When the heat of reaction at constant pressure is -2.5 × 10^3 cal and entropy change is 7.4 cal deg^ -1 at 25^ C , the reaction is predicted as

Correct answer: B. spontaneous For the reaction at constant pressure, aligned & Δ H=-2.5 × 10^3 cal & Δ S=7.4 cal deg^ -1 & Δ G=Δ H-T Δ S=(-) 2.5 × 10^3-298 × 7.4 & =(-) 2.5 × 10^3-2205.2 & Δ G=-4705 cal & aligned Reaction is spontaneous.

JIPMER 2017

Exam: JIPMER
Year: 2017
Subject: Chemistry
Chapter: Thermodynamics (C)

JIPMER 2017 · Chemistry · Thermodynamics (C)

When the heat of reaction at constant pressure is -2.5 × 10^3 cal...

When the heat of reaction at constant pressure is

-2.5 \times 10^3 \mathrm{cal}

and entropy change is

7.4

\mathrm{cal} \mathrm{deg}^{-1}

25^{\circ} \mathrm{C}

, the reaction is predicted as

Options

Areversible
Bspontaneous
Cnon-spontaneous
Dirreversible

Correct answer

B. spontaneous

Step-by-step solution

For the reaction at constant pressure,

\begin{aligned} & \Delta H=-2.5 \times 10^3 \mathrm{cal} \\ & \Delta S=7.4 \mathrm{cal} \mathrm{deg}^{-1} \\ & \therefore \Delta G=\Delta H-T \Delta S=(-) 2.5 \times 10^3-298 \times 7.4 \\ & =(-) 2.5 \times 10^3-2205.2 \\ & \Delta G=-4705 \mathrm{cal} \\ &\end{aligned}

\therefore$ Reaction is spontaneous.

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