Question

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not affect the reaction rate. What is the rate law for this reaction? A. rate = k[B] B. rate = k[A]2 C. rate = k[A][B] D. rate = k[A]

Answer 1

Answer:

rate = k[A]

Explanation:

The equation that relate reaction rate with reactant concentrations is known as the rate law.

for a reaction:

• A + B  → C

the rate law can be expressed as:  

• Rate = k[A]ᵃ[B]ᵇ

The proportionality constant, k, is known as the rate constant, the powers a and b is the reaction order with respect to reactants A and B, respectively.

for this reaction doubling the concentration of A doubles the reaction rate that means

Rate₂ = 2 *Rate₁     and     [A]₂ = 2 [A]₁

• Rate₁ = k[A]₁ᵃ[B]ᵇ    → eq. 1

• Rate₂ = k[A]₂ᵃ[B]ᵇ    → eq. 2

Dividing eq. 2 by eq. 1 one can get

• (Rate₂ / Rate₁) = (k [A]₂ᵃ[B]ᵇ) / (k[A]₁ᵃ[B]ᵇ)

using

• Rate₂ = 2 *Rate₁     and     [A]₂ = 2 [A]₁

∴ (2 Rate₁ / Rate₁) = ( k [2]ᵃ[B]ᵇ) / (k[1]ᵃ[B]ᵇ)

• (2) = (2)ᵃ
• taking log of both sides
• log (2) = a Log (2)
• 0.693 = a * 0.693
• a =1  

∴ order of reaction with respect to A is first (=1)        →     (1)

Doubling the concentration of B does not affect the reaction rate.

that means

Rate₂ = Rate₁     and     [B]₂ = 2 [B]₁

• Rate₁ = k[A]ᵃ[B]₁ᵇ    → eq. 1

• Rate₂ = k[A]ᵃ[B]₂ᵇ    → eq. 2

Dividing eq. 2 by eq. 1 one can get

• (Rate₂ / Rate₁) = (k [A]ᵃ[B]₂ᵇ) / (k[A]ᵃ[B]₁ᵇ)

using

• Rate₂ = Rate₁     and   [B]₂ = 2 [B]₁

∴ (Rate₁ / Rate₁) = ( k [A]ᵃ[2]ᵇ) / (k[A]ᵃ[1]ᵇ)

• (1) = (2)ᵇ
• taking log of both sides
• log (1) = b Log (2)
• 0 = 0.693 * b
• b = 0

∴ order of reaction with respect to B is zero         →     (2)

So, from 1 and 2  the right choice is rate = k[A]¹[B]⁰= k[A]