Question

Using the diagram given below, the relation between k1 and k2 for the reaction A → C is: A. k1 = k2. B. k2 <<< k1. C. k1 ≤ k2. D. k1 <<< k

Answer 1

if k2 << k1, then B → C is the rate-determining step of the reaction, and the reaction rate depends mainly on k2. B. A k. ⎯→. ⎯ 1 dt. Bd dt

Answer 2

Answer:

The correct answer is Option A

Explanation:

The mass action-based model has steady-state concentrations $a and b$ characterized by

$k_{1} a=k_{2} b$

where conservation gives $a+b=T$. The steady-state concentrations are then

$a=\frac{k_{2} T}{k_{1}+k_{2}} \quad \text { and } \quad b=\frac{k_{1} T}{k_{1}+k_{2}}$

The expected (mean) abundance of A in the probability distribution (7.31) is

$\begin{aligned}E\left(N_{A}\right) &=2\left(\frac{k_{2}^{2}}{\left(k_{1}+k_{2}\right)^{2}}\right)+1\left(\frac{2 k_{1} k_{2}}{\left(k_{1}+k_{2}\right)^{2}}\right)+0\left(\frac{k_{1}^{2}}{\left(k_{1}+k_{2}\right)^{2}}\right) \\&=\frac{2 k_{2}^{2}+2 k_{1} k_{2}}{\left(k_{1}+k_{2}\right)^{2}}=\frac{2 k_{2}\left(k_{2}+k_{1}\right)}{\left(k_{1}+k_{2}\right)^{2}}=\frac{2 k_{2}}{\left(k_{1}+k_{2}\right)}\end{aligned}$

Likewise

$\begin{aligned}E\left(N_{B}\right) &=0\left(\frac{k_{2}^{2}}{\left(k_{1}+k_{2}\right)^{2}}\right)+1\left(\frac{2 k_{1} k_{2}}{\left(k_{1}+k_{2}\right)^{2}}\right)+2\left(\frac{k_{1}^{2}}{\left(k_{1}+k_{2}\right)^{2}}\right) \\&=\frac{2 k_{1}}{\left(k_{1}+k_{2}\right)} .\end{aligned}$

With molecule count T = 2, these expected values correspond to the deterministic description.