Question

i. \\(\text{h}_2(\text{g}) + \text{i}_2(\text{g}) \
ightleftharpoons 2\text{hi}(\text{g})\\)
\\(k = 54\\) @ \\(700\text{ k}\\)

ii. \\(\text{n}_2(\text{g}) + 3\text{h}_2(\text{g}) \
ightleftharpoons 2\text{nh}_3(\text{g})\\)
\\(k = 1.04 \times 10^{-4}\\) @ \\(700\text{ k}\\)

what is the \\(k\\) 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})\\)

\\(k = ? \times 10^{?}\\)

Explanation:

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<reasoning>

Identify the target reaction and given equations

Using the Chemical Equilibrium and Equilibrium Constant knowledge points
Given reactions:

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

ightleftharpoons 2\text{HI}(g)\) with \(K_1 = 54\)

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

ightleftharpoons 2\text{NH}_3(g)\) with \(K_2 = 1.04 \times 10^{-4}\)

Target reaction:
\(\text{2NH}_3(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g) + \text{N}_2(g)\) with \(K_{\text{target}}\)

Manipulate given equations to match target

Using the Chemical Equilibrium and Equilibrium Constant knowledge points

• Reverse reaction (2) to place \(\text{2NH}_3(g)\) on the reactant side:

\(2\text{NH}_3(g)
ightleftharpoons \text{N}_2(g) + 3\text{H}_2(g)\)
\(K_2' = \frac{1}{K_2} = \frac{1}{1.04 \times 10^{-4}}\)

• Multiply reaction (1) by 3 to get \(3\text{I}_2(g)\) and \(6\text{HI}(g)\):

\(3\text{H}_2(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g)\)
\(K_1' = (K_1)^3 = 54^3\)

Combine reactions and calculate target K

Using the Chemical Equilibrium and Equilibrium Constant knowledge points
Adding the two manipulated equations yields the target equation:
\(2\text{NH}_3(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g) + \text{N}_2(g)\)

\[

$$\begin{aligned} K_{\text{target}} &= K_1' \times K_2' = \frac{54^3}{1.04 \times 10^{-4}} \\ K_{\text{target}} &= \frac{157464}{1.04 \times 10^{-4}} \approx 1.514 \times 10^9 \end{aligned}$$

\]

Format the final answer

Using the Chemical Equilibrium and Equilibrium Constant knowledge points
The target value is expressed in scientific notation as \(K = [?] \times 10^{[?]}\).
Comparing \(1.51 \times 10^9\) to the format \(K = [?] \times 10^{[?]}\):
The coefficient is \(1.51\) (or \(1.5\)) and the exponent is \(9\).
</reasoning>

<answer>
What is the K value for the reaction below?
\(2\text{NH}_3(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g) + \text{N}_2(g)\)
\(K =\) <blank>\(1.5\)</blank> \(\times 10^{\text{}9\text{}}\)
</answer>

<post_analysis>
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"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Natural Science",
"Chemistry",
"Chemical Equilibrium"
]
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</post_analysis>

Answer:

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"Equilibrium Constant",
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</pre_analysis>

<reasoning>

Identify the target reaction and given equations

Using the Chemical Equilibrium and Equilibrium Constant knowledge points
Given reactions:

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

ightleftharpoons 2\text{HI}(g)\) with \(K_1 = 54\)

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

ightleftharpoons 2\text{NH}_3(g)\) with \(K_2 = 1.04 \times 10^{-4}\)

Target reaction:
\(\text{2NH}_3(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g) + \text{N}_2(g)\) with \(K_{\text{target}}\)

Manipulate given equations to match target

Using the Chemical Equilibrium and Equilibrium Constant knowledge points

• Reverse reaction (2) to place \(\text{2NH}_3(g)\) on the reactant side:

\(2\text{NH}_3(g)
ightleftharpoons \text{N}_2(g) + 3\text{H}_2(g)\)
\(K_2' = \frac{1}{K_2} = \frac{1}{1.04 \times 10^{-4}}\)

• Multiply reaction (1) by 3 to get \(3\text{I}_2(g)\) and \(6\text{HI}(g)\):

\(3\text{H}_2(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g)\)
\(K_1' = (K_1)^3 = 54^3\)

Combine reactions and calculate target K

Using the Chemical Equilibrium and Equilibrium Constant knowledge points
Adding the two manipulated equations yields the target equation:
\(2\text{NH}_3(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g) + \text{N}_2(g)\)

\[

$$\begin{aligned} K_{\text{target}} &= K_1' \times K_2' = \frac{54^3}{1.04 \times 10^{-4}} \\ K_{\text{target}} &= \frac{157464}{1.04 \times 10^{-4}} \approx 1.514 \times 10^9 \end{aligned}$$

\]

Format the final answer

Using the Chemical Equilibrium and Equilibrium Constant knowledge points
The target value is expressed in scientific notation as \(K = [?] \times 10^{[?]}\).
Comparing \(1.51 \times 10^9\) to the format \(K = [?] \times 10^{[?]}\):
The coefficient is \(1.51\) (or \(1.5\)) and the exponent is \(9\).
</reasoning>

<answer>
What is the K value for the reaction below?
\(2\text{NH}_3(g) + 3\text{I}_2(g)
ightleftharpoons 6\text{HI}(g) + \text{N}_2(g)\)
\(K =\) <blank>\(1.5\)</blank> \(\times 10^{\text{}9\text{}}\)
</answer>

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"question_type": "Fill-in-the-blank",
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"Natural Science",
"Chemistry",
"Chemical Equilibrium"
]
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</post_analysis>