Reaction Mechanism - Molecularity, Difference Between Order and Molecularity, Elementary and Complex Reactions, Practice Problems and FAQs

Imagine there’s a car company which produces cars. There is a separate unit for each part of the car. One unit produces 25 bodies in a day. Another unit produces 16 tyres in a day. While another unit produces 3 engines per day. So can you tell how many cars will be produced a day? Three! The number of cars that can be made in a day depends upon the number of engines made in a day. So the rate of formation of cars depends on the slowest step.

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Similarly in the reaction, the rate depends on the slowest step. And that the slowest step determines the order of the reaction.

So let's study the order of the reaction in detail!

Table of content:

Introduction
Molecularity of reactions

Difference between order and molecularity
Elementary Reaction
Complex reaction
Practice problems
Frequently Asked Questions-FAQs

Introduction:

$2 N O + 2 H_{2} \to N_{2} + {2 H}_{2} O$

According to law of mass action, the rate of this reaction is given as:

$R a t e = k [N O]^{2} [H_{2}]^{2}$

But experimentally it has been found that this follows third order kinetics.

$R a t e = k' [N O]^{2} [H_{2}]^{}$

It is the experimentally observed rate law.

So, the law of mass action does not always support experimental facts. This made us wonder that reactions do not occur according to the chemical reaction as written. That is the reaction is not always elementary but sometimes complicated. In order to explain the observed rate law, we need to go deep into its mechanism.

Molecularity of reactions:

It is defined as the number of reacting molecules that collide simultaneously to produce a chemical reaction. In other terms, it is defined as the number of reactant molecules involved in an elementary reaction.

Consider the reaction, $H_{2} + I_{2} \to 2 H I$

In this reaction, the molecularity is two.

Note: Reactions having molecularity greater than three are not observed.

Difference between order and molecularity:

Molecularity gives an idea about how many molecules take part in the RDS, whereas Order gives an idea about how sensitive those molecules are toward the reaction rate.

Molecularity	Order
1. Molecularity refers to the quantity of ions or molecules involved in the rate-determining step.	1. The order of the reaction is defined as the sum of powers to which the reactant concentrations are increased in the rate law equation.
2. It is always a whole number	2. It can be a whole number or a fraction.
3. It can be determined from the balanced chemical equation	3. It can be determined only experimentally.

Elementary reactions:

Reactions that occur in a single step are known as elementary reactions. Example-

Decomposition of ammonium nitrate, ${N H_{4} N O_{3}}^{} \to N_{2} O + {2 H}_{2} O$

It is a unimolecular reaction as only one reactant is involved in the given elementary reaction.

$H_{2} + I_{2} \to 2 H I$

It is a bimolecular reaction, as two reactants are involved in the given elementary reaction.

$2 N O + {O_{2}}^{} \to 2 N O_{2}$

It is a trimolecular reaction, as three reactants are involved in the given elementary reaction.

Complex reactions:

Reactions that occur in a multistep, that is, in a series of elementary steps are known as complex reactions.

For a complex reaction, order of reaction may or may not be equal to the molecularity of the slowest step in a reaction, known as rate determining step (RDS).

Example:

1. $2 H_{2} O_{2} \to O_{2} + 2 H_{2} O$

Alkaline medium

This reaction occurs following in two steps:

$H_{2} O_{2} + I^{-} \to H_{2} O + I O^{-}$ (slow)

$H_{2} O_{2} + {I O}^{-} \to H_{2} O + I^{-} + O_{2}$ (fast)

It has been observed that step 1 is the slowest step, RDS.

$R a t e = k [H_{2} O_{2}$ ][I^-], where k is the rate constant.

So, the order of this reaction will be 2.

2. $2 N O + 2 H_{2} \to N_{2} + {2 H}_{2} O$

$N O + N O ⇌ N_{2} O_{2} (f a s t a n d r e v e r s i b l e) N_{2} O_{2} + H_{2} \to N_{2} O + H_{2} O (s l o w) N_{2} O + H_{2} \to N_{2} + H_{2} O (f a s t)$

Step II being the slowest step is the rate determining step. So the rate of reaction will be:

$R a t e = k [N_{2} O_{2}] [H_{2}] \dots \dots \dots (i)$ , where k is the rate constant of this step.

N2O2 act as the intermediate(species that appears and disappears in the reaction mechanism). Since intermediate cannot appear in the rate law, it has to be substituted by other reactants.

Using the equilibrium reaction.

$K_{C} = \frac{[N_{2} O_{2}]}{[N O]^{2}} {[N}_{2} O_{2}] = {K_{C} [N O]^{2}}_{}$

Substituting the value of N₂O₂ in the equation (i)

$R a t e = k K_{C} [N O]^{2} [H_{2}]$

$R a t e = k' [N O]^{2} {[H_{2}],}^{}$ ,where $k^{'} = K_{C} k$

This is the same rate law as observed experimentally.

