A) Forward
B) Backward
C) May be forward or backward
D) Reaction is in equilibrium
Correct Answer: B
Solution :
[b]:\[C{{H}_{4(g)}}+2{{H}_{2}}{{S}_{(g)}}C{{S}_{2(g)}}+4{{H}_{2(g)}}\] \[K=\frac{[C{{S}_{2}}]{{[{{H}_{2}}]}^{4}}}{[C{{H}_{4}}]{{[{{H}_{2}}S]}^{2}}}\] ?(i) \[[C{{S}_{2}}]=\left( \frac{3}{10} \right);[{{H}_{2}}]=\left( \frac{3}{10} \right);\] \[[C{{H}_{4}}]=\left( \frac{2}{10} \right);[{{H}_{2}}S]=\left( \frac{4}{10} \right)\] Substituting all the concentration values in equation (i), we get \[K=\frac{\left( \frac{3}{10} \right){{\left( \frac{3}{10} \right)}^{4}}}{\left( \frac{2}{10} \right){{\left( \frac{4}{10} \right)}^{2}}}=\frac{{{3}^{5}}}{{{10}^{5}}}\times \frac{{{10}^{3}}}{2\times {{4}^{2}}}=\frac{243}{3200}\Rightarrow 0.076\]Therefore, \[K>2.5\times {{10}^{-3}}({{K}_{c}})\]. Hence, the reaction will proceed in backward direction.You need to login to perform this action.
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