Question

The decomposition of a gaseous substance (A) to yield gaseous products (B) and (C) follows first order kinetics. If initially only (A) is present and after 10 minutes the reaction pressure of (A) is $200$ mm Hg and the overall pressure of the mixture is $300$ mm Hg, the rate constant for the reaction $2A\to B+3C$ is:

• A. $\frac{1}{600}\ln 1.25{\text{s}}^{-1}$
• B. $\frac{2.303}{10}\text{log}1.5{\text{min}}^{-1}$
• C. $\frac{1}{10}\ln 1.25{\text{s}}^{-1}$
• D. None of these

Answer 1

The correct option is B $\frac{2.303}{10}\text{log}1.5{\text{min}}^{-1}$

The initial and final concentrations can be expressed in terms of pressure.
Given:
Let the initial pressure be $P_0$
Pressure at time $(t=10\text{min})=200\text{mm Hg}$
Final pressure after completion $\left(t_{\infty }\right)=300\text{mm Hg}$
$2A\to B+3C$
$\begin{array}{cccc}t=0& P_0& 0& 0\\ t=10& P_0-2P^{\prime }& P^{\prime }& 3P^{\prime }\\ t=t_{\infty }& 0& P_0/2& 3P^0/2\end{array}$

From above,
Given: $P_0-2P^{\prime }=200\text{mm Hg}$
$2P_0=300\text{mm Hg}$
So, $\overline{P_0=150}$
& $P^{\prime }=25$

Integrated rate expression,
$k=\frac{2.303}{t}\text{log}\left(\frac{P_0}{P_0-2P^{\prime }}\right)$
$k=\frac{2.303}{t}\text{log}\left(\frac{150}{100}\right)$
$k=\frac{2.303}{10}\text{log}1.5$