AT `300K`, the standard enthalpies of formation of `C_(6)H_(5)COOH(s), CO_(2)(g)`, and `H_(2)O(l)` are `-408, 393`, and `-286 kJ mol^(-1)`, respectively. Calculate the heat of combustion of benzoic acid at (a) constant pressure and (b) constant volume. `(R = 8.31 J mol^(-1)K^(-1))`

Options

A. `+3201` kJ
B. `+3199.75` kJ
C. `-3201` kJ
D. `-3199.75` kJ

Answer

D

Explanation

`C_(6)H_(5)COOH_((s))+(15)/(2)O_(2(g))rarr 7CO_(2(g))+3H_(2)O_((l))`
`Delta H=Sigma (Delta H_(f))_("Products")-Sigma (Delta H_(f))_("Reactants")`
`Delta H=[7xx(-393)+3xx(-286)]-[1xx(-408)+0]`
`Delta H=-3201 kJ`
now, `Delta H=Delta E+Delta n_(g)RT`
`Delta E=Delta H-Delta n_(g)RT`
`Delta E=-3201-(-(1)/(2))xx8.314xx10^(-3)xx300`
`therefore Delta E =-3199.75 kJ`