State whether the entropy change is positive or negative for 2H_2(g) + O_2(g) → 2H_2O(l), and why. [2 marks]

A reaction has H = +20 kJ mol^-1 and S = +50 J K^-1mol^-1. Find the temperature above which it is feasible. [2 marks]

State what a negative value of G tells you, and what it does not tell you, about a reaction. [2 marks]

SQA SQA AH 2019-style practice: For the decomposition MgCO_3(s) → MgO(s) + CO_2(g), H = +100.7 kJ mol^-1 and S = +175.0 J K^-1mol^-1. (a) Calculate the temperature above which the reaction becomes feasible. (b) Explain why S is positive.

Markers reward setting G = 0, the unit conversion, the temperature, and the entropy reasoning. (a) The reaction just becomes feasible when G = 0, so T = H S. Converting H to joules: $T = (100700)/(175.0) = 575 K$ Above about 575 K the decomposition is feasible. (b) S is positive because a gas (CO_2) is produced from a solid, greatly increasing the disorder of the system (gases have far higher entropy than solids).

SQA SQA AH specimen-style practice: A reaction has H = -56 kJ mol^-1 and S = -125 J K^-1mol^-1. Calculate G at 298 K and state whether the reaction is feasible at this temperature.

The answer must use the Gibbs equation with consistent units and judge feasibility from the sign. Convert S to kilojoules: S = -0.125 kJ K^-1mol^-1. $G = H - T S = -56 - (298 × -0.125) = -56 + 37.3 = -18.7 kJ mol^-1$ G is negative, so the reaction is feasible at 298 K. Markers reward the unit conversion, the value of G, and the feasibility judgement from its sign.

ScotlandChemistrySyllabus dot point

Why do some reactions happen spontaneously and others do not?

Standard enthalpy and entropy changes, the second law of thermodynamics, and the Gibbs free energy relationship delta G = delta H - T delta S used to decide reaction feasibility, find the temperature of feasibility, and interpret Ellingham diagrams.

An SQA Advanced Higher Chemistry answer on reaction feasibility, covering standard enthalpy change, entropy and the second law of thermodynamics, the Gibbs free energy relationship delta G = delta H - T delta S, calculating the temperature at which a reaction becomes feasible, and interpreting Ellingham diagrams for metal extraction.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Entropy $S$ measures the disorder or dispersal of energy in a system, with gases far higher than liquids, which are higher than solids; a reaction that produces more gas has a positive entropy change. The second law of thermodynamics states that the total entropy of the universe increases in any spontaneous change. The Gibbs free energy change combines enthalpy and entropy: $\Delta G = \Delta H - T\Delta S$ . A reaction is feasible when $\Delta G$ is negative (or zero). Setting $\Delta G = 0$ gives the temperature of feasibility $T = \Delta H/\Delta S$ . Watch units, because $\Delta H$ is usually in kilojoules and $\Delta S$ in joules. Ellingham diagrams plot $\Delta G$ against temperature for metal oxides and show which reducing agent can extract a metal at a given temperature. Feasibility says a reaction can happen, not how fast it goes.

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What this key area is asking
Enthalpy, entropy and the second law
Gibbs free energy
Feasibility versus rate
Ellingham diagrams
Examples in context
Try this

What this key area is asking

The SQA wants you to use standard enthalpy and entropy changes, to state the second law of thermodynamics, and to use $\Delta G = \Delta H - T\Delta S$ to decide whether a reaction is feasible, to find the temperature at which it becomes feasible, and to interpret Ellingham diagrams. The free-energy calculation and the temperature-of-feasibility calculation are reliable exam earners.

Enthalpy, entropy and the second law

The standard enthalpy change $\Delta H^{\circ}$ is the heat energy change at constant pressure under standard conditions, found from enthalpies of formation, combustion or Hess's law. The second law of thermodynamics states that the total entropy of the universe (system plus surroundings) increases in every spontaneous process.

