Consider the reaction given below. a + b → products the following data are obtained at a certain temperature. what is the rate constant

K =  0.167 s⁻¹

Explanation:

1) Rate law, at a given temperature:

Since all the data are obtained at the same temperature, the equilibrium constant is the same.Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

2) Use the data from the table

Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

3) Write the rate law

r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

4) Use any set of data to find K

With the first set of data

r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

The correct answer is b

The answer would be