$\Delta_r H_T^{\circ}=\sum a_i H_{m,T}^{\circ}-\sum b_i H_{m,T}^{\circ}$

$\Delta_r H_T^{\circ}$=$\sum a_i\Delta_fH_{i,T}^{\circ}$-$\sum b_i\Delta_fH_{i,T}^{\circ}$

Standard enthalpy of formation

Standard enthalpy of phase change

Hess's Law

A balanced chemical reaction along with the value $\Delta_rH^0(T)$

stoichiometric coefficient ⇒ no. of moles

$Δ_rΗ^0(T) ⇒ extensive$

Reverse reaction $-\Delta _rH^0(T)$

Enthalpy of different types of processes/reactions

Standard enthalpy of combustion $\Delta_cH^0(T)$

$R (1mole) + O_2 \rightarrow Co_2(g) + H_2O(l)$

All reactants and products are in their standard states at temp T

$\Delta rH^\circ (T) \equiv \Delta_c H^\circ (T)$

Glucose

• $C_6 H_{12} O_6(s) (l) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6H_2O(l)$

• $\Delta_c H^0_g {glucose}(298k) = -2802kJ mol^ {-1}$

Solution

$\Delta_c H^0_{C_6H_6} = - 3267 KJ mol^{-1}$

$\Delta_f H^0_{CO_2}(g) = -393.5 KJ mol^{-1}$

$\Delta_f H^0_{H_2O}(l) = -285.8 KJ mol^{-1}$

$\Delta_f H^0_{C_6H_6}(g)= ?$

$C_6H_6(l) + \frac{15}{2}O_2(g)\rightarrow 6CO_2(g) + 3H_2O(l)$

$\Delta_c H^0_{c_6H_6}(l) = \Delta_rH^o$ = $∑a_i\Delta_f H^0_i(products) - ∑b_i\Delta_f H^0_i(reactants)$

$-3267 = 6×(-393.5) + 3×(-285.8) - \Delta_f H^0_{c_6H_6}(l)-\frac{15}{2}\Delta_f H^0O{_2}(g) = 0$

$\Delta_f C_6H_6 (l) = 49.8kJ mol^{-1}$