Liquid benzene is oxidised at constant pressure according to following equation.
`C_6H_6 (l) + 15/2 O_2 (s) to 6CO_2(s) + 3H_2O (l)`
Enthalpy change `(Delta H)` for the reaction at 300K is - 3728 kJ Calculate the internal energy charge.

Explanation

The reaction is
`C_6H_6(l) + 15/2 O_2(g) to 6CO_2(g) + 3H_2O (l)`
` Delta n` = number of mol of gaseous products - number of mol of gaseous reactants
` Delta n = 6 - 15/2 = - 3/2 `
`Delta H = Delta U + RT (Delta n)`
` - 3728 = Delta U + (-3/2) xx 8.314 xx 10^(-3) xx 300 `
` Delta = - 3724.25 kJ "mol"^(-1)`