Question

For a reaction scheme \({\rm{A}}\mathop \to \limits^{{\rm{\;}}{{\rm{K}}_1}{\rm{\;}}} {\rm{B}}\mathop \to \limits^{{\rm{\;}}{{\rm{K}}_2}{\rm{\;}}} {\rm{C}}\), if the rate of formation of B is set to be zero then the concentration of B is given by:
1. (k₁ – k₂)[A]
2. (k₁ k₂) [A]
3. (k₁ k₂) [A]
4. \(\frac{{{k_1}}}{{{k_2}}}{\rm{\;}}\left[ A \right]\)

Answer 1

Correct Answer - Option 4 : \(\frac{{{k_1}}}{{{k_2}}}{\rm{\;}}\left[ A \right]\)

Concept:

Steady-state approximation:

When a reaction mechanism has several steps of comparable rates, the rate-determining step is often not obvious. However, there is an intermediate in some of the steps. An intermediate is a species that is neither one of the reactants, nor one of the products. The steady-state approximation is a method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated. Its concentration remains the same in a duration of the reaction.

The steady state is a situation in which all state variables are constant in spite of ongoing processes that strive to change them.

Calculation:

Given equation,

\({\rm{A}}\mathop \to \limits^{{\rm{\;}}{{\rm{K}}_1}{\rm{\;}}} {\rm{B}}\mathop \to \limits^{{\rm{\;}}{{\rm{K}}_2}{\rm{\;}}} {\rm{C}}\)

By using steady state approximation, we can find the concentration of B.

Let us consider that equation is differentiated with respect to B is 0, then

\(\frac{{d\left[ B \right]}}{{dt}} = {k_1}\left[ A \right] - {k_2}\left[ B \right] = 0\)

Where k₁ and k₂ are rates of the reaction

A is an initial reactant

B is an intermediate reactant

C is the final product.

\(\frac{{d\left[ B \right]}}{{dt}} = 0\)

⇒ 0 = k₁ [A] – k₂ [B]

\(\Rightarrow \left[ B \right] = \frac{{{k_1}}}{{{k_2}}}\left[ A \right]\)