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MHT-CET Chemistry · Chemical Thermodynamics and Energetics

Entropy and the Second Law

Entropy measures disorder; a spontaneous process raises the entropy of the universe, ΔS_total = ΔS_sys + ΔS_surr > 0, where the surroundings' share is −ΔH_sys/T.

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Why this matters

13 PYQs, none HARD. Two shapes: pick the reaction whose entropy falls (gas moles decrease, or gas becomes liquid) or rises (solid dissolves, gas made), and compute ΔS_surr = −ΔH/T or ΔS_total with the units matched — ΔH in kJ, ΔS in J K⁻¹. One conversion and one sign.

Concept 1 of 2: Predicting the Sign of ΔS

Gas is far more disordered than liquid, liquid than solid. So count gas molecules on each side and watch phase changes: more gas or a solid dissolving means entropy rises; fewer gas molecules or a gas condensing means it falls.

Definition

  • Sgas≫Sliquid>SsolidS_{\text{gas}} \gg S_{\text{liquid}} > S_{\text{solid}}. Melting, vaporisation, sublimation: ΔS > 0. Freezing, condensation, deposition: ΔS < 0.
  • Gas moles up → ΔS > 0: CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g); H2→2H\text{H}_2 \to 2\text{H}; 2H2O2(l)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g).
  • Gas moles down → ΔS < 0: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3; 2H2+O2→2H2O(l)2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}(l) (three gas moles to a liquid); CaO+CO2→CaCO3\text{CaO} + \text{CO}_2 \to \text{CaCO}_3.
  • Dissolving an ionic solid: NaNO3(s)→Na++NO3−\text{NaNO}_3(s) \to \text{Na}^+ + \text{NO}_3^-, ΔS > 0. Crystallising from solution: ΔS < 0.

Rule of thumb

Δng>0⇒ΔS>0;Δng<0 or gas→liquid/solid⇒ΔS<0\Delta n_g > 0 \Rightarrow \Delta S > 0;\qquad \Delta n_g < 0 \text{ or gas} \to \text{liquid/solid} \Rightarrow \Delta S < 0

Worked example

Order by entropy change, most positive first: I2(s)→I2(g)\text{I}_2(s) \to \text{I}_2(g); 2SO2+O2→2SO32\text{SO}_2 + \text{O}_2 \to 2\text{SO}_3; H2O(l)→H2O(s)\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(s).
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q98Moderate

Example 1 · Chemical Thermodynamics and Energetics · Entropy and Second Law

Which of the following reactions exhibits decrease in entropy?

Counting moles without looking at phases

2H2O2(l)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g) has 2 → 3 moles overall, but what matters is that a GAS appears from liquids: ΔS > 0. Conversely 3 gas moles to 2 moles of LIQUID water is a large decrease.
Drill 2 more on predicting the sign of δs

Concept 2 of 2: ΔS = q_rev/T, ΔS_surr = −ΔH/T and ΔS_total

Heat flowing reversibly at temperature T changes entropy by q/T — at a phase change that is ΔH/T. The surroundings receive whatever heat the system gives out, so their entropy change is −ΔH_sys/T, and the total is the sum. Watch the units: ΔH arrives in kJ, ΔS in J K⁻¹.

Definition

  • Phase change: ΔS=ΔHT\Delta S = \dfrac{\Delta H}{T}. Melting 1 g ice, 80 J g⁻¹ at 273 K: 0.2930.293 J g⁻¹ K⁻¹.
  • ΔSsurr=−ΔHsysT\Delta S_{\text{surr}} = -\dfrac{\Delta H_{\text{sys}}}{T}. Ice melting, +7 kJ at 300 K: −7000/300=−23.3-7000/300 = -23.3 J K⁻¹. Water forming, −525 kJ at 300 K: +1750+1750 J K⁻¹.
  • ΔStotal=ΔSsys+ΔSsurr=ΔSsys−ΔHT\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = \Delta S_{\text{sys}} - \dfrac{\Delta H}{T}. NH₄NO₃ dissolving, 28.1 kJ, 108.7 J K⁻¹, 300 K: 108.7−93.7=15.1108.7 - 93.7 = 15.1 J K⁻¹ (spontaneous though endothermic).
  • ΔH = −150 kJ, ΔS = 32 J K⁻¹, 300 K: 32+500=53232 + 500 = 532. ΔH° = −208.6 kJ, ΔS° = −36 J K⁻¹, 298 K: −36+700=664-36 + 700 = 664.
  • Second law: ΔStotal>0\Delta S_{\text{total}} > 0 spontaneous, =0= 0 equilibrium, <0< 0 non-spontaneous.

Entropy of surroundings and total

ΔSsurr=−ΔHsysT,ΔStotal=ΔSsys−ΔHsysT\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T},\qquad \Delta S_{\text{total}} = \Delta S_{\text{sys}} - \frac{\Delta H_{\text{sys}}}{T}

Worked example

A reaction has ΔH = −90 kJ and ΔS_sys = −120 J K⁻¹ at 300 K. Find ΔS_total and say whether it is spontaneous.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q95Moderate

Example 2 · Chemical Thermodynamics and Energetics · Entropy and Second Law

Calculate ΔStotal\Delta S_\text{total} for the following reaction at 300 K. NH4NO3(s)→NH4+(aq)+NO3−(aq)\text{NH}_4\text{NO}_3(s) \to \text{NH}_4^+(aq) + \text{NO}_3^-(aq); ΔH=28.1 kJ mol−1\Delta H = 28.1\text{ kJ mol}^{-1}, ΔSsys=108.7 J K−1mol−1\Delta S_\text{sys} = 108.7\text{ J K}^{-1}\text{mol}^{-1}

Dividing kilojoules by kelvin

28.1/300 = 0.094 is in kJ K⁻¹ and cannot be added to 108.7 J K⁻¹. Convert ΔH to joules first: 28100/300 = 93.7 J K⁻¹.
Drill 5 more on δs = q_rev/t, δs_surr = −δh/t and δs_total

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Predicting the Sign of ΔS

    Rule of thumb

    Δng>0⇒ΔS>0;Δng<0 or gas→liquid/solid⇒ΔS<0\Delta n_g > 0 \Rightarrow \Delta S > 0;\qquad \Delta n_g < 0 \text{ or gas} \to \text{liquid/solid} \Rightarrow \Delta S < 0
  • ΔS = q_rev/T, ΔS_surr = −ΔH/T and ΔS_total

    Entropy of surroundings and total

    ΔSsurr=−ΔHsysT,ΔStotal=ΔSsys−ΔHsysT\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T},\qquad \Delta S_{\text{total}} = \Delta S_{\text{sys}} - \frac{\Delta H_{\text{sys}}}{T}

Watch out for (2)

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