Question

The heat liberated on complete combustion of 7.8 g benzene is 327 kJ. This heat has been measured at a constant volume and at $27{}^{\circ }\text{C}$. Calculate the heat of combustion of benzene at constant pressure.

• A. -327.37 kJ
• B. 327.37 kJ
• C. 32.737 kJ
• D. -32.737 kJ

Answer 1

The correct option is A -327.37 kJ

The required chemical equation is:
$\begin{array}{ccccccc}{C}_6{H}_6\left(l\right)& +& \frac{15}{2}{O}_2\left(g\right)& \to & 6C{O}_2\left(g\right)& +& 3H_{2}O\left(l\right)\\ \text{1 mol}& & \text{7.5 mol}& & \text{6 mol}& & \text{3 mol}\\ \text{0.1 mol}& & \text{0.75 mol}& & \text{0.6 mol}& & \text{0.3 mol}\end{array}$
Moles of benzene taken for combustion $=\frac{7.8}{78}=0.1\text{mol}$
$\because$ Heat is liberated, $\Delta E=-327\text{kJ}$
Here, $\Delta E=-327\text{kJ}$
$R=8.3\times {10}^{-3}{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
$T=27+273=300K$
$\Delta n=(0.6-0.75)=-0.15$
We know that,
$\Delta H=\Delta E+\Delta nRT$
$=-327+(-0.15)(8.3\times {10}^{-3})\left(300\right)$
$=-327-0.3735=-327.3735\text{kJ}$