InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 `-286kJ mol^(-1)` respectively. Calculate the heat of combustion of benzoic acid at `(i)` constant pressure, `(ii)` constant volume. `(R=8.31J mol^(-1)K^(-1))` |
|
Answer» Correct Answer - `-3199.75 kj` Given, `7C (s) + 3H_(2) (g) + O_(2) (g) rarr C_(6) H_(5) COOH (s), Delta H^(0) = -408 kJ` (i) (i) `C(s) + O_(2) (g) rarr CO_(2) (g)` `Delta H^(0) = -393 kJ`...(ii) `H_(2) (g) + (1)/(2) O_(2) (g) rarr H_(2)O (l)` `Delta H^(0) = -286 kJ`...(iii) `C_(6)H_(5)COOH (s) + (15)/(2) O_(2) (g) rarr 7 CO_(2) (g) + 3H_(2) O (l) Delta H = ?`.....(iv) By (ii) `xx 7 + (iii) xx 3 - (i)` `C_(6)H_(5)COOH (s) + (15)/(2) O_(2) (g) rarr 7CO_(2) (g) + 3H_(2) O (l)` `Delta H^(0) = [-393 xx 7 - 286 xx 3 + 408]` `:. Delta H^(0) = - 3201 kJ` Also `Delta H = Delta U + Delta n_(g) RT` `:gt = 3201 = Delta U + (-0.5) xx 8.314 xx 10^(-3) xx 300` `:. Delta U = - 3201 + 1.2471 = -3199.75 kJ` |
|