At a given temperature, the elementary reaction a > b in the forward direction is first order in a with a rate constant of 1.60*10^2 s^-1. the reverse reaction is first order in b and the rate constant is 9.30*10^-2 s^-1what is the value of the equilibrium constant for the reaction a > b at this temperature? what is the value of equilibrium constant for the reaction b--> a at this temperature?

The value of the equilibrium constant for the reaction A ⇒ B is Kc = 1.72 × 10³.

The value of the equilibrium constant for the reaction B ⇒ A is K'c = 5.81 × 10⁻⁴.

Explanation:

For the reaction A ⇒ B, the equilibrium constant (Kc) is equal to the forward rate constant (kf) divided by the reverse rate constant (ki).

If we consider the inverse reaction B ⇒ A, its equilibrium constant (K'c) is the inverse of the forward reaction equilibrium constant.

it's filtration, but this is an assumption so my apologies if i got this

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