Enthalpy and entropy

In a nutshell

Enthalpy alone cannot tell you whether a reaction can happen: some endothermic reactions go on their own. The missing piece is entropy, S, a measure of how spread out energy is.

This subtopic shows you how to predict and calculate the entropy change, ΔS, and then combine it with ΔH and temperature in the Gibbs' equation, ΔG = ΔH - TΔS, to decide whether a process is feasible. It ends with the big limitation: ΔG says nothing about rate.

Assumed knowledge: Enthalpy changes; Reaction rates.

Core content

Entropy: a measure of the dispersal of energy

Entropy, S, is a measure of the dispersal of energy in a system, which is greater the more disordered a system is.

Particles store energy in their movement (vibrating, rotating, moving from place to place). The more freely particles can move and arrange themselves, the more ways the energy can be shared out, so the more dispersed the energy is and the higher the entropy.

Facts about entropy you need:

• Units: J K^-1 mol^-1. Note joules, not kilojoules.
• Standard entropies, S^⦵, are quoted for one mole of a substance under standard conditions (100 kPa and 298 K).
• Every substance has a positive standard entropy. Only a perfect crystal at 0 K would have zero entropy.
• Elements do not have zero entropy. (This is different from Δ_fH^⦵, which is zero for an element in its standard state.)

Entropy, together with enthalpy, is the model chemists use to explain thermodynamic stability: whether a substance or a reaction mixture tends to change on its own.

Still don't get it? · entropy as the dispersal of energy

Imagine sharing out 10 sweets. If only 2 friends are allowed to hold sweets, there are only a few ways to share them. If 50 friends can hold sweets, there are a huge number of ways, and the sweets end up spread out among lots of people.

Now rebuild the chemistry one step at a time:

1. The "sweets" are small packets of energy. The "friends" are all the different ways particles can move and be arranged.
2. In a solid, particles are locked in place and can only vibrate. Few ways to hold the energy, so the energy is concentrated.
3. In a liquid, particles can also slide past each other. More ways, so the energy is more spread out.
4. In a gas, particles move freely and randomly through the whole container. By far the most ways, so the energy is most spread out.
5. More ways to arrange particles and energy means more disorder, and more disorder means higher entropy.

Exam version: entropy is a measure of the dispersal of energy in a system, which is greater the more disordered a system. Gases have the highest entropy because their particles have the most freedom of movement and are the most disordered.

Entropy: solids, liquids and gases

For the same substance, entropy increases in the order solid < liquid < gas.

• Solid: particles vibrate about fixed positions, so the system is ordered and entropy is low.
• Liquid: particles move around each other, so there is more disorder and entropy is higher.
• Gas: particles move quickly and randomly with lots of space between them, so there is most disorder and entropy is much higher.

The jump on boiling is much bigger than the jump on melting, because turning a liquid into a gas gives the particles far more freedom than turning a solid into a liquid.

Reading the graph:

• Entropy rises gradually within each state as temperature rises, because particles move more and energy spreads out more.
• It jumps vertically at the melting point and, by much more, at the boiling point.

Real standard entropies at 298 K show the same pattern. The biggest values belong to gases:

Notice water: 69.9 J K^-1 mol^-1 as a liquid, but 188.8 J K^-1 mol^-1 as a gas.

Entropy: reactions that change the number of gaseous molecules

Because gases have so much more entropy than solids and liquids, the number of gaseous molecules usually decides the sign of ΔS for a reaction.

• More moles of gas in the products than the reactants: disorder increases, so ΔS is positive.
• Fewer moles of gas in the products: disorder decreases, so ΔS is negative.
• No change in moles of gas: ΔS is usually small, and you cannot confidently predict its sign from the equation alone.

To earn the mark, count the moles of gas on each side and quote the numbers:

A change of state works the same way. H₂O(l) → H₂O(g) has a positive ΔS because a liquid becomes a gas.

