A note on Gibbs free energy

The change in Gibbs free energy, ΔG, is a convenient way to determine whether a reaction is spontaneous or feasible without having to consider the entropy change of the surroundings, ΔS_surr.  This means that ΔG can be calculated from the properties of the system (which is, of course, our reaction).

Let’s figure out how all this is possible …

• We know that for a spontaneous reaction ΔS_total  >  0 according to the second law of thermodynamics and that ΔS_total  =  ΔS_surr  +  ΔS_sys

• We also know that we can calculate a value for ΔS_surr if we know the enthalpy change of the reaction, ΔH (remember that exothermic reactions release energy to the surroundings, increasing their entropy, and vice versa for endothermic reactions) and the temperature (we assume that T_sys  =  T_surr)

• So we can write an expression for ΔS_total as 

• If we multiply both sides of the expression by -T we get

• which, if we compare this to the expression for ΔG, means that ΔG is the exact same thing as -T ΔS_total

Finally let’s consider the contribution of the enthalpy and entropy terms to ΔG.  

ΔG    =    ΔH   –   T ΔS_sys

In general, a negative value for ΔH (exothermic reaction) makes a favourable contribution, as does a positive value for ΔS_sys because of the minus sign in the (-TΔS_sys) term.  

However, the temperature is often the deciding factor.  

• At low temperatures the  -TΔS_sys  term will be small and the feasibility of the reaction will be mainly determined by ΔH.  Endothermic reactions will become feasible on heating.  

• At high temperatures the  -TΔS_sys  term plays a more significant role and the feasibility of the reaction will depend on ΔS_sys.  Reactions in which there is a decrease in entropy will become feasible on cooling. 

Practice questions

1.     (a)  Suggest why dissolving potassium chloride in water is accompanied by a positive change in entropy (ΔS^⦵ = +75.0 JK^-1mol^-1).

(b)  The enthalpy change of solution for dissolving potassium chloride in water is endothermic (Δ_solH^⦵ = +17.2 kJ mol^-1).  Determine by calculation whether potassium chloride is soluble at 25°C.

2. Calculate ΔG^⦵ for the reaction between ammonia and hydrogen chloride at 25°C

NH_3(g)    +    HCl_(g)   ⇾   NH₄Cl_(s)

Substance	ΔfH⦵ / kJ mol-1	S⦵ / JK-1mol-1
NH3(g)	-46.0	192.5
HCl(g)	-92.3	186.7
NH4Cl(s)	-315.5	94.6

3. (a)  Determine by calculation whether the following reaction is feasible at 1000K

2NO_(g)   +  O_2(g)   ⇾    N₂O_4(g)   Δ_rH^⦵  =  -171.6 kJ mol^-1 ΔS^⦵  =  -322 JK^-1 mol^-1

(b)  Explain under what conditions, if any, reactions in which there is a decrease in entropy are feasible.

Answers

1. (a)  On dissolving, one particle becomes two and as there are more ways of arranging two particles than one (as well as more ways of distributing the energy), there is an increase in entropy.

KCl_(s) ⇾  K⁺_(aq)  +  Cl^–_(aq)

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