In this context, we learn about rate law, the order of reaction, and their importance in chemical kinetics. The rate law explains the dependence of the rate of reaction on the concentration of species in the reaction. This dependence is very complex as it changes with a change in the concentration of specie. We only study the simplest form of independence. There is only one method to explain or find the dependence is experimental.
What does the Rate Law Expression?
Let us a general reaction having “A” and “B” reactants and gives the product as given,
2A + B → Product
We conduct the experiment,
- If we double the concentration of “A”, the rate of reaction will be double. (concentration of B remains constant)
[A] → Double then ROR → Double
ROR ∝ [A]1
- If we double the concentration of “B”, the rate of reaction will be a fourth time.
[B] → Double then ROR → Increases 4 times
ROR ∝ [B]2
The overall rate of reaction will be:
ROR ∝ [A]1 [B]2
ROR = K [A]1 [B]2
Where K is the rate constant and this expression is called rate law expression. This rate law expression tells the independence of the rate of reaction on the concentration of species involved in the reaction. Now we write the general form of rate law expression:
2A + B → Product
By experiment,
ROR (r) ∝ [A]x
ROR (r) ∝ [B]y
ROR ∝ [A]x [B]y
ROR = K [A]x [B]y
Where,
- [A] = concentration of A
- [B] = concentration of B
- x = order of reaction with respect to “A”
- y = order of reaction with respect to “B”
- x+y = overall order of reaction (n)
- K = rate constant or velocity constant or specific reaction rate
If the concentration of all species in the rate law expression is unity ([A] = [B] = 1), then
ROR = K
Unit of Rate Constant K
ROR = K [A]x [B]y
K = ROR / [A]x [B]y
As, ROR = Conc / time
K = (mol/L) s-1 / (mol/L)x (mol/L)y
K = (mol/L) s-1 / (mol/L)x+y
K = (mol/L) s-1 / (mol/L)n
K = (mol/L)1-n s-1
The unit of “K” depends upon the order of the reaction.
Relation Between Order of Reaction and Unit of Rate constant K
| Order of Reaction | Unit of K |
|---|---|
| Zero-order reaction | (mol/L)1 s-1 |
| First-order reaction | s-1 |
| Second-order reaction | (mol/L)-1 s-1 |
| Third-order reaction | (mol/L)-2 s-1 |
Difference Between Molecularity and Order of Reaction?
If K = 3×10-3 l2mol-2s-1. Find the Order of Reaction.
Unit of K = (mol/L)1-n s-1
Unit of K is compared with the given unit
(mol/L)1-n s-1 = (mol/L)-2 s-1
1-n = -2
n = 1+2
n = 3
n = 3 so it is the third-order reaction.
The reaction 2A + 3B → Products. The order of reaction with respect to “A” is 2 and the order of reaction with respect to “B” is -1. With rate law expression and also write calculate the order of the reaction. What is the effect on the rate when the concentration of “A” is double alone, the concentration of “B” is halved alone, the concentration of “A” and “B” is double, and the volume of the container increases three times?
Rate law expression =?
ROR = K [A]x [B]y
ROR = K [A]2 [B]-1
Order of reaction =?
Order of reaction = x + y
Order of reaction = (2) + (-1)
Order of reaction = 1
1) When the concentration of “A” is double alone:
Initial rate = r = K [A]2 [B]-1 —— (1)
Final rate = r’ = K [2A]2 [B]-1 ——(2)
Divide (2) by (1)
r’ / r = K [2A]2 [B]-1 / K [A]2 [B]-1
r’ / r = (2)2
r’ = 4r
rate increses 4 times
2) When the concentration of “B” is halved alone:
Initial rate = r = K [A]2 [B]-1 —— (1)
Final rate = r’ = K [A]2 [B/2]-1 ——(2)
Divide (2) by (1)
r’ / r = K [A]2 [B/2]-1 / K [A]2 [B]-1
r’ / r = (1/2)-1
r’ = 2r
rate increses 2 times
3) When the concentration of both “A” and “B” is double:
Initial rate = r = K [A]2 [B]-1 —— (1)
Final rate = r’ = K [2A]2 [2B]-1 ——(2)
Divide (2) by (1)
r’ / r = K [2A]2 [2B]-1 / K [A]2 [B]-1
r’ / r = (2)2 (2)-1
r’ / r = (4) (1/2)
r’ = 2r
rate increses 2 times
4) When the volume of the container increases three times:
Concentration ∝ 1/volume
If we increase the volume, the concentration is decreased. So, if we increase the volume 3 times, the concentration will also decrease 3 times.