Entropy: calculating the entropy change of a reaction

You are given the standard entropies of each reactant and product. Then:

$\Delta S = \sum S(\text{products}) - \sum S(\text{reactants})$

Three rules make this work:

1. Multiply each entropy by its coefficient in the balanced equation.
2. Include every element. Elements have non-zero entropies, so they never drop out.
3. Give ΔS in J K^-1 mol^-1 with a sign (+ or -).

"Related quantities" means you can also be asked to rearrange the same equation, for example to find the unknown entropy of one substance when you are given ΔS and the other entropies (Worked example 2).

Always sense-check the sign against the moles of gas. If the number of gas moles rises and you get a negative ΔS, you have subtracted the wrong way round.

Free energy: feasibility depends on ΔH and TΔS

A feasible process is one that can take place on its own, without continuous outside help. Whether a process is feasible depends on two things:

• the enthalpy change of the system, ΔH: a negative (exothermic) ΔH favours feasibility;
• the entropy change of the system multiplied by temperature, TΔS: a positive ΔS favours feasibility, and the higher the temperature, the more the entropy term counts.

That is why the sign of ΔH alone is not enough. An endothermic process can still be feasible if ΔS is positive and the temperature is high enough. Dissolving ammonium nitrate in water is an everyday example: it is endothermic (the solution gets cold), yet it happens on its own.

The four combinations of signs:

Still don't get it? · why temperature can flip a reaction from not feasible to feasible

Think of a tug of war between two teams. Team "Enthalpy" always pulls with the same strength. Team "Entropy" gets stronger and stronger the hotter it gets. At a low temperature Team Enthalpy wins; heat things up enough and Team Entropy takes over.

Now rebuild it with calcium carbonate, CaCO₃(s) → CaO(s) + CO₂(g), where ΔH = +178 kJ mol^-1 and ΔS = +0.161 kJ K^-1 mol^-1:

1. ΔH is positive, which works against feasibility. It is the same at every temperature.
2. ΔS is positive (a gas is made), which works for feasibility. Its effect is the term TΔS, so it grows as T grows.
3. At 298 K, TΔS is only about 48 kJ mol^-1. That is smaller than 178, so ΔG is positive: not feasible. Limestone does not fall apart at room temperature.
4. As T rises, TΔS gets bigger. At about 1106 K it exactly equals 178, so ΔG = 0. That is the balance point.
5. Above that temperature TΔS is bigger than ΔH, so ΔG is negative: feasible. This is why limestone is decomposed in very hot kilns.

Exam version: the feasibility of a process depends on the entropy change and temperature, TΔS, and the enthalpy change, ΔH. A process is feasible when ΔG = ΔH - TΔS is negative, and the minimum temperature for feasibility comes from setting ΔG = 0, so T = ΔH/ΔS.

Free energy: the Gibbs' equation

The free energy change, ΔG, combines both effects into one number using the Gibbs' equation:

$\Delta G = \Delta H - T\Delta S$

A process is feasible when ΔG has a negative value (ΔG < 0).

Get the units right before you substitute, because this is where most marks go:

The temperature where feasibility changes. At the boundary, ΔG = 0, so:

$0 = \Delta H - T\Delta S \quad \Rightarrow \quad T = \frac{\Delta H}{\Delta S}$

Then decide the direction from the signs (see the table above): above this temperature if ΔH and ΔS are both positive, below it if both are negative.

The same idea finds the temperature of a change of state. At the boiling point under standard pressure, liquid and gas are in equilibrium, so ΔG = 0 and T = ΔH/ΔS gives the boiling point (Worked example 6).

ΔG against temperature is a straight line. If ΔH and ΔS are assumed not to change with temperature, ΔG = ΔH - TΔS has the form y = c + mx:

• intercept on the ΔG axis (at T = 0) = ΔH
• gradient = -ΔS
• the line crosses ΔG = 0 at the temperature where feasibility changes

Reading the two lines:

• CaCO₃ decomposition slopes down (ΔS positive). ΔG becomes negative above 1106 K.
• Ammonia formation slopes up (ΔS negative). ΔG is negative only below 462 K.

