Question

The heat of combination of napthalene, $C_{10}{H}_8\left(s\right)$, at constant volume at ${25}^{o}C$ was found to be $-5133kJmo{l}^{-1}$. Calculate type value of enthalpy change at constant pressure at the same temperature.

Answer 1

${C_{10}{H}_8}_{\left(s\right)}+12{O_2}_{\left(g\right)}⟶10{C{O}_2}_{\left(g\right)}+4{H_{2}O}_{\left(l\right)}$

For the above reaction-

$\Delta n_g=n_P-n_R=10-12=-2$

$\Delta E=-5133kJ/mol\left(\text{Given}\right)$

$T=25℃=(25+273)K=298K$

Now, as we know that

$\Delta H=\Delta E+\Delta n_{g}RT$

$\Rightarrow \Delta H=-5133+(-2\times 8.314\times {10}^{-3}\times 298)$

$\Rightarrow \Delta H=-5133-4.95=-5137.95kJ/mol$

Hence the enthalpy change at constant pressure is $-5137.95kJ/mol$.