Question

For the reaction \[2NOCl(g) \rightleftharpoons 2NO(g) + C{l_2}(g)\] calculate the standard equilibrium constant at 298K. given that the values of \[\Delta {H^o}\] and \[\Delta {S^o}\] of the reaction at 298K are \[77.2kJmo{l^{ - 1}}\] and \[122JKmo{l^{ - 1}}\] .

Answer 1

Hint: There are certain things we need to keep in mind to solve this question. These are: \[\Delta {G^o}\] is used to calculate the equilibrium constant for the reaction \[K_c^o\] by using the formula \[\Delta {G^o} = - 2.303RT\log K_c^o\] where, \[\Delta {G^o}\] is the free energy change of the reaction which further affects the enthalpy and the entropy of the reaction.
Formulas used:
  \[\Delta {G^o} = - 2.303RT\log K_c^o\]

Complete step by step answer:
Gibbs free energy change can be defined as change in free energy of a system as it goes from some initial state such as reactants to some other final state or products. The value thus obtained for this thermodynamic variable tells us the maximum energy released which can be used or absorbed in going from initial to final state.
 \[2NOCl(g) \rightleftharpoons 2NO(g) + C{l_2}(g)\]
We can write Gibbs free energy change for this reaction in terms of enthalpy and entropy change.
 \[\Delta {G^o} = \Delta {H^o} - T\Delta {S^o}\]
We are provided with the values of enthalpy change as \[77.2kJmo{l^{ - 1}}\] , entropy change as \[122JKmo{l^{ - 1}}\] and temperature as 298K. Let us substitute these values in the above formula and get the value of \[\Delta {G^o}\] .
 \[
   = 77200 - 298 \times 122 \\
   = 40844Jmo{l^{ - 1}} \\
 \]
Now, we know the value of \[\Delta {G^o}\] , so using its other formula we can calculate the value of equilibrium constant at standard conditions. The formula is \[\Delta {G^o} = - 2.303RT\log K_c^o\]
Substituting the values, R=8.314, T=298K, \[\Delta {G^o} = 40844Jmo{l^{ - 1}}\] , we get
 \[
  \log K_c^o = - \dfrac{{\Delta {G^o}}}{{2.303 \times 8.314 \times 298}} \\
   = - \dfrac{{40844}}{{2.303 \times 8.314 \times 298}} \\
   = - 7.158 \\
 \]
Taking antilog of this value, we obtain
 \[K_c^o = 6.95 \times {10^{ - 8}}\]
Hence, For the reaction \[2NOCl(g) \rightleftharpoons 2NO(g) + C{l_2}(g)\] , the standard equilibrium constant at 298K is \[6.95 \times {10^{ - 8}}\] given that the values of \[\Delta {H^o}\] and \[\Delta {S^o}\] of the reaction at 298K are \[77.2kJmo{l^{ - 1}}\] and \[122JKmo{l^{ - 1}}\] .

Note:
For a reaction to proceed spontaneously the value of Gibbs free energy for that reaction must be negative. Entropy of a system always increases for a spontaneous reaction and enthalpy change depends upon whether the reaction is exothermic or endothermic i.e. negative value if exothermic and positive value if endothermic.