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MHTCET 25 Apr 2025 Shift 1 Paper

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Question : 55 of 150

Marks: +1, -0

Calculate

\Delta \mathrm{H}^{\circ}

for a reaction if

\Delta \mathrm{S}^{\circ}=120 \mathrm{JK}^{-1}

\Delta \mathrm{G}^{\circ}=28000 \mathrm{~J}

15.94 kJ
31.83 kJ
94.12 kJ
63.76 kJ

Solution:

Concept:

Use the Gibbs free energy equation to relate

\Delta G^\circ

,

\Delta H^\circ

, and

\Delta S^\circ

at standard temperature.

Explanation:

The relation is

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

.

Given

\Delta S^\circ = 120\ \mathrm{J\,K^{-1}}

and

\Delta G^\circ = 28000\ \mathrm{J}

.

For standard conditions,

T = 298\ \mathrm{K}

is assumed.

First calculate

T\Delta S^\circ = 298 \times 120 = 35760\ \mathrm{J}

.

Rearrange the equation:

\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ

.

Substitute values:

\Delta H^\circ = 28000 + 35760 = 63760\ \mathrm{J}

.

Convert to kJ:

63760\ \mathrm{J} = 63.76\ \mathrm{kJ}

.

Answer:

\Delta H^\circ = 63.76\ \mathrm{kJ}

, which is option D.

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