\( \left(\mathrm{H}_{4}(g)+\mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{HCH}_{(g)}+3 \mathrm{H}_{2}(\mathrm{~g}) \Delta H=\right. \) ? Mermbdemical Eqns: \[ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H=-9 \cdot 8 \mathrm{lg} \] (cs)t \( 2 H_{2}(g) \rightarrow C H 4(g) \quad \Delta A=-74.9 \mathrm{ks} \) \( H_{2}(y)+2 C(s)+N_{2}(y) \rightarrow 2 H(H \) cs) \( D H=+270.3 \mathrm{~kJ} \)

To find the enthalpy change (\( \Delta H \)) for the reaction \[ \mathrm{H}_{4}(g) + \mathrm{NH}_{3}(g) \rightarrow \mathrm{HCH}(g) + 3 \mathrm{H}_{2}(g), \] we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction. ### Given Reactions and Their Enthalpy Changes 1. \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H = -9.8 \, \text{kJ} \) 2. \( 2 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g) \quad \Delta H = -74.9 \, \text{kJ} \) 3. \( \mathrm{H}_{2}(g) + 2 \mathrm{C}(s) + \mathrm{N}_{2}(g) \rightarrow 2 \mathrm{H}(g) \quad \Delta H = +270.3 \, \text{kJ} \) ### Steps to Find \( \Delta H \) 1. **Reverse the second reaction** to express \( \mathrm{H}_{4}(g) \) in terms of \( \mathrm{H}_{2}(g) \): \[ \mathrm{CH}_{4}(g) \rightarrow 2 \mathrm{H}_{2}(g) \quad \Delta H = +74.9 \, \text{kJ} \] 2. **Use the first reaction** as is to form \( \mathrm{NH}_{3}(g) \): \[ \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H = -9.8 \, \text{kJ} \] 3. **Use the third reaction** as is to form \( \mathrm{H}(g) \): \[ \mathrm{H}_{2}(g) + 2 \mathrm{C}(s) + \mathrm{N}_{2}(g) \rightarrow 2 \mathrm{H}(g) \quad \Delta H = +270.3 \, \text{kJ} \] ### Combining the Reactions Now we will combine these reactions to find the overall reaction and its enthalpy change. 1. Start with the third reaction: \[ \mathrm{H}_{2}(g) + 2 \mathrm{C}(s) + \mathrm{N}_{2}(g) \rightarrow 2 \mathrm{H}(g) \quad \Delta H = +270.3 \, \text{kJ} \] 2. Add the first reaction: \[ \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H = -9.8 \, \text{kJ} \] 3. Add the reversed second reaction: \[ \mathrm{CH}_{4}(g) \rightarrow 2 \mathrm{H}_{2}(g) \quad \Delta H = +74.9 \, \text{kJ} \] ### Calculate the Total \( \Delta H \) Now we can sum the enthalpy changes: \[ \Delta H = +270.3 \, \text{kJ} - 9.8 \, \text{kJ} + 74.9 \, \text{kJ} \] Let's calculate this: \[ \Delta H = 270.3 - 9.8 + 74.9 \] Now, I will perform the calculation. Calculate the value by following steps: - step0: Calculate: \(270.3-9.8+74.9\) - step1: Calculate: \(335.4\) The total enthalpy change (\( \Delta H \)) for the reaction \[ \mathrm{H}_{4}(g) + \mathrm{NH}_{3}(g) \rightarrow \mathrm{HCH}(g) + 3 \mathrm{H}_{2}(g) \] is \[ \Delta H = 335.4 \, \text{kJ}. \] Thus, the final answer is: \[ \Delta H = 335.4 \, \text{kJ}. \]