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The combustion of 1 mole of benzene takes placeat 298K and 1 atm. After combustion, CO_(2)(g) and H_(2)O(l) are produced and 3267.0kJ of heat is liberated. Calculate the standard enthalpy of formation, Delta _(f) H^(@) of benzene. Standard enthalpies of formation of CO_(2)(g) and H_(2)O(l) are -393.5kJ mol^(-1) and -258.83kJ mol^(_1) respectively |
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Answer» Solution :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 requiredthermochemical equation, multiply Eq. (ii) 6 and Eq. (iii) by 3 and subtract Eq. (i) from THEIRSUM, 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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