Question

Consider a reaction involving two reactants (A and B) in which the reaction is first-order in reactant A and second-order in reactant B.
a. write the rate law for this equation.

b. what is the overall order of the reaction?

c. identify how the reaction rate would change if...


i. [A] is doubled and [B] held constant.

ii. [A] is held constant and [B] is doubled.

iii. [A] is tripled and [B] is doubled

iv. [A] is doubled and [B] is halved.

Answer 1

Answer:

(a) R = k [A]¹ [B]²

(b) The given chemical reaction is a third order reaction

(c)

1. [A] is doubled and [B] held constant: the reaction rate doubles.
2. [A] is held constant and [B] is doubled: the reaction rate becomes 4 times.
3. [A] is tripled and [B] is doubled : the reaction rate becomes 12 times.
4. [A] is doubled and [B] is halved: the reaction rate becomes half.            

Explanation:

Rate law is the equation that defines the rate of a given chemical reaction and depends on the concentration of the reactants, raised to the power partial orders of reaction.

The overall order of the given chemical reaction is equal to the sum of partial orders of reaction.

Given: Partial order of reaction of reactant A: a = 1,

Partial order of reaction of reactant B: b = 2

(a) Therefore, the rate law equation of the given reaction is given by

R = k [A]ᵃ [B]ᵇ = k [A]¹ [B]²                          ....equation 1

here k is the rate constant

(b) The overall order of the reaction = a + b = 1 + 2 = 3

Therefore, the given chemical reaction is a third order reaction.

(c) Since, the rate of a reaction is directly proportional to the reactant concentration. Therefore, when

i. [A] is doubled and [B] held constant.

⇒ Concentration of reactant A becomes 2[A]

The new rate law is:

R' = k {2[A]}¹ [B]² = 2 {k [A]¹ [B]²}                   ....equation 2

Comparing equations 1 and 2, we get

R' = 2 R  ⇒ the reaction rate doubles.

ii. [A] is held constant and [B] is doubled.

⇒ Concentration of reactant B becomes 2[B]

The new rate law is:

R' = k [A]¹ {2[B]}² = 4 {k [A]¹ [B]²}                   ....equation 3

Comparing equations 1 and 3, we get

R' = 4 R  ⇒ the reaction rate becomes 4 times.

 

iii. [A] is tripled and [B] is doubled

⇒ Concentration of reactant A becomes 3[A], Concentration of reactant B becomes 2[B]

The new rate law is:

R' = k {3[A]}¹ {2[B]}² = 12 {k [A]¹ [B]²}                   ....equation 4

Comparing equations 1 and 4, we get

R' = 12 R  ⇒ the reaction rate becomes 12 times.

iv. [A] is doubled and [B] is halved.

⇒ Concentration of reactant A becomes 2[A], Concentration of reactant B becomes 1/2 [B]

The new rate law is:

R' = k {2[A]}¹ {1/2[B]²} = 1/2 {k [A]¹ [B]²}                   ....equation 5

Comparing equations 1 and 5, we get

R' = 1/2 R  ⇒ the reaction rate becomes half.