## 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 50^{0}c.

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 Ag_{2}O (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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