For the reaction SnO_{2(s)} + 2H_{2(g)} right | vedclass

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(A) The reaction is $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$.Since $SnO_2$ and $Sn$ are in solid and liquid states respectiv

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$\frac{9}{4}$

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(A) The reaction is $SnO_{2(s)} + 2H_{2(g)} ightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$.Since $SnO_2$ and $Sn$ are in solid and liquid states respectively,they do not appear in the $K_p$ expression.Let the total moles of gaseous mixture at equilibrium be $n_T = 1$.Given that the mixture contains $40\%$ $H_2$ by volume,the mole fraction of $H_2$ is $X_{H_2} = 0.4$.Therefore,the mole fraction of $H_2O$ is $X_{H_2O} = 1 - 0.4 = 0.6$.$K_p = \frac{(P_{H_2O})^2}{(P_{H_2})^2} = \frac{(X_{H_2O} 	imes P_T)^2}{(X_{H_2} 	imes P_T)^2} = \frac{(0.6)^2}{(0.4)^2}$.$K_p = \frac{0.36}{0.16} = \frac{36}{16} = \frac{9}{4}$.

For the reaction $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$,at equilibrium,the mixture of steam and hydrogen contains $40\%$ $H_2$ by volume. Find $K_p$ for the reaction.

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