3 years ago

The least massive of all the alkaline earth metals. Who am I?

1 answer:

5 0

Answer:argon

Explanation:

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Answer : The equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

Explanation :

The given main chemical reaction is:

$N_2(g)+2O_2(g)\rightarrow N_2O_4(g)$ ; $K$

The intermediate reactions are:

(1) $N_2O_4(g)\rightarrow 2NO_2(g)$ ; $K_a$

(2) $\frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g)$ ; $K_b$

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) $2NO_2(g)\rightarrow N_2O_4(g)$ ; $\frac{1}{K_a}$

(2) $N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$ ; $(K_b)^2$

Thus, the equilibrium constant for this reaction will be:

$K=\frac{1}{K_a}\times (K_b)^2$

$K=\frac{(K_b)^2}{K_a}$

Thus, the equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

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3 years ago