5.4.7 The Feasibility of a Reaction

5.4.7 The Feasibility of a Reaction: Use the total entropy change to decide whether a reaction is thermodynamically feasible, and explain why feasible reactions may still not occur.

A reaction is feasible if the total entropy of system and surroundings increases.

Second law idea

A process can happen spontaneously only if ΔS(total) is positive, since the universe becomes more disordered.

Surroundings

ΔS(surr) = −ΔH / T, so an exothermic reaction (negative ΔH) increases the entropy of the surroundings.

Total

ΔS(total) = ΔS(system) + ΔS(surr), and both terms must be in the same units.

Units

ΔH is usually in kJ mol⁻¹, so convert it to J mol⁻¹ by multiplying by 1000 before dividing by T in kelvin.

The sign of ΔS(total) decides feasibility at a given temperature.

Calculate ΔS(system) from standard entropy data and ΔH from enthalpy data.

Calculate ΔS(surr) = −ΔH × 1000 / T, with T in kelvin.

Add the two values to obtain ΔS(total) in J K⁻¹ mol⁻¹.

If ΔS(total) is positive the reaction is feasible, and if it is negative it is not feasible at that temperature.

Temperature changes ΔS(surr), so a reaction that is not feasible at room temperature may become feasible when heated.

Feasible Does Not Mean Fast

Kinetic stability

A reaction with a high activation energy may not occur at a measurable rate even though it is thermodynamically feasible.

Examples

Diamond turning into graphite and the reaction of hydrogen and oxygen at room temperature are feasible but extremely slow.

Catalyst role

A catalyst can make a feasible reaction occur at a useful speed, but cannot make an unfeasible reaction proceed.

Conditions

Standard values are given at 298 K, so results at other temperatures need the temperature-dependent term recalculated.

Question: For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS(system) = +160.4 J K⁻¹ mol⁻¹. Is it feasible at (a) 298 K and (b) 1500 K? Find the minimum temperature at which it becomes feasible.

(a) ΔS(surr) = −178 000 ÷ 298 = −597 J K⁻¹ mol⁻¹, so ΔS(total) = 160.4 − 597 = −437 J K⁻¹ mol⁻¹: not feasible.
(b) ΔS(surr) = −178 000 ÷ 1500 = −118.7 J K⁻¹ mol⁻¹, so ΔS(total) = 160.4 − 118.7 = +41.7 J K⁻¹ mol⁻¹: feasible.
At the limit, ΔS(total) = 0, so T = ΔH ÷ ΔS(system) = 178 000 ÷ 160.4 = 1110 K.

Answer: Not feasible at 298 K, feasible at 1500 K, minimum temperature about 1110 K.

5.4.7 – Key Points for the Exam

Feasible reaction: One for which the total entropy change is positive, so it is thermodynamically possible.
ΔS(surroundings): −ΔH/T, the entropy change of the surroundings.
ΔS(total): The sum of the entropy change of the system and surroundings.
Kinetic stability: Lack of reaction at a noticeable rate because of a high activation energy.

Common Mistakes

Leaving ΔH in kJ mol⁻¹ while ΔS is in J K⁻¹ mol⁻¹.
Using temperature in °C in the expression −ΔH/T.
Concluding that a feasible reaction must be fast, when feasibility says nothing about rate.
Forgetting the minus sign in ΔS(surr) = −ΔH/T.