In a reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,the initial concentrations of $A$ and $B$ are $a$ and $b$ in moles per litre,and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively. Express $x$ in terms of $k_1, k_2, a,$ and $b$.

  • 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
    $\frac{k_1 a + k_2 b}{k_1 + k_2}$

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