Chemical EquilibriumHard

Question

For a reversible reaction $A\overset{\quad K_{1}K_{2}\quad}{\leftleftarrows}B$, the initial molar concentration of A and B are a M and b M, respectively. If x M of A is reacted till the achievement of equilibrium, then x is

Options

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.$\frac{K_{1}a + K_{2}b}{K_{1} + K_{2}}$

Solution

A $\rightleftharpoons$ B

Initial a M b M

Equilibrium (a – x) M (b + x) M

Now, $K_{eq} = \frac{K_{1}}{K_{2}} = \frac{b + x}{a - x}$

$\therefore x = \frac{K_{1}a - K_{2}b}{K_{1} + K_{2}}$

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