Question

Consider the two gaseous equilibria: The values of the equilibrium constants K 1 and K 2 are related by

Answer 1

The question is missing parts. The complete question is as follows.

Consider the two gaseous equilibria involving SO2 and the corresponding equilibrium constants at 298K:

$SO_{2}_{(g)} + \frac{1}{2}O_{2}$ ⇔ $SO_{3}_{(g)}$; $K_{1}$

$2SO_{3}_{(g)}$ ⇔ $2SO_{2}_{(g)}+O_{2}_{(g)}; K_{2}$

The values of the equilibrium constants are related by:

a) $K_{1}$ = $K_{2}$

b) $K_{2} = K_{1}^{2}$

c) $K_{2} = \frac{1}{K_{1}^{2}}$

d) $K_{2}=\frac{1}{K_{1}}$

Answer: c) $K_{2} = \frac{1}{K_{1}^{2}}$

Explanation: Equilibrium constant is a value in which the rate of the reaction going towards the right is the same rate as the reaction going towards the left. It is represented by letter K and is calculated as:

$K=\frac{[products]^{n}}{[reagents]^{m}}$

The concentration of each product divided by the concentration of each reagent. The indices, m and n, represent the coefficient of each product and each reagent.

The equilibrium constants of each reaction are:

$SO_{2}_{(g)} + \frac{1}{2}O_{2}$ ⇔ $SO_{3}_{(g)}$

$K_{1}=\frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$

$2SO_{3}_{(g)}$ ⇔ $2SO_{2}_{(g)}+O_{2}_{(g)}$

$K_{2}=\frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$

Now, analysing each constant, it is easy to see that $K_{1}$ is the inverse of $K_{2}$.

If you doubled the first reaction, it will have the same coefficients of the second reaction. Since coefficients are "transformed" in power for the constant, the relationship is:

$K_{2}=\frac{1}{K_{1}^{2}}$