Free energy: the limitations of ΔG predictions

A negative ΔG tells you a reaction can happen. It does not tell you how fast it happens.

The limitation, in the words that earn the mark:

• The reaction may have a high activation energy.
• So the rate may be too slow for any reaction to be seen, even though ΔG is negative.

Examples you can quote:

• Diamond → graphite has a negative ΔG at room temperature, but diamonds do not visibly turn into graphite. Diamond is thermodynamically unstable but kinetically stable.
• N₂(g) + 3H₂(g) → 2NH₃(g) has ΔG of about -33 kJ mol^-1 at 298 K, yet a mixture of nitrogen and hydrogen at room temperature does not react at a noticeable rate.
• A mixture of methane and air has a very negative ΔG for combustion, but nothing happens until a spark or flame supplies the activation energy.

So ΔG is a thermodynamic prediction (is it possible?), and kinetics (how fast?) has to be considered separately.

Worked examples

Data are given in each question, as in the exam. Answers are given to the significant figures of the least precise data.

1. Entropy change of a reaction from standard entropies

Calculate ΔS for N₂(g) + 3H₂(g) → 2NH₃(g).

S / J K^-1 mol^-1: N₂(g) 192; H₂(g) 131; NH₃(g) 193.

$\sum S(\text{products}) = 2 \times 193 = 386 \text{ J K}^{-1}\text{ mol}^{-1}$ $\sum S(\text{reactants}) = 192 + (3 \times 131) = 585 \text{ J K}^{-1}\text{ mol}^{-1}$ $\Delta S = 386 - 585 = -199 \text{ J K}^{-1}\text{ mol}^{-1}$

Sense check: 4 moles of gas become 2, so ΔS should be negative. It is.

Where marks go: forgetting the 3 in front of H₂ gives 386 - 323 = +63, which has the wrong sign. Leaving out N₂ because "it is an element" gives the wrong answer too.

2. Finding an unknown standard entropy

For CaCO₃(s) → CaO(s) + CO₂(g), ΔS = +161 J K^-1 mol^-1.

S / J K^-1 mol^-1: CaCO₃(s) 93; CO₂(g) 214. Calculate the standard entropy of CaO(s).

$\Delta S = [S(\text{CaO}) + S(\text{CO}_2)] - S(\text{CaCO}_3)$ $+161 = S(\text{CaO}) + 214 - 93$ $S(\text{CaO}) = 161 - 214 + 93 = +40 \text{ J K}^{-1}\text{ mol}^{-1}$

Sense check: a standard entropy can never be negative, and a solid should be low. +40 fits.

3. ΔH, ΔS and ΔG at a given temperature

For CaCO₃(s) → CaO(s) + CO₂(g), use the data to calculate ΔG at 298 K and at 1500 K, and say whether the reaction is feasible at each.

Step 1: ΔH from enthalpies of formation.

$\Delta H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$ $\Delta H = (-635 + (-394)) - (-1207) = +178 \text{ kJ mol}^{-1}$

Step 2: ΔS, then convert to kJ.

$\Delta S = (40 + 214) - 93 = +161 \text{ J K}^{-1}\text{ mol}^{-1}$ $\Delta S = \frac{+161}{1000} = +0.161 \text{ kJ K}^{-1}\text{ mol}^{-1}$

Step 3: ΔG at 298 K.

$\Delta G = \Delta H - T\Delta S = 178 - (298 \times 0.161)$ $\Delta G = 178 - 47.978 = +130 \text{ kJ mol}^{-1}$

ΔG is positive, so the reaction is not feasible at 298 K.

Step 4: ΔG at 1500 K.

