A reversible reaction has a forward rate constant of 0.412 mol/L/s and a reverse reaction rate constant of 0.827 mol/L/s. What's the equilibrium constant for this reaction?

0.498

Explanation:

Given the following :

Forward rate constant (Kf) = 0.412 mol/L/s

Reverse rate constant (Kr) = 0.827 mol/L/s

The ratio of the rate constant for the forward reaction and the rate constant for the reverse reaction produces the equilibrium constant.

Therefore,

Keq = (forward rate constant (Kf) / Reverse rate constant)

Keq = 0.412mol/L/s / 0.827mol/L/s

Keq = 0.498