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If the equilibrium constant for the reaction $2 A B \rightleftharpoons A_2+B_2$ is 49 , what is the equilibrium constant for $A B \rightleftharpoons \frac{1}{2} A_2+\frac{1}{2} B_2 ?$
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The correct answer is:
7
$\begin{array}{r}2 A B \rightleftharpoons A_2+B_2 \\ K_c=\frac{\left[A_2\right]\left[B_2\right]}{[A B]^2}\end{array}$
again, $\quad \begin{aligned} A B & \rightleftharpoons \frac{1}{2} A_2+\frac{1}{2} B_2 \\ K_c{ }^{\prime} & =\frac{\left[A_2\right]^{1 / 2}\left[B_2\right]^{1 / 2}}{[A B]} \\ K_c & =\left(K_c{ }^{\prime}\right)^2\end{aligned}$
or $\quad K_c{ }^{\prime}=\sqrt{K_c}=\sqrt{49}=7$
again, $\quad \begin{aligned} A B & \rightleftharpoons \frac{1}{2} A_2+\frac{1}{2} B_2 \\ K_c{ }^{\prime} & =\frac{\left[A_2\right]^{1 / 2}\left[B_2\right]^{1 / 2}}{[A B]} \\ K_c & =\left(K_c{ }^{\prime}\right)^2\end{aligned}$
or $\quad K_c{ }^{\prime}=\sqrt{K_c}=\sqrt{49}=7$
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