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Consider a reversible reaction, $A \underset{k_{2}}{\overset{k_{1}}{⇌}} B$ which is hi first order in both directions, which statement about this reaction is correct?
(P) The reaction will never reach equilibrium
(Q) If $\frac{k_{1}}{k_{2}} = 2,$ then at $t = \infty; [A] = \frac{2 a}{3}$ dinitially only A was present and at t = 0, [A] = a
(R) At any instant, if $[A] = \frac{a}{5}$ then $[B] = \frac{2 a}{5}$ if $\frac{k_{1}}{k_{2}} = 2$

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Detailed Solution

(II)

$\underset{a - x_{eq}}{A} ⇌ \underset{x_{eq}}{B} \begin{aligned} k_{1} (a - x_{eq}) = k_{2} (x_{eq}) \\ 2 a - 2 x_{eq} = x_{eq} \\ x_{eq} = \frac{2 a}{3} \\ [A] = a - \frac{2 a}{3} = \frac{a}{3} \end{aligned}$
(III)

$[A] = a - x_{e q} = \frac{a}{5} \begin{array}{l} \Rightarrow x = \frac{4 a}{5} \\ \Rightarrow [B] = \frac{4 a}{5} \end{array}$

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