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Enthalpy

For the combu...

Question

For the combustion of 1 mole of liquid benzene at $25^{o} C,$ the heat of reaction at constant pressure is given by
$C_{6} H_{6} (l) + 7 \frac{1}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l); △ U = - 780980 c a l$
What would be the heat of reaction at constant volume?

A

- 780 k c a l

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B

780 k c a l

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C

- 580 k c a l

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D

580 k c a l

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Solution

The correct option is A $- 780 k c a l$
$△ H$ =Heat of reaction
We have, $△ H = △ U + △ n_{g} R T$
Here, $△ n_{g} = 6 - 7.5 = - 1.5$
Thus, $△ H = △ U + △ n_{g} R T$
$△ H = 780980 + (- 1.5) \times 2 \times 298$
$△ H = - 780090 c a l = - 780.090 k c a l$

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