MCQ
For a reaction; $A + B \rightleftharpoons C + D$ , the initial concentration of $A$ and $B$ are equal but the equilibrium concentration of $C$ is twice that of equilibrium concentration of $A$. The $K_c$ is
- ✓$4$
- B$9$
- C$0.25$
- D$1/9$
$x=2(a-x)$
$\therefore \mathrm{x}=\frac{2 \mathrm{a}}{3}$
$\therefore[\mathrm{A}]=\frac{\mathrm{a}}{3}=[\mathrm{B}]$ and $[\mathrm{C}]=[\mathrm{D}]=\frac{2 \mathrm{a}}{3}$
$\therefore K_{c}=\frac{\frac{2 a}{3} \times \frac{2 a}{3}}{\frac{a}{3} \times \frac{a}{3}}=4$
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