Order Integrated Rate Law Graph Slope 0 [A] = - kt + [A] subscript 0 [A] vs. t -

Question

Order            Integrated Rate Law                    Graph    Slope

0       [A] = - kt + [A] subscript 0        [A] vs. t          -k

1      ln[A] = - kt + ln[A] subscript 0         ln[A] vs. t         -k

2       1/[A] = kt + 1/[A] subscript 0           1/[A] vs. t          k

*****PART A********

******PART B********

******PART C*******

The reactant concentration in a first-order reaction was .055M after 30.0s and .075M after 85.0s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

*******PART D********

Explanation / Answer

1) a-(a-x) = kt for zero order

k = (0.09-0.035)/(400-120) = 0.000196 M/s

B) inital raectant conc = 0.09 M

C) k = (1/dt) ln(a/a-x) for 1st order

k =(1/85-30) ln( 0.075/0.055) = 0.00564 s-1

D) k = (1/dt) (1/a-x -1/a) = (1/800-300)(1/0.068-1/0.48)

      = 0.02524 M-1*s-1

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