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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2020 · Chemistry · Thermodynamics (C)

For the combustion reaction at 298 K2 Ag(s)+1 / 2 O_ 2 ( g) 2 Ag_ 2...

For the combustion reaction at 298K2Ag(s)+1/2O2(g)⟶2Ag2O(s)298 \mathrm{ K}2 \mathrm{Ag}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{ g}) \longrightarrow 2 \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) Which of the following alternatives is correct?

Options

  1. AΔH=ΔU\Delta \mathrm{H}=\Delta \mathrm{U}
  2. BΔH>ΔU\Delta \mathrm{H} > \Delta \mathrm{U}
  3. CΔH<ΔU\Delta \mathrm{H}< \Delta \mathrm{U}
  4. DΔH\Delta \mathrm{H} and ΔU\Delta \mathrm{U} has no relation with each other

Correct answer

C. ΔH<ΔU\Delta \mathrm{H}< \Delta \mathrm{U}

Step-by-step solution

ΔH=ΔU+ΔnRTΔ\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{nRT}\Delta\mathrm{n}=\mathrm{n}p−nR_{\mathrm{p}}-\mathrm{n}_{\mathrm{R}} Now, Δn=2−52=−12∴ΔH=ΔU−12RT\Delta \mathrm{n}=2-\frac{5}{2}=-\frac{1}{2}\therefore \Delta \mathrm{H}=\Delta \mathrm{U}-\frac{1}{2} \mathrm{RT} Thus, ΔU=ΔH+12RT∴ΔU>ΔH\Delta \mathrm{U}=\Delta \mathrm{H}+\frac{1}{2} \mathrm{RT}\therefore \Delta \mathrm{U} > \Delta \mathrm{H}
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