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).
[tex]Kc=\frac{kf}{ki} =\frac{1.60 \times 10^{2} s^{-1} }{ 9.30 \times 10^{-2} s^{-1}} =1.72 \times 10^{3}[/tex]
If we consider the inverse reaction B ⇒ A, its equilibrium constant (K'c) is the inverse of the forward reaction equilibrium constant.
[tex]K'c=\frac{1}{Kc} =\frac{1}{1.72 \times 10^{3} } =5.81 \times 10^{-4}[/tex]
Since the forward and reverse reactions are first order;
The equilibrium constant, Kc of a reaction is the ratio of the concentration of the products to that of the reactants.
The rate constant of the forward reaction = 1.60*10^2 s^-1.
The rate constant of the backward reaction = 9.30*10^-2 s^-1
Since the forward and reverse reaction is first order, the equilibrium constant, Kc of the reaction A ---> B is given as:
Kc = 1.60*10^2 s^-1 / 9.30*10^-2 s^-1
Kc = 1.72 × 10^3
The equilibrium constant, Kc of the reaction B ---> A is the inverse of Kc of A ---> B and is is given as:
Kc = 1/1.72 × 10^3
Kc = 5.81 × 10^-4
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