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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    At 1400 K, \[{{K}_{c}}=2.5\times {{10}^{-3}}\]for the reaction : \[C{{H}_{4(g)}}+2{{H}_{2}}{{S}_{(g)}}C{{S}_{2(g)}}+4{{H}_{2(g)}}\] A 10 L reaction vessel at 1400 K contains 2.0 mol of \[C{{H}_{4}},3.0\] mol of \[C{{S}_{2}},3.0\]mol of\[{{H}_{2}}\] and 4.0 mol of \[{{H}_{2}}S\]. In which direction does the reaction proceed to reach equilibrium?

    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.


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