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For a bimolecular association reaction A + B ⇌ A.B at 20 °C, the experimentally measured on rate

is 1.0 x108 M-1.s-1, while the off rate is 640 s-1.
.
What is the equilibrium association constant for the reaction (M-1)?

User Kizz
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1 Answer

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Final answer:

The equilibrium association constant for the reaction A + B ⇌ A.B at 20 °C is 1.56 x 10⁵ M⁻¹.

Step-by-step explanation:

The equilibrium association constant (Ka) can be calculated using the equation Ka = kon/koff

Where

kon is the on rate

koff is the off rate.

In this case, the on rate is given as 1.0 x 10⁸ M⁻¹.s⁻¹ and the off rate is given as 640 s⁻¹.

Substituting the values into the equation,

Ka = (1.0 x 10⁸ M⁻¹.s⁻¹)/(640 s⁻¹) = 1.56 x 10⁵ M⁻¹.

Therefore, the equilibrium association constant for the reaction at 20 °C is 1.56 x 105 M⁻¹.

User Jan Werkhoven
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