1. Tardigrade
  2. Question
  3. Chemistry
  4. At 25 °C , for the combustion of 1 mole of liquid benzene, the heat of reaction at constant pressure is given by C6H6 (l) +(7/2) O2(g) → 6CO2(g) + 3H2O(l) , Δ H= 780980 cal Calculate the heat of reaction at constant volume

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Q. At $ 25 ^{\circ}C $ , for the combustion of $1$ mole of liquid benzene, the heat of reaction at constant pressure is given by
$ C_{6}H_{6} (l) +\frac{7}{2} O_{2}(g) \to 6CO_{2}(g) + 3H_{2}O(l) $ , $ \Delta H= 780980\, cal $
Calculate the heat of reaction at constant volume

AMUAMU 2014Thermodynamics

Solution:

$\Delta H =\Delta U+\Delta n_{g} RT$
(where, $\Delta H$= heat of reaction at constant pressure
$\Delta U$= heat of reaction at constant volume)
heat of reaction at constant
$\Delta n_{g}=6-\frac{7}{2}=\frac{5}{2}=2.5$
$\therefore \Delta U=\Delta H - \Delta n_{g} RT$
$=780980-2.5\times2\times298$
$=780980-1490=779490$
$=779.49\, kcal$