If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The $$|\Delta H_f°|$$ of ethanol at 298.15 K is _________ $$\times 10^2$$ kJ mol$$^{-1}$$. (Nearest integer)

Given: Standard enthalpy for combustion of graphite = $$-393.5$$ kJ mol$$^{-1}$$

Standard enthalpy of formation of water (l) = $$-285.8$$ kJ mol$$^{-1}$$

Molar mass in g mol$$^{-1}$$ of C, H, O are 12, 1 and 16 respectively

The combustion of liquid ethanol at $$298.15\,$$K is

$$\text{C}_2\text{H}_5\text{OH}(l) + 3\,\text{O}_2(g) \longrightarrow 2\,\text{CO}_2(g) + 3\,\text{H}_2\text{O}(l)$$

1. Moles of ethanol burnt:
$$n = \frac{m}{M} = \frac{3.365\;\text{g}}{46.07\;\text{g mol}^{-1}} = 0.07304\;\text{mol}$$

2. Heat evolved in the bomb (constant-volume) ⇒ change in internal energy:
$$\Delta U_{\text c}^{\circ} = \frac{-99.472\;\text{kJ}}{0.07304\;\text{mol}} = -1.362\times 10^{3}\;\text{kJ mol}^{-1}$$

3. Convert $$\Delta U$$ to $$\Delta H$$ using $$\Delta H = \Delta U + \Delta n_{\text{g}}RT$$.
For the balanced equation, gaseous moles change:
$$\Delta n_{\text g} = (2)_{\text{CO}_2} - (3)_{\text{O}_2} = -1$$
$$RT = (8.314\times10^{-3}\;\text{kJ mol}^{-1}\text{K}^{-1})(298.15\;\text{K}) = 2.48\;\text{kJ mol}^{-1}$$
$$\Delta H_{\text c}^{\circ} = -1.362\times10^{3} - 1(2.48) = -1.365\times10^{3}\;\text{kJ mol}^{-1}$$

4. Apply Hess’s law:
$$\Delta H_{\text c}^{\circ} = \big[2\,\Delta H_f^{\circ}(\text{CO}_2) + 3\,\Delta H_f^{\circ}(\text{H}_2\text O)\big] - \big[\Delta H_f^{\circ}(\text{C}_2\text H_5\text{OH}) + 3\,\Delta H_f^{\circ}(\text O_2)\big]$$
Because $$\Delta H_f^{\circ}(\text O_2)=0$$,

$$-1365 = \big[2(-393.5) + 3(-285.8)\big] - \Delta H_f^{\circ}(\text{ethanol})$$

$$-1365 = (-787.0 - 857.4) - \Delta H_f^{\circ}(\text{ethanol})$$

$$-1365 = -1644.4 - \Delta H_f^{\circ}(\text{ethanol})$$

$$\Delta H_f^{\circ}(\text{ethanol}) = -1644.4 + 1365 = -279.4\;\text{kJ mol}^{-1}$$

5. Magnitude (as asked) rounded to two significant figures:
$$|\Delta H_f^{\circ}(\text{ethanol})| = 2.80 \times 10^{2}\;\text{kJ mol}^{-1}$$

Answer: 2.80