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 (\( \Delta H \)) 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 chemical reaction is the sum of the enthalpy changes for the individual steps. We have the following reactions and their enthalpy changes: 1. Combustion of propane: \[ \mathrm{C}_{3} \mathrm{H}_{8}(g) + 5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2}O(g) \quad \Delta H = -2043 \, \text{kJ} \] 2. Formation of carbon dioxide: \[ \mathrm{C}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \quad \Delta H = -393.5 \, \text{kJ} \] 3. Formation of water: \[ 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}O(g) \quad \Delta H = -483.6 \, \text{kJ} \] Now, we need to reverse the first reaction to find the enthalpy change for the formation of propane from carbon and hydrogen. Reversing the reaction changes the sign of \( \Delta H \): \[ 3 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2}O(g) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(g) + 5 \mathrm{O}_{2}(g) \quad \Delta H = +2043 \, \text{kJ} \] Next, we will scale the second and third equations as necessary. From the second reaction, for 3 moles of carbon dioxide formed, we have: \[ 3 \times (\mathrm{C}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)) = 3 \mathrm{C}(s) + 3 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g) \quad \Delta H = 3 \times (-393.5) = -1180.5 \, \text{kJ} \] From the third reaction, for 4 moles of water formed, we have: \[ 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}O(g) \quad \text{(since we need 4 water, multiply by 2)} \] Thus, we scale it: \[ 4 \mathrm{H}_{2}(g) + 2 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{H}_{2}O(g) \quad \Delta H = 2 \times (-483.6) = -967.2 \, \text{kJ} \] Now, let's consider the overall enthalpy change: \[ \Delta H = +2043 \, \text{kJ} - 1180.5 \, \text{kJ} - 967.2 \, \text{kJ} \] Calculating gives: \[ \Delta H = 2043 - 1180.5 - 967.2 = -104.7 \, \text{kJ} \] Thus, the standard enthalpy change for the reaction \[ 3 \mathrm{C}(s) + 4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(g) \] is \[ \Delta H = -104.7 \, \text{kJ} \]