Question

For the combustion of 1 mole of liquid benzene at ${25}^{\circ }\text{C}$, the heat of reaction at constant pressure is -780.9 kcal
${\text{C}}_6{\text{H}}_{6\left(1\right)}+7\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to 6\text{C}{\text{O}}_{2\left(\text{g}\right)}+3{\text{H}}_2{\text{O}}_{\left(1\right)}$
The heat of reaction at constant volume

• A. -781.8 kcal
• B. -780.0 kcal
• C. 781.8 kcal
• D. 780 kcal

Answer 1

The correct option is A -781.8 kcal

${\text{C}}_6{\text{H}}_{6\left(l\right)}+\frac{7}{2}{\text{O}}_{2\left(g\right)}\to 6\text{C}{\text{O}}_{2\left(g\right)}+3{\text{H}}_2{\text{O}}_{\left(l\right)}$
$\Delta \text{ng}$ = (number of moles of gaseous product) - (number of moles of gaseous reactant)
$\Delta \text{ng}\text{=}6-\frac{7}{2}=\frac{5}{2}$
$\Delta \text{H}=\Delta \text{U}+\Delta \text{ngRT}$
$\text{T}={25}^{\circ }\text{C}=25+273\text{K}=298\text{K}$
$\text{R}=2\text{Cal}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}=2\times {10}^{-3}\text{K}\text{cal}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}$
$\underset{\left(\text{Heat of reaction at constant volume}\right)}{\Delta \text{U}}$ $=\Delta \text{H}-\Delta \text{ng}\text{RT}$
$\Delta \text{U}=-780.\text{9K}\text{cal -}\frac{5}{2}\times 2\times {10}^{-3}\times 298$
$\Delta \text{U}=-782.\text{K}\text{cal}$
Thus, correct option is (A)