MCQ
In a reversible reaction $A\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}B,$ the initial concentration of $A$ and  $B$ are $a$ and $b$ and the equilibrium concentration are $(a -x)$ and $(b + x)$ respectively. Express $"x"$ in terms of $k_1, k_2,$ $a$ and $b$
  • $\frac{{{k_1}a - {k_2}b}}{{{k_1}\, + \,{k_2}}}$
  • 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

Correct option: A.
$\frac{{{k_1}a - {k_2}b}}{{{k_1}\, + \,{k_2}}}$
a
$\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}=\frac{[\mathrm{B}]}{[\mathrm{A}]}=\frac{(\mathrm{b}+\mathrm{x})}{\mathrm{a}-\mathrm{x}}$

$\mathrm{K}_{1} \mathrm{a}-\mathrm{K}_{1} \mathrm{x}=\mathrm{K}_{2} \mathrm{b}+\mathrm{K}_{2} \mathrm{x}$

$\mathrm{K}_{1} \mathrm{a}-\mathrm{K}_{2} \mathrm{b}=\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) \mathrm{x}$

$\mathrm{x}=\frac{\mathrm{K}_{1} \mathrm{a}-\mathrm{K}_{2} \mathrm{b}}{\mathrm{K}_{1}+\mathrm{K}_{2}}$

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