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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