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NCERT Exemplar · Q53

Q.Match the following parameters with description for spontaneity. The parameters are ΔrH°, ΔrS° and ΔrG° (in that order).
Parameters

(i) ΔrH° = +, ΔrS° = -, ΔrG° = +
(ii) ΔrH° = -, ΔrS° = -, ΔrG° = + at high T
(iii) ΔrH° = -, ΔrS° = +, ΔrG° = -
Description
(a) Non-spontaneous at high temperature.
(b) Spontaneous at all temperatures
(c) Non-spontaneous at all temperatures
Rajasthan RbseShort· 2mImportance★★★★★est
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Spontaneity is governed by the Gibbs equation ΔrG°=ΔrH°−TΔrS°\Delta_r G° = \Delta_r H° - T\Delta_r S°. Match the sign combinations to predict when reactions are spontaneous (ΔrG°<0\Delta_r G° < 0), non-spontaneous (ΔrG°>0\Delta_r G° > 0), or temperature-dependent.

(i) → (c), (ii) → (a), (iii) → (b)

The Gibbs free energy change ΔrG°\Delta_r G° is the ultimate arbiter of spontaneity at constant temperature and pressure. A reaction proceeds spontaneously when ΔrG°<0\Delta_r G° < 0, is at equilibrium when ΔrG°=0\Delta_r G° = 0, and is non-spontaneous when ΔrG°>0\Delta_r G° > 0.

The relationship between enthalpy, entropy, and free energy is captured in the fundamental equation:

ΔrG°=ΔrH°−TΔrS°\Delta_r G° = \Delta_r H° - T\Delta_r S°

This tells us that spontaneity depends on the interplay between the enthalpy change (energy released or absorbed) and the entropy change (disorder created or destroyed), weighted by temperature. The temperature acts as a lever: at high TT, the entropy term dominates; at low TT, the enthalpy term dominates.

Let's analyze each parameter set systematically.

Analysis of each case

1. Case (i): ΔrH°=+\Delta_r H° = +, ΔrS°=−\Delta_r S° = -, ΔrG°=+\Delta_r G° = +

The reaction is endothermic (absorbs heat, positive ΔrH°\Delta_r H°) and decreases disorder (negative ΔrS°\Delta_r S°). Substituting into the Gibbs equation:

ΔrG°=(+)−T×(−)=(+)+T×(+)\Delta_r G° = (+) - T \times (-) = (+) + T \times (+)

Both terms are positive. The enthalpy term is positive, and subtracting a negative entropy contribution (which means adding a positive term) makes ΔrG°\Delta_r G° even more positive. At any temperature, ΔrG°>0\Delta_r G° > 0.

This reaction is non-spontaneous at all temperatures. It fights both thermodynamic driving forces: it requires energy input and decreases disorder.

Match: (i) → (c)

2. Case (ii): ΔrH°=−\Delta_r H° = -, ΔrS°=−\Delta_r S° = -, ΔrG°=+\Delta_r G° = + at high TT

The reaction is exothermic (releases heat, negative ΔrH°\Delta_r H°) but decreases disorder (negative ΔrS°\Delta_r S°). The Gibbs equation becomes:

ΔrG°=(−)−T×(−)=(−)+T×(+)\Delta_r G° = (-) - T \times (-) = (-) + T \times (+)

At low temperatures, the enthalpy term (negative, favorable) dominates, so ΔrG°<0\Delta_r G° < 0 and the reaction is spontaneous. As temperature increases, the positive term T∣ΔrS°∣T|\Delta_r S°| grows larger. Eventually, at high enough temperature, this positive contribution overwhelms the negative enthalpy term, making ΔrG°>0\Delta_r G° > 0.

The problem states that ΔrG°=+\Delta_r G° = + at high TT, confirming the reaction becomes non-spontaneous at high temperature.

Match: (ii) → (a)

Tip

When ΔrH°\Delta_r H° and ΔrS°\Delta_r S° have opposite signs, there's always a crossover temperature Tc=ΔrH°∣ΔrS°∣T_c = \frac{\Delta_r H°}{|\Delta_r S°|} where spontaneity flips. If ΔrH°<0\Delta_r H° < 0 and ΔrS°<0\Delta_r S° < 0, the reaction is spontaneous below TcT_c and non-spontaneous above it.

3. Case (iii): ΔrH°=−\Delta_r H° = -, ΔrS°=+\Delta_r S° = +, ΔrG°=−\Delta_r G° = - …

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