At a particular temperature, the equilibrium constants for the following reaction is:
2NO(g) + 2H₂(g) ⇄ N₂(g) + 2H₂O(g) Kc = 6.7x10²
So, the equilibrium constants for the given reactions are:
a. NO(g) + H₂(g) ⇄ 1/2N₂(g) + H₂O(g) Kc = 2.6x10
b. 2N₂(g) + 4H₂O(g) ⇄ 4NO(g) + 4H₂(g) Kc = 2.2x10⁻⁶
The first reaction is:
2NO(g) + 2H₂(g) ⇄ N₂(g) + 2H₂O(g) (1)
The equilibrium constant of the above reaction is:
[tex]K_{c_{1}} = \frac{[N_{2}][H_{2}O]^{2}}{[NO]^{2}[H_{2}]^{2}} = 6.7 \cdot 10^{2}[/tex] (2)
Now, let's find the equilibrium constant for the given reactions.
a) The reaction is:
NO(g) + H₂(g) ⇄ 1/2N₂(g) + H₂O(g) (3)
So, the constant for the reaction (3) is:
[tex] K_{c_{2}} = \frac{[N_{2}]^{1/2}[H_{2}O]}{[NO][H_{2}]} [/tex] (4)
When comparing equations (2) and (4), we can see that:
[tex] \frac{[N_{2}]^{1/2}[H_{2}O]}{[NO][H_{2}]} = (\frac{[N_{2}][H_{2}O]^{2}}{[NO]^{2}[H_{2}]^{2}})^{1/2} [/tex]
So, the equilibrium constant of the reaction (3) is:
[tex] K_{c_{2}} = (K_{c_{1}})^{1/2} = (6.7 \cdot 10^{2})^{1/2} = 2.6 \cdot 10 [/tex]
Hence, the equilibrium constant for reaction (3) is 2.6x10.
b) The given reaction is:
2N₂(g) + 4H₂O(g) ⇄ 4NO(g) + 4H₂(g) (5)
And the equilibrium constant is:
[tex] K_{c_{3}} = \frac{[NO]^{4}[H_{2}]^{4}}{[N_{2}]^{2}[H_{2}O]^{4}} [/tex] (6)
From equations (2) and (6), we have first:
[tex] \frac{[NO]^{4}[H_{2}]^{4}}{[N_{2}]^{2}[H_{2}O]^{4}} = \left(\frac{[N_{2}][H_{2}O]^{2}}{[NO]^{2}[H_{2}]^{2}}\right)^{-1} [/tex]
and second:
[tex] \frac{[NO]^{4}[H_{2}]^{4}}{[N_{2}]^{2}[H_{2}O]^{4}} = \left(\frac{[NO]^{2}[H_{2}]^{2}}{[N_{2}][H_{2}O]^{2}}\right)^{2} [/tex]
Hence, the equilibrium constant of the reaction (5) is:
[tex] K_{c_{3}} = (K_{c_{1}})^{-2} = (6.7 \cdot 10^{2})^{-2} = 2.2 \cdot 10^{-6} [/tex]
Therefore, the equilibrium constant for this reaction is 2.2x10⁻⁶.
You can learn more about equilibrium constants here:
- https://brainly.com/question/9173805?referrer=searchResults
- https://brainly.com/question/1619133?referrer=searchResults
I hope it helps you!
