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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2007 · Paper 1 · Chemistry · Thermodynamics (C)

The value of _ 10 ~K for a reaction A leftharpoons B is (Given: Δ_...

The value of log⁡10 K\log _{10} \mathrm{~K} for a reaction A⇌BA \rightleftharpoons B is (Given: ΔrH298 K∘=−54.07 kJ mol−1,ΔrS298 K∘=10JK−1 mol−1\Delta_{\mathrm{r}} H_{298 \mathrm{~K}}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{r}} S_{298 \mathrm{~K}}^{\circ}=10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} and R=8.314JK−1 mol−1;2.303×8.314×298=5705R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} ; 2.303 \times 8.314 \times 298=5705 )

Options

  1. A5
  2. B10
  3. C95
  4. D100

Correct answer

B. 10

Step-by-step solution

A⇌BΔG∘=ΔH∘−TΔS∘ΔG∘=−2303RTlog⁡Klog⁡K=ΔH∘−TΔS∘−2303RT=−54.07×103−298×10−2303×8.314×298=−57050−5705=10\begin{aligned} A \rightleftharpoons B \\ \Delta G^{{}^{\circ}} & =\Delta H^{{}^{\circ}}-T \Delta S^{{}^{\circ}} \\ \Delta G^{{}^{\circ}} & =-2303 R T \log K \\ \log K & =\frac{\Delta H^{{}^{\circ}}-T \Delta S^{{}^{\circ}}}{-2303 R T}\\ & =\frac{-54.07 \times 10^3-298 \times 10}{-2303 \times 8.314 \times 298}\\ & =\frac{-57050}{-5705}=10 \end{aligned}
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