Gibbs Free Energy

A2 · 12 min

Whether a reaction can happen on its own depends on two things: the enthalpy change (energy released to or taken from the surroundings) and the entropy change of the system. The Gibbs free energy change, $\Delta G$ , combines them into one number whose sign tells you whether a reaction is feasible. Paper 4 regularly asks you to calculate $\Delta G^{\ominus}$ , decide feasibility, and find the temperature at which a reaction becomes feasible, usually as part of a longer question on thermal decomposition or industrial processes.

The Gibbs equation

Key result

\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

$\Delta G^{\ominus}$ and $\Delta H^{\ominus}$ in $\text{kJ mol}^{-1}$
$T$ in kelvin ( $T/\text{K} = \theta/^\circ\text{C} + 273$ )
$\Delta S^{\ominus}$ must be converted to $\text{kJ K}^{-1}\ \text{mol}^{-1}$ (divide by 1000) before substituting

Definition

A reaction or process is feasible if $\Delta G$ is negative. If $\Delta G$ is positive, the reaction is not feasible under those conditions (the reverse reaction is feasible). When $\Delta G = 0$ , the system is at the boundary between feasible and not feasible.

Where the equation comes from

You do not need to derive it, but the idea makes it easier to use. The total entropy change of the universe must be positive for a process to happen spontaneously. The surroundings gain entropy when they absorb heat released by the reaction: $\Delta S_{\text{surr}} = -\Delta H/T$ . Adding the system and surroundings and multiplying by $-T$ gives

-T\Delta S_{\text{total}} = \Delta H - T\Delta S_{\text{system}} = \Delta G

So a negative $\Delta G$ is the same as a positive total entropy change. The $\Delta H$ term accounts for the surroundings; the $T\Delta S$ term accounts for the system.

The four combinations

Because $T$ is always positive, the signs of $\Delta H$ and $\Delta S$ decide how feasibility depends on temperature.

Feasibility and temperature

$\Delta H$	$\Delta S$	$\Delta G = \Delta H - T\Delta S$	Feasible?
negative	positive	always negative	at all temperatures
positive	negative	always positive	at no temperature
negative	negative	negative at low $T$ , positive at high $T$	only below $T = \Delta H/\Delta S$
positive	positive	positive at low $T$ , negative at high $T$	only above $T = \Delta H/\Delta S$

The temperature at which feasibility changes is found by setting $\Delta G = 0$ :

T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}

with $\Delta S$ in $\text{kJ K}^{-1}\ \text{mol}^{-1}$ so that the units cancel to give kelvin.

Seeing it on a graph

Since $\Delta H$ and $\Delta S$ change only slightly with temperature, a graph of $\Delta G$ against $T$ is approximately a straight line with intercept $\Delta H$ (at $T = 0$ ) and gradient $-\Delta S$ .

The first graph is for the decomposition of calcium carbonate ( $\Delta H^{\ominus} = +178\ \text{kJ mol}^{-1}$ , $\Delta S^{\ominus} = +160.4\ \text{J K}^{-1}\ \text{mol}^{-1}$ ), plotted as $\Delta G$ in kJ mol⁻¹ against $T$ in K. The line falls and crosses zero at about 1110 K: above this, decomposition is feasible.

y = 178 - 0.1604x (1110, -100) -- (1110, 200)

The second graph is for the Haber process ( $\Delta H^{\ominus} = -92\ \text{kJ mol}^{-1}$ , $\Delta S^{\ominus} = -198.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ ). The line rises and crosses zero at about 463 K: the forward reaction is feasible only below this temperature.

y = -92 + 0.1988x (463, -100) -- (463, 100)

A positive gradient means $\Delta S$ is negative; a negative gradient means $\Delta S$ is positive. Exam questions sometimes give you the graph and ask you to read off $\Delta H$ (intercept) or $\Delta S$ (minus the gradient).

Feasible does not mean fast

$\Delta G$ tells you whether a reaction is thermodynamically possible. It says nothing about the rate. A reaction with a very negative $\Delta G$ may be immeasurably slow at room temperature if its activation energy is high.

The conversion of diamond into graphite has $\Delta G^{\ominus} \approx -3\ \text{kJ mol}^{-1}$ at 298 K, but diamonds do not turn into graphite, because breaking the strong covalent network has an enormous activation energy.
Methane and oxygen can be mixed safely ( $\Delta G^{\ominus}$ about $-818\ \text{kJ mol}^{-1}$ ) until a spark supplies the activation energy.

