For a first order reaction A( g ) \to 2B( g ) + C( g ) 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_1/3 is the time required for the partial pressure of A to reach 1/3^rd of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases)

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Advanced 2018 · Shift 2 · Q11 with Solution

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Correct answer: A, D

For $A(g) \rightarrow 2B(g)+C(g)$ at constant volume and temperature, let initial moles of $A$ be $a$ .

If $x$ moles of $A$ react at time $t$ , then:

So total moles at time $t$ are $n_t=(a-x)+2x+x=a+2x$

Initially, total moles are just $a$ .

Since $P \propto n$ at constant $V,T$ , $\frac{P_1}{P_0}=\frac{a+2x}{a}$

Hence, $P_1=P_0\left(1+\frac{2x}{a}\right)$

Partial pressure of $A$ at time $t$ is $p_A=\frac{n_A RT}{V}=\frac{(a-x)RT}{V}$

Initially, $P_0=\frac{aRT}{V}$

Therefore, $p_A=P_0\left(1-\frac{x}{a}\right)$

From the total-pressure relation, $\frac{P_1}{P_0}=1+2\frac{x}{a}$ Rightarrow $\frac{x}{a}=\frac{P_1/P_0-1}{2}=\frac{P_1-P_0}{2P_0}$

Substitute into $p_A$ :

=\frac{3P_0-P_1}{2}$$ So one correct relation is $$\boxed{p_A=\frac{3P_0-P_1}{2}}$$ 3. **Use first-order kinetics** For a first-order reaction, $$[A]_t=[A]_0 e^{-kt}$$ Since partial pressure of a gas at constant $T,V$ is proportional to concentration, $$p_A=p_{A0}e^{-kt}=P_0 e^{-kt}$$ Thus, $$e^{-kt}=\frac{p_A}{P_0}$$ Using $p_A=\dfrac{3P_0-P_1}{2}$, $$e^{-kt}=\frac{3P_0-P_1}{2P_0}$$ Hence, $$\boxed{k=\frac{1}{t}\ln\left(\frac{2P_0}{3P_0-P_1}\right)}$$ 4. **Find $t_{1/3}$** Given $t_{1/3}$ is the time when partial pressure of $A$ becomes one-third of initial: $$p_A=\frac{P_0}{3}$$ For first order decay, $$\frac{p_A}{P_0}=e^{-kt_{1/3}}=\frac13$$ So, $$kt_{1/3}=\ln 3$$ $$\boxed{t_{1/3}=\frac{\ln 3}{k}}$$ 5. **Useful pressure result at $t=t_{1/3}$** When $p_A=P_0/3$, then $$1-\frac{x}{a}=\frac13$$ Rightarrow $$\frac{x}{a}=\frac23$$ Therefore, $$P_1=P_0\left(1+2\cdot\frac23\right)=P_0\left(\frac73\right)$$ So at $t_{1/3}$, $$\boxed{P_1=\frac{7P_0}{3}}$$ 6. **Conclusion** The correct options are the ones corresponding to: - $$p_A=\frac{3P_0-P_1}{2}$$ - $$t_{1/3}=\frac{\ln 3}{k}$$ - equivalently, any option stating $$P_1=\frac{7P_0}{3}$$ at $t=t_{1/3}$ - and/or $$k=\frac{1}{t}\ln\left(\frac{2P_0}{3P_0-P_1}\right)$$ Matching with the stored answer, the correct choices are **A and D**.