Equilibrium constants for the following react | vedclass

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(D) The given reactions are:$(1) \ 2 \ H_2O_{(g)} ightleftharpoons 2 \ H_{2(g)} + O_{2(g)}, \ K_1 = 6.4 	imes 10^{-8}$$(2) \ 2 \ CO_{2(g)} ightle

Vedclass · Accepted Answer

$5$

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(D) The given reactions are:$(1) \ 2 \ H_2O_{(g)} ightleftharpoons 2 \ H_{2(g)} + O_{2(g)}, \ K_1 = 6.4 	imes 10^{-8}$$(2) \ 2 \ CO_{2(g)} ightleftharpoons 2 \ CO_{(g)} + O_{2(g)}, \ K_2 = 1.6 	imes 10^{-6}$We need the equilibrium constant $(K)$ for the reaction: $H_{2(g)} + CO_{2(g)} ightleftharpoons CO_{(g)} + H_2O_{(g)}$Divide reaction $(1)$ by $2$ and reverse it: $H_{2(g)} + \frac{1}{2} O_{2(g)} ightleftharpoons H_2O_{(g)}$, $K_3 = \frac{1}{\sqrt{K_1}}$Divide reaction $(2)$ by $2$: $CO_{2(g)} ightleftharpoons CO_{(g)} + \frac{1}{2} O_{2(g)}$, $K_4 = \sqrt{K_2}$Adding these two reactions gives the target reaction, so $K = K_3 	imes K_4 = \frac{\sqrt{K_2}}{\sqrt{K_1}}$$K = \sqrt{\frac{1.6 	imes 10^{-6}}{6.4 	imes 10^{-8}}} = \sqrt{\frac{16 	imes 10^{-7}}{6.4 	imes 10^{-8}}} = \sqrt{\frac{160}{6.4}} = \sqrt{25} = 5$

Similar Questions

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,which of the following relations is correct?

For the reaction $2HI \rightleftharpoons H_2 + I_2$,which of the following is true?

$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$
$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$
Then the equilibrium constant for the following reaction is ....
$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

For which of the following equilibria are $K_P$ and $K_C$ different?

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at equilibrium,the partial pressures of $PCl_5, PCl_3$ and $Cl_2$ are $2 \ atm, 4 \ atm$ and $4 \ atm$ respectively. The value of $K_C$ at $27^{\circ}C$ is (approx):-

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