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

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

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