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, ΔSsurr. 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 ΔStotal > 0 according to the second law of thermodynamics and that ΔStotal = ΔSsurr + ΔSsys
- We also know that we can calculate a value for ΔSsurr 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 Tsys = Tsurr)
- So we can write an expression for ΔStotal 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 ΔStotal
Finally let’s consider the contribution of the enthalpy and entropy terms to ΔG.
ΔG = ΔH – T ΔSsys
In general, a negative value for ΔH (exothermic reaction) makes a favourable contribution, as does a positive value for ΔSsys because of the minus sign in the (-TΔSsys) term.
However, the temperature is often the deciding factor.
- At low temperatures the -TΔSsys 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 -TΔSsys term plays a more significant role and the feasibility of the reaction will depend on ΔSsys. Reactions in which there is a decrease in entropy will become feasible on cooling.
Practice questions
- (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 (ΔsolH⦵ = +17.2 kJ mol-1). Determine by calculation whether potassium chloride is soluble at 25°C.
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2. Calculate ΔG⦵ for the reaction between ammonia and hydrogen chloride at 25°C
NH3(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 |
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3. (a) Determine by calculation whether the following reaction is feasible at 1000K
2NO(g) + O2(g) ⇾ N2O4(g) ΔrH⦵ = -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.
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4. (a) Determine by calculation whether the following reaction is feasible at 1000K
H2O(l) ⇾ H2O(g) ΔG⦵ = +8.6 kJ mol-1 at 298K
ΔS⦵ = +119 JK-1mol-1
(b) Explain under what conditions, if any, endothermic reactions are spontaneous.
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5. A reaction has a positive value for the enthalpy change, ΔrH⦵, and a negative value for the entropy change, ΔS⦵.
(a) Predict whether this reaction will be feasible at 25°C
(b) Explain why, with reference to the equation for calculating ΔG⦵, changing the temperature will not affect your answer to part (a).
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6. Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide.
CaCO3(s) ⇾ CaO(s) + CO2(g) ΔrH⦵ = +177.9 kJ mol-1
ΔS⦵= +160.4 JK-1mol-1
(b) Determine the temperature at which this reaction becomes feasible.
(a) Calculate ΔG⦵ for this reaction at 25°C.
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7. If you investigate the relationship between Gibbs free energy and temperature, you will get the following results:
Explain the significance of the gradient, the point at which the line intercepts the x- axis and the point at which the line intercepts the y-axis.
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)
4. (a) ΔG = ΔH – TΔS
(we can calculate a value for ΔH because we know the value of ΔG at 298K)
ΔH = ΔG + TΔS
= 8.6 + (298 x (119/1000)) = +44.1 kJ mol-1
ΔG (at 1000K) = +44.1 – (1000 x 0.119) = -74.9 kJ mol-1
The reaction is feasible at 1000K because ΔG is negative.
(b) ΔG = ΔH – TΔS
ΔH is positive in endothermic reactions, so these reactions may become feasible / spontaneous at high temperatures (as long as the reaction has an increase in entropy) as the -TΔS term is more significant and ΔG relies more upon the entropy change than the enthalpy change.
5. (a) ΔG = ΔH – TΔS
If ΔH is positive / endothermic and ΔS is negative then ΔG will be positive and the reaction his not feasible.
(b) ΔH is positive; -TΔS is also positive regardless of the temperature so the reaction will never be spontaneous or feasible.
6. (a) ΔG = ΔH – TΔS
= +177.9 – (298 x (160.4/1000))
= +130.1 kJ mol-1
EXAM TIP: you won’t get the marks if you forget to include the sign in your final answer)
(b) A reaction becomes feasible when ΔG = 0
0 = ΔH – TΔS
T = 177.9 / 0.1604 = 1109K
EXAM TIP: you can convert 160.4J to kJ by dividing by 1000 or multiplying by 10-3.
7. The relationship between ΔG and T is given by ΔG = TΔS.
Graphically this can be represented in the form y = mx + c, where y is ΔG, m (the gradient) is -ΔS for the reaction, x is T and ΔH is the intercept of the y-axis.
The point of intercept on the x-axis, where ΔG = 0, gives us the temperature at which this reaction becomes feasible.