Question

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

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

Answer 1

The correct option is B

-7.483 kJ

Heat of combustion determined by bomb calorimeter is at constant volume. i.e. $\Delta U.$

$2C_6{H}_{6\left(l\right)}+15O_{2\left(g\right)}\to 12C{O}_{2\left(g\right)}+6H_2{O}_{l}Hence,\Delta H-\Delta U=\Delta nRT=(12-15)\times 8.315\times 300=-7.483kJ$