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 the solid/liquid state,their acti

Vedclass · Accepted Answer

$1.49$

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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 the solid/liquid state,their activities are taken as $1$.Thus,the equilibrium constant expression is $K_p = \frac{(p_{H_2O})^2}{(p_{H_2})^2}$.Given that the mixture is $45 \% \ H_2$ by volume,the mole fraction of $H_2$ is $0.45$ and the mole fraction of $H_2O$ is $0.55$.Since partial pressure is proportional to mole fraction,$p_{H_2} \propto 0.45$ and $p_{H_2O} \propto 0.55$.Substituting these values,$K_p = \frac{(0.55)^2}{(0.45)^2} = (\frac{55}{45})^2 = (1.222)^2 \approx 1.49$.

For the reaction $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$,calculate $K_p$ at $900 \ K$,where the equilibrium steam-hydrogen mixture was $45 \%$ $H_2$ by volume.

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