Steady state approximation
Supplementary notes for the course “Chemistry for Physicists”
Course coordinator: Prof. Dr. Mathias Nest
Teaching assistant: Dr. Raghunathan Ramakrishnan
contact: [email protected], [email protected]
Department Chemie
Technische Universität München
1 When is steady state approximation useful?
We apply steady state approximation for reactions where the concentration of the reaction intermediate can be considered to be constant. Steady state means an chemical equilibrium, i.e. the rate of formation equals the rate of consumption.
2 Why do we need steady state approximation?
We need steady state approximation to simplify the derivation of rate laws of many step reac-tions.
3 A general reaction to show the applicability of the steady state
approximation
Consider the following reaction
A → B (1)
Let us say that the reaction occurs in two steps
A → Ik1 (2)
I → Bk2 (3)
The differential rate laws are given by
dct(A) dt = −k1ct(A) (4) dct(I) dt = k1ct(A) − k2ct(I) (5) dct(B) dt = k2ct(I) (6)
The integrated rate laws which are obtained by solving the coupled differential equations are given by ct(A) = c0(A)e−k1t (7) ct(I) = c0(A)k1k−k1 2 e−k!t− e−k2t ; k16= k2 c0(A)k1te−k1t; k1 = k2 (8)
ct(B) = c0(A) − ct(A) − ct(I) (9)
The concentrations of the reactant, intermediate and the product for the cases i) k1 k2, ii) k1= k2, and iii) k1 k2 are shown below
0.0 0.5 1.0 0.0 2.0 4.0 6.0 8.0 10.0 concentration k1>> k2 A I B 0.0 0.5 1.0 0.0 2.0 4.0 6.0 8.0 10.0 concentration k1= k2 A I B 0.0 0.5 1.0 0.0 10.0 20.0 30.0 40.0 50.0 concentration k1<< k2 A I B
Clearly the steady state approximation is applicable for the case k1 k2 where the concen-tration of the intermediate is small and it varies slowly and can be considered to be constant most of the time during the reaction (i.e. to a good approximation, dct(I)
dt = 0). The situation k1 k2 also means that the intermediate is very reactive and this step is very fast. Also note that the first step is relatively very slow hence the first step determines the rate of the entire reaction. Now let us apply the steady state approximation for this case to derive the rate law.
In most of the cases, we look for an expression for ct(I) in terms of the rate constants and the concentration of reactants. Let us apply dct(I)
dt = 0 which is the steady state approximation to eq. 5 dct(I) dt = 0 ⇒ ct(I) = k1 k2 ct(A) (10)
Using the above equation we can rewrite the rate laws (differential and integral) as follows
dct(A) dt = −k1ct(A) (11) dct(I) dt = 0 (12) dct(B) dt = k1ct(A) (13) and ct(A) = c0(A)e−k1t (14) ct(I) = constant (15) ct(B) = c0(A) − ct(A) − k1 k2 ct(A) = c0(A) 1 − c0(A)e−k1t− k1 k2 c0(A)e−k1t = c0(A) n 1 − c0(A)e−k1t o (16)
where in the last equation we have used k1
k2 = 0 because k2 is large.
4 In this course we use steady state approximation only to derive
the differential rate laws of many step reactions
Let us try an example (p. 260 of Chemie by Mortimer, Müller 9. Auflage). PLEASE READ THIS SECTION IN THE BOOK (p.260) BEFORE READING THE FOLLOWING. The complete reaction is given by
H3COH + H++ Br−→ H3CBr + H2O (17)
For this reaction the experimentally derived rate law is given by
v = k · c (H3COH) · c H+ · c Br−
(18)
The proposed mechanism has three steps
step 2 : H3COH+2→ Hk2
3COH + H+ fast (20)
step 3 : Br−+ H3COH+2 → Hk3 3CBr + H2O slow (21) Since the third step is the rate determining step, we can write the rate of the reaction as (remember, the slowest step determines the rate of the entire reaction)
v = v3 = k3· c H3COH+2 · c Br−
(22)
This expression involves the concentration of the intermediate. Recall that we usually write the rate law in terms of the concentration of the reactant only. Let us use the stationary state approximation and see if we can eliminate c H3COH+2 in eq. 22. Let us equate the net rate of formation of the intermediate to zero
dc H3COH+2 dt = k1c (H3COH) c H + − k 2c H3COH+2 − k3c H3COH+2 c Br− = 0 (23) ⇒ k1c (H3COH) c H+ − c H3COH+2 k2+ k3c Br− = 0 (24) ⇒ c H3COH+2 = k1c (H3COH) c (H +) k2+ k3c (Br−) (25)
Note that the third step is the rate determining step, hence k3 k2 or k2+ k3c (Br−) ≈ k2
c H3COH+2 ≈
k1c (H3COH) c (H+) k2
(26)
By substituting the above equation in eq. 22, we arrive at the experimental rate law
v = v3 = k1k3 k2 · c (H3COH) · c H+ · c Br− (27) = k · c (H3COH) · c H+ · c Br− (28)
Now let us do a bit of analysis. For the two-step reaction given in page 2 of this document we saw that steady state approximation is valid if the intermediate is very reactive, right?. But in this three-step reaction the third step is a slow step, i.e., in the third step the intermediate reacts slowly. Then why did we use steady state approximation here? The reason is that we assume that k2 is very large, i.e., the intermediate undergoes a very rapid reverse reaction as in step 2. In this example if k2 k1 k3 the concentration of the intermediate increases very rapidly at the beginning of the reaction, for a very short time and then for most of the reaction it varies very slowly, i.e., dc(I)dt ≈ 0. This is illustrated in the following figure. Note that in the bottom figure (short time) the concentration of the intermediate increases very rapidly, but it stays essentially constant through out the reaction.