Practice Problems:

Q1. Mechanism of a hypothetical reaction

$X_{2} + Y_{2} \to 2 X Y$ is given below :

$(i) X_{2} \to X + X (f a s t) (i i) X + Y_{2} ⇌ X Y + Y (s l o w) (i i i) X + Y \to X Y (f a s t)$

The overall order of the reaction will be

(A)1 (B) 2

Answer (D)

Solution: The solution of this equation is given by assuming step (i) to be reversible which is not given in question

Overall rate = Rate of slowest step (ii)

$= k [X] [Y_{2}] \dots (1)$

k = rate constant of step (ii)

Assuming step (i) to be reversible, its equilibrium constant,

$k_{e q} = \frac{[X]^{2}}{[X_{2}]} \Rightarrow [X] = {k_{e q}}^{\frac{1}{2}} {[X_{2}]_{}}^{\frac{1}{2}} \dots (2)$

Put (2) in (1)

Rate = $k {k_{e q^{}}}^{\frac{1}{2}} [X_{2}]^{\frac{1}{2}} [Y_{2}]$

Overall order $= \frac{1}{2} + 1 = \frac{3}{2}$

Q2. For the formation of phosgene from CO(g) and chlorine,

$C O (g) + C l_{2} (g) \to C O C l_{2} (g)$

The experimentally determined rate equation is,

$\frac{d [C O C l_{2}]}{d t} = k [C O] [C l_{2}]^{\frac{3}{2}}$

Is the following mechanism consistent with the rate equation?

$(i) C l_{2} ⇌ 2 C l (f a s t) (i i) C l + C O ⇌ C O C l (f a s t) (i i i) C O C l_{} + C l_{2} ⇌ C O C l_{2} + C l (s l o w)$

Solution: Multiplying equation (ii) by 2 and adding (i), we get :

$C l_{2} + 2 C O ⇌ 2 C O C l K = \frac{[C O C l]^{2}}{[C l_{2}] [C O]^{2}} [C O C l] = (K)^{\frac{1}{2}} [C l_{2}]^{\frac{1}{2}} [C O] \dots . . (i)$

Slowest step is rate determining, hence

Rate = $k [C O C l] [C l_{2}] \dots . . (i i)$

From (i) and (ii), we get

Rate = $k K^{\frac{1}{2}} [C l_{2}]^{\frac{1}{2}} [C l_{2}] [C O]$

Rate = $k' [C l_{2}]^{\frac{3}{2}} [C O] [k' = k . k^{\frac{1}{2}}]$

Thus, rate law is in accordance with the mechanism.

Q3. Rate law for ozone layer depletion is

$\frac{d [O_{3}]}{d t} = \frac{K [O_{3}]^{2}}{[O_{2}]}$

Give the probable mechanism of reaction?

Solution: $O_{3} ⇌ O_{2} + O (f a s t r e a c t i o n)$

$O_{3} + O \to 2 O_{2} (s l o w r e a c t i o n, r a t e c o n s t a n t k^{'})$

Rate = $k^{'} [O_{3}] [O]$

$K_{c} \frac{[O_{2}] [O]}{[O_{3}]}$ or $[O] = K_{c} \frac{[O_{3}]}{[O_{2}]},$ , (equilibrium constant Kc)

putting the value in (1)

From (1) Rate = $k^{'} . [O_{3}] . K_{c} \frac{[O_{3}]}{[O_{2}]} = k^{'} . K_{c} . \frac{[{O_{3}}^{2}]}{[O]_{2}]} = k . \frac{[O_{3}]^{2}}{[O_{2}]}$

$k = k^{'} \times K_{c}$

Q4. What is the molecularity of the following reaction?

$H_{2} + I_{2} \to 2 H I$

(A) 1

(B) 2

(D) 1.5

Answer (B)

Solution: Molecularity is defined as the number of reacting molecules that collide simultaneously to produce a chemical reaction. In other terms, it is defined as the number of reactant molecules involved in an elementary reaction. In this reaction, the molecularity is two.

Frequently Asked Questions-FAQs:

Q1: Why are reactions with molecularity greater than three not observed?
Answer: The probability that more than three molecules collide and react simultaneously is very small. Hence the molecularity greater than three is not observed.

Q2: What is the rate determining step and how can we determine this step?
Answer: In the complex reaction which occurs in multiple steps, the step which is slowest of all, is known as rate determining step. It can only be determined experimentally and not just looking at the complex reactions.

Q3: Is it always true that order of the reaction is always equal to molecularity of the reaction of the rate determining step, in complex reactions?
Answer: No, order of the reaction may or may not be equal to the molecularity of the reaction of the rate determining step, in multistep reactions. Therefore, order of the reaction is calculated only after derivation of rate law.

Q4: Can the molecularity of a reaction be zero?
Answer: No, molecularity of a reaction can never be zero. We know that for any reaction to occur, collision is a necessary condition. So it will never happen that a product is formed with no two molecules colliding.