Given the following thermochemical equations, $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-1299.5 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & & \Delta H=-393.5 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-285.8 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the decomposition of one mole of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), to its elements in their stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

First, we need to determine the reaction for the decomposition of one mole of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), into its component elements in their standard states at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). Carbon is a solid, hydrogen is a gas, and oxygen is a gas. The balanced reaction is: $$\mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{C}(s) + \mathrm{H}_{2}(g)$$

Now we need to use the given reactions to construct this decomposition reaction. 1. In the combustion of acetylene (\(\Delta H_1 = -1299.5\text{ kJ}\)), the acetylene gas is on the reactant side, which is what we want. However, we need to reverse the reaction to get the carbon and hydrogen as products. So, reverse the first reaction: $$2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g) + \frac{5}{2}\mathrm{O}_{2}(g)$$ 2. In the formation of CO2 from carbon (\(\Delta H_2 = -393.5\text{ kJ}\)), we need to have 2 moles of carbon solid as a product. To do that, multiply the second reaction by 2: $$2\mathrm{C}(s) + 2\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CO}_{2}(g)$$ 3. In the formation of water from H2 gas (\(\Delta H_3 = -285.8\text{ kJ}\)), the reaction is already correctly representing the formation of one mole of water from one mole of H2 gas, so we don't need adjustments here.

Now, add the manipulated reactions obtained in step 2: $$\begin{aligned} 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) &\longrightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g) + \frac{5}{2}\mathrm{O}_{2}(g) \\ 2\mathrm{C}(s) + 2\mathrm{O}_{2}(g) &\longrightarrow 2\mathrm{CO}_{2}(g) \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) &\longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \end{aligned}$$ Notice that some species cancel out: \(2\mathrm{CO}_{2}\), \(\mathrm{H}_{2} \mathrm{O}\) and \(\frac{3}{2}\mathrm{O}_{2}\). The summed reaction becomes: $$\mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{C}(s) + \mathrm{H}_{2}(g) $$ Which is exactly the decomposition reaction we wanted to find the enthalpy change for.

Now we can calculate the enthalpy change for the decomposition reaction by summing the enthalpy changes for the manipulated reactions in step 2. Remember to adjust the sign and multiply the values when needed: $$\Delta H_\text{decomposition} = \Delta H_1^\text{reversed} + 2\Delta H_2 + \Delta H_3 = (1)(1299.5\text{ kJ}) + 2(-393.5\text{ kJ}) + (-285.8\text{ kJ})$$ Calculate the value: $$\Delta H_\text{decomposition} = 1299.5\text{ kJ} - 787\text{ kJ} - 285.8\text{ kJ} = 226.7\text{ kJ}$$ The enthalpy change for the decomposition of one mole of acetylene to its elements in their stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is 226.7 kJ.