2.8: Steady-State Approximation

Learning Objectives

By the end of this section, you should be able to:

• Select the rate-determining step given the consecutive reactions and overall rate law of the reaction

Derive the rate law of a reaction given the consecutive reactions

Derive the rate law under pre-equilibria conditions

\[A\xrightarrow{\text{ka}}I\xrightarrow{\text{kb}}P\]

This approximation assumes intermediate, I, is in a low constant concentration,ie:

\[\frac{d[I]}{dt}=0\]

Generally the approximation is more accurate when \(k_{b}>>k_{a}\) (\(k_{b}\) is much greater than \(k_{a}\)). The bigger the difference, the more accurate the assumption.

As we saw before with first order reactions: \(\frac{d[A]}{dt}=-k_{a}[A]\) leads to \([A]=[A]_{0}e^{-k_{a}t}\)

Now: \(\frac{d[I]}{dt}=k_{a}[A]-k_{b}[I]≈0\) so \([I]=\frac{k_{a}}{k_{b}}[A]\)

For product: \(\frac{d[P]}{dt}=k_{b}[I]≈k_{a}[A]\) if we solved this, we would find: \([P]=[A]_{0}(1-e^{-k_{a}t})\)

Detailed proof: This is not required for the course, but might be useful for understanding.

For reactant:

\[\frac{d[A]}{dt}=-k_{a}[A]\]

\[[A]\]

\[[A]=constant*e^{-k_{a}t}\]

\[[A]=[A]_{0}*e^{-k_{a}t}\]

For intermediate:

\[\frac{d[I]}{dt}=k_{a}[A]-k_{b}[I]≈0\]

For product:

\[\frac{d[P]}{dt}=k_{b}[I]≈k_{a}[A]\]

\begin{align*}
\frac{d[P]}{dt} & = k_{a}[A] \\
\frac{d[P]}{dt} & = k_{a}*[A]_{0}*e^{-k_{a}t}\\
[P] & = \int k_{a}*[A]_{0}*e^{-k_{a}t}\\
[P] & = -[A]_{0}*e^{-k_{a}t} + constant
\end{align*}

To find the constant, we know that there is no product in the system, initially. At t=0, [P]=0. Subsitute t=0, the exponential term is equal to 1, so we are left with \(0=-[A]_{0}+constant\), so \(constant=[A]_{0}\), Therefore:

\[[P]= -[A]_{0}*e^{-k_{a}t} + [A]_{0}=[A]_{0}(1-e^{-k_{a}t})\]

Taking arbitrary values of \(k_{a}=1\) and \(k_{b}=10\), we can plot the concentration vs. time for all the components:

Exercise: Using Steady-state Approximation to Find Rate Law Expressions

To derive the overall rate law for this reaction:

\[2N_{2}O_{5(g)}→4NO_{2(g)}+O_{2(g)}\]

Assume the reaction follows the following three-step mechanism:

The final expression of the rate law should only contain \([N_{2}O_{5}]\) and the rate constants of the elementary steps.\(^{[1]}\)

Solution

Add example text here.

Step 1: Find the intermediates and express their rate of formation. Equate their rate of formation to 0 due to steady-state approximation.

\[\frac{d[NO]}{dt}=k_{2}[NO_{3}][NO_{2}]-k_{3}[NO_{3}][NO]=0\;\;\;\;\;(1)\]

\[\frac{d[NO_{3}]}{dt}=k_{f}[N_{2}O_{5}]-k_{b}[NO_{2}][NO_{3}]-k_{2}[NO_{3}][NO_{2}]-k_{3}[NO_{3}][NO]=0\;\;\;\;\;(2)\]

Step 2: Express the overall rate law. The easiest way is to express \(\frac{d[O_{2}]}{dt}\) because it is a product that has stoichiometric coefficient of 1.

\[\frac{d[O_{2}]}{dt}=k_{2}[NO_{3}][NO_{2}]\]

Step 3: From the expression of \(\frac{d[O_{2}]}{dt}\), we can see that to find the overall rate law, we need to find \([NO_{3}][NO_{2}]\) as an expression of the concentrations of reactants or products (no intermediates) by manipulating (1) and (2).

From (1):

\[k_{2}[NO_{3}][NO_{2}]=k_{3}[NO_{3}][NO]\]

Substitute into (2):
\begin{align*}
k_{f}[N_{2}O_{5}]-k_{b}[NO_{2}][NO_{3}]-k_{2}[NO_{3}][NO_{2}]-k_{2}[NO_{3}][NO_{2}] & = 0\\
k_{b}[NO_{2}][NO_{3}]+2k_{2}[NO_{3}][NO_{2}] & = k_{f}[N_{2}O_{5}]\\
[NO_{3}][NO_{2}] & =\frac{k_{f}[N_{2}O_{5}]}{k_{b}+2k_{2}}
\end{align*}

Step 4: We can substitute the expression for \([NO_{3}][NO_{2}]\) we just found into the expression for \(\frac{d[O_{2}]}{dt}\), which is the same as the overall rate law.

\[r=\frac{d[O_{2}]}{dt}=\frac{k_{2}k_{f}[N_{2}O_{5}]}{k_{b}+2k_{2}}\]

---