Question

For a reversible reaction $A\stackrel{k_1}{\underset{k_2}{\rightleftharpoons }}B$ the initial molar concentration of $A$ and $B$ are ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ M respectively. If ${}^{\prime }{x}^{\prime }$ mol/litre of $A$ is reacted till the achievement of equilibrium, then ${}^{\prime }{x}^{\prime }$ is:

($K_1$ and $K_2$ are rate constants)

• A. $\frac{K_{1}a-K_{2}b}{K_1+K_2}$
• B. $\frac{K_{1}a-K_{2}b}{K_1-K_2}$
• C. $\frac{K_{1}a-K_{2}b}{K_1{K}_2}$
• D. $\frac{K_{1}a+K_{2}b}{K_1+K_2}$

Answer 1

The correct option is A $\frac{K_{1}a-K_{2}b}{K_1+K_2}$

$A\underset{K_2}{\stackrel{K_1}{\rightleftharpoons }}B$

Initially $a$ $b$ $R_1=K_1\left[A\right],R_2{K}_2\left[B\right]$

At $Eq{b}^m$ $a+x$ $b+x$ at $Eq{b}^m$ $R_1=R_2$

$K_{eq}=\frac{b+x}{a-x}$ as, $K_{eq}=\frac{K_1}{K_2}=\frac{\left[B\right]}{\left[A\right]}$

$\Rightarrow \frac{K_1}{K_2}=\frac{b+x}{a-x}$

$\Rightarrow K_1(a-x)=K_2(b+x)$

$\Rightarrow a{K}_1-x_{K_1}=b{K}_2+x{K}_2$

$\Rightarrow a{K}_1-b_{K_2}=x(K_1+K_2)$

$\Rightarrow x=\frac{K_{1}a-K_{2}b}{K_1+K_2}$