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

BITSAT 2019 · Chemistry · Thermodynamics (C)

The heat of reaction for C _ 10 H _ 8 (s) + 12 O _ 2 (g) ⟶ 10 CO _...

The heat of reaction for

C10H8(s)+12O2(g)⟶10CO2(g)+4H2O(l){C}_{10}{H}_{8}\left(s\right) + 12{O}_{2}\left(g\right) ⟶ 10{CO}_{2}\left(g\right) + 4{H}_{2}O ( l ) at constant volume is −1228.2kcal- 1228. 2kcal at 250C{25}^{0}C. The heat of reaction at constant pressure and same temperature is

Options

  1. A−1228.2kcal- 1228. 2kcal
  2. B−1229.3kcal- 1229. 3kcal
  3. C−1232.9kcal- 1232. 9kcal
  4. D−1242.6kcal- 1242. 6kcal

Correct answer

B. −1229.3kcal- 1229. 3kcal

Step-by-step solution

Heat of reaction at constant pressure is represented by& ΔH

Heat of reaction at constant volume is represented by& ΔU

C10H8(s)+12O2(g)⟶10CO2(g)+4H2O{C}_{10}{H}_{8}\left(s\right) + 12{O}_{2}\left(g\right) ⟶ 10{CO}_{2}\left(g\right) + 4{H}_{2}O

We known that ΔH=ΔE+ΔngRT\Delta H = \Delta E + {Δn}_{g}RT

Δng=10−12=−2{Δn}_{g} = 10 - 12 = - 2

ΔH=−1228.2×103+(−2)×2×298\Delta H = - 1228. 2 \times {10}^{3} + \left(- 2\right) \times 2 \times 298

=−1229393Cal= - 1229393Cal

=−1229.39kcal= - 1229. 39kcal

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