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\( \left(\mathrm{H}_{4}(g)+\mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{HCH}_{(g)}+3 \mathrm{H}_{2}(\mathrm{~g}) \Delta H=\right. \) ? 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) \( D H=+270.3 \mathrm{~kJ} \)

Ask by Gray Spencer. in the Philippines
Feb 25,2025

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The enthalpy change (\( \Delta H \)) for the reaction is \( 335.4 \, \text{kJ} \).

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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: \[ \Delta H = 335.4 \, \text{kJ}. \]

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Beyond the Answer

Did you know that the reactions involving nitrogen and hydrogen to produce ammonia, known as the Haber process, were a game-changer in the early 20th century? This revolutionary method not only helped to circumvent the limitations of natural nitrogen sources but also paved the way for increased agricultural yield through synthetic fertilizers. It’s a fascinating intersection of chemistry and history that transformed food production globally! If you're looking to tackle thermochemical equations, a common mistake is to overlook the direction of the reaction when interpreting enthalpy changes. Always remember that the sign of ΔH indicates whether a reaction is exothermic (releases heat, negative ΔH) or endothermic (absorbs heat, positive ΔH). Mapping out a reaction cycle can help visualize these transitions and aid in avoiding confusion, ensuring you get those calculations right!

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