Question

For a first order reaction $A\left(g\right)\to 2B\left(g\right)+C\left(g\right)$ at constant volume and $300K$, the total pressure at the beginning $(t=0)$ and at time $t$ are$P_0$ and $P_1$ respectively. Initially, only $A$ is present with concentration $[A{]}_0$ and $t_{\frac{1}{3}}$ is the time required for the partial pressure of $A$ to reach ${\frac{1}{3}}^{rd}$ of its initial value. The correct option (s) is/are:
(Assume that all these gases behave as ideal gases)

• A.
• B.
• C. $t_{\frac{1}{3}}$ is independent of initial concentration
• D.

Answer 1

Answer: (A), (C), (D)

Given, $V=constant,T=300K$
$\begin{array}{cccc}& A\left(g\right)& \stackrel{\text{first order}}{\to }2B\left(g\right)+& C\left(g\right)\\ t=0& P_0& & \\ t=t_{\frac{1}{3}}& (P_0-\frac{2P_0}{3})& \frac{4P_0}{3}& \frac{2P_0}{3}\\ & =\frac{P_0}{3}& & \\ t=t& P_0-x& 2x& x\end{array}So,P_t=P_0-x+2x+x=P_0+2x\Rightarrow 2x=P_t-P_{0}t=\frac{1}{k}ln\frac{P_0}{(P_0-x)}\Rightarrow t=\frac{1}{k}ln\frac{P_0}{P_0-\frac{(P_t-P_0)}{2}}=\frac{1}{k}ln\frac{2P_0}{2P_0-P_1+P_0}\Rightarrow kt=ln\frac{2P_0}{3P_0-+P_t},kt=ln2P_0-ln(3P_0-P_t)\Rightarrow ln(3P_0-P_t)=-kt+ln2P_0$
Graph between in $(3P_0-P_t)$ vs ${}^{\prime }{t}^{\prime }$ is a straight line with negative slope. So,(a) is correct option.
$t_{\frac{1}{3}}=\frac{1}{k}ln\frac{P_0}{P_0/3}=\frac{1}{k}ln3$
It is independent of initial concentration.
So (B) is wrong option.
As rate constant is a constant quantity and independent of initial concentration.
So graph (D) is correct.