For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g) H° = -45.6 kJ and S° = -125.7 J/K The equilibrium constant for this reaction at 262.0 K is

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Question:

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)
H° = -45.6 kJ and S° = -125.7 J/K
The equilibrium constant for this reaction at 262.0 K is

Answers

Answer : The value of equilibrium constant for this reaction at 262.0 K is [tex]3.35\times 10^{2}[/tex]Explanation :As we know that,[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]where,[tex]\Delta G^o[/tex] = standard Gibbs free energy  = ?[tex]\Delta H^o[/tex] = standard enthalpy = -45.6 kJ = -45600 J[tex]\Delta S^o[/tex] = standard entropy = -125.7 J/KT = temperature of reaction = 262.0 KNow put all the given values in the above formula, we get:[tex]\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)[/tex][tex]\Delta G^o=-12666.6J=-12.7kJ[/tex]The relation between the equilibrium constant and standard Gibbs free energy is:[tex]\Delta G^o=-RT\times \ln k[/tex]where,[tex]\Delta G^o[/tex] = standard Gibbs free energy  = -12666.6 JR = gas constant  = 8.314 J/K.molT = temperature  = 262.0 KK = equilibrium constant = ?Now put all the given values in the above formula, we get:[tex]-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k[/tex][tex]k=3.35\times 10^{2}[/tex]Therefore, the value of equilibrium constant for this reaction at 262.0 K is [tex]3.35\times 10^{2}[/tex]

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