At $\mathrm{T}(\mathrm{K})$, the equilibrium constant for the reaction $\mathrm{a} \mathrm{A}(\mathrm{g})…

At $\mathrm{T}(\mathrm{K})$, the equilibrium constant for the reaction $\mathrm{a} \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{b} \mathrm{B}(\mathrm{g})$ is $\mathrm{K}_{\mathrm{c}}$. If the reaction takes place in the following form $2 \mathrm{aA}(\mathrm{g}) \rightleftharpoons 2 \mathrm{bB}(\mathrm{g})$ its equilibrium constant is $\mathrm{K}_{\mathrm{c}}^{\prime}$. The correct relationship between $\mathrm{K}_{\mathrm{c}}$ and $\mathrm{K}_{\mathrm{c}}$ is
  1. $\mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^2$
  2. $\mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^{\frac{1}{2}}$
  3. $\mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^{-1}$
  4. $\mathrm{K}_{\mathrm{c}}^{\prime}=\mathrm{K}_{\mathrm{c}}$

Solution

$\begin{aligned} & \text {} \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{B}]^{\mathrm{b}}}{[\mathrm{A}]^{\mathrm{a}}} \text { and } \mathrm{K}_{\mathrm{c}}^{\prime}=\frac{[\mathrm{B}]^{2 \mathrm{~b}}}{[\mathrm{~A}]^{2 \mathrm{a}}} \\ & \Rightarrow \mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^2\end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 2)

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