Answer:
(a) R = k [A]¹ [B]²
(b) The given chemical reaction is a <u>third order reaction</u>
(c)
- [A] is doubled and [B] held constant: <u>the reaction rate doubles.</u>
-
[A] is held constant and [B] is doubled: <u>the reaction rate becomes 4 times.</u>
-
[A] is tripled and [B] is doubled
: <u>the reaction rate becomes 12 times.</u>
- [A] is doubled and [B] is halved: <u>the reaction rate becomes half.</u>
Explanation:
<em>Rate law</em><em> 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.</em>
<em>The </em><em>overall order</em><em> of the given chemical reaction is equal to the sum of partial orders of reaction. </em>
Given: Partial order of reaction of reactant A: a = 1,
Partial order of reaction of reactant B: b = 2
(a) <u>Therefore, the rate law equation of the given reaction is given by</u>
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
<u>Therefore, the given chemical reaction is a </u><u>third order reaction</u><u>.</u>
(c) <em>Since, the rate of a reaction is directly proportional to the reactant concentration. Therefore, when</em>
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 ⇒ <u>the reaction rate doubles.</u>
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 ⇒ <u>the reaction rate becomes 4 times.</u>
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 ⇒ <u>the reaction rate becomes 12 times.</u>
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 ⇒ <u>the reaction rate becomes half.</u>