Ch 3 · Entropy change, ΔS · AS & A Level Chemistry 9701

Topic 3 of 4 in Chemical energetics — AS & A Level Chemistry 9701 — 56 lessons.

Introduction

Enthalpy alone cannot explain why some endothermic processes, such as ice melting or limestone decomposing, happen spontaneously. A second thermodynamic quantity, entropy, measures how disordered a system is, and it works alongside enthalpy to determine whether a change is thermodynamically feasible.

Key ideas — Defining entropy

Entropy, S, is a measure of the number of possible ways in which the particles of a system, and the energy they possess, can be arranged. A system with more possible arrangements of particles and energy has higher entropy. Entropy has units of mathrmJ\ K^-1\ mol^-1, and standard entropies, S^ominus, are tabulated for elements and compounds at 298 K and 1 bar; unlike enthalpies of formation, the standard entropy of an element is not zero.

🧑‍🏫 Teacher's explanation

Imagine you have a box of 100 distinct colored marbles. If all the marbles are sorted perfectly by color, there is only one way to arrange them. This is a state of low entropy. Now, imagine you shake the box; the marbles mix up, and there are millions of different ways they could be positioned. This state of disorder has high entropy. In chemistry, we aren't just talking about the position of atoms, but also the ways energy can be distributed among them.

📖 Definitions

🔍 Worked example — Standard

Predict the sign (positive or negative) of Δ S for the following process: 2H_2O(l) arrow 2H_2(g) + O_2(g). Explain your reasoning.

🔍 Worked example — Exam-style

A reaction has Δ H = -150 kJ mol^-1 and Δ S = -200 J K^-1 mol^-1. Calculate the Gibbs free energy change (Δ G) at 298 K. (Use Δ G = Δ H - TΔ S)

⚠️ Common errors

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🎯 Exam question

Key ideas — Entropy changes on a change of state or temperature change

Entropy increases in the order solid < liquid < gas, since particles gain increasing freedom to move and increasing numbers of ways to arrange their positions and energy. Melting (Δ S > 0), boiling (Δ S > 0, and much larger than melting) and sublimation all increase entropy, while freezing and condensing decrease it. Raising the temperature of a single phase also increases entropy (Δ S > 0 as T increases), since particles gain access to a wider range of vibrational, rotational and translational energy levels; lowering temperature decreases entropy.

🧑‍🏫 Teacher's explanation

Imagine you have a deck of cards. If the cards are neatly stacked in a box, there is only one way to arrange them (low entropy). If you throw them into the air, they can land in millions of different patterns (high entropy). In chemistry, entropy is a measure of this 'disorder' or, more accurately, the number of ways particles can be arranged in space and energy.

📖 Definitions

🔍 Worked example — Easy

Predict the sign of the entropy change (Δ S) for the following process: H_2O(l) arrow H_2O(g)

🔍 Worked example — Exam-style

A reaction involves the decomposition of calcium carbonate: CaCO_3(s) arrow CaO(s) + CO_2(g). Predict the sign of Δ S and explain why, considering the change in the number of moles of gas.

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🎯 Exam question

Key ideas — Entropy changes on a change in the number of gaseous molecules

Because gas particles have far more positional and energetic freedom than particles in a solid or liquid, the entropy of a reaction is dominated by any change in the number of moles of gas. If the number of moles of gaseous product exceeds the number of moles of gaseous reactant, Δ S is positive (more possible arrangements are created); if the number of moles of gas decreases, Δ S is negative. For example, CaCO_3(s) arrow CaO(s) + CO_2(g) has Δ S > 0, since one mole of gas is created from none, while N_2(g) + 3H_2(g) arrow 2NH_3(g) has Δ S < 0, since four moles of gas become two.

🧑‍🏫 Teacher's explanation

Think of entropy as a measure of 'chaos' or, more accurately, the number of different ways you can arrange the particles in a system. Imagine a classroom full of students. If every student is sitting perfectly still in their assigned desk, there are very few ways to arrange them. If every student is running around randomly, there are millions of possible positions they could occupy. The 'running around' state has much higher entropy.

📖 Definitions

🔍 Worked example — Easy

Predict the sign of Δ S for the following reaction: CaCO_3(s) arrow CaO(s) + CO_2(g)

🔍 Worked example — Standard

Calculate the Δ S^° for the reaction N_2(g) + 3H_2(g) arrow 2NH_3(g) given: S^°(N_2) = 202 J K^-1 mol^-1, S^°(H_2) = 118 J K^-1 mol^-1, S^°(NH_3) = 168 J K^-1 mol^-1.

🔍 Worked example — Exam-style

For the reaction 2H_2O(l) arrow 2H_2(g) + O_2(g), calculate Δ S^° if S^°(H_2O, l) = 70 J K^-1 mol^-1, S^°(H_2) = 131 J K^-1 mol^-1, and S^°(O_2) = 205 J K^-1 mol^-1.

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🎯 Exam question

Key ideas — Calculating entropy change for a reaction

The standard entropy change of a reaction is calculated, like an enthalpy change of reaction, from standard molar entropies of reactants and products:

🧑‍🏫 Teacher's explanation

Imagine you are organizing a messy bedroom. If all your clothes are neatly folded in a drawer, there are very few ways they can be arranged. This is low entropy. If you throw them all on the floor, there are millions of ways they could land, and many more ways they could be 'messy'. Entropy is essentially a measure of this 'disorder' or, more accurately, the number of different ways the particles in a system can be arranged in space and energy.

📖 Definitions

🔍 Worked example — Easy

Calculate the entropy change for the reaction: CaCO_3(s) ightarrow CaO(s) + CO_2(g). Given S^ominus(CaCO_3) = 92.3 J K^-1 mol^-1, S^ominus(CaO) = 39.7 J K^-1 mol^-1, and S^ominus(CO_2) = 213.6 J K^-1 mol^-1.

🔍 Worked example — Standard

Calculate Δ S^ominus for N_2(g) + 3H_2(g) ightarrow 2NH_3(g). Given S^ominus(N_2) = 191.6 J K^-1 mol^-1, S^ominus(H_2) = 130.7 J K^-1 mol^-1, and S^ominus(NH_3) = 192.8 J K^-1 mol^-1.

🔍 Worked example — Exam-style

A reaction A(g) + B(g) ightarrow C(g) + D(l) has Δ S^ominus = +45.0 J K^-1 mol^-1. If S^ominus(A) = 200.0 J K^-1 mol^-1 and S^ominus(B) = 210.0 J K^-1 mol^-1, calculate S^ominus(C) assuming S^ominus(D) = 50.0 J K^-1 mol^-1.

⚠️ Common errors

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🎯 Exam question

Key ideas — Why increasing temperature increases entropy

At a higher temperature, particles have more kinetic energy, which can be distributed among a much larger number of accessible vibrational, rotational and translational energy levels. Since entropy counts the number of ways energy (as well as position) can be arranged, spreading a fixed quantity of energy over more available levels, and adding more total energy to spread, increases the number of possible arrangements, so entropy rises as temperature increases.

🧑‍🏫 Teacher's explanation

Imagine you are organizing a massive party in a ballroom. If the party is small and everyone is sitting in assigned seats, there are very few ways to arrange the guests. This is low entropy. Now, imagine the music turns up and everyone starts dancing wildly around the room. There are suddenly thousands of different positions and movements the guests could be in at any given second. This represents high entropy.

Ch 3 · Entropy change, ΔS · AS & A Level Chemistry 9701

All chapters in Chemical energetics