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The combusition of 1 mol of benzene (C(6...

The combusition of 1 mol of benzene `(C_(6) H_(6))` takes place at `298 K` and 1 bar pressure. After combustion, `CO_(2) (g)` and `H_(2) O (l)` are produced and `3267 kJ` of heat is liberated. Calculate the standard enthaply of formation, `Delta_(f) H^(@)` of benzene. Standard enthapies of formation of `CO_(2) (g)` and `H_(2) O (l)` are `- 393.5 kJ mol^(-1)` and `- 258.83 kJ mol^(-1)`, respectively.
Strategy : Apply Eq. the mathematical form of Hesis's law, to the combustion reaction of 1 mol of benzene. Remember `Delta_(f) H^(@)` for `O_(2) (g)` is zero by convention. We are give `Delta_(1) H^(@)` and `Delta_(f) H^(@)` values for all substance except `C_(6) H_(6) (1)`. We can solve for this unknown.

Text Solution

Verified by Experts

Combustion of 1 mol of benzene takes place as :
`C_(6)H_(6)(l) + (15)/(2)O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l)`
`Delta_(c)H^(@) = -3267.0 kJ mol^(-1)`
`Delta_(r)H^(@) = Sigma Delta_(f)H^(@)("products") - Sigma Delta_(f)H^(@)("reactants")`
`6[ Delta_(f)H^(@){CO_(2)(g)}] + 3[ Delta_(f)H^(@){H_(2)O(l)}]`
` - [Delta_(f)H^(@){C_(6)H_(6)(l)}] - (15)/(2)[Delta_(f)H^(@){O_(2)(g)}]`
`Delta_(f)H^(@) {CO_(2)(g)} = -393.5 kJ mol^(-1), Delta_(f)H^(@){H_(2O(l))} = -285.83 kJ mol^(-1)`
`Delta_(f)H^(@) {C_(6)H_(6)(l)} = ? Delta_(f)H^(@){O_(2)(g)} = 0`
`therefore - 3267.0 = 6(-393.5) + 3(-285.3) -Delta_(f)H^(@)[C_(6)H_(6)(l)] -(15)/(2)(0)`
`-3267.0 = - 2361.0 - 85749 - Delta_(f)H^(@){C_(6)H_(6)(l)} -0`
`therefore Delta_(f)H^(@){C_(6)H_(6)(l)} = - 2361.0 - 857.49 + 3267.0`
= `48.51 kJ mol^(-1)`
Standard enthalpy of formation of `C_(6)H_(6) = 48.51 kJ mol^(-1)`
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