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

$\mathrm{K}_{\mathrm{c}}$ for the reaction $\mathrm{A}_2(\mathrm{~g}) \stackrel{\mathrm{T}(\mathrm{~K})}{\rightleftharpoons} \mathrm{B}_2(\mathrm{~g})$ is 39.0. In a closed one litre flask, one mole of $\mathrm{A}_2(\mathrm{~g})$ was heated to $\mathrm{T}(\mathrm{K})$. What are the concentrations of $\mathrm{A}_2(\mathrm{~g})$ and $\mathrm{B}_2(\mathrm{~g})$ (in $\mathrm{mol}^{-1}$ ) respectively at equilibrium?
  1. $0.025,0.975$
  2. $0.975,0.025$
  3. $0.05,0.95$
  4. $0.02,0.98$

Solution


$\begin{aligned} & \text { Initial } \\ & \begin{array}{l}\text { Equilibrium } \\ \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{B}_2\right]}{\left[\mathrm{A}_2\right]} \\ 39=\frac{\left[\frac{\mathrm{x}}{1}\right]}{\left[\frac{1-\mathrm{x}}{1}\right]} \\ \text { or, } 39-39 \mathrm{x}=\mathrm{x} \\ \Rightarrow 40 \mathrm{x}=39 \\ \Rightarrow \mathrm{x}=\frac{39}{40} \\ \mathrm{x}=0.975 \\ \text { Concentration of } \mathrm{B}_2=0.975 \\ \text { Concentration of } \mathrm{A}_2=1-0.975 \\ \end{array} \text { - } 1-0.025\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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