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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2024 · Paper 2 · Chemistry · Thermodynamics (C)

The standard Gibbs energy (Δ G^0) for the following reaction is...

The standard Gibbs energy (ΔG0)\left(\Delta G^0\right) for the following reaction is A(s)+B2+(aq)→A2+(aq)+B(s),Kc=1012at25∘C\begin{aligned} & \mathrm{A}(\mathrm{s})+\mathrm{B}^{2+}(\mathrm{aq}) \rightarrow \mathrm{A}^{2+}(\mathrm{aq})+\mathrm{B}(\mathrm{s}), \mathrm{K}_{\mathrm{c}}=10^{12} \mathrm{at} \\ & 25^{\circ} \mathrm{C}\end{aligned}\left(\mathrm{K}_c=\right.$ equilibrium constant )

Options

  1. A−150 kJ-150 \mathrm{~kJ}
  2. B−96.80 kJ-96.80 \mathrm{~kJ}
  3. C−68.47 kJ-68.47 \mathrm{~kJ}
  4. D-100 kJ

Correct answer

C. −68.47 kJ-68.47 \mathrm{~kJ}

Step-by-step solution

A+B2+⇌A2++B\mathrm{A}+\mathrm{B}^{2+} \rightleftharpoons \mathrm{A}^{2+}+\mathrm{B} G∘=−RTln⁡Kc\mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}} =−8.314×298×2.303×log⁡1012=68.47 kJ/=-8.314 \times 298 \times 2.303 \times \log 10^{12}=68.47 \mathrm{~kJ} / mol
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