Such systems are described as kinetically stable (or kinetically inert) but thermodynamically unstable.

Watch out

Never write "the reaction happens quickly because $\Delta G$ is negative". A negative $\Delta G$ means the reaction is feasible; whether it is observed depends on the activation energy and the rate.

Gibbs calculations

If not given, calculate $\Delta H^{\ominus}$ (for example from enthalpies of formation) and $\Delta S^{\ominus}$ (from standard entropies).
Convert $\Delta S^{\ominus}$ from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000.
Convert the temperature to kelvin.
Substitute into $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ and state the sign and units.
For a minimum or maximum temperature, set $\Delta G = 0$ and calculate $T = \Delta H/\Delta S$ . State whether the reaction is feasible above or below it, using the signs.

Worked examples

Decomposition of calcium carbonate

For $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ , $\Delta H^{\ominus} = +178\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = +160.4\ \text{J K}^{-1}\ \text{mol}^{-1}$ .

(a) Calculate $\Delta G^{\ominus}$ at 298 K and comment. (b) Calculate the minimum temperature at which the decomposition becomes feasible.

Solution

(a)

\Delta G^{\ominus} = 178 - 298 \times 0.1604 = 178 - 47.8 = +130\ \text{kJ mol}^{-1}

Positive, so not feasible at room temperature: limestone is stable.

(b)

T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} = \frac{178}{0.1604} = 1110\ \text{K}

Since $\Delta S$ is positive, $-T\Delta S$ becomes more negative as $T$ rises, so the reaction is feasible above 1110 K (about $837\ ^\circ\text{C}$ ). Lime kilns operate above this temperature.

The Haber process

For $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ , $\Delta H^{\ominus} = -92\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = -198.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ .

(a) Show that the reaction is feasible at 298 K. (b) Calculate the temperature above which it is no longer feasible. (c) Suggest why the industrial process operates at about 700 K anyway.

Solution

(a)

\Delta G^{\ominus} = -92 - 298 \times (-0.1988) = -92 + 59.2 = -32.8\ \text{kJ mol}^{-1}

Negative, so feasible.

(b)

T = \frac{-92}{-0.1988} = 463\ \text{K}

Above 463 K, $-T\Delta S$ (which is positive) outweighs $\Delta H$ , so $\Delta G^{\ominus}$ is positive.

(c) At low temperature the rate is far too slow, even with a catalyst. A compromise temperature gives an acceptable rate. (Feasibility under standard conditions is not the whole story: high pressure and removal of ammonia shift the position of equilibrium in favour of ammonia, so a useful yield is still obtained.)

Melting ice

For $\ce{H2O(s) -> H2O(l)}$ , $\Delta H^{\ominus} = +6.01\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = +22.0\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Calculate $\Delta G$ at 263 K and at 283 K, and the temperature at which $\Delta G = 0$ .

Solution

\Delta G_{263} = 6.01 - 263 \times 0.0220 = 6.01 - 5.786 = +0.22\ \text{kJ mol}^{-1}

\Delta G_{283} = 6.01 - 283 \times 0.0220 = 6.01 - 6.226 = -0.22\ \text{kJ mol}^{-1}

T = \frac{6.01}{0.0220} = 273\ \text{K}

Ice does not melt at $-10\ ^\circ\text{C}$ but does at $+10\ ^\circ\text{C}$ . At 273 K, $\Delta G = 0$ : ice and water coexist in equilibrium at the melting point.

From enthalpies of formation and entropies

Sodium hydrogencarbonate decomposes on heating:

\ce{2NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}

Substance	$\Delta H^{\ominus}_{\text{f}}$ / kJ mol⁻¹	$S^{\ominus}$ / J K⁻¹ mol⁻¹
$\ce{NaHCO3(s)}$	–950.8	101.7
$\ce{Na2CO3(s)}$	–1130.7	135.0
$\ce{H2O(g)}$	–241.8	188.7
$\ce{CO2(g)}$	–393.5	213.6

Calculate the minimum temperature for decomposition.

