Final answer:
The rate law for the reaction is rate = k[B]. The integrated rate law for this reaction is [A] = [A]₀ - kt. The rate constant is k = -3.40×10² L mol⁻¹ s⁻¹.
Step-by-step explanation:
The rate law for the given reaction is rate = k[A]ⁿ[B]ᵐ, where n and m are the orders of the reaction with respect to reactants A and B, respectively. To determine the rate law, we can use the given information that 1/[A] versus time resulted in a straight line with a slope value of +3.40×10² L mol⁻¹ s⁻¹. Since 1/[A] is directly proportional to time, the reaction is zeroth order with respect to A. Therefore, the rate law for the reaction is rate = k[A]⁰[B]¹ = k[B].
The integrated rate law for a zeroth-order reaction is [A] = [A]₀ - kt, where [A] is the concentration of A at a given time, [A]₀ is the initial concentration of A, k is the rate constant, and t is the time. By rearranging the equation, we get [A]₀ = [A] + kt. Since the plot of 1/[A] versus time resulted in a straight line with a positive slope, we can conclude that the reaction is zeroth order. Therefore, the integrated rate law for this reaction is [A] = [A]₀ - kt.
To determine the value of the rate constant, we can use the slope of the graph, which is equal to -k. From the given data, the slope is +3.40×10² L mol⁻¹ s⁻¹. Therefore, the rate constant is k = -3.40×10² L mol⁻¹ s⁻¹.