$\Delta G = 178 - (1500 \times 0.161) = 178 - 241.5 = -63.5 \text{ kJ mol}^{-1}$

ΔG is negative, so the reaction is feasible at 1500 K. (Check: this is the point at 1500 K on the CaCO₃ line of the ΔG graph.)

Where marks go: using ΔS = 161 without dividing by 1000 gives 178 - 47 978 = about -47 800 kJ mol^-1, which is absurd. If your ΔG is tens of thousands of kJ, you forgot to convert.

4. Minimum temperature for a reaction to become feasible

Using ΔH = +178 kJ mol^-1 and ΔS = +161 J K^-1 mol^-1 for the decomposition of CaCO₃, calculate the minimum temperature at which the reaction is feasible.

$\text{At the boundary, } \Delta G = 0 \text{, so } T = \frac{\Delta H}{\Delta S}$ $T = \frac{178}{0.161} = 1106 \text{ K} \quad (1110 \text{ K to 3 s.f.})$

Both ΔH and ΔS are positive, so the reaction is feasible above 1106 K. This matches where the CaCO₃ line crosses ΔG = 0 on the graph.

Where marks go: dividing 178 by 161 (not 0.161) gives 1.11 K, which is impossible for limestone decomposition.

5. Maximum temperature for a reaction to be feasible

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = -92.0 kJ mol^-1 and ΔS = -199 J K^-1 mol^-1. Find the temperature range over which the reaction is feasible.

$T = \frac{\Delta H}{\Delta S} = \frac{-92.0}{-0.199} = 462 \text{ K}$

Both values are negative, so -TΔS is positive and grows with temperature. The reaction is feasible only below 462 K.

Where marks go: students often get the number right but write "above 462 K". Always state the direction, and check it against the sign table.

6. Boiling point from ΔH and entropies (ΔG = 0)

For Br₂(l) → Br₂(g), ΔH = +30.9 kJ mol^-1. S / J K^-1 mol^-1: Br₂(l) 152.2; Br₂(g) 245.5. Estimate the boiling point of bromine.

$\Delta S = 245.5 - 152.2 = +93.3 \text{ J K}^{-1}\text{ mol}^{-1} = +0.0933 \text{ kJ K}^{-1}\text{ mol}^{-1}$ $\text{At the boiling point, } \Delta G = 0 \text{, so } T = \frac{\Delta H}{\Delta S} = \frac{30.9}{0.0933}$ T = 331 \text{ K} \quad (58\,^{\circ}\text{C})

This is very close to the measured boiling point of bromine (59 °C), which shows the method works.

Where marks go: ΔG = 0 at the boiling point because liquid and gas are in equilibrium. Students who set ΔG to a negative value cannot get a temperature at all.

7. Finding ΔH and ΔS from a ΔG against temperature graph

Use the ammonia line on the ΔG graph in Core content to find ΔH and ΔS.

ΔH is the intercept at T = 0:

$\Delta H = -92 \text{ kJ mol}^{-1}$

The gradient is -ΔS. Read two points: (0 K, -92) and (1000 K, +107).

$\text{gradient} = \frac{107 - (-92)}{1000 - 0} = \frac{199}{1000} = +0.199 \text{ kJ K}^{-1}\text{ mol}^{-1}$ $\Delta S = -\text{gradient} = -0.199 \text{ kJ K}^{-1}\text{ mol}^{-1} = -199 \text{ J K}^{-1}\text{ mol}^{-1}$

Where marks go: taking the gradient as ΔS gives the wrong sign. An upward-sloping line always means ΔS is negative.

