A) 4.20
B) 0.42
C) 0.21
D) 0.042
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} 0.2\,mol \\ (0.2-x) \end{smallmatrix}}{\mathop{{{H}_{2}}(g)}}\,+\underset{\text{1}\,\text{mol}}{\mathop{S(s)}}\,\underset{\begin{smallmatrix} 0\,mol \\ x\,mol \end{smallmatrix}}{\mathop{{{H}_{2}}S(g)}}\,\,\underset{\begin{smallmatrix} initially \\ at\,equilibrium \end{smallmatrix}}{\mathop{K=6.8\times {{10}^{-2}}}}\,\] \[K=\frac{[{{H}_{2}}S]}{[{{H}_{2}}]}=\frac{x}{0.2-x}=6.8\times {{10}^{-2}}\] \[\therefore \] \[x=1.273\times {{10}^{-2}}\] \[{{p}_{{{H}_{2}}s}}=\frac{n}{V}\times R\times T\] \[=\frac{1.273\times {{10}^{-2}}\times 0.082\times 363}{1}\] \[=0.38\approx 0.42\]
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