Solution

\Delta H^{\ominus} = [-1130.7 + (-241.8) + (-393.5)] - 2(-950.8) = -1766.0 + 1901.6 = +135.6\ \text{kJ mol}^{-1}

\Delta S^{\ominus} = (135.0 + 188.7 + 213.6) - 2(101.7) = +333.9\ \text{J K}^{-1}\ \text{mol}^{-1}

T = \frac{135.6}{0.3339} = 406\ \text{K}

Both $\Delta H$ and $\Delta S$ are positive, so decomposition is feasible above 406 K (about $133\ ^\circ\text{C}$ ). This is why baking powder releases carbon dioxide in a hot oven.

Extracting ΔH and ΔS from two values of ΔG (exam-hard)

For a reaction, $\Delta G = +20.0\ \text{kJ mol}^{-1}$ at 300 K and $-4.0\ \text{kJ mol}^{-1}$ at 500 K. Assuming $\Delta H$ and $\Delta S$ do not change with temperature, calculate $\Delta H$ , $\Delta S$ and the temperature at which the reaction becomes feasible.

Solution

Write the Gibbs equation at each temperature:

20.0 = \Delta H - 300\Delta S \qquad -4.0 = \Delta H - 500\Delta S

Subtract the second from the first:

24.0 = 200\Delta S \quad\Rightarrow\quad \Delta S = 0.120\ \text{kJ K}^{-1}\ \text{mol}^{-1} = +120\ \text{J K}^{-1}\ \text{mol}^{-1}

\Delta H = 20.0 + 300 \times 0.120 = +56.0\ \text{kJ mol}^{-1}

T = \frac{56.0}{0.120} = 467\ \text{K}

Feasible above 467 K. (This is the graph method in algebra: the gradient between the two points is $-\Delta S$ .)

Common mistakes

Unit mismatch. Substituting $\Delta S = 160.4$ (in J) with $\Delta H = 178$ (in kJ). Always divide $\Delta S$ by 1000.
Celsius. $T$ must be in kelvin.
Wrong direction. Saying the Haber process is feasible "above 463 K". Check the signs: negative $\Delta S$ means higher temperatures make $\Delta G$ more positive.
Confusing feasibility with rate. A feasible reaction may not occur at a measurable rate.
Losing signs when $\Delta S$ is negative. $-T \times (-0.1988)$ is $+0.1988T$ . Use brackets.

Exam technique

Exam tip

"State whether the reaction is feasible" needs a reason: " $\Delta G$ is negative (or positive), so it is (or is not) feasible".
When asked to "predict the effect of temperature on feasibility", refer to the sign of $\Delta S$ and the size of the $T\Delta S$ term compared with $\Delta H$ .
When calculating $T$ at which a reaction becomes feasible, state "above" or "below" it. Without this the final mark is usually lost.
Assumptions examiners like: " $\Delta H$ and $\Delta S$ do not vary with temperature" and "the reaction occurs under standard conditions".
If a reaction is feasible but does not occur, the expected explanation is a high activation energy, so the rate is very slow.

Summary

$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ , with $\Delta S$ in kJ K⁻¹ mol⁻¹ and $T$ in K.
A reaction is feasible when $\Delta G$ is negative.
$\Delta H < 0$ , $\Delta S > 0$ : always feasible. $\Delta H > 0$ , $\Delta S < 0$ : never feasible.
When $\Delta H$ and $\Delta S$ have the same sign, feasibility switches at $T = \Delta H / \Delta S$ : positive $\Delta S$ means feasible above it; negative $\Delta S$ means feasible below it.
$\Delta G$ against $T$ is a straight line with intercept $\Delta H$ and gradient $-\Delta S$ .
Feasibility says nothing about rate; high activation energy can make a feasible reaction too slow to observe.

