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:

Identify target reaction and given equations

Using the Manipulating Equilibrium Constants knowledge point
Given:

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

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

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

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

Target:

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

Manipulate given equations to match target

Using the Manipulating Equilibrium Constants knowledge point
Multiply reaction (1) by 3:

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

Reverse reaction (2):

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

Combine equations and calculate final constant

Using the Manipulating Equilibrium Constants knowledge point
Add the manipulated equations:

$$ K = K_1' \times K_2' = \frac{54^3}{1.04 \times 10^{-4}} $$

$$ K = \frac{157464}{1.04 \times 10^{-4}} \approx 1.514 \times 10^9 $$

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> \(\times 10^{\text{9}}\)