The enthalpy of combustion of propane, graphite and dihydrogen at $$298 \text{ K}$$ are: $$-2220.0 \text{ kJ mol}^{-1}$$, $$-393.5 \text{ kJ mol}^{-1}$$ and $$-285.8 \text{ kJ mol}^{-1}$$ respectively. The magnitude of enthalpy of formation of propane ($$C_3H_8$$) is ______ $$\text{kJ mol}^{-1}$$. (Nearest integer)

We need to find the enthalpy of formation of propane ($$C_3H_8$$).

(i) $$C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$$; $$\Delta H_1 = -2220.0 \text{ kJ mol}^{-1}$$

(ii) $$C(s) + O_2(g) \rightarrow CO_2(g)$$; $$\Delta H_2 = -393.5 \text{ kJ mol}^{-1}$$

(iii) $$H_2(g) + \dfrac{1}{2}O_2(g) \rightarrow H_2O(l)$$; $$\Delta H_3 = -285.8 \text{ kJ mol}^{-1}$$

$$3C(s) + 4H_2(g) \rightarrow C_3H_8(g)$$; $$\Delta H_f = ?$$

Using Hess's law: $$\Delta H_f = 3 \times \Delta H_2 + 4 \times \Delta H_3 - \Delta H_1$$

$$\Delta H_f = 3(-393.5) + 4(-285.8) - (-2220.0)$$

$$= -1180.5 + (-1143.2) + 2220.0$$

$$= -1180.5 - 1143.2 + 2220.0$$

$$= -2323.7 + 2220.0$$

$$= -103.7 \text{ kJ mol}^{-1}$$

Magnitude of $$\Delta H_f = |-103.7| \approx 104 \text{ kJ mol}^{-1}$$

Hence, the answer is 104.