IB Colourful Solutions in Chemistry

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 reference 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
Temperature and equilibrium
The relationship between ΔG and the equilibrium constant, k.
Worked examples

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 ΔH^o = +210 kJ and ΔS^o= 216 JK^-1. What is the value of ΔG^o for this process at 298K?

ΔS^o = 216 JK^-1, therefore: ΔS^o = 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
S^o water(l) = 71 JK^-1
S^o 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:

is spontaneous at 300 K and becomes non-spontaneous at higher temperatures.
is spontaneous at 300 K and becomes non-spontaneous at lower temperatures.
is non-spontaneous at 300 K and becomes spontaneous at higher temperatures.
is non-spontaneous at 300 K and becomes spontaneous at lower temperatures.

Answer

To calculate spontaneity Gibbs free energy equation must be used.

G^o = H^o - T S^o

Substituting the values into the equation

G^o = -597.3 - 300 x (-0.33)

G^o = -498.3 kJ

As G^o has a negative value then the reaction is spontaneous under these conditions… and as the temperature rises the term (- T S^o) 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)

ΔH^o = -9.83 kJmol^-1 and ΔS^o = -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)
ΔHf^o/kJ mol^-1	-1219	-394	-558
S^o /JK^-1 mol^-1	+112	+214	+70

Calculate the value of ΔG^o 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 ΔH^o = +210kJ and ΔS^o = 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:

	ΔH^of / kJmol^-1	S^o / J K^-1 mol^-1
CuSO₄.5H₂O (s)	-2278	305
CuSO₄.H₂O (s)	-1084	150
H₂O (g)	-242	189

Calculate the value of ΔH^o for the reaction above and state what information the sign of ΔH^o provides about this reaction.
Calculate ΔS^o for the reaction and state the meaning of the sign of ΔS^o 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 25^oC

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 ΔS^o for the reaction and explain the sign of ΔS^o

Calculate ΔG^o 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, H^of, 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) - H^of( butanoic acid)

∴ H^of( butanoic acid) = 4(-393.5) + 4(-285.9) + 2183.5 = -534 kJ

Substance	Standard Enthalpy of Formation, H^of /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, S^of, 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, S^of = 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, G^of, 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, ΔH^o 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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