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

Where:

• ΔG = Gibbs free energy change (J mol^-1)
• ΔH = enthalpy change (J mol^-1)
• T = temperature (K)
• ΔS = entropy change (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 = +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

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 = ∆H

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

Where:

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

∆S = 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

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

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

Calculating enthalpy changes from temperature changes

Sometimes, ∆H needs to be determined experimentally by measuring the temperature change in a known mass of water or solution where the reaction occurs.

The formula to calculate the energy change (q) is as follows:

q = mcΔT

Where:

• q = energy change (J)
• m = mass of water or solution (g)
• c = specific heat capacity (4.18 J g^-1 K^-1 for water)
• ΔT = temperature change (K or °C)

To find ΔH (energy change per mole), divide q by the number of moles of reactant. Once ΔH is known, you can calculate ∆G.

Worked example 3 - Calculating ∆G for the dissolution of sodium thiosulfate in water

Calculate the free energy change for the dissolution of sodium thiosulfate (Na₂S₂O₃) in water, when 50.0 g of Na₂S₂O₃ dissolves and the solution temperature decreases from 298 K to 295 K:

Na₂S₂O_3(s) ➔ 2Na⁺_(aq) + S₂O₃^2-_(aq)

The total solution mass is 1,000 g, and the specific heat capacity of the solution is assumed to be 4.18 J g^-1 K^-1.

The standard molar entropies are:

Step 1: Calculate q

$\text{q }=\text{mc}\mathrm{\Delta }\text{T }=\mathrm{1,000}\times 4.18\times (295-298)=-\mathrm{12,540}\text{ J}$

Step 2: Calculate number of moles of Na₂S₂O₃

$\text{n }=\frac{\text{m}}{{\text{M}}_{\text{r}}}=\frac{50.0}{158.2}=0.316\text{ mol}$

Step 3: Calculate ΔH

$\mathrm{\Delta }\text{H }=\frac{\text{q}}{\text{n}}=\frac{-\mathrm{12,540}}{0.316}=\mathrm{39,676.56}{\text{ J mol}}^{-1}$

Step 4: Calculate ΔS

ΔS = S^⦵_products $-$ S^⦵_reactants = ((2 x 59.0) + 248.2) $-$ 150.4 = 215.8 J K^-1 mol^-1

Step 5: Calculate ΔG at 298 K

$\mathrm{\Delta }\text{G }=\mathrm{\Delta }\text{H}-\text{T}\mathrm{\Delta }\text{S }=-\mathrm{39,676.56}-(298\times 215.8)=-\mathrm{104,000}{\text{ J mol}}^{-1}$

Given that ΔG is negative, the dissolution of sodium thiosulfate in water is thermodynamically possible at 25°C.

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:

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

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