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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2024 · Paper 1 · Chemistry · Thermodynamics (C)

The combustion of benzene (L) gives CO_2(~g) and H_2 O(L) . Given...

The combustion of benzene (L) gives CO2( g)\mathrm{CO}_2(\mathrm{~g}) and H2O(L)\mathrm{H}_2 \mathrm{O}(L). Given that heat of combustion of benzene at constant volume is −3263.9 kJ mol−1-3263.9 \mathrm{~kJ} \mathrm{~mol}^{-}{ }^1 at 25∘C25^{{}^{\circ}} \mathrm{C}; heat of combustion (in kJmol−1\mathrm{kJ} \mathrm{mol}{ }^{-1} ) of benzene at constant pressure will be : (R=8.314JK−1 mol−1)\left(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)

Options

  1. A4152.64152.6
  2. B452.46452.46
  3. C32603260
  4. D−3267.6-3267.6

Correct answer

D. −3267.6-3267.6

Step-by-step solution

C6H6(\mathrm{C}_6 \mathrm{H}_6(\mathrm{I})+152O2(g)→6CO2(g)+3H2O(l)Δng=6−152=−32ΔH=ΔU+ΔngRT=−3263.9+(−32)×8.314×10−3×298=−3263.9+(−3.71)=−3267.6kJmol−1)+\frac{15}{2} \mathrm{O}_2(\mathrm{ g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{ g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\Delta n_g=6-\frac{15}{2}=-\frac{3}{2}\Delta H=\Delta U+\Delta n_{\mathrm{g}} R T=-3263.9+\left(-\frac{3}{2}\right) \times 8.314 \times 10^{-3} \times 298=-3263.9+(-3.71)=-3267.6 \mathrm{ kJ} \mathrm{ mol}^{-1}
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