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Question: If \[\Delta G^\circ \] for the reaction given below is \[1.7{\text{ }}kJ\], the equilibrium constant...

If ΔG∘\Delta G^\circ for the reaction given below is 1.7 kJ1.7{\text{ }}kJ, the equilibrium constant for the reaction,2HI(g)⇆H2(g)+I2(g)2HI(g)\overset {} \leftrightarrows {H_2}(g) + {I_2}(g). at 25 ∘C25{\text{ }}^\circ C, is:
A.0.5A.0.5
B.2.0B.2.0
C.3.9C.3.9
D.24.0D.24.0

Explanation

Solution

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 ∘C25{\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 KK is the ratio of product concentration and reactant concentration which can also be expressed in terms of free energy i.e.ΔG\Delta G.
The mathematical equation relating Gibbs free energy and equilibrium constant is:
ΔG=ΔG∘+RTln⁡K\Delta G = \Delta G^\circ + RT\ln K
where ΔG\Delta G is the difference of free energy change between products and reactants, ΔG∘\Delta G^\circ is the Gibbs free energy change of system under 1  atm1\;atmpressure and 298 K298{\text{ }}K.
For a reaction at equilibrium the difference in energy change of reactant and product is zero. ThusΔG=0\Delta G = 0.
For the above reaction, given ΔG∘=1.7kJ=1700 J\Delta G^\circ = 1.7kJ = 1700{\text{ }}J.
R = 8.314 Jmol−1K−1, T = 25 ∘C = 298 KR{\text{ }} = {\text{ }}8.314{\text{ }}Jmo{l^{ - 1}}{K^{ - 1}},{\text{ }}T{\text{ }} = {\text{ }}25{\text{ }}^\circ C{\text{ }} = {\text{ }}298{\text{ }}K.
As ΔG=ΔG∘+RTln⁡K\Delta G = \Delta G^\circ + RT\ln K
0=ΔG∘+RTln⁡K0 = \Delta G^\circ + RT\ln K
ΔG∘=−RTln⁡K\Delta G^\circ = - RT\ln K
Now we can write the above equation as
ΔG∘=−2.303RTlog⁡K\Delta G^\circ = - 2.303RT\log K
And hence on substituting the values ,we have
1700=−2.303∗8.314∗298∗log⁡K1700 = - 2.303*8.314*298*\log K
log⁡K=−17002.303∗8.314∗298\log K = - \dfrac{{1700}}{{2.303*8.314*298}}
And hence on doing further simplification we have
log⁡K=−17005705.85\log K = - \dfrac{{1700}}{{5705.85}}
log⁡K=−0.297\log K = - 0.297
K=10−0.297=0.5K = {10^{ - 0.297}} = 0.5

Hence option A is correct. The equilibrium constant for the reaction at 25 ∘C25{\text{ }}^\circ C in 0.50.5.

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