Entropy Change of the Surroundings

Introduction

The second law concerns a system together with its surroundings. A reaction can make the system more ordered yet proceed because the heat it releases increases the surroundings' entropy. For a large isothermal thermal reservoir, its entropy change can often be estimated from the system's enthalpy change as ΔS surr = −ΔH sys/T under suitable constant-pressure conditions.

Core explanation

Suppose a system releases heat q sys < 0 to a large surrounding reservoir at fixed temperature T. Energy conservation gives q surr = −q sys > 0 when other heat leaks are negligible. For the reservoir, a reversible heat-transfer reference at its essentially constant temperature gives ΔS surr = q surr/T. If the system undergoes a constant-pressure process with only P–V work, q sys = ΔH sys, so ΔS surr = −ΔH sys/T. Each link has an assumption: a defined boundary, equal-and-opposite heat, constant reservoir T and appropriate relation between system heat and ΔH.

For an exothermic reaction, ΔH sys < 0, so ΔS surr > 0. The reservoir receives energy and gains entropy. For an endothermic reaction, ΔH sys > 0 and the surroundings lose heat, so ΔS surr < 0. The sign is intuitive when system and surroundings are kept separate. It does not by itself decide spontaneity, because ΔS sys may have either sign and must be added.

Temperature affects the magnitude. Releasing 10 kJ to a reservoir at 300 K increases its entropy by about 33.3 J K⁻¹; the same heat released at 600 K gives about 16.7 J K⁻¹. A given energy transfer has a larger entropy effect at lower temperature. This dependence helps explain why an exothermic ordering process may be favorable at low temperature but not at high temperature when the system's entropy decreases.

The formula is not universal for every process. If the surroundings temperature changes substantially, one must integrate δq rev/T over its thermal path or use heat capacity data. If pressure is not constant or non-P–V work is important, q sys may not equal ΔH sys. If matter flows across the boundary, a simple heat-only relation may omit entropy carried by matter. State the idealized thermal-reservoir conditions before applying the shortcut.

Even if the actual heat transfer occurs across a finite temperature difference and is irreversible, the reservoir's entropy change can be evaluated from its own initial and final states. The reservoir approximation makes q surr/T useful because T is effectively fixed. The system's entropy change should be calculated from state data or a reversible path, not by dividing its actual irreversible q by T without justification.

At constant T and P, adding ΔS sys and ΔS surr gives ΔS total = ΔS sys − ΔH sys/T. Multiplying by −T yields ΔH sys − TΔS sys, the Gibbs-energy change. This derivation shows why ΔG is a convenient system-only criterion under these constraints.

Step-by-step reasoning

1. Define the reacting system and thermal surroundings. 2. Check constant pressure, P–V-only work and large constant-T reservoir assumptions. 3. Use q surr = −q sys = −ΔH sys. 4. Divide by positive absolute T in kelvin. 5. Add ΔS sys before deciding total-entropy direction.

Visual explanation

Draw a reaction box releasing a red heat arrow to a large thermal bath labelled T. Write ΔH sys < 0 on the box and ΔS surr = + ΔH sys /T on the bath. A separate arrow from system entropy and bath entropy converges on ΔS total.

Real-world analogy

One person's financial loss can be another's gain, but the significance of a fixed payment may depend on the recipient's overall scale. Heat leaves one defined system and enters another; dividing the reservoir's receipt by its temperature gives its entropy contribution under the model.

Real-world example

Water freezing below its melting point reduces the water system's entropy, but the released latent heat raises the surroundings' entropy. At sufficiently low temperature, the surroundings gain can outweigh the system loss, allowing freezing to proceed spontaneously.

Why?

Why does heat release give positive surroundings entropy? The surroundings receive energy at a positive temperature, giving q surr/T > 0. The system's heat has the opposite sign.

Common misconception

“An exothermic reaction always increases the system's entropy.” Exothermic describes heat flow and surroundings entropy contribution; the system's entropy may increase or decrease independently.

Worked example

A reaction at constant pressure releases 12.0 kJ to a large bath at 300 K. For the system, ΔH sys = −12.0 kJ and q sys ≈ −12.0 kJ. The bath receives +12,000 J, so ΔS surr = 12,000/300 = +40.0 J K⁻¹ for the stated reaction extent. If ΔS sys = −25 J K⁻¹, total entropy change is +15 J K⁻¹, favoring the forward direction under the stated isolated-total model.

Quick check

1. What sign is ΔS surr for an endothermic system at constant T and P under the reservoir model? Answer: Negative, because heat is taken from the surroundings.

Exam focus

Convert kJ to J when entropy is in J K⁻¹, use absolute T and show the minus sign between system heat and surroundings heat. State assumptions rather than applying −ΔH/T to any arbitrary process.

Advanced insight

The reservoir formula is a bridge from the total-entropy form of the second law to Gibbs energy. It is particularly useful for chemical reactions at fixed temperature and pressure, but composition-dependent actual reaction Gibbs energy requires care beyond standard-state quantities.

Summary

For a constant-temperature reservoir receiving the heat of a suitable constant-pressure system process, ΔS surr = −ΔH sys/T. Exothermic system changes raise surroundings entropy and endothermic ones lower it. Total entropy, not either contribution alone, determines the second-law direction.

Practice questions

1. A reaction absorbs 5.0 kJ from a 250 K reservoir. Find ΔS surr. Answer: −5000/250 = −20 J K⁻¹. 2. A system has ΔS sys = −10 J K⁻¹ and ΔS surr = +30 J K⁻¹. Find ΔS total. Answer: +20 J K⁻¹ for the combined system and surroundings. 3. Why is −ΔH sys/T unsafe if the bath temperature changes greatly? Answer: The reservoir is not isothermal; its entropy change requires a temperature-dependent heat integration or other state calculation.