Colourful Solutions > Entropy and spontaneity > Gibbs free energy calculations

The Mad Science Lab

Higher-level only

The calculated value of Gibbs free energy is useful for prediction of reaction spontaneity. There are few complications in the actual calculation procedure once you have the units correctly matched.

Syllabus ref: R1.4.3

Reactivity 1.4.3 - At constant pressure, a change is spontaneous if the change in Gibbs energy, ΔG, is negative. (HL)

  • Interpret the sign of ΔG calculated from thermodynamic data.
  • Determine the temperature at which a reaction becomes spontaneous.

Guidance

  • ΔG takes into account the direct entropy change resulting from the transformation of the chemicals and the indirect entropy change of the surroundings resulting from the transfer of heat energy.

Tools and links

  • Reactivity 3.2 - How can electrochemical data also be used to predict the spontaneity of a reaction?

Gibbs free energy calculations

Gibbs free energy is calculated using the relationship:

ΔG = ΔH - TΔS

Note that the units of enthalpy change are usually quoted in kiloJoules, whereas the units for entropy are given in Joules per Kelvin. This means that you must convert the entropy values to kiloJoules per Kelvin before using the Gibbs free energy relationship.

Example: Methanol can be made from synthesis gas, produced by the reaction:

CH4(g) + H2O(g) → 3H2(g) + CO(g)

For this reaction ΔHo = +210 kJ and ΔSo= 216 JK-1. What is the value of ΔGo for this process at 298K?

ΔSo = 216 JK-1, therefore: ΔSo = 0.216 kJ K-1

ΔG = ΔH - TΔS

∴ ΔG = 210 - (298 x 0.216) = 210 - 64 = 146 kJ

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Temperature and equilibrium

The condition for equilibrium in a process is that Gibbs free energy is zero.

ΔH - TΔS = 0

The temperature at which equilibrium is established may be calculated if the enthalpy and entropy changes for the system are known. This is the case for changes of state. If the enthalpy value for the change is known as are the absolute entropies of the starting and finishing materials, then the equilibrium temperature can be calculated.

Example: Use the following data to find the temperature at which water boils.

  • ΔH(vaporisation) = 44 kJ
  • So water(l) = 71 JK-1
  • So water(g) = 189 JK-1

ΔS = 189 - 71 JK-1 = 118 JK-1 = 0.118 kJ K-1

T = ΔH/ΔS = 44/0.118 = 373 (3 sig figs)

Therefore the boiling point of water = 373K

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Worked examples

Q463-01 For the reaction;

3HC≡CH(g) → C6H6(g)          H = - 597.3 kJ and S = - 0.33 kJ K-1.

This reaction:

  1. is spontaneous at 300 K and becomes non-spontaneous at higher temperatures.
  2. is spontaneous at 300 K and becomes non-spontaneous at lower temperatures.
  3. is non-spontaneous at 300 K and becomes spontaneous at higher temperatures.
  4. is non-spontaneous at 300 K and becomes spontaneous at lower temperatures.
Answer

To calculate spontaneity Gibbs free energy equation must be used.

Go = Ho - T So

Substituting the values into the equation

Go = -597.3 - 300 x (-0.33)

Go = -498.3 kJ

As Go has a negative value then the reaction is spontaneous under these conditions… and as the temperature rises the term (- T So) becomes more positive, i.e. the reaction gets LESS spontaneous. At a high enough temperature it will not be feasible.
Q463-02 For the process:

C6H6(l) → C6H6(s)

ΔHo = -9.83 kJmol-1 and ΔSo = -35.2 J K-1 mol-1. Predict and explain the effect of an increase in temperature of the spontaneity of the process.

Calculate the temperature (in ºC) at which ΔG = 0 for the above process and explain the significance of this temperature.

Answer

Using the Gibbs free energy equation:

ΔG = ΔH - TΔS

When ΔG is negative the process is spontaneous. ΔS is negative making the term - TΔS positive. Hence, high temperatures produce a large positive term in the free energy equation. As ΔH is negative the process will be spontaneous when the term TΔS has no importance, i.e. when T is very small.

At low temperature the process is spontaneous and at high temperature it is non-spontaneous.

When ΔG = 0 the system is at equilibrium.

ΔH = TΔS (not forgetting to change the units of entropy to kJ K-1)

∴ T = ΔH/ΔS = 279 K = 6ºC

The equilibrium for the above process is the temperature at which both liquid and solid co-exist, i.e. the melting point.
Q463-03 Decomposition of solid barium carbonate is given by the following equation:

BaCO3(s) → BaO(s) + CO2(g)

compound
BaCO3(s)
CO2(g)
 BaO(s)
ΔHfo/kJ mol-1
-1219
-394
-558
So /JK-1 mol-1
+112
+214
 +70

Calculate the value of ΔGo in kJ mol-1 at 25ºC
State with a reason whether the reaction is spontaneous at 25ºC
Determine the minimum temperature above which the reaction is spontaneous.

