Free energy change predicts reaction feasibility

The Gibbs free energy change (ΔG) represents the overall change in energy during a chemical reaction. It is used to determine if a chemical reaction is thermodynamically feasible under certain conditions.

It considers two important thermodynamic properties:

1. Enthalpy change (ΔH) - This is the heat energy absorbed or given off by the reaction at constant pressure.
2. Entropy change (ΔS) - This represents the change in disorder or randomness within the system.

For a reaction to be thermodynamically feasible or to occur spontaneously, the free energy change must be negative (ΔG < 0) or exactly zero (ΔG = 0).

If ΔG is > 0, the reaction is not feasible without an external energy source being supplied to drive the reaction forward.

The free energy equation

The formula to calculate free energy change is as follows:

ΔG = ΔH $-$ TΔS_system

Where:

• ΔG = Gibbs free energy change (J mol^-1)
• ΔH = enthalpy change (J mol^-1)
• T = temperature (K)
• ΔS_system = entropy change of the system (J K^-1 mol^-1)

Worked example 1 - Calculating ΔG for the dissolution of ammonium chloride in water

Calculate the free energy change for the dissolution of ammonium chloride (NH₄Cl) in water at 25.0°C:

NH₄Cl_(s) ➔ NH₄⁺_(aq) + Cl^-_(aq)

Given:

• ΔH = +14.6 kJ mol^-1
• ΔS_system = +75.3 J K^-1 mol^-1

Step 1: Conversion of °C into K

To convert from °C to K, add 273

25.0°C = 298.0 K

Step 2: Conversion of kJ mol^-1 into J mol^-1

To convert from kJ mol^-1 into J mol^-1, multiply by 1,000

14.6 kJ mol^-1 = 14,600 J mol^-1

Step 3: Equation

ΔG = ΔH $-$ TΔS_system

Step 4: Substitution and correct evaluation

$\mathrm{\Delta }\text{G }=\mathrm{14,600}-(298\times 75.3)$

$\mathrm{\Delta }\text{G }=-\mathrm{7,839.4}{\text{ J mol}}^{-1}$

Since ∆G is negative, dissolving NH₄Cl in water is thermodynamically feasible at 25.0°C.

Temperature affects reaction feasibility

The possibility of some reactions happening depends on the temperature.

Based on the signs of ∆H and ∆S, there are four scenarios:

1. If ΔH is negative (exothermic) and ΔS is positive, ΔG is always negative, making the reaction feasible at any temperature.

2. If ΔH is positive (endothermic) and ΔS is negative, ΔG is always positive, and the reaction is not feasible at any temperature.

3. If both ΔH and ΔS are positive, the reaction is only feasible above a specific temperature. Example - The thermal decomposition of calcium carbonate:

CaCO_3(s) ➔ CaO_(s) + CO_2(g) ΔH = +10 kJ mol^-1, ΔS = +10 J K^-1 mol^-1

• At 300 K: ΔG = +7,000 J mol^-1 (not feasible)
• At 1,200 K: ΔG = $-$2,000 J mol^-1 (feasible)

1. If ∆H is negative (exothermic) and ∆S is negative, the reaction is feasible below a certain temperature. Example - The freezing of water:

H₂O_(l) ➔ H₂O_(s) ΔH = $-$10 kJ mol^-1, ΔS = $-$10 J K^-1 mol^-1

• At 300 K: ΔG = $-$7,000 J mol^-1 (feasible)
• At 1,200 K: ΔG = +2,000 J mol^-1 (not feasible)

These scenarios are summarised in the table below:

Calculating the temperature at which a reaction becomes feasible

To find out at what temperature a reaction just becomes feasible (∆G = 0), we can rearrange the free energy equation:

When ΔG = 0, TΔS_system = ∆H

$\text{Thus, T }=\frac{\mathrm{\Delta }\text{H}}{\mathrm{\Delta }{\text{S}}_{\text{system}}}$

Where:

• T = temperature at which the reaction becomes feasible (K)
• ΔH = enthalpy change (J mol^-1)
• ΔS_system = entropy change of the system (J K^-1 mol^-1)

Worked example 2 - Calculating the temperature at which a reaction becomes feasible

Calculate the minimum temperature at which the decomposition of calcium carbonate to calcium oxide and carbon dioxide becomes feasible:

CaCO_3(s) ➔ CaO_(s) + CO_2(g) ΔH = +178 kJ mol^-1

Given entropy values:

Step 1: Conversion of kJ mol^-1 into J mol^-1

To convert from kJ mol^-1 into J mol^-1, multiply by 1,000

178 kJ mol^-1 = 178,000 J mol^-1

Step 2: Calculate ∆S_system

∆S_system = S^⦵_products $-$ S^⦵_reactants = (39.8 + 213.7) $-$ 92.9 = 160.6 J K^-1 mol^-1

Step 3: Calculate minimum temperature when reaction becomes feasible

$\text{T }=\frac{\mathrm{\Delta }\text{H}}{\mathrm{\Delta }{\text{S}}_{\text{system}}}=\frac{\mathrm{178,000}}{160.6}=\mathrm{1,108}\text{ K}$

Thus, the decomposition of calcium carbonate into calcium oxide and carbon dioxide is feasible above 1,108 K.

