Represent the potential energy/enthalpy change in the following processes graphically. (a) Throwing a stone fr

NCERT Exemplar · Q47

Q. Represent the potential energy/enthalpy change in the following processes graphically.

(a) Throwing a stone from the ground to roof.

(b) (1/2) H2(g) + (1/2) Cl2(g) ⇌ HCl(g) ΔrH° = -92.32 kJ mol^-1
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?

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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 key idea is that spontaneity depends on the total entropy change of the universe, not just energy change. For process (a), potential energy change is the main driver; for process (b), enthalpy change (a form of potential energy) is a contributing factor, but entropy also matters.

Let’s understand the core concept first. Spontaneity of a process is decided by the second law of thermodynamics: a process is spontaneous if the total entropy of the universe increases. This total entropy change is the sum of the entropy change of the system and the entropy change of the surroundings. The surroundings’ entropy change is directly linked to the heat exchanged — and for constant pressure processes, that heat is the enthalpy change ( $\Delta H$ ). So, a negative $\Delta H$ (exothermic) increases the entropy of the surroundings, favouring spontaneity. But a positive $\Delta H$ can still be spontaneous if the system’s entropy increase is large enough. In short, both energy and entropy matter, but their relative importance varies.

Now, let’s work through each process.

Process (a): Throwing a stone from ground to roof

This is a mechanical process, not a chemical reaction. The stone gains gravitational potential energy as it rises. The driving force here is the external work done by the thrower — the stone itself does not spontaneously move upward. If we consider the stone alone as the system, its potential energy increases. But spontaneity in thermodynamics is about whether a process occurs without external intervention. Here, the stone does not spontaneously go up; it requires an external agent. So, the potential energy change is not a contributing factor to spontaneity in the thermodynamic sense — the process is non-spontaneous for the stone alone.

Watch out

A common mistake is to think that because the stone’s potential energy increases, this energy change drives the process. In reality, the stone’s upward motion is forced by external work; the stone’s own energy change is a result, not a cause.
Process (b): $\frac{1}{2} \text{H}_2(g) + \frac{1}{2} \text{Cl}_2(g) \rightleftharpoons \text{HCl}(g)$ , $\Delta_r H^\circ = -92.32 \text{ kJ mol}^{-1}$ …

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Q.Represent the potential energy/enthalpy change in the following processes graphically.

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