For the combustion of 1 mole of liquid benzene at 27° C, the heat of reaction at constant pressure is given by :
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (l)$
$Δ H = - 78 kcal$
What would the be heat of reaction at constant volume?

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$- 78.9 k c a l$

$816.0 k c a l$

$- 77.1 k c a l$

$- 78.0 k c a l$

answer is C.

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Detailed Solution

The change in the enthalpy is,

$Δ H = Δ U + (Δ n_{g}) R T$

$Δ U = Δ H - (Δ n_{g}) R T$

$= - 78 - (- 1.5) \times \frac{2}{1000} \times 300$

$= - 78 \frac{3}{2} \times \frac{2}{1000} \times 300 = - 78 + 0.9$

$= - 77.1 kcal$

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