For a first order reaction $A_{(g)} \rightarrow 2 B_{(g)}+C_{(g)}$ at constant volume and 300 K , th

Question

For a first order reaction $A_{(g)} ightarrow 2 B_{(g)}+C_{(g)}$ at constant volume and 300 K , the total pressure at the beginning $(t=0)$ and at time $t$ are $P_{0}$ and $P_{t}$, respectively. Initially, only $A$ is present with concentration $[A]_{0}$, and $t_{1 / 3}$ is the time required for the partial pressure of $A$ to reach $1 / 3^{	ext {rd }}$ of its initial value. The correct option(s) is (are)\ (Assume that all these gases behave as ideal gases)\

Professor PC Thomas Classes · Answer

$(\mathrm{a}, \mathrm{d}): A_{(\mathrm{g})} \xrightarrow{	ext { First order }} 2 B_{(\mathrm{g})}+C_{(g)}$; \$\$ \$\$ So, $P_{t}=P_{0}-x+2 x+x=P_{0}+2 x$\ or $2 x=P_{t}-P_{0}$\ $t=\frac{1}{k} \ln \frac{P_{0}}{\left(P_{0}-xight)}$\ or $t=\frac{1}{k} \ln \frac{P_{0}}{P_{0}-\frac{\left(P_{t}-P_{0}ight)}{2}}=\frac{1}{k} \ln \frac{2 P_{0}}{2 P_{0}-P_{t}+P_{0}}$\ or $k t=\ln \frac{2 P_{0}}{3 P_{0}-P_{t}}$,\ $k t=\ln 2 P_{0}-\ln \left(3 P_{0}-P_{t}ight)$\ or $\ln \left(3 P_{0}-P_{t}ight)=-k t+\ln 2 P_{0}$\ Comparing the above equation with general straight line equation we get, slope $=-k$, intercept $=\ln 2 P_{0}$\ So, (a) is correct option.\ Now, $t_{1 / 3}=\frac{1}{k} \ln \frac{P_{0}}{\left(P_{0} / 3ight)}=\frac{1}{k} \ln 3$\ $\Rightarrow$ It is independent of initial concentration.\ So, (b) is wrong option.\ For first order reaction, rate constant is independent of initial concentration.\ So, graph (d) is correct.\