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Find the equilibrium constants, Kp, for the following equilibria, (i) NO(g) + ½ O2(g) ⇄ NO2(g), Kp = ? (ii) NO2(g) ⇄ NO(g) + ½ O2(g), Kp = ?, given the equilibrium constant, Kp, for the reaction: 2NO (g) + O2(g) ⇄ 2NO2(g) Kp= 100 at the same temperature

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Answer :

(i) The value of equilibrium constants for this reaction is, 10

(ii) The value of equilibrium constants for this reaction is, 0.1

Explanation :

The given equilibrium reaction is,


2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)
K_p=100

Now we have to determine the equilibrium constants for the following equilibrium reactions.

(i)
NO(g)+(1)/(2)O_2(g)\rightleftharpoons NO_2(g)
K_p_1=?

From the given reaction we conclude that, the reaction (i) will takes place when the given main reaction will be multiplied by half (1/2). That means when reaction will be half then the equilibrium constant will be:


K_p_1=(K_p)^{(1)/(2)}


K_p_1=(100)^{(1)/(2)}


K_p_1=10

The value of equilibrium constants for this reaction is, 10

(ii)
NO_2(g)\rightleftharpoons NO(g)+(1)/(2)O_2(g)
K_p_2=?

From the given reaction we conclude that, the reaction (ii) will takes place when the reaction (i) will be reverse. That means when reaction will be reverse then the equilibrium constant will be:


K_p_2=(1)/((K_p_1))


K_p_2=(1)/((10))


K_p_2=0.1

The value of equilibrium constants for this reaction is, 0.1

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