Question

For the combustion of 1 mole of a liquid benzene at ${25}^{o}C$, the enthalpy is given by:
$C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(l\right)$ $\Delta H=-780980$ cal
What would be the internal energy of the reaction?

• A. $-780kcal$
• B. $780kcal$
• C. $-580kcal$
• D. $580kcal$

Answer 1

The correct option is A $-780kcal$

We know. $\Delta H=\Delta U+n_{g}RT$
where H and U are the enthalpy and the internal energy of a reaction.
$\Delta n_g$ = difference in moles of gaseous products and reactants = $6-\frac{15}{2}$ = $-\frac{3}{2}$ = -1.5
$\Delta U=\Delta H-n_{g}RT$ = $-780980$$-(-1.5)\times 2\times 298$
$\Delta U=-780086$ cal = $-780$ kcal