Question

The gaseous decomposition reaction, $A\left(g\right)⟶2B\left(g\right)+C\left(g\right)$ is observed to first order the excess of liquid water at ${25}^{o}C$. It is found that after $10$ minutes the total pressure of system is $188$ torr and after very long time it is $388$ torr. The rate constant of the reaction (in $h{r}^{-1}$) is:
[Given: vapour pressure of $H_{2}O$ at ${25}^{o}C$ is $28$ torr. ($\ln 2=0.7,\ln 3=1.1,\ln 10=2.3$)]

• A. $0.02$
• B. $1.2$
• C. $0.2$
• D. none of these

Answer 1

Answer: (B)

$1.2$

By putting values in equation $k=\frac{2.303}{t}log\frac{P_0}{(P_0-x)}=\frac{2.303}{t}log\frac{P_{\infty }/3}{(P_{\infty }/3-(P_t-P_0\left)\right)}$

We will get answer as $k=1.2h{r}^{-1}$

$(A,B,C)$ let the initial pressure of $A$ be $P_s$ mm of $Hg$ and the pressure of $A$ decreases in 10 minute be $x$ unit.

$\begin{array}{cccccc}& & & & & \\ & A\left(g\right)& ⟶& 2B\left(g\right)& +& C\left(g\right)\\ \text{Initially}& P_0& & 0& & 0\\ \text{At time 10 min}& P_0-x& & 2x& & x\\ \text{At}\infty \text{time}& 0& & 2P_0& & P_0\end{array}$

After long time interval, $P_{\infty }=2P_0+P_0=3P_0$

After $t$ time $P_t=P_0+2x$

For a first order reaction, the rate constant expression would be

$k=\frac{2.303}{t}\log \frac{P_0}{P_0-x}$