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

answer by yourhope:

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: )

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