The combustion of `1 mol` of benzene takes place at `298 K` and `1 atm`. After combustion, `CO_(2)(g)` and `H_(2)O(l)` are produced and `3267.0 kJ` of heat is librated. Calculate the standard entalpy of formation, `Delta_(f)H^(Θ)` of benzene
Given: `Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1)`
`Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1)`.

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

The combustion of one mole of benzene takes place at 298 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. triangle_f H^@ benzene. Standard enthalpies of formation of CO_2(g) and H_20(l) are -393,5 kJ mol ""^(-1) and "-285.83 kJ mol"^(-1) respectively.

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 (1) 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 (1) 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.

Estimate the standard enthalpy of combustion of acetylene from the following data: Delta_fH^@ (CO_2) =-393.5 kj mol^(-1) Delta_fH^@ (H_2O) =-285.8 kj mol^(-1) Delta_fH^@ (C_2H_2) = + 227.3 kj mol^(-1)