Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b. Order Integrated Rate Law Graph Slope 0 [A]=−kt+[A]0 [A] vs. t −k 1 ln[A]=−kt+ln[A]0 ln[A] vs. t −k 2 1[A]= kt+1[A]0 1[A] vs. t k Part A The reactant concentration in a zero-order reaction was 0.100 M after 200 s and 4.00×10−2 M after 310 s . What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. k0th = SubmitHintsMy AnswersGive UpReview Part Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. [A]0 = SubmitHintsMy AnswersGive UpReview Part Part C The reactant concentration in a first-order reaction was 6.70×10−2 M after 50.0 s and 3.20×10−3 M after 60.0 s . What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. k1st = SubmitHintsMy AnswersGive UpReview Part Part D The reactant concentration in a second-order reaction was 0.210 M after 120 s and 6.20×10−2 M after 850 s . What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.