Common exam mistakes

• Not dividing ΔS by 1000. ΔS is in J K^-1 mol^-1 but ΔH is in kJ mol^-1. Substituting 161 instead of 0.161 into ΔG = ΔH - TΔS wrecks the answer.
• Using temperature in °C. T in the Gibbs' equation must be in kelvin (add 273).
• Giving ΔS in the wrong units, such as kJ mol^-1 or J mol^-1. It needs J K^-1 mol^-1 (the K^-1 is part of the unit).
• Forgetting the coefficients when summing entropies: 3H₂ means 3 × S(H₂).
• Treating elements as having zero entropy. That is true of Δ_fH^⦵, not of S. O₂, H₂, N₂ and solid metals all have positive entropies.
• Subtracting the wrong way round. It is products minus reactants. Check your sign against the change in moles of gas.
• Vague explanations of ΔS. "It gets more disordered" is not enough. Quote the number of moles of gas on each side (for example, "4 mol of gas form 2 mol of gas, so disorder decreases").
• Defining entropy only as "disorder". The spec wording is a measure of the dispersal of energy in a system, which is greater the more disordered the system.
• Saying a reaction is feasible when ΔG is positive, or mixing up the direction. Feasible means ΔG negative.
• Giving T = ΔH/ΔS without the direction. Say above that temperature (both positive) or below it (both negative).
• Sign slips with -TΔS. When ΔS is negative, -TΔS is positive: -(298 × -0.199) = +59.3, not -59.3.
• Claiming a negative ΔG means the reaction "will happen quickly" or "will definitely happen". ΔG says nothing about rate. The limitation needs high activation energy and slow rate.
• Writing "activation energy" on its own for the limitation. Say the activation energy may be high, so the rate is very slow.
• Rounding too early. Keep 47.978 in your calculator, and round only the final ΔG.
• Leaving off the sign of ΔS or ΔG. Write "+161" or "-63.5", never just "161" or "63.5".

Key definitions

• Entropy, S: a measure of the dispersal of energy in a system, which is greater the more disordered a system.
• Standard entropy, S^⦵: the entropy of one mole of a substance under standard conditions (100 kPa and 298 K); units J K^-1 mol^-1; always positive.
• Entropy change of a reaction: ΔS = ΣS(products) - ΣS(reactants).
• Gibbs' equation: ΔG = ΔH - TΔS (ΔG and ΔH in kJ mol^-1, T in K, ΔS in kJ K^-1 mol^-1).
• Feasibility: a process is feasible when ΔG has a negative value.
• Temperature at which feasibility changes: found by setting ΔG = 0, so T = ΔH/ΔS.
• Limitation of ΔG: a reaction with a negative ΔG may not take place because it has a high activation energy, so the rate is too slow.

Specification

• I can explain that entropy is a measure of the dispersal of energy in a system, which is greater the more disordered the system. (5.2.2 a)
• I can explain why entropy increases from solid to liquid to gas. (5.2.2 b i)
• I can explain the sign of ΔS for a reaction from the change in the number of gaseous molecules. (5.2.2 b ii)
• I can calculate ΔS from the entropies of reactants and products, and work back to an unknown entropy. (5.2.2 c)
• I can explain that feasibility depends on ΔH and on TΔS, and use the signs of ΔH and ΔS to say how temperature affects feasibility. (5.2.2 d)
• I can use ΔG = ΔH - TΔS to calculate ΔG, state that a process is feasible when ΔG is negative, and calculate the temperature at which feasibility changes. (5.2.2 e)
• I can find ΔH and ΔS from a graph of ΔG against temperature. (5.2.2 e)
• I can explain why a reaction with a negative ΔG may not happen, in terms of a high activation energy and a slow rate. (5.2.2 f)

Related notes

• Enthalpy changes: ΔH, standard conditions and Hess' law with Δ_fH^⦵, used in every ΔG calculation
• Reaction rates: activation energy and the Boltzmann distribution, behind the kinetic limitation of ΔG
• How far?: equilibrium constants, the other way of describing how far a reaction goes
• Lattice enthalpy: enthalpy changes of solution, where entropy explains why some endothermic dissolving still happens
• Redox and electrode potentials: predicting feasibility from E values, with the same kinetic limitation