A) At equilibrium, the concentrations of \[C{{O}_{2}}(g)\] and \[{{H}_{2}}O(l)\] are not equal
B) The equilibrium constant for the reaction is given by \[{{K}_{p}}=\frac{[C{{O}_{2}}]}{[C{{H}_{4}}][{{O}_{2}}]}\]
C) Addition of \[C{{H}_{4}}(g)\] or \[{{O}_{2}}(g)\] at equilibrium will cause a shift to the right
D) The reaction is exothermic
Correct Answer: B
Solution :
For the reaction, \[C{{H}_{4}}(g)+2{{O}_{2}}(g)\rightleftharpoons C{{O}_{2}}(g)+2{{H}_{2}}O\,(\ell )\] \[\Delta {{H}_{r}}=-170.8\,kJ\,mo{{l}^{-1}}\] This equilibrium is an example of heterogeneous chemical equilibrium. Hence, for it \[{{K}_{c}}=\frac{[C{{O}_{2}}]}{[C{{H}_{4}}]\,{{[{{O}_{2}}]}^{2}}}....(i)\] (Equilibrium constant on the basis of conc.) and \[{{K}_{p}}=\frac{{{P}_{C{{O}_{2}}}}}{{{P}_{C{{H}_{4}}}}+{{P}_{O_{2}^{2}}}}...(ii)\] (Equilibrium constant according to partial pressure) Thus in it concentration of \[C{{O}_{2}}(g)\] and \[{{H}_{2}}O(l)\] are not equal at equilibrium The equilibrium constant \[({{K}_{p}})=\frac{[C{{O}_{2}}]}{[C{{H}_{4}}]\,\,[{{O}_{2}}]}\] not correct expression In addition of \[C{{H}_{4}}(g)\] or \[{{O}_{2}}(g)\] at equilibrium \[{{K}_{c}}\] will be decreased according to expression (i) but \[{{K}_{c}}\] remains constant at constant temperature for a reaction, so for maintaining the constant value of \[{{K}_{c}}\], the concentration of \[C{{O}_{2}}\] will increased in same order. Hence, on addition of \[C{{H}_{4}}\] or \[{{O}_{2}}\] equilibrium will cause to the right. This reaction is an example of exothermic reaction.
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