Question

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

• A. - 2.494 kJ
• B. 2.494 kJ
• C. - 7.483 kJ
• D. 7.483 kJ

Answer 1

Answer: (C)

- 7.483 kJ

Answer 2

By bomb calorimeter we get $\Delta E$.
$2 C _{6} H _{6}( l )+15 O _{2}( g ) \longrightarrow 12 CO _{2}( g )+6 H _{2} O (l)$
$\Delta H -\Delta E =\Delta nRT$
$=(12-15) \times 8.314 \times 300=-7.483\, kJ$