Chemistry XII Content





Criteria for spontaneity in terms of H, S and G

ΔG,  ΔH and ΔS is related by Gibb’s – He/mhottz equation as,

            ΔG = ΔH  – ΔS 

The conditions under which different process are spontaneous is discussed as follows.

a. Exothermic reaction with increase in entropy

            for exothermic reaction,   ΔH = -ve

            for increase in entropy,  ΔS = +ve

            For such process, the sign of   G is always negative. So such process are spontaneous at all temperature.

b. Exothermic reaction with decrease in entropy

            for exothermic reaction,   ΔH = -ve

            for decrease in entropy,  ΔS = -ve

            For such process, the value of   S can be +ve  or -ve. So the process is spontaneous at low temperature when  H > TΔS.

 

c. Endothermic reaction with increase in entropy

            for endothermic r× n,  Δ H = +ve

            for increase in entropy,  ΔS = +ve

            For such process, the value of   G can be +ve  or -ve. The process is spontaneous at high temperature when,  TΔS > H.

d. Endothermic reaction with decrease in entropy

            for endothermic r× n,   ΔH = +ve

            for increase in entropy, ΔS = -ve

            For such process, the value of   G is always positive. So, this type of process is non-spontaneous at all temperature. 

ΔH ΔS ΔG Remarks
-ve +ve -ve Spontaneous at all temperature
-ve -ve +ve or -ve Spontaneous at low temperature, when  ΔH > TΔ S
+ve +ve +ve or -ve Spontaneous at high temperature, when  TΔ S >Δ H
+ve -ve +ve Non Spontaneous at all temperature

Numerical
For a process the change in enthalpy and entropy is 40.82 KJ mole-1 and 109.4 JK-1 mole-1 respectively will the process be spontaneous at 500c.
Soln Given,
Δ S = 109.4 JK-1mol-1
ΔH = 40.82 KJ mol-1 × 1000J
= 40820 Jmol-1
ΔG = ΔH  – T Δ S
= 40820 – 323 × 109.4

For a chemical reaction,
2 Ag2O (s)                4 Ag (s) + O2 (g).         H = 61.2 KJmol-1
Calculate the temperature above which the reaction is spontaneous.
Here,
ΔH°  = 61.2 × 1000 Jmol-1 = 61200 Jmol-1
ΔS°  = 132 JK-1mol-1
TΔ S >Δ H°
T > 61200/132
T > 463.63K

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