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:

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<reasoning>

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:
\[

$$\begin{aligned} 2\text{NH}_3(g) & ightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \\ +\quad 3\text{H}_2(g) + 3\text{I}_2(g) & ightleftharpoons 6\text{HI}(g) \\ \hline 2\text{NH}_3(g) + 3\text{I}_2(g) & ightleftharpoons 6\text{HI}(g) + \text{N}_2(g) \end{aligned}$$

\]
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.
</reasoning>

<answer>
<mcq-correct>1. \(\Delta H = 3(\Delta H_1) + -1(\Delta H_2)\)</mcq-correct>
<mcq-option>2. \(\Delta H = (\Delta H_1)^3 \times -1(\Delta H_2)\)</mcq-option>
<mcq-option>3. \(\Delta H = (\Delta H_1)^3 + \frac{1}{\Delta H_2}\)</mcq-option>
<mcq-option>4. \(\Delta H = 3(\Delta H_1) \times \frac{1}{\Delta H_2}\)</mcq-option>
</answer>

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"Hess's Law"
]
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Answer:

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<reasoning>

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:
\[

$$\begin{aligned} 2\text{NH}_3(g) & ightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \\ +\quad 3\text{H}_2(g) + 3\text{I}_2(g) & ightleftharpoons 6\text{HI}(g) \\ \hline 2\text{NH}_3(g) + 3\text{I}_2(g) & ightleftharpoons 6\text{HI}(g) + \text{N}_2(g) \end{aligned}$$

\]
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.
</reasoning>

<answer>
<mcq-correct>1. \(\Delta H = 3(\Delta H_1) + -1(\Delta H_2)\)</mcq-correct>
<mcq-option>2. \(\Delta H = (\Delta H_1)^3 \times -1(\Delta H_2)\)</mcq-option>
<mcq-option>3. \(\Delta H = (\Delta H_1)^3 + \frac{1}{\Delta H_2}\)</mcq-option>
<mcq-option>4. \(\Delta H = 3(\Delta H_1) \times \frac{1}{\Delta H_2}\)</mcq-option>
</answer>

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