Answer

Firstly we must calculate ΔH and ΔS for the reaction:

Formation enthalpy of reactants

BaCO3(s) = -1219 kJ

Formation enthalpy of products

BaO(s)= --558 kJ
CO2(g) = -394 kJ
Total = -952 kJ
Reaction enthalpy ΔH = ΔHf products - ΔHf reactants = -952 + 1219 = +267 kJ

Entropy of reactants

BaCO3(s) = 112 J K-1

Entropy of products

BaO(s)= 70 J K-1
CO2(g) = 214 J K-1
Total = 284 J K-1

Entropy change ΔS = ΔS products - ΔS reactants = 284 - 112 = +172 J K-1

This is a positive entropy change as the entropy increases from reactants to products

Spontaneity of reaction can be predicted using Gibb's free energy change, ΔG. If ΔG is negative then the reaction is spontaneous.

ΔG = ΔH - TΔS

at 25ºC (298K) ΔG = +267 - 298 (+0.172)

∴ ΔG = +267 - 51 = +216 kJ

The reaction is non-spontaneous at 25ºC

The temperature at which it becomes spontaneous is when ΔG = 0 and ΔH = TΔS

Therefore T = ΔH/ΔS = +267/0.172 = 1552 K

Above 1552 K the reaction becomes spontaneous.
Q463-04 Methanol can be made from synthesis gas, produced by the reaction:

CH4(g) + H2O(g) → 3H2(g) + CO(g)

For this reaction ΔHo = +210kJ and ΔSo = 216 JK-1

Use these values to explain why this reaction is not spontaneous at 298K. Calculate the temperature at which it becomes spontaneous.

Answer

Using the Gibbs free energy equation:

ΔG = ΔH - TΔS

At 298K: ΔG = +210 - 298(+0.216) = 210 - 64.4 = +145.6 kJ

This value is positive meaning that the process is non-spontaneous at this temperature.

For spontaneity the value of ΔG must be negative. The limiting value is ΔG = 0.

ΔG = 0 when ΔH = TΔS

∴ T = ΔH/ΔS = 210/0.216 = 972 K

Above 972 K the reaction becomes spontaneous.
Q463-05 When solid blue copper(II) sulfate pentahydrate CuSO4.5H2O loses water, the white solid copper(II) sulfate monohydrate CuSO4.H2O is produced as represented by the equation:

CuSO4.5H2O(s) → CuSO4.H2O(s) + 4H2O(g)

The thermodynamic data for the substances involved in the reversible process are:

 
ΔHof / kJmol-1
So / J K-1 mol-1
CuSO4.5H2O (s)
-2278
305
CuSO4.H2O (s)
-1084
150
H2O (g)
-242
189

Calculate the value of ΔHo for the reaction above and state what information the sign of ΔHo provides about this reaction.
Calculate ΔSo for the reaction and state the meaning of the sign of ΔSo obtained.
Identify a thermodynamic function that can be used to predict the reaction spontaneity and state its units.
Use the values obtained in the above to determine whether the reaction is spontaneous or non-spontaneous at 25ºC.

Answer

Formation enthalpy of reactants

CuSO4.5H2O(s) = -2278 kJ

Formation enthalpy of products

CuSO4.H2O(s) = -1084 kJ
4 x H2O(g) = 4 x -242 = -968kJ
Total = -2052 kJ
Reaction enthalpy ΔH = ΔHf products - ΔHf reactants = -2052 + 2278 = +226 kJ (endothermic)

Entropy of reactants

CuSO4.5H2O(s) = 305 J K-1

Entropy of products

CuSO4.H2O(s) = 150 J K-1
4 x H2O(g) = 4 x 189 = 756 J K-1
Total = 906 J K-1

Entropy change ΔS = ΔS products - ΔS reactants = 906 - 305 = +601 J K-1

This is a positive entropy change as the entropy increases from reactants to products

Spontaneity of reaction can be predicted using Gibbs free energy change, ΔG. If ΔG is negative then the reaction is spontaneous.

ΔG = ΔH - TΔS

at 25ºC (298K) ΔG = 226 - 298 (+0.601)

∴ ΔG = 226 - 179 = +47 kJ

The reaction is non-spontaneous at 25oC
Q463-06 Using the answers in question 7 above, identify which compound CuSO4.5H2O(s), or CuSO4.H2O (s) is more stable at 25ºC.
Use the values obtained in question 7 to determine the centigrade temperature above which the other compound is more stable.
Answer

As the reaction is shown to be non-spontaneous at 25ºC, the most stable component is CuSO4.5H2O(s). This is also borne out by its much lower enthalpy of formation.

The temperature above which CuSO4.H2O(s) is more stable is when the Gibbs Free energy = negative. The limit of this is when ΔG = 0.

When ΔG = 0, ΔH = TΔS.

Therefore T = ΔH /ΔS = +226/0.601 = 376 K = 103 ºC
Q463-07 Consider the following reaction:

N2(g) + 3H2(g) → 2NH3(g)

Using average bond enthalpies values below, calculate the standard enthalpy change for this reaction.

Bond enthalpies (kJ mol-1) : N≡N, 942; H-H, 432; N-H, 386.

The absolute entropy values, S, at 300K for N2(g), H2(g) and NH3(g) are 193, 131 and 192 JK-1 mol-1 respectively.

