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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Chemistry
Chapter
Thermodynamics (C)

JIPMER 2019 · Chemistry · Thermodynamics (C)

If equilibrium constant is 2.6 × 10^8 at 298 K then find the value...

If equilibrium constant is 2.6×1082.6 \times 10^8 at 298 K then find the value of ΔG∘\Delta \mathrm{G}^{\circ}.

Options

  1. A-48.14 kJ
  2. B-63.2 kJ
  3. C-23.3 kJ
  4. D+40 kJ

Correct answer

A. -48.14 kJ

Step-by-step solution

ΔG∘\Delta G^{\circ} is related to thermodynamic equilibrium constant as ΔG∘=−2303RTlog⁡K\Delta \mathrm{G}^{\circ}=-2303 \mathrm{RT} \log \mathrm{K}=−2303×8.314×298log⁡(26×108)=−48014.50 J=−48.14 kJ\begin{aligned} & =-2303 \times 8.314 \times 298 \log \left(26 \times 10^8\right) \\ & =-48014.50 \mathrm{~J}=-48.14 \mathrm{~kJ} \end{aligned}$$
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