Question

The steady state concentration of the activated molecule [A] in the following sequence of steps is given by:

A + A $\overset{\quad K_{1}\quad}{\rightarrow}$A + A*

A* + A$\overset{\quad K_{2}\quad}{\rightarrow}$ 2A

• A. $\frac{K_{2}\lbrack A\rbrack}{K_{1}}$
• B. $\frac{K_{1}\lbrack A\rbrack}{K_{2}}$
• C. K1K2[A]
• D. $\frac{K_{1}K_{2}}{\lbrack A\rbrack}$

Answer 1

Answer: (B)

$\frac{K_{1}\lbrack A\rbrack}{K_{2}}$

Answer 2

$A + A\overset{\quad K_{1}\quad}{\rightarrow}A + A* A* + A\overset{\quad K_{2}\quad}{\rightarrow}2A$Rate of

Decomposition $\text{A = }\text{K}_{1}\lbrack A\rbrack^{2}$

Rate of formation $\text{A = }\text{K}_{2}\lbrack A\rbrack\ \lbrack\text{A*}\rbrack$

At equilibrium rate of decomposition = rate of formation

$\lbrack A*\rbrack = \frac{K_{1}}{K_{2}}\lbrack A\rbrack$