$\mathrm{K}_{\mathrm{c}}$ for the following reaction is 99.0 $\mathrm{A}_2(\mathrm{~g})…
$\mathrm{K}_{\mathrm{c}}$ for the following reaction is 99.0
$\mathrm{A}_2(\mathrm{~g}) \stackrel{\mathrm{T}(\mathrm{~K})}{\rightleftharpoons} \mathrm{B}_2(\mathrm{~g})$
In a one litre flask, 2 moles of $\mathrm{A}_2$ was heated to $\mathrm{T}(\mathrm{K})$ and the above equilibrium is reached. The concentrations at equilibrium of $\mathrm{A}_2$ and $\mathrm{B}_2$ are $\mathrm{C}_1\left(\mathrm{~A}_2\right)$ and $\mathrm{C}_2\left(\mathrm{~B}_2\right)$ respectively. Now, one mole of $\mathrm{A}_2$ was added to flask and heated to $\mathrm{T}(\mathrm{K})$ to establish the equilibrium again. The concentrations of $\mathrm{A}_2$ and $\mathrm{B}_2$ are $\mathrm{C}_3\left(\mathrm{~A}_2\right)$ and $\mathrm{C}_4\left(\mathrm{~B}_2\right)$ respectively. What is the value of $\mathrm{C}_3\left(\mathrm{~A}_2\right)$ in $\mathrm{mol}^{-1}$ ?
1.98
0.01
0.03
2.97
Solution
Given, $\mathrm{K}_{\mathrm{c}}=99.0$
On adding one mole of $\mathrm{A}_2$ Hence, $\mathrm{C}_3\left(\mathrm{~A}_2\right)=0.03$ moles.
