Question

i. \\(\text{h}_2(\text{g}) + \text{i}_2(\text{g}) \
ightleftharpoons 2\text{hi}(\text{g})\\) \\(\delta h_1\\)
ii. \\(\text{n}_2(\text{g}) + 3\text{h}_2(\text{g}) \
ightleftharpoons 2\text{nh}_3(\text{g})\\) \\(\delta h_2\\)

what is the \\(\delta h\\) value for the reaction below?

\\(2\text{nh}_3(\text{g}) + 3\text{i}_2(\text{g}) \
ightleftharpoons 6\text{hi}(\text{g}) + \text{n}_2(\text{g})\\)

enter the answer choice number.

1. \\(\delta h = 3(\delta h_1) + -1(\delta h_2)\\)
2. \\(\delta h = (\delta h_1)^3 \times -1(\delta h_2)\\)
3. \\(\delta h = (\delta h_1)^3 + \frac{1}{\delta h_2}\\)
4. \\(\delta h = 3(\delta h_1) \times \frac{1}{\delta h_2}\\)

Explanation:

Analyze the target chemical reaction

Using the Hesss Law for Equilibrium knowledge point
We need to find the enthalpy change \(\Delta H\) for the target reaction:

$$ 2\text{NH}_3(g) + 3\text{I}_2(g) ightleftharpoons 6\text{HI}(g) + \text{N}_2(g) $$

Express target reaction using given steps

Using the Hesss Law for Equilibrium knowledge point
We are given two reference reactions with their respective enthalpy changes:

1. \(\text{H}_2(g) + \text{I}_2(g)

ightleftharpoons 2\text{HI}(g) \quad \Delta H_1\)

1. \(\text{N}_2(g) + 3\text{H}_2(g)

ightleftharpoons 2\text{NH}_3(g) \quad \Delta H_2\)

To construct the target reaction:

• We need \(2\text{NH}_3(g)\) on the reactant side. We reverse reaction II:

$$ 2\text{NH}_3(g) ightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \quad \Delta H = -\Delta H_2 $$

• We need \(6\text{HI}(g)\) on the product side. We multiply reaction I by 3:

$$ 3\text{H}_2(g) + 3\text{I}_2(g) ightleftharpoons 6\text{HI}(g) \quad \Delta H = 3(\Delta H_1) $$

Combine the modified reactions

Using the Hesss Law for Equilibrium knowledge point
Adding the two modified equations:

$$
LATEXBLOCK0
$$

The overall enthalpy change \(\Delta H\) is the sum of the modified enthalpies:

$$ \Delta H = 3(\Delta H_1) + (-1)(\Delta H_2) $$

This matches option 1.

Answer:

• 1. \(\Delta H = 3(\Delta H_1) + -1(\Delta H_2)\) (Correct answer)
• 2. \(\Delta H = (\Delta H_1)^3 \times -1(\Delta H_2)\)
• 3. \(\Delta H = (\Delta H_1)^3 + \frac{1}{\Delta H_2}\)
• 4. \(\Delta H = 3(\Delta H_1) \times \frac{1}{\Delta H_2}\)