The heat of combustion of napthalene, $C_{10} H_{8}$ , at constant volume at $25^{o} C$ was found to be $- 5133 k J m o l^{- 1}$ . Calaculate the value of enthalpy change at constant pressure at the same temperature.

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Solution

$Δ H = Δ U + Δ n_{g} R T$

The balanced equation for the combustion of naphthalene is as follows:
$C_{10} H_{8} + 12 O_{2} \to 10 C O_{2} + 4 H_{2} O$

Here $Δ n_{g} = (10 - 12) = - 2$
on substituting the above value in the equation of enthalpy , we get
$Δ H = Δ U - 2 R T$
which means $Δ H = - 5133 - 2 \times R T$
$Δ H = - 5133 - 4.96$ KJ/mol
$Δ H = - 5137.96$ KJ/mol

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