The enthalpy of combustion of 2 moles of benzene at 27^ ^ C differs from the value determined in bomb calorimeter by

Correct answer: C. -7.483 kJ By bomb calorimeter we get Δ E . array l 2 C_ 6 H_ 6 (l)+15 O_ 2 ( g) 12 CO_ 2 ( g)+6 H_ 2 O(1) Δ H-Δ E=Δ nRT =(12-15) ×8.314 ×300=-7.483 kJ array

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Chemistry
Chapter: Thermodynamics (C)

BITSAT 2013 · Chemistry · Thermodynamics (C)

The enthalpy of combustion of 2 moles of benzene at 27^ ^ C differs...

The enthalpy of combustion of 2 moles of benzene at

27^{{}^{\circ}} \mathrm{C}

differs from the value determined in bomb calorimeter by

Options

A $-2.494 \mathrm{ kJ}$
B $2.494 \mathrm{ kJ}$
C $-7.483 \mathrm{ kJ}$
D $7.483 \mathrm{ kJ}$

Correct answer

C. $-7.483 \mathrm{ kJ}$

Step-by-step solution

By bomb calorimeter we get

\Delta \mathrm{E}

\begin{array}{l}2 \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})+15 \mathrm{O}_{2}(\mathrm{ g}) \longrightarrow\\ 12 \mathrm{CO}_{2}(\mathrm{ g})+6 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta \mathrm{H}-\Delta \mathrm{E}=\Delta \mathrm{nRT}\\ =(12-15) \times8.314 \times300=-7.483 \mathrm{ kJ} \end{array}

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