Question

For the second-order reaction below, the concentration of the product, B, after 535 seconds is 0.0784 M. If no B is initially present, and the initial concentration of A is 0.579 M, what is the rate constant? 2 A B Rate = k[A]2

A. 1.20×10-3 M-1s-1

B. 8.30×10-4 M-1s-1

C. 8.39×103 M-1s-1

D. 5.90×10-4 M-1s-1

E. 83.9 M-1s-1

Answer 1

the rate constant is k = 1.2* 10⁻³ M⁻¹/s

Step-by-step explanation:

Assuming a reaction of the type 2*A → B with no side reactions , then denoting the concentration of a [A]= Cᵃ , and assuming that the volume of reaction is not affected , then

Rate = k[A]² → -dCᵃ/dt= kCᵃ²

-∫dCᵃ/Cᵃ² = ∫k*dt

1/Cᵃ-1/Cᵃ₀ = k*t

also from the reaction stoichiometry

(Cᵇ- Cᵇ₀)/νᵇ=(Cᵃ- Cᵃ₀)/νᵃ

where νᵇ and νᵃ are stoichiometric coefficients , thus

νᵃ = 2 , νᵇ= -1

then since also Cᵇ₀=0 ( no B initially present)

0 - Cᵇ = (Cᵃ- Cᵃ₀)/2 → Cᵃ = Cᵃ₀ -2*Cᵇ

thus

1/(Cᵃ₀ -2*Cᵇ)-1/Cᵃ₀ = k*t

k = (1/(Cᵃ₀ -2*Cᵇ)-1/Cᵃ₀)/t =2*Cᵇ/[(Cᵃ₀ -2*Cᵇ)*Cᵃ₀*t]

k =2*Cᵇ/[(Cᵃ₀ -2*Cᵇ)*Cᵃ₀*t]

replacing values

k =2*Cᵇ/[(Cᵃ₀ -2*Cᵇ)*Cᵃ₀*t] = 2*0.0784 M/[(0.579 M-2*0.0784 M)*0.579 M*535 s]

k = 1.2* 10⁻³ M⁻¹/s