Q9 / 125 · BITSAT 2013 · Chemistry
The enthalpy of combustion of 2 moles of benzene at $27^{\circ} \mathrm{C}$ differs from the value determined in bomb calorimeter by
- A$-2.494 \mathrm{~kJ}$
- B$2.494 \mathrm{~kJ}$
- Correct answerC$-7.483 \mathrm{~kJ}$
- D$7.483 \mathrm{~kJ}$
Solution
By bomb calorimeter we get $\Delta \mathrm{E}$.
$$
\begin{array}{l}
2 \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})+15 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \\
12 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(1) \\
\Delta \mathrm{H}-\Delta \mathrm{E}=\Delta \mathrm{nRT} \\
=(12-15) \times 8.314 \times 300=-7.483 \mathrm{~kJ}
\end{array}
$$
Hence, the correct answer is option: 'C'.