Free energy change and equilibrium constants

The equilibrium constant (K) is the ratio of product to reactant concentrations at equilibrium for a reversible reaction at a specific temperature.

Feasible reactions with negative ΔG have equilibrium constants greater than 1. Unfeasible reactions with positive ΔG have equilibrium constants less than 1.

The relationship between ΔG and K is given by:

ΔG= $-RTlnK$

Where:

• ΔG = free energy change (kJ mol^-1)
• R = gas constant (8.31 J K^-1 mol^-1).
• T = temperature (K)
• K = equilibrium constant

Rearranging for K yields:

$K=e^{\frac{-\mathrm{\Delta }{G}^{\ominus }}{RT}}$

Using K and ΔG^⦵ to predict reaction favourability

The magnitude of the equilibrium constant (K) indicates whether reactants or products are favoured at equilibrium. This can also be determined from the sign and magnitude of the standard Gibbs energy change (ΔG^⦵).

By calculating K from ΔG^⦵ (or vice versa), you can predict the spontaneity and favourability of a reversible reaction at a given temperature.

Worked example 3 - Calculating Gibbs energy change

Consider the following reversible reaction at 600 K:

2SO_2(g)_ + O_2(g)_ ⇌ 2SO_3(g)_

Given that the equilibrium constant (K) is 19.6 at 600 K, calculate the Gibbs energy change (ΔG), in kJ mol^-1, and comment on the spontaneity of the forward reaction.

The gas constant, R = 8.31 J K^-1 mol^-1

Step 1: Equation

ΔG = $-RTlnK$

Step 2: Substitution and correct evaluation

ΔG = $-$8.31 x 600 x ln(19.6) = $-$14,800 J mol^-1

Step 3: Conversion of J mol^-1 into kJ mol^-1

To convert from J mol^-1 into kJ mol^-1, divide by 1,000

$-$14,800 J mol^-1 = $-$14.8 kJ mol^-1

Step 4: Interpretation of ΔG

Since ΔG < 0, the forward reaction is spontaneous at this point.

Worked example 4 - Determining the equilibrium constant

The following reversible reaction has a Gibbs energy change (ΔG) of 6.40 kJ mol^-1.

H_2(g)_ + I_2(g)_ ⇌ 2HI_(g)_

Determine the value of the equilibrium constant (K) for this reaction at a temperature of 400 K, and use your result to comment on the favourability of the forward reaction at this temperature.

The gas constant, R = 8.31 J K^-1 mol^-1

Step 1: Conversion of kJ mol^-1 into J mol^-1

To convert kJ mol^-1 into J mol^-1, multiply by 1,000

6.40 kJ mol^-1 = 6,400 J mol^-1

Step 2: Equation

ΔG = $-RTlnK$

Step 3: Rearrange equation

ln K = $\frac{-\mathrm{\Delta }G}{RT}$

$K=e^{\frac{-\mathrm{\Delta }G}{RT}}$

Step 4: Substitution and correct evaluation

ln K = $\frac{-\mathrm{6,400}}{(8.31\times 400)}=-1.93$

K = e^-1.93^ = 0.146 (unitless)

Step 5: Interpretation of K

The small K value indicates the forward reaction (formation of HI) is not favourable at 400 K.

Limitations of using ∆G to predict reaction feasibility

Although a negative ∆G indicates a reaction is thermodynamically feasible, it doesn't account for the reaction's rate or kinetic barriers. Some reactions with a negative ∆G may proceed very slowly or require significant activation energy, making them practically unobservable under normal conditions.

For instance, hydrogen gas combustion is highly exothermic and entropically favourable at 298 K:

H_2(g) + ½O_2(g) ➔ H₂O_(g) ∆H^⦵ = $-$242 kJ mol^-1, ∆S^⦵ = $-$44.4 J K^-1 mol^-1

At 298 K:

∆G^⦵= ∆H^⦵ $-$ T∆S^⦵_system = $-\mathrm{242,000}-(298\times -44.4)=-\mathrm{229,000}{\text{ J mol}}^{-1}$

Despite the favourable ∆G, this reaction requires a spark to initiate due to its high activation energy.