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

Vedclass · Accepted Answer

$1.49$

Vedclass · Answer

(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 respectively,their activities are taken as $1$.Thus,the expression for the equilibrium constant $K_p$ 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 $1 - 0.45 = 0.55$.Since partial pressure is proportional to mole fraction,$K_p = \frac{(0.55)^2}{(0.45)^2} = \left(\frac{55}{45}\right)^2 = (1.222)^2 \approx 1.49$.

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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