At $\mathrm{T}(\mathrm{K}), \mathrm{K}_{\mathrm{c}}$ for the reaction. $\mathrm{A}_2(\mathrm{~g})…

At $\mathrm{T}(\mathrm{K}), \mathrm{K}_{\mathrm{c}}$ for the reaction. $\mathrm{A}_2(\mathrm{~g}) \rightleftharpoons \mathrm{B}_2(\mathrm{~g})$ is 99.0 . Two moles of $A_2(\mathrm{~g})$ was heated to $T(K)$ in a $1 L$ closed flask to reach the above equilibrium. What are the concentrations (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of $\mathrm{A}_2(\mathrm{~g})$ and $\mathrm{B}_2(\mathrm{~g})$ respectively at equilibrium?
  1. $1.86,0.0187$
  2. $1.98,0.02$
  3. $0.0187,1.86$
  4. $0.02,1.98$

Solution

given $K_c=99.0$ $\mathrm{n}=2$ moles of $\mathrm{A}_2$ $\mathrm{v}=1 \mathrm{~L}$

$\begin{aligned} & \mathrm{K}_{\mathrm{c}}=\frac{[\text { Product }]}{[\text { Reactant }]} \\ & \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{x} / \mathrm{l}]}{\left[\frac{2-\mathrm{x}}{1}\right]}\end{aligned}$ $\begin{aligned} & 99=\frac{x}{2-x} \\ & 198-99 x=x \\ & \therefore x=\frac{198}{100}=1.98\end{aligned}$ concentration of $\mathrm{B}_2$ at equilibrium is $1.98 \mathrm{~mol} \mathrm{~L}^{-1}$ concentration of $\mathrm{A}_2$ at equilibrium is $(2-198)$ $=0.02 \mathrm{~mol}^{-1}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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