Question

For combustion of 1 mole of benzene at $25^{o}C$, the heat of reaction at constant pressure is $- 780.9$ kcal. What will be the heat of reaction at constant volume?

$C_{6}H_{6(l)} + 7\frac{1}{2}O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_{2}O_{(l)}$

• A. $- 781.8$ kcal
• B. $- 780.0$ kcal
• C. $+ 781.8$kcal
• D. $+ 780.0$kcal

Answer 1

Answer: (B)

$- 780.0$ kcal

Answer 2

: $q_{P} = q_{v} + \Delta n_{g}RT,q_{v} = q_{P} - \Delta n_{g}RT$

($\Delta H = \Delta E = \Delta n_{g}RT$)

$q_{v} = - 780.9 - ( - 1.5 \times 2 \times 298 \times 10^{- 3} = - 780kCal$