Calculate the entropy change in surroundings when 1.00 mol of $H_2O(l)$ is formed under standard conditions. $\Delta_f H = -286$ kJ mol$^{-1}$.

Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. Δf H

Exercises · 5.22

H_2O(l)

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Concept understanding — Entropy Change Prediction

Entropy Change Prediction

Imagine you have a box of marbles — all neatly arranged, reds on one side, blues on the other. Now shake the box. What happens? The colours mix. They never spontaneously unmix. That tendency — for things to go from ordered to disordered — is what entropy measures. Entropy is a measure of disorder or randomness in a system.

When you predict an entropy change, you're asking: Will this process make the system more disordered or less disordered? And by how much?

The Core Intuition

Entropy change ( $\Delta S$ ) is positive when disorder increases, negative when disorder decreases. Three things drive this:

Volume change — More space means more positions for particles → more disorder. A gas expanding into vacuum has $\Delta S > 0$ .
Temperature change — Higher temperature means particles move faster, explore more states → more disorder. Heating something increases entropy.
Phase change — Solid → liquid → gas is a ladder of increasing disorder. Melting ice increases entropy; freezing water decreases it.

Note

Entropy always increases for spontaneous processes in an isolated system (Second Law of Thermodynamics). But for a non-isolated system, entropy can decrease locally — as long as the surroundings' entropy increases enough to compensate.

The Precise Statement

For a reversible process at constant temperature, the entropy change is:

\Delta S = \frac{Q_{\text{rev}}}{T}

where $Q_{\text{rev}}$ is the heat transferred reversibly, and $T$ is the absolute temperature (in Kelvin).

For an irreversible process (which is what actually happens), you calculate $\Delta S$ by imagining a reversible path between the same initial and final states — because entropy is a state function. It depends only on where you start and end, not how you get there.

\Delta S = \int \frac{dQ_{\text{rev}}}{T}

For common cases, you use these:

Process	Formula	Sign intuition
Isothermal expansion/compression (ideal gas)	$\Delta S = nR \ln\frac{V_2}{V_1}$	Expanding → more volume → $\Delta S > 0$
Heating/cooling (constant pressure)	$\Delta S = nC_p \ln\frac{T_2}{T_1}$	Heating → higher T → $\Delta S > 0$
Phase change (melting, boiling)	$\Delta S = \frac{\Delta H_{\text{phase}}}{T_{\text{phase}}}$	Melting/boiling → more disorder → $\Delta S > 0$

How to Predict Without Calculation

You don't always need numbers. Ask these questions in order:

Is there a phase change? Solid → liquid or liquid → gas always increases entropy. Reverse decreases it.
Is the number of gas molecules changing? In a reaction, more gas molecules means more disorder. $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ (gas → liquid) has $\Delta S < 0$ because 3 gas molecules become 2 liquid molecules.
Is temperature increasing or decreasing? Higher temperature → higher entropy.
Is volume increasing? More space → more entropy. …

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The entropy change of the surroundings is determined by the heat transferred from the system at constant pressure, divided by the temperature. For the formation of 1.00 mol of liquid water under standard conditions, $\Delta S_{\text{surr}} = +960 \text{ J K}^{-1} \text{mol}^{-1}$ .

Why Gibbs Free Energy isn’t the direct tool here — but the concept of reversibility is

You might be tempted to reach for $\Delta G = \Delta H - T\Delta S$ when you see formation enthalpy and entropy. But the question asks specifically for the entropy change of the surroundings, not the system. The surroundings are the rest of the universe at constant temperature and pressure. When the system undergoes a process at constant pressure, the heat exchanged with the surroundings is exactly $-\Delta H_{\text{sys}}$ . And if that heat transfer occurs reversibly (which we assume for a standard state change), the entropy change of the surroundings is simply $Q_{\text{surr}}/T$ .

The key insight: the surroundings are so large that they absorb or release heat isothermally and reversibly, no matter what the system does. So we don’t need to know anything about the system’s entropy — only the enthalpy change of the reaction.

Step-by-step reasoning

1. Identify the heat flow from the system to the surroundings

The reaction is:

\text{H}_2(g) + \frac12 \text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)

with $\Delta_f H = -286 \text{ kJ mol}^{-1}$ at 298 K (standard conditions). The negative sign means the system releases 286 kJ of heat per mole to the surroundings at constant pressure.

So the heat absorbed by the surroundings is:

Q_{\text{surr}} = -\Delta H_{\text{sys}} = +286 \text{ kJ mol}^{-1}

2. Apply the definition of entropy change for the surroundings

For a reversible heat transfer at constant temperature:

\Delta S_{\text{surr}} = \frac{Q_{\text{surr}}}{T}

Standard temperature is $T = 298 \text{ K}$ .

3. Plug in the numbers — watch the units

\Delta S_{\text{surr}} = \frac{+286 \times 10^3 \text{ J mol}^{-1}}{298 \text{ K}}

Watch out

A common mistake is to forget converting kJ to J. Entropy is typically expressed in J K $^{-1}$ , so always convert enthalpy from kJ to J before dividing.

4. Calculate …

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Q.Calculate the entropy change in surroundings when 1.00 mol of H2O(l)H_2O(l)H2​O(l) is formed under standard conditions. ΔfH=−286\Delta_f H = -286Δf​H=−286 kJ mol−1^{-1}−1.

Why Gibbs Free Energy isn’t the direct tool here — but the concept of reversibility is

Step-by-step reasoning

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Q.Calculate the entropy change in surroundings when 1.00 mol of $H_2O(l)$ is formed under standard conditions. $\Delta_f H = -286$ kJ mol $^{-1}$ .

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