Question

Equilibrium constants K₁ and K₂ for the following equilibria: NO_(g)+\(\bigg(\frac{1}{2}\bigg)\)O₂ \(^{K_1}_\rightleftharpoons\) NO₂(g) and 2NO_2(g) \(^{K_2}_\rightleftharpoons\) 2NO_(g)+O_2(g) are related as:

• A. \(K_2\) = \(\frac{1}{K_1}\)
• B. K₂ = \(K^2_1\)
• C. K₂ = \(K^1_2\)
• D. K₂ = \(\frac{1}{K^2_1}\)

Answer 1

The Correct Option is D

To solve the problem of determining the relationship between the equilibrium constants \(K_1\) and \(K_2\) for the given chemical equilibria, we need to understand how these constants relate to each other based on the provided reactions.

The first equilibrium is given by:

NO_{(g)} + \bigg(\frac{1}{2}\bigg)O_{2(g)} \rightleftharpoons NO_{2(g)}\quad \text{with equilibrium constant}\, K_1

The second equilibrium is given by:

2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)} \quad \text{with equilibrium constant}\, K_2

To find the relationship between \(K_1\) and \(K_2\), we need to analyze how the second reaction is related to the first one. Notice that the second reaction is essentially the reverse of the first reaction, but with coefficients multiplied by 2:

The reverse of the first reaction with doubled coefficients is:

2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}\quad \text{which is the same as the second reaction given}

The relationship between the equilibrium constants for the reverse reactions is:

• For a reaction occurring in reverse, the equilibrium constant is the reciprocal; hence, the equilibrium constant for reversing the first reaction, doubled, would be:

(K_1)^{-1} when reversed and squared when doubled (as coefficients are doubled):

K_2 = \frac{1}{K_1^2}

Thus, the relationship is established as:

K_2 = \frac{1}{K_1^2}

This aligns with the given correct answer.