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.

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.