How do you calculate ΔSsystem from standard entropies?

ΣS(products) − ΣS(reactants), with each value multiplied by its coefficient.

Give the equation for ΔSsurroundings.

ΔSsurroundings = −ΔH/T, with ΔH in J mol⁻¹ and T in kelvin.

What must you do to ΔH (in kJ mol⁻¹) before calculating ΔSsurroundings?

Multiply by 1000 to convert to J mol⁻¹.

What is the sign of ΔSsurroundings for an exothermic reaction?

Positive, because heat released to the surroundings increases their entropy.

What value of ΔStotal makes a reaction feasible?

A positive value (zero is the limit of feasibility).

Can an endothermic reaction be spontaneous?

Yes, if a large positive ΔSsystem outweighs the negative ΔSsurroundings so that ΔStotal is positive.

Can a reaction be feasible if ΔSsystem is negative?

Yes, if ΔSsurroundings is positive and larger in magnitude, so ΔStotal is positive.

How does raising the temperature affect the magnitude of ΔSsurroundings?

It decreases, because T is in the denominator of −ΔH/T.

How do you find the minimum temperature at which a reaction becomes feasible?

Set ΔStotal = 0, so T = ΔH/ΔSsystem (ΔH in J mol⁻¹).

Write the Gibbs equation relating ΔG, ΔH and ΔSsystem.

ΔG = ΔH − TΔSsystem; the reaction is feasible if ΔG is negative.

What does a positive ΔStotal not tell you?

How fast the reaction is; the rate depends on the activation energy.

Define kinetic stability.

A substance that could react (ΔStotal positive) but does not, because a high activation energy makes the rate negligible.

Why is the standard molar entropy of a gas larger than that of the solid?

Gas particles are free to move randomly, so there are many more ways to arrange the particles and energy.

Total entropy change and feasibilityEdexcel International A Level Chemistry: Flashcards

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Write the equation for ΔStotal.

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Write the equation for ΔStotal.: ΔStotal = ΔSsystem + ΔSsurroundings
How do you calculate ΔSsystem from standard entropies?: ΣS(products) − ΣS(reactants), with each value multiplied by its coefficient.
Give the equation for ΔSsurroundings.: ΔSsurroundings = −ΔH/T, with ΔH in J mol⁻¹ and T in kelvin.
What must you do to ΔH (in kJ mol⁻¹) before calculating ΔSsurroundings?: Multiply by 1000 to convert to J mol⁻¹.
What is the sign of ΔSsurroundings for an exothermic reaction?: Positive, because heat released to the surroundings increases their entropy.
What value of ΔStotal makes a reaction feasible?: A positive value (zero is the limit of feasibility).
Can an endothermic reaction be spontaneous?: Yes, if a large positive ΔSsystem outweighs the negative ΔSsurroundings so that ΔStotal is positive.
Can a reaction be feasible if ΔSsystem is negative?: Yes, if ΔSsurroundings is positive and larger in magnitude, so ΔStotal is positive.
How does raising the temperature affect the magnitude of ΔSsurroundings?: It decreases, because T is in the denominator of −ΔH/T.
How do you find the minimum temperature at which a reaction becomes feasible?: Set ΔStotal = 0, so T = ΔH/ΔSsystem (ΔH in J mol⁻¹).
Write the Gibbs equation relating ΔG, ΔH and ΔSsystem.: ΔG = ΔH − TΔSsystem; the reaction is feasible if ΔG is negative.
What does a positive ΔStotal not tell you?: How fast the reaction is; the rate depends on the activation energy.
Define kinetic stability.: A substance that could react (ΔStotal positive) but does not, because a high activation energy makes the rate negligible.
Why is the standard molar entropy of a gas larger than that of the solid?: Gas particles are free to move randomly, so there are many more ways to arrange the particles and energy.

Exam questions on Total entropy change and feasibility

Instant cold packs contain solid ammonium nitrate and a sealed pouch of water. When the pouch is burst, the solid dissolves: NH₄NO₃(s) + aq → NH₄⁺(aq) + NO₃⁻(aq). The enthalpy change of solution is +25.7 kJ mol⁻¹ and the entropy change of the system is +108.5 J K⁻¹ mol⁻¹. The pack is used at 298 K and becomes noticeably cold as the solid dissolves.
Calculate the total entropy change for the dissolving at 298 K and state what it shows about feasibility.2 marks
In the Haber process, nitrogen and hydrogen react to make ammonia: N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.0 kJ mol⁻¹. Standard molar entropies at 298 K are: N₂(g) 191.6, H₂(g) 130.6 and NH₃(g) 192.3 J K⁻¹ mol⁻¹.
Explain, with a calculation of ΔSsurroundings at 298 K, why this reaction is feasible even though ΔSsystem is negative.2 marks
Limestone is heated in a kiln to make quicklime: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹. Standard molar entropies at 298 K are: CaCO₃(s) 92.9, CaO(s) 39.7 and CO₂(g) 213.6 J K⁻¹ mol⁻¹. Assume that ΔH and the entropy values do not change with temperature.
Calculate ΔSsystem and ΔSsurroundings at 298 K for this reaction, and use them to show that the decomposition is not feasible at 298 K.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).