Gibbs free energy

Temperature at which a reaction becomes feasible

For $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ , $\Delta H = +178 \text{ kJ mol}^{-1}$ and $\Delta S = +161 \text{ J K}^{-1}\text{mol}^{-1}$ . Find the minimum temperature for feasibility.

step 1: Set the feasibility boundary

The reaction just becomes feasible when $\Delta G = 0$ , so $0 = \Delta H - T\Delta S$ , giving $T = \Delta H/\Delta S$ .

step 2: Make the units consistent

Convert $\Delta H$ to joules: $178 \text{ kJ mol}^{-1} = 178000 \text{ J mol}^{-1}$ . (Or convert $\Delta S$ to kilojoules; either works provided both match.)

step 3: Calculate the temperature

T = \frac{178000}{161} = 1106 \text{ K} \ (\text{about } 1110 \text{ K})

step 4: Interpret

Above about $1106 \text{ K}$ , $\Delta G$ becomes negative and the decomposition is feasible, which is why a lime kiln must reach a high temperature.

Feasibility versus rate

Ellingham diagrams

An Ellingham diagram plots the free energy change $\Delta G$ for the formation of metal oxides against temperature. Each line slopes upward because $\Delta S$ for forming an oxide from a metal and oxygen gas is usually negative. A metal whose oxide line lies below another's can reduce that other oxide at a given temperature, because the overall $\Delta G$ for the reduction is negative. Carbon is widely used to reduce metal oxides because its line slopes downward and crosses below most metal oxide lines at high temperature, which is why carbon (coke) extracts iron from its oxide in a blast furnace.

Examples in context

Free energy explains everyday spontaneity. Ice melts above $273 \text{ K}$ because melting is endothermic but increases disorder, so above the melting point the $T\Delta S$ term outweighs $\Delta H$ and $\Delta G$ becomes negative. The thermal decomposition of carbonates needs high temperatures because it is endothermic with a positive entropy change. In metallurgy, Ellingham diagrams set the temperature at which carbon can reduce a given oxide, governing how iron, zinc and other metals are extracted economically. The distinction between feasibility and rate explains why petrol is stable in air at room temperature (the combustion is highly feasible but kinetically slow without a spark) despite a strongly negative $\Delta G$ .

Try this

Q1. State whether the entropy change is positive or negative for $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$ , and why. [2 marks]

Cue. Negative, because three moles of gas become two moles of liquid, decreasing disorder.

Q2. A reaction has $\Delta H = +20 \text{ kJ mol}^{-1}$ and $\Delta S = +50 \text{ J K}^{-1}\text{mol}^{-1}$ . Find the temperature above which it is feasible. [2 marks]

Cue. $T = \Delta H/\Delta S = 20000/50 = 400 \text{ K}$ .

Q3. State what a negative value of $\Delta G$ tells you, and what it does not tell you, about a reaction. [2 marks]

Cue. It tells you the reaction is thermodynamically feasible, but not how fast it will go.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH 20194 marksFor the decomposition

\text{MgCO}_3(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g)

\Delta H = +100.7 \text{ kJ mol}^{-1}

and

\Delta S = +175.0 \text{ J K}^{-1}\text{mol}^{-1}

. (a) Calculate the temperature above which the reaction becomes feasible. (b) Explain why

\Delta S

is positive.

Show worked answer →

Markers reward setting $\Delta G = 0$ , the unit conversion, the temperature, and the entropy reasoning.

(a) The reaction just becomes feasible when $\Delta G = 0$ , so $T = \dfrac{\Delta H}{\Delta S}$ . Converting $\Delta H$ to joules:

T = \frac{100700}{175.0} = 575 \text{ K}

Above about $575 \text{ K}$ the decomposition is feasible.

(b) $\Delta S$ is positive because a gas ( $\text{CO}_2$ ) is produced from a solid, greatly increasing the disorder of the system (gases have far higher entropy than solids).

SQA AH specimen3 marksA reaction has

\Delta H = -56 \text{ kJ mol}^{-1}

and

\Delta S = -125 \text{ J K}^{-1}\text{mol}^{-1}

. Calculate

\Delta G

298 \text{ K}

and state whether the reaction is feasible at this temperature.

Show worked answer →

The answer must use the Gibbs equation with consistent units and judge feasibility from the sign.

Convert $\Delta S$ to kilojoules: $\Delta S = -0.125 \text{ kJ K}^{-1}\text{mol}^{-1}$ .

\Delta G = \Delta H - T\Delta S = -56 - (298 \times -0.125) = -56 + 37.3 = -18.7 \text{ kJ mol}^{-1}

$\Delta G$ is negative, so the reaction is feasible at $298 \text{ K}$ . Markers reward the unit conversion, the value of $\Delta G$ , and the feasibility judgement from its sign.

Related dot points

Sources & how we know this

SQA Advanced Higher Chemistry Course Specification — SQA (2019)