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]

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]