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Question: $K_1$, $K_2$ and $K_3$ are the equilibrium constants of the following reactions (I), (II) and (III) ...

K1K_1, K2K_2 and K3K_3 are the equilibrium constants of the following reactions (I), (II) and (III) respectively

I) N2+2O2⇌2NO2N_2 + 2O_2 \rightleftharpoons 2NO_2 II) 2NO2⇌N2+2O22NO_2 \rightleftharpoons N_2 + 2O_2 III) NO2⇌12N2+O2NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2 The correct relation from the following is

A

K1=1K2=1K3K_1 = \frac{1}{K_2} = \frac{1}{K_3}

B

K1=1K2=1(K3)2K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}

C

K1=K2=K3K_1 = \sqrt{K_2} = K_3

D

K1=1K2=K3K_1 = \frac{1}{K_2} = K_3

Answer

K1=1K2=1(K3)2K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}

Explanation

Solution

To determine the correct relation between the equilibrium constants K1K_1, K2K_2, and K3K_3 for the given reactions, we will write down the expression for each constant and then establish the relationships.

The given reactions are: I) N2+2O2⇌2NO2N_2 + 2O_2 \rightleftharpoons 2NO_2 II) 2NO2⇌N2+2O22NO_2 \rightleftharpoons N_2 + 2O_2 III) NO2⇌12N2+O2NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2

The equilibrium constant expressions are: For Reaction I: K1=[NO2]2[N2][O2]2K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} For Reaction II: K2=[N2][O2]2[NO2]2K_2 = \frac{[N_2][O_2]^2}{[NO_2]^2} For Reaction III: K3=[N2]1/2[O2][NO2]K_3 = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}

1. Relationship between K1K_1 and K2K_2:

Reaction II is the reverse of Reaction I. When a reaction is reversed, its equilibrium constant becomes the reciprocal of the original equilibrium constant. Therefore, K2=1K1K_2 = \frac{1}{K_1}, which implies K1=1K2K_1 = \frac{1}{K_2}.

2. Relationship between K1K_1 and K3K_3:

Let's manipulate Reaction III to obtain Reaction I. Reaction III: NO2⇌12N2+O2NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2 (with equilibrium constant K3K_3)

First, reverse Reaction III: 12N2+O2⇌NO2\frac{1}{2}N_2 + O_2 \rightleftharpoons NO_2 When a reaction is reversed, its equilibrium constant becomes the reciprocal. So, the equilibrium constant for this reversed reaction is 1K3\frac{1}{K_3}.

Next, multiply the reversed reaction by 2: 2×(12N2+O2⇌NO2)2 \times (\frac{1}{2}N_2 + O_2 \rightleftharpoons NO_2) This gives: N2+2O2⇌2NO2N_2 + 2O_2 \rightleftharpoons 2NO_2, which is Reaction I. When a reaction is multiplied by a factor 'n', its equilibrium constant is raised to the power 'n'. Here, n=2. So, the equilibrium constant for Reaction I (K1K_1) will be (1K3)2\left(\frac{1}{K_3}\right)^2. K1=1(K3)2K_1 = \frac{1}{(K_3)^2}

Combining the relationships:

We have found two relationships:

  1. K1=1K2K_1 = \frac{1}{K_2}
  2. K1=1(K3)2K_1 = \frac{1}{(K_3)^2}

Therefore, the correct combined relation is K1=1K2=1(K3)2K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}.