One mole of $\mathrm{A}(\mathrm{g})$ is heated to $\mathrm{T}(\mathrm{K})$ till the following equilibrium is…
One mole of $\mathrm{A}(\mathrm{g})$ is heated to $\mathrm{T}(\mathrm{K})$ till the following equilibrium is obtained
$\mathrm{A}(\mathrm{g}) \stackrel{\mathrm{T}(\mathrm{K})}{\rightleftharpoons} \mathrm{B}(\mathrm{g})$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium, 0.5 moles of $\mathrm{A}(\mathrm{g})$ is added and heated. The equilibrium is again established. The value of $\frac{[\mathrm{A}]}{[\mathrm{B}]}$ is
$10^{-1}$
10
$10^{-2}$
100
Solution
$\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{B}]}{[\mathrm{A}]}=10^{-1}$ or $\frac{1}{10}$
Since $\mathrm{K}_{\mathrm{C}}$ remains constant at a given temperature, the value of $\frac{[\mathrm{B}]}{[\mathrm{A}]}$ will still be the same.
$\Rightarrow \frac{[\mathrm{B}]}{[\mathrm{A}]}=\frac{1}{10} \Rightarrow \frac{[\mathrm{A}]}{[\mathrm{B}]}=10$