Question

\(K_1\) and \(K_2\) are equilibrium constant for reactions (i) and (ii) 
(i) \(N_2 ( g ) + O_2 ( g ) \leftrightharpoons 2 NO ( g)\)
(ii) \(NO ( g ) \leftrightharpoons \frac{1}{2} N_2 ( g ) + \frac{ 1}{ 2} O_2 ( g )\).Then,

• A. $ K_1 = \bigg [ \frac{1}{ K_2 } \bigg ] ^2 $
• B. $ K_1 = K_2^2$
• C. $ K_1 = \frac{1}{ K_2 } $
• D. $ K_1 = ( K_2 )^0 $

Answer 1

The Correct Option is A

We are given two reactions with their equilibrium constants \(K_1\) and \(K_2\):

1. Reaction (i): \(N_2(g) + O_2(g) \leftrightharpoons 2NO(g)\) with equilibrium constant \(K_1\).
2. Reaction (ii): \(NO(g) \leftrightharpoons \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g)\) with equilibrium constant \(K_2\).

To determine the relationship between \(K_1\) and \(K_2\), let us analyze each reaction and their corresponding equilibrium expressions.

Step-by-step Analysis

1. Equilibrium expression for Reaction (i)

The balanced chemical reaction is:

N_2(g) + O_2(g) \leftrightharpoons 2NO(g)

The equilibrium constant \(K_1\) is given by:

K_1 = \frac{[NO]^2}{[N_2][O_2]}

2. Equilibrium expression for Reaction (ii)

The balanced chemical reaction is:

NO(g) \leftrightharpoons \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g)

The equilibrium constant \(K_2\) is given by:

K_2 = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}

3. Relating \(K_1\) and \(K_2\)

Note that Reaction (ii) is the reverse of Reaction (i) but halved. Thus, the relationship between \(K_1\) and \(K_2\) can be derived from the fact that reversing a reaction and changing stoichiometry impacts the equilibrium constant.

The operation of halving the coefficients of the reverse reaction and thus squaring the equilibrium expression gives:

K_1 = \left( \frac{1}{K_2} \right)^2

Conclusion

Therefore, the correct relationship between the equilibrium constants \(K_1\) and \(K_2\) is:

Correct Answer: K_1 = \bigg [ \frac{1}{ K_2 } \bigg ] ^2