Entropy and Gibbs free energy — predict spontaneity of chemical reactions

Question

For a reaction, $\Delta H = -10$ kJ/mol and $\Delta S = -40$ J/(mol K). Determine whether the reaction is spontaneous at (a) 200 K and (b) 300 K using the Gibbs free energy equation.

(NCERT Class 11, Chapter 6 — Thermodynamics)

Solution — Step by Step

\Delta G = \Delta H - T\Delta S

The spontaneity criterion:

$\Delta G < 0$ → spontaneous (reaction proceeds forward)
$\Delta G > 0$ → non-spontaneous
$\Delta G = 0$ → equilibrium

$\Delta H = -10$ kJ/mol $= -10000$ J/mol

$\Delta S = -40$ J/(mol K)

Both must be in the same unit (J) before substituting. Mixing kJ and J is a guaranteed wrong answer.

\Delta G = -10000 - (200)(-40) = -10000 + 8000 = -2000 \text{ J/mol}

\boxed{\Delta G = -2 \text{ kJ/mol (spontaneous at 200 K)}}

\Delta G = -10000 - (300)(-40) = -10000 + 12000 = +2000 \text{ J/mol}

\boxed{\Delta G = +2 \text{ kJ/mol (non-spontaneous at 300 K)}}

At the boundary, $\Delta G = 0$ :

0 = \Delta H - T\Delta S

T = \frac{\Delta H}{\Delta S} = \frac{-10000}{-40} = 250 \text{ K}

Below 250 K → spontaneous. Above 250 K → non-spontaneous.

Why This Works

Spontaneity depends on two competing factors: enthalpy (energy change) and entropy (disorder change). A reaction with $\Delta H < 0$ (exothermic) wants to happen because it releases energy. But if $\Delta S < 0$ (decrease in disorder), nature resists it.

At low temperatures, the enthalpy term dominates (the $T\Delta S$ term is small), so the exothermic nature wins — the reaction is spontaneous. At high temperatures, the $T\Delta S$ term becomes large enough to override the enthalpy advantage, and the reaction becomes non-spontaneous.

This is why some exothermic reactions only occur at lower temperatures. The Gibbs equation captures this temperature dependence precisely.

Alternative Method — Four-case analysis

$\Delta H$	$\Delta S$	Spontaneity
$-$ (exo)	$+$	Always spontaneous ( $\Delta G < 0$ at all $T$ )
$+$ (endo)	$-$	Never spontaneous ( $\Delta G > 0$ at all $T$ )
$-$ (exo)	$-$	Spontaneous at low $T$ (our case)
$+$ (endo)	$+$	Spontaneous at high $T$

For NEET, memorise this table. Most MCQs give you signs of $\Delta H$ and $\Delta S$ and ask “at what temperature is the reaction spontaneous?” You don’t need to calculate — just match the sign pattern. Our case ( $\Delta H < 0, \Delta S < 0$ ) = spontaneous at low temperature.

Common Mistake

The unit mismatch between $\Delta H$ (usually given in kJ) and $\Delta S$ (usually given in J/K) catches students every time. Always convert to the same unit before substituting into $\Delta G = \Delta H - T\Delta S$ . If you keep $\Delta H$ in kJ and $\Delta S$ in J/K, your answer will be off by a factor of 1000.

Question

Solution — Step by Step

Why This Works

Alternative Method — Four-case analysis

Common Mistake

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