Question

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?

Answer 1

Answer:

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).

$Kc=\frac{kf}{ki} =\frac{1.60 \times 10^{2} s^{-1} }{ 9.30 \times 10^{-2} s^{-1}} =1.72 \times 10^{3}$

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

$K'c=\frac{1}{Kc} =\frac{1}{1.72 \times 10^{3} } =5.81 \times 10^{-4}$