Practice questions

Question

State the Gibbs equation and the units of each term.
For a reaction, $\Delta H^{\ominus} = -46\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = -99\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Calculate $\Delta G^{\ominus}$ at 298 K and state whether the reaction is feasible.
For the decomposition of magnesium carbonate, $\Delta H^{\ominus} = +117\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = +175\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Calculate the minimum temperature for decomposition.
Explain why a reaction with $\Delta H$ positive and $\Delta S$ negative can never be feasible.
For $\ce{2SO2(g) + O2(g) -> 2SO3(g)}$ , $\Delta H^{\ominus} = -197.8\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = -187.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Calculate $\Delta G^{\ominus}$ at 298 K and the temperature above which the reaction is not feasible.
Ammonium nitrate dissolves with $\Delta H^{\ominus} = +25.7\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = +108.7\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Show that dissolving is feasible at 298 K.
The conversion of diamond to graphite has a negative $\Delta G^{\ominus}$ at 298 K. Explain why diamonds do not change into graphite.
A graph of $\Delta G$ against $T$ for a reaction is a straight line passing through $(0\ \text{K}, +80\ \text{kJ mol}^{-1})$ and $(800\ \text{K}, -16\ \text{kJ mol}^{-1})$ . Find $\Delta H$ and $\Delta S$ .
Iron(III) oxide is reduced by carbon: $\ce{Fe2O3(s) + 3C(s) -> 2Fe(s) + 3CO(g)}$ . Data: $\Delta H^{\ominus}_{\text{f}}$ / kJ mol⁻¹: $\ce{Fe2O3}$ –824.2, $\ce{CO}$ –110.5. $S^{\ominus}$ / J K⁻¹ mol⁻¹: $\ce{Fe2O3}$ 87.4, C 5.7, Fe 27.3, CO 197.7. Calculate the minimum temperature at which the reduction is feasible.
For a reaction, $\Delta G = +20.0\ \text{kJ mol}^{-1}$ at 300 K and $-4.0\ \text{kJ mol}^{-1}$ at 500 K. Without calculating, explain how you can tell the sign of $\Delta S$ , and then state whether $\Delta H$ must be positive or negative.

Answers

$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ ; $\Delta G$ and $\Delta H$ in kJ mol⁻¹, $T$ in K, $\Delta S$ in kJ K⁻¹ mol⁻¹ (data are usually given in J K⁻¹ mol⁻¹ and must be divided by 1000).
$\Delta G^{\ominus} = -46 - 298(-0.099) = -46 + 29.5 = -16.5\ \text{kJ mol}^{-1}$ . Negative, so feasible.
$T = 117 / 0.175 = 669\ \text{K}$ . $\Delta S$ is positive, so decomposition is feasible above 669 K.
$\Delta H$ is positive and $-T\Delta S$ is also positive (since $\Delta S$ is negative and $T$ is positive), so $\Delta G$ is positive at every temperature.
$\Delta G^{\ominus} = -197.8 + 298(0.1878) = -197.8 + 56.0 = -141.8\ \text{kJ mol}^{-1}$ (feasible). $T = 197.8/0.1878 = 1053\ \text{K}$ ; above this the reaction is not feasible.
$\Delta G^{\ominus} = 25.7 - 298(0.1087) = 25.7 - 32.4 = -6.7\ \text{kJ mol}^{-1}$ . Negative, so feasible: the large entropy increase outweighs the endothermic enthalpy change.
The reaction has a very high activation energy because many strong covalent bonds in the giant diamond lattice must be broken. The rate at room temperature is negligible, so diamond is kinetically stable even though it is thermodynamically unstable relative to graphite.
Intercept: $\Delta H = +80\ \text{kJ mol}^{-1}$ . Gradient $= (-16 - 80)/800 = -0.120\ \text{kJ K}^{-1}\ \text{mol}^{-1} = -\Delta S$ , so $\Delta S = +0.120\ \text{kJ K}^{-1}\ \text{mol}^{-1} = +120\ \text{J K}^{-1}\ \text{mol}^{-1}$ .
$\Delta H^{\ominus} = 3(-110.5) - (-824.2) = -331.5 + 824.2 = +492.7\ \text{kJ mol}^{-1}$ . $\Delta S^{\ominus} = [2(27.3) + 3(197.7)] - [87.4 + 3(5.7)] = 647.7 - 104.5 = +543.2\ \text{J K}^{-1}\ \text{mol}^{-1}$ . $T = 492.7 / 0.5432 = 907\ \text{K}$ ; reduction is feasible above about 907 K.
$\Delta G$ becomes more negative as $T$ increases, so $-T\Delta S$ must be becoming more negative: $\Delta S$ is positive. At 300 K, $\Delta G$ is positive even though $-T\Delta S$ is negative, so $\Delta H$ must be positive (and larger than $300\Delta S$ ).

The Gibbs equation

Where the equation comes from

The four combinations

Seeing it on a graph

Feasible does not mean fast

Worked examples

Exam technique

Summary

Practice questions

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