Question

The gaseous decomposition reaction:$A\left(g\right)\to 2B\left(g\right)+C\left(g\right)$, is observed to be first order over the excess of liquid water at ${25}^{o}C$ at constant volume. It is found that after 10 minutes, the total pressure of the system is $188torr$ and at completion of reaction the pressure is $388torr$. Calculate the rate constant of the reaction in $h{r}^{-1}$. The vapour pressure of $H_{2}O$ at ${25}^{o}C$ is $28torr$.
Take
$ln1.2\approx 0.18$

• A. 0.018
• B. 1.08
• C. 0.24
• D. 0.12

Answer 1

The correct option is B 1.08

The gaseous decomposition reaction is occurs over the excess of liquid water. So the vapour pressure of water at that temperature will also contribute to the total pressure of the system.

$A\left(g\right)\to 2B\left(g\right)+C\left(g\right)$
Let,
$\text{At time, t}=0P_{0}00$
$\text{After 10 mins,}(P_0-x)2xx$
$Att=\infty 0{2P}_0{P}_0$
Now,
At time, t = 10 min
$\text{Total pressure}=(P_0-x)+2x+x+\text{vapour pressure of}{H}_{2}O$
$188=P_0+2x+28$
$P_0+2x=\mathrm{160......}eqn\left(1\right)$

At time, $t=\infty$
$\text{Total pressure}=3P_0+\text{vapour pressure of}{H}_{2}O$
$3P_0+28=388$
So,
$P_0=120torr$
Substituting $P_0$ in equation (1), we get,
$x=20torr$
For $1^{st}$ order reaction,
$k=\frac{1}{t}ln\frac{a}{a-x}\text{(In terms of concentration})$

$k=\frac{1}{t}ln\left(\frac{P_0}{P_0-x}\right)\text{(In terms of pressure})$
$k=\frac{1}{10}ln\left(\frac{120}{120-20}\right)\approx 0.018{min}^{-1}=1.08h{r}^{-1}$