Calculate ΔSo for the reaction and explain the sign of ΔSo

Calculate ΔGo for the reaction at 300K.

Answer

Bonds broken (endothermic)

1 x N≡N = 1 x 942 = 942 kJ
3 x H-H = 3 x 432 = 1296 kJ
Total = 2238 kJ

Bonds formed (exothermic)

6 x N-H = 6 x -386 = -2316 kJ
Reaction enthalpy ΔH = 2238 - 2316 = -78 kJ

Initial entropy

N2 = 1 x 193 = 193 J K-1
3 x H2 = 3 x 131 = 393 J K-1
Total = 586 J K-1

Product entropy

2x NH3 = 2 x 192 = 384 J K-1

The entropy change, ΔS = 384 - 586 = -202 J K-1

The sign of the entropy change is negative as the overall entropy decreases.

Using the Gibbs free energy equation, and not forgetting to convert the Joules per Kelvin of the entropy change to kiloJoules per Kelvin (divide by 1000):

ΔG = ΔH - TΔS

ΔG = -78 - 300 x (-0.202)

ΔG = -78 + 60.6 = -17.4 kJ

Q463-08 The enthalpy change for the combustion of butanoic acid at 25ºC is - 2183.5 kJ mol-1. The combustion reaction is:

C3H7COOH(l) + 5O2(g) → 4CO2(g) + 4H2O(l)


Write the balanced equation for the formation of butanoic acid from its elements. ANS
4C(s) + 4H2(g) + O2(g) → C3H7COOH(l)

Using the data below, calculate the standard enthalpy of formation, Hof, for butanoic acid. ANS

Reaction enthalpy for the combustion of butanoic acid = formation enthalpy of products - formation enthalpy of reactants

Combustion enthalpy, - 2183.5 = 4(-393.5) + 4(-285.9) - Hof( butanoic acid)

∴ Hof( butanoic acid) = 4(-393.5) + 4(-285.9) + 2183.5 = -534 kJ

Substance Standard Enthalpy of Formation, Hof /kJ mol-1 Absolute Entropy, S /J mol-1 K-1
C(s) 0 5.7
CO2(g) -393.5 213.6
H2(g) 0 130.6
H2O(l) -285.9 60.9
O2(g) 0 205.0
C3H7COOH(l)   226.3

Calculate the standard entropy change, Sof, for the formation of butanoic acid at 25ºC ANS

Initial entropy

4 x C(s) = 4 x 5.7 = 21.8 J K-1
4 x H2 = 4 x 130.6 = 522.4 J K-1
1 x O2(g) = 1 x 205.0 = 205.0 J K-1
Total = 750.2 J K-1

Product entropy

1 x C3H7COOH(l) = 1 x 226.3 = 226.3 J K-1

The standard entropy change, Sof = 750.2 - 226.3 = -523.9 J K-1

The sign of the entropy change is negative as the overall entropy decreases.

Calculate the standard free energy of formation, Gof, for butanoic acid at 25ºC ANS

Using the Gibbs free energy equation, and not forgetting to convert the Joules per Kelvin of the entropy change to kiloJoules per Kelvin (divide by 1000):

ΔG = ΔH - TΔS

ΔG = -534 - 298 x (-0.524)

ΔG = -534 + 156 = -378 kJ


Is this reaction spontaneous at 25ºC? Explain your answer ANS
The reaction is spontaneous at 25ºC as the value for Gibbs free energy is negative.
Q463-09 Consider the following reaction:

N2(g) + 3H2(g) → 2NH3(g)

The enthalpy change, ΔHo for this reaction = -92 kJ. The magnitude of the entropy change ΔS, at 27ºC for the reaction is -202 J K-1 mol-1. Calculate ΔG for this reaction at 27ºC and determine whether this reaction is spontaneous at this temperature.

Answer

Spontaneity of reaction can be predicted using Gibbs free energy change, ΔG. If ΔG is negative then the reaction is spontaneous.

ΔG = ΔH - TΔS

at 27ºC: ΔG = -92 - 300 x (-0.202)

∴ ΔG = -92 + 60.6 = -31.4 kJ

The reaction is spontaneous at 27ºC

REM Although this answer shows that the reaction is spontaneous at low temperatures, the kinetics of the process cause it to be too slow for practical purposes. In industry the temperature is elevated to 450-500ºC to increase the rate of formation of ammonia.
Q463-10 The equation for the decomposition of calcium carbonate is given below:

CaCO3(s) → CaO(s) + CO2(g)

At 500K, ΔH for this reaction is +177 kJ mol-1 and ΔS is 161 JK-1 mol-1. Calculate the value of ΔG at 500K and determine, giving a reason, whether or not the reaction is spontaneous.

Answer

Spontaneity of reaction can be predicted using Gibbs free energy change, ΔG. If ΔG is negative then the reaction is spontaneous.

ΔG = ΔH - TΔS

at 500K: ΔG = +177 - 500 (+0.161)

∴ ΔG = +177 - 80.5 = +96.5 kJ

The reaction is non-spontaneous at 500K

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