Initial rate = r = K [A]2 [B]-1 —— (1)
Final rate = r’ = K [A/3]2 [B/3]-1 ——(2)
Divide (2) by (1)
r’ / r = [1/3]2 [1/3]-1
r’ / r = 1/9 × 3
r’ = r/3
rate decreased 3 times
The reaction 2A + B → Products. When the concentration of “A” is double alone, the rate of reaction increases 2 times. When the concentration of both reactants “A” and “B” are double, the rate of reaction increases 8 times. Write the rate law expression. Find the order of reaction.
As we know,
Initial rate = r = K [A]x [B]y —— (1)
- If [A] will be doubled, the ROR increases 2 times
2r = K [2A]x [B]y —— (2)
- If [A] and [B] will be doubled, the ROR increases 8 times
8r = K [2A]x [2B]y —— (3)
Divide (2) by (1)
2r / r = K [2A]x [B]y / K [A]x [B]y
2 = 2x
x = 1
Divide (3) by (1)
8r / r = K [2A]x [2B]y / K [A]x [B]y
8 = (2)x (2)y
8 = 2x+y
23 = 2x+y
x + y = 3
put x = 1
1 + y = 3
y = 2
1) The rate law expression:
ROR = K [A]x [B]y
ROR = K [A]1 [B]2
2) The order of reaction:
order of reaction = n = x + y = 1 + 2 = 3
The reaction A + 2B + 4C → Products. When the volume of the container is double, the rate of reaction decreases 8 times. When the concentration of “A” is double and the concentration of “B” is halved, the rate of reaction increases 4 times. When the concentration of “C” increases 4 times alone, the rate of reaction increases 64 times. Write the rate law expression and find the order of the reaction.
As we know,
Initial rate = r = K [A]x [B]y [C]z —— (1)
- When the volume is double, the ROR decreases 8 times.
concentration = mol/v
C ∝ 1 / V
r/8 = K [A/2]x [B/2]y [C/2]z —— (2)
- When [A] is double and [B] is halved, the ROR increases 4 times.
4r = K [2A]x [B/2]y [C]z —— (3)
- When [C] is increasing 4 times alone, the ROR increases 64 times.
64 r = K [A]x [B]y [4C]z —— (4)
Divide (1) by (2)
r / r/8 = K [A]x [B]y [C]z / K [A/2]x [B/2]y [C/2]z
8 = 1 / (1/2)x (1/2)y (1/2)z
8 = 2x 2y 2z
8 = 2x+y+z
23 = 2x+y+z
x + y + z = 3 —— (5)
Divide (4) by (1)
64r / r = K [A]x [B]y [4C]z / K [A]x [B]y [C]z
64 = 4z
43 = 4z
z = 3
Divide (3) by (1)
4r / r = K [2A]x [B/2]y [C]z / K [A]x [B]y [C]z
4 = [2]x [1/2]y
4 = 2x-y
22 = 2x-y
x-y = 2 —— (6)
Put the value of z=3 in eq (5)
x + y + z = 3
x + y = 0 —— (7)
For the value of “x” add eq (6) and (7)
after adding the x + y = 0 and x – y = 2, we get
x = 1
For the value of y put x=1 in eq (7)
x + y = 0
1 + y = 0
y = -1
1) The rate law expression:
ROR = K [A]x [B]y [C]z
ROR = K [A]1 [B]-1 [C]3
The order of reaction:
order of reaction = n = x+y+z = (1) + (-1) + (3)
n = 3
Some Important Points About Order of Reaction
- Stoichiometric coefficients have nothing to do with order.
2N2O5 → 4NO2 + O2
ROR = K [N2O5]1 (experimentally)
- In rate law, the concentration term of products can be present.
2O3 → 3O2
ROR = K [O3]2 [O2]-1 (experimentally)
- In rate law, the concentration term of the reactant may be absent.
NO2 + CO → NO + CO2
ROR = K [NO]2 [CO]0 (experimentally)
- In rate law, the concentration term of the catalyst may be present.
2SO2 + O2 → 2SO3
In this reaction, NO is used as a catalyst and the rate of reaction is given experimentally.
ROR = K [O2]2 [NO]1
Simple or Elementary Single Step Reactions
Gulberg and Waage law, the law of mass action is true only for simple or elementary single-step reactions.
H2 + I2 → 2HI (simple reaction)
ROR = K [H2]1 [I2]1
H2 + CI2 → 2HCI (Not a single-step reaction)
According to the law of mass action,
ROR = K [H2]1 [CI2]1
But it is wrong, by experiment the rate of reaction will be:
ROR = K [H2]0 [CI2]0
Click here to know what the factors that affect the rate of reaction are.
Leave a Reply