Question

Heat of combustion of ${CH}_4$ at ${25}^{0}C$ and 1 atm constant pressure is -210 kcal ${mol}^{-1}$

${CH}_4\left(g\right)+{20}_2\left(g\right)\to {CO}_2\left(g\right)+{2H}_{2}O\left(l\right)$

Hence, heat of combustion at constant volume is :

• A. -210 kcal ${mol}^{-1}$
• B. -198 kcal ${mol}^{-1}$
• C. -208.8 kcal ${mol}^{-1}$
• D. -211.2 kcal ${mol}^{-1}$

Answer 1

The correct option is C -208.8 kcal ${mol}^{-1}$

$C{H}_4\left(g\right)+2O_2\left(g\right)⟶C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

$\Delta ng=1-3=-2$

$Q_p=-210$ $kcal$ $mo{l}^{-1}$

$Q_p=Q_v+\Delta n_{g}RT$

$-210=Q_v+\frac{(-2\times 2\times 298)}{1000}$

$Q_v=-208.8$ $kcal$ $mo{l}^{-1}$