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The combustion of 1 mole of benzene take...

The combustion of 1 mole of benzene takes place at 298K and 1 atm. After combustion, `CO_(2(g)) and H_(2)O_((l))` are produced and 3267.0 KJ of heat is liberated. Calculate the standard enthalpy of formation, `Delta_(f)H^(0)` of benene. Standard enthalpies of formation of `CO_(2(g))andH_(2)O_((l))` are -393.5 KJ `mol^(-1)` and - 285.83 KJ `mol^(-1)` respectively

Text Solution

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Aim `: 6C(s) + 3H_(2)(g) rarr C_(6)H_(6)(l),DeltaH= ?`
Given `: (i) C_(6) H_(6)(l) + (15)/(2) O_(2)(g) rarr 6 CO_(2)(g) + 3H_(2)O(l), Delta H = - 3267.0 kJ mol^(-1)`
(ii) `C(s) + O_(2)(g) rarr CO_(2)(g) , DeltaH = -393.5 kJ mol^(-1)`
(iii) `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(l) , Delta H = -285.83 kJ mol^(_1)`
In order to get th required thermochemical equation, multiply Eq. (ii) 6 and Eq. (iii) by 3 and subtract Eq. (i) from their sum, i.e., operating 6 `xx` Eqn. (ii) `+ 3 xx `Eqn (iii) - Eqn (i), we get
`6 C(s) + 3H_(2) (g) rarr C_(6) H_(6)(l) , DeltaH = 6(-393.5) + 3(-285.83) - ( -3267.0)`
`= - 2361- 857.49 + 3267.0`
`= - 48.51kJ mol^(-1)`
Thus, the enthalpy of formation of benzene is `Delta _(f) H= - 48.51kJ mol^(-1)`
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