Calculate the standard enthalpy change for the reaction: \[ 3 \mathrm{C}(s)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(g) \] Given that: \[ \begin{array}{cl} \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) & \Delta H_{r x n}^{\circ}=-2043 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) & \Delta H_{r x n}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) & \Delta H_{r x n}^{\circ}=-483.6 \mathrm{~kJ} / \mathrm{mol} \end{array} \]

To calculate the standard enthalpy change (\( \Delta H^\circ \)) for the reaction: \[ 3 \mathrm{C}(s) + 4 \mathrm{H}_2(g) \rightarrow \mathrm{C}_3\mathrm{H}_8(g) \] we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction, regardless of the pathway taken. ### Given Data: 1. **Combustion of Propane (\( \mathrm{C}_3\mathrm{H}_8 \))**: \[ \mathrm{C}_3\mathrm{H}_8(g) + 5 \mathrm{O}_2(g) \rightarrow 3 \mathrm{CO}_2(g) + 4 \mathrm{H}_2\mathrm{O}(g) \quad \Delta H^\circ = -2043 \ \mathrm{kJ/mol} \] 2. **Combustion of Carbon (\( \mathrm{C}(s) \))**: \[ \mathrm{C}(s) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) \quad \Delta H^\circ = -393.5 \ \mathrm{kJ/mol} \] 3. **Combustion of Hydrogen (\( \mathrm{H}_2(g) \))**: \[ 2 \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2\mathrm{O}(g) \quad \Delta H^\circ = -483.6 \ \mathrm{kJ/mol} \] ### Steps to Calculate \( \Delta H^\circ \) for the Formation of \( \mathrm{C}_3\mathrm{H}_8 \): 1. **Determine the Enthalpy of Formation for \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \)**: - \( \Delta H^\circ_f(\mathrm{CO}_2) = -393.5 \ \mathrm{kJ/mol} \) - \( \Delta H^\circ_f(\mathrm{H}_2\mathrm{O}) = \frac{-483.6 \ \mathrm{kJ/mol}}{2} = -241.8 \ \mathrm{kJ/mol} \) 2. **Apply Hess's Law to the Combustion of Propane**: \[ \Delta H^\circ_{\text{combustion}} = \left[ 3 \Delta H^\circ_f(\mathrm{CO}_2) + 4 \Delta H^\circ_f(\mathrm{H}_2\mathrm{O}) \right] - \Delta H^\circ_f(\mathrm{C}_3\mathrm{H}_8) \] Plugging in the known values: \[ -2043 \ \mathrm{kJ} = \left[ 3(-393.5) + 4(-241.8) \right] - \Delta H^\circ_f(\mathrm{C}_3\mathrm{H}_8) \] \[ -2043 \ \mathrm{kJ} = (-1180.5 - 967.2) \ \mathrm{kJ} - \Delta H^\circ_f(\mathrm{C}_3\mathrm{H}_8) \] \[ -2043 \ \mathrm{kJ} = -2147.7 \ \mathrm{kJ} - \Delta H^\circ_f(\mathrm{C}_3\mathrm{H}_8) \] 3. **Solve for \( \Delta H^\circ_f(\mathrm{C}_3\mathrm{H}_8) \)**: \[ \Delta H^\circ_f(\mathrm{C}_3\mathrm{H}_8) = -2147.7 \ \mathrm{kJ} + 2043 \ \mathrm{kJ} = -104.7 \ \mathrm{kJ/mol} \] ### Conclusion: The standard enthalpy change for the formation of propane (\( \mathrm{C}_3\mathrm{H}_8 \)) from its elements is: \[ \Delta H^\circ = -104.7 \ \mathrm{kJ/mol} \] Rounded to an appropriate number of significant figures based on the given data, the standard enthalpy change is: \[ \Delta H^\circ = -105 \ \mathrm{kJ/mol} \] **Final Answer:** The standard enthalpy change is –104.7 kJ mol⁻¹ for the formation of propane.