Calculate the equilibrium constant for the re | vedclass

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(B) For $2H_2O_{(g)} \rightleftharpoons 2H_{2(g)} + O_{2(g)}; K_1 = \frac{[H_2]^2[O_2]}{[H_2O]^2} \dots (1)$For $2CO_{2(g)} \rightleftharpoons 2CO_{(g

Vedclass · Accepted Answer

$2.58$

Vedclass · Answer

(B) For $2H_2O_{(g)} ightleftharpoons 2H_{2(g)} + O_{2(g)}; K_1 = \frac{[H_2]^2[O_2]}{[H_2O]^2} \dots (1)$For $2CO_{2(g)} ightleftharpoons 2CO_{(g)} + O_{2(g)}; K_2 = \frac{[CO]^2[O_2]}{[CO_2]^2} \dots (2)$For the target reaction $H_{2(g)} + CO_{2(g)} ightleftharpoons H_2O_{(g)} + CO_{(g)}; K = \frac{[H_2O][CO]}{[CO_2][H_2]} \dots (3)$By subtracting equation $(1)$ from equation $(2)$ and dividing by $2$ (or manipulating the expressions):$\frac{K_2}{K_1} = \frac{[CO]^2[O_2]}{[CO_2]^2} 	imes \frac{[H_2O]^2}{[H_2]^2[O_2]} = \frac{[CO]^2[H_2O]^2}{[CO_2]^2[H_2]^2} = K^2$Therefore,$K = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{1.4 	imes 10^{-12}}{2.1 	imes 10^{-13}}} = \sqrt{\frac{14}{2.1}} = \sqrt{6.666...} \approx 2.58$

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