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Thermodynamics question

2010 · Shift 0 · Q8

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Thermodynamics
2010 · Shift 0 · Q8

Thermodynamics question

2010 · Shift 0 · Q8

JEE MainChemistryThermodynamicsMCQ+4 / −1

For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when :

A
Te > T
B
T > Te
C
Te is 5 times T
D
T = Te

View written solutionFree

Correct answer: B

For spontaneity, the Gibbs free energy change must be negative:

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

A reaction is spontaneous when:

$\Delta G < 0$

Given:

$\Delta H > 0, \quad \Delta S > 0$

So spontaneity condition becomes:

$\Delta H - T\Delta S < 0$

$T\Delta S > \Delta H$

$T > \frac{\Delta H}{\Delta S}$

At equilibrium, by definition:

$\Delta G = 0$

So the equilibrium temperature $T_e$ is:

$\Delta H - T_e\Delta S = 0$

$T_e = \frac{\Delta H}{\Delta S}$

Therefore, the spontaneity condition becomes:

$T > T_e$

So the reaction is spontaneous when the actual temperature is greater than the equilibrium temperature.

Checking options:

A: $T_e > T$ → not spontaneous
B: $T > T_e$ → correct
C: $T_e$ is 5 times $T$ → not generally true
D: $T = T_e$ → equilibrium, not spontaneous driving force

Hence, the correct option is B.

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