Module 5: Physical Chemistry and Transition Elements · Year 13
5.2.2 Enthalpy and Entropy
Use entropy and Gibbs calculations to decide whether a process is thermodynamically feasible, and see how temperature can change the answer.
What you need to know
Open a line for a quick recap. If it feels obvious, move straight to the linked practice.
5.2.2(a) Explain entropy in terms of energy dispersal and the number of possible arrangements. Quick revision
Think of entropy as a measure of how widely energy is dispersed through a system. A system has greater entropy when its particles and energy can be arranged in more possible ways.
That is why OCR links greater entropy with greater disorder. I would keep the energy-dispersal idea in your explanation, because “more disordered” on its own does not say what entropy is measuring.
5.2.2(b)(i) Compare the entropy of solids, liquids and gases and explain the differences. Quick revision
If you compare the same substance in different states, entropy normally increases solid → liquid → gas. In a solid the particles are restricted to positions in a lattice; in a liquid they can move around one another; in a gas they have far more freedom of position and movement.
So the gas has many more possible arrangements and its energy is more dispersed. That gives it the largest entropy.
5.2.2(b)(ii) Predict the sign of an entropy change from changes in the number of gas particles. Quick revision
When a reaction makes more moles of gas, entropy will usually increase, so ΔS is positive. If the number of gas particles falls, entropy will usually decrease and ΔS is negative.
For N₂(g) + 3H₂(g) → 2NH₃(g), four moles of gas become two, so you should predict a negative ΔS before doing any calculation.
5.2.2(c) Calculate ΔS for a reaction from the entropies of reactants and products. Quick revision
For a reaction, add the entropy values for the products and subtract the total for the reactants. Include every coefficient from the balanced equation.
I would write the product total and reactant total separately before subtracting. It makes reversed subtraction and missed coefficients much easier to spot.
For N₂(g) + 3H₂(g) → 2NH₃(g), take S values of 192, 131 and 193 J K⁻¹ mol⁻¹ respectively.
- ΣS(products) = 2 × 193 = 386 J K⁻¹ mol⁻¹
- ΣS(reactants) = 192 + (3 × 131) = 585 J K⁻¹ mol⁻¹
- ΔS = 386 − 585 = −199 J K⁻¹ mol⁻¹
Answer ΔS = −199 J K⁻¹ mol⁻¹, which matches the prediction from the decrease in gas particles.
5.2.2(d) Explain qualitatively how enthalpy, entropy and temperature affect feasibility. Quick revision
Feasibility depends on the competition between ΔH and TΔS in the Gibbs equation. A negative ΔH helps make ΔG negative; a positive ΔS also helps because you subtract TΔS.
Temperature decides how much weight the entropy term has. If ΔH and ΔS favour opposite outcomes, changing T can therefore change whether the process is thermodynamically feasible.
- ΔH < 0 and ΔS > 0 → feasible at all positive temperatures
- ΔH > 0 and ΔS < 0 → not feasible at any positive temperature
- ΔH < 0 and ΔS < 0 → favoured at lower temperatures
- ΔH > 0 and ΔS > 0 → favoured at higher temperatures
5.2.2(e) Calculate ΔG using consistent units in the Gibbs equation. Quick revision
Use ΔG = ΔH − TΔS, with T in kelvin. The trap is that ΔH is usually in kJ mol⁻¹ while ΔS is usually in J K⁻¹ mol⁻¹, so put them in the same energy unit before you subtract.
I normally convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000, then calculate TΔS.
At 298 K, ΔH = +40.0 kJ mol⁻¹ and ΔS = +150 J K⁻¹ mol⁻¹.
- ΔS = 0.150 kJ K⁻¹ mol⁻¹
- TΔS = 298 × 0.150 = 44.7 kJ mol⁻¹
- ΔG = 40.0 − 44.7 = −4.7 kJ mol⁻¹
Answer ΔG = −4.7 kJ mol⁻¹.
5.2.2(e) Use the sign of ΔG to judge thermodynamic feasibility. Quick revision
For OCR, a process is thermodynamically feasible in the forward direction when ΔG is negative at the temperature you are considering.
A positive ΔG means the forward process is not thermodynamically feasible under those conditions. If ΔG = 0, you are at the boundary where feasibility changes.
5.2.2(e) Calculate the temperature at which ΔG becomes zero and feasibility changes. Quick revision
At the crossover temperature, set ΔG = 0. The Gibbs equation then becomes ΔH = TΔS, so T = ΔH/ΔS.
Match the units before dividing, and keep T in kelvin. Once you have the crossover temperature, use the signs of ΔH and ΔS to decide which side of it gives negative ΔG.
A process has ΔH = +50.0 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹.
- ΔS = 0.100 kJ K⁻¹ mol⁻¹
- T = 50.0 ÷ 0.100 = 500 K
- Here ΔH and ΔS are both positive, so the process is feasible above 500 K.
Answer The crossover temperature is 500 K.
5.2.2(f) Explain why a thermodynamically feasible reaction may still be too slow to observe. Quick revision
A negative ΔG tells you that a reaction is thermodynamically feasible; it does not tell you how quickly it will happen.
If the activation energy is large, very few collisions have enough energy to react, so the reaction may be extremely slow even though ΔG is negative. This is the kinetic limitation of using ΔG to predict what you will actually observe.