10 \Rightarrow x=0.05$
$k=\frac{2.303}{10} log \frac{0.4}{0.4-0.05}=\frac{2.303}{10} log \frac{0.40}{0.35}$
$=\frac{2.303}{10} log \frac{40}{35}=\frac{2.303}{10}\left[1.6021-1.5441\right]$
$=\frac{2.303}{10}\times0.0580=\frac{0.1335}{10}=0.01335 min^{-1}$
$=0.01335\times60=0.8 s^{-1}$