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+by=mx+b.

 Order Integrated Rate Law Graph Slope 0 [A]=−kt+[A]0[A]=−kt+[A]0 [A] vs. t[A] vs. t −k 1 ln[A]=−kt+ln[A]0ln⁡[A]=−kt+ln⁡[A]0 ln[A] vs. tln[A] vs. t −k 2 1[A]= kt+1[A]01[A]= kt+1[A]0 1[A] vs. t1[A] vs. t k

Part A

The reactant concentration in a zero-order reaction was 8.00×10−2 MM after 130 ss and 4.00×10−2 MM after 380 ss . 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.

Part C

The reactant concentration in a first-order reaction was 8.00×10−2 MM after 25.0 ss and 4.90×10−3 MM after 60.0 ss . 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.

Part D

The reactant concentration in a second-order reaction was 0.460 MM after 300 ss and 3.60×10−2 MM after 705 ss . 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.   #### Earn Coins

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