Question

In the following chapter, enzyme catalysis reactions will be extensively reviewed. The first step in these reactions involves the binding of a reactant molecule (referred to as a substrate) to a binding site on the enzyme. If this binding is extremely efficient (that is, equilibrium strongly favors the enzyme–substrate complex over separate enzyme and substrate) and the formation of product rapid, then the rate of catalysis could be diffusion limited. Estimate the expected rate constant for a diffusion controlled reaction using typical values for an enzyme ( and Å) and a small molecular substrate ( and Å).

Answer 1

Complete Question

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Answer:

The rate constant is
`k_d = 3.44*10^(10) \ L \cdot mol^(-1) s^(-1)`

Step-by-step explanation:

From the question we are told that

The values for an enzyme is given as

`D_1 = 1.00 *10^(-7) \ cm^2 s^(-1)`

`r_1 = 40.0 \r A = 40*10^(-8) cm`

The values of a small molecular substrate is

`D_2 = 1.00 *10^(-5) \ cm^2 s^(-1)`

`r_2 = 5.00 \r A = 5.00*10^(-8) \ cm`

The equation relating the rate constant is

`k_d = 4 \pi N_A (D_1 +D_2) (r_1 +r_2)`

substituting values

`k_d = 4 \pi (6.022 *10^(23))(1 *10^(-7) * 1*10^(-5) (40*10^(-8) + 5*10^(-8)))`

`k_d = 3.44*10^(10) \ L \cdot mol^(-1) s^(-1)`