Question

If \[\Delta G^\circ \] for the reaction given below is \[1.7{\text{ }}kJ\], the equilibrium constant for the reaction,\[2HI(g)\overset {} \leftrightarrows {H_2}(g) + {I_2}(g)\]. at \[25{\text{ }}^\circ C\], is:
\[A.0.5\]
\[B.2.0\]
\[C.3.9\]
\[D.24.0\]

Answer 1

Hint: The relation between Gibbs free energy and equilibrium constant of a reaction follows a mathematical relationship. The reaction is at equilibrium at normal condition i.e \[25{\text{ }}^\circ C\] means the rate of formation of product is equal to the rate of formation of reactants.

Complete step by step answer:
In thermodynamics, the energy change in a reaction is related in terms of free energy change of the reacting substances. The equilibrium constant \[K\] is the ratio of product concentration and reactant concentration which can also be expressed in terms of free energy i.e.\[\Delta G\].
The mathematical equation relating Gibbs free energy and equilibrium constant is:
\[\Delta G = \Delta G^\circ + RT\ln K\]
where \[\Delta G\] is the difference of free energy change between products and reactants, \[\Delta G^\circ \]is the Gibbs free energy change of system under \[1\;atm\]pressure and \[298{\text{ }}K\].
For a reaction at equilibrium the difference in energy change of reactant and product is zero. Thus\[\Delta G = 0\].
For the above reaction, given \[\Delta G^\circ = 1.7kJ = 1700{\text{ }}J\].
\[R{\text{ }} = {\text{ }}8.314{\text{ }}Jmo{l^{ - 1}}{K^{ - 1}},{\text{ }}T{\text{ }} = {\text{ }}25{\text{ }}^\circ C{\text{ }} = {\text{ }}298{\text{ }}K\].
As \[\Delta G = \Delta G^\circ + RT\ln K\]
\[0 = \Delta G^\circ + RT\ln K\]
\[\Delta G^\circ = - RT\ln K\]
Now we can write the above equation as
\[\Delta G^\circ = - 2.303RT\log K\]
And hence on substituting the values ,we have
\[1700 = - 2.303*8.314*298*\log K\]
\[\log K = - \dfrac{{1700}}{{2.303*8.314*298}}\]
And hence on doing further simplification we have
\[\log K = - \dfrac{{1700}}{{5705.85}}\]
\[\log K = - 0.297\]
\[K = {10^{ - 0.297}} = 0.5\]

Hence option A is correct. The equilibrium constant for the reaction at \[25{\text{ }}^\circ C\] in \[0.5\].


Note:
Actually the value of \[\Delta G^\circ \] gives information regarding the spontaneity of a reaction. If \[\Delta G^\circ > 1,K < 1\] and the reaction favours reactants at equilibrium. If \[\Delta G^\circ < 1,K > 1\] and the reaction favours products at equilibrium. If \[\Delta G^\circ = 0,K = 1\] which indicates neither product nor reactant are favored at equilibrium.