Suppose that the gas-phase reactions A→B and B→A are both elementary processes with rate constants of 4.7*10-3s-1 and 5.8*10-1s-1, respectively.

What is the value of the equilibrium constant for the equilibrium A (g) ⇌B (g) ?

K = 8.1 x 10⁻³

Explanation:

We are told here that these gas phase reactions are both elementary processes, thus the reactions forward and reverse are both first order:

A→B Rate (forward) = k (forward) x [A]

and for

B→A Rate (reverse) = k (reverse) x [B]

At equilibrium we know the rates of the forward and reverse reaction are equal, so

k (forward) x [A] = k (reverse) x [B] for A (g) ⇌B (g)

⇒ k (forward) / k (reverse) = [B] / [A] = K

4.7 x 10⁻³ s⁻1 / 5.8 x 10⁻¹ s⁻¹ = 8.1 x 10⁻³ = K

Notice how this answer is logical : the rate of the reverse reaction is greater than the forward reaction (a factor of approximately 120 times), and will be expecting a number for the equilibrium constant, K, smaller than one where the reactant concentration, [A], will prevail.

It is worth to mention that this is only valid for reactions which are single, elementary processes and not true for other equilibria.