Calculate the heat of combustion of ethene at constant pressure, given the standard enthalpy of combustion is -1411 kJ/mol at 17°C.

Combustion of ethene with oxygen produces carbon dioxide and water. Since water is specified to be in the liquid state, the equation is:

$$\text{C}_2\text{H}_4(\text{g}) + 3\text{O}_2(\text{g}) \longrightarrow 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$$

$\Delta n_g$ is the moles of gaseous products minus the moles of gaseous reactants.

$$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$$

From the balanced equation:
Gaseous products: 2 moles of $\text{CO}_2(\text{g})$
Gaseous reactants: 1 mole of $\text{C}_2\text{H}_4(\text{g}) + 3$ moles of $\text{O}_2(\text{g}) = 4$ moles

$$\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2 \text{ mol}$$

Given temperature $T = 17^\circ\text{C}$.

$$T(\text{K}) = 17 + 273.15 = 290.15 \text{ K}$$

Given gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.

$$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = 0.008314 \text{ kJ K}^{-1} \text{ mol}^{-1}$$

The heat of combustion at constant volume ($\Delta U$) is given as $-1389.55 \text{ kJ}$.
The formula relating $\Delta H$ and $\Delta U$ is:

$$\Delta H = \Delta U + \Delta n_g RT$$

Substitute the values:

$$\Delta H = -1389.55 \text{ kJ} + (-2 \text{ mol}) \times (0.008314 \text{ kJ K}^{-1} \text{ mol}^{-1}) \times (290.15 \text{ K})$$ $$\Delta H = -1389.55 \text{ kJ} - (2 \times 0.008314 \times 290.15) \text{ kJ}$$ $$\Delta H = -1389.55 \text{ kJ} - 4.8248 \text{ kJ}$$ $$\Delta H = -1394.3748 \text{ kJ}$$

Rounding to two decimal places:

$$\Delta H = \boxed{-1394.37 \text{ kJ}}$$