The integrated rate laws for zero-, first-, and second-order reaction may be arr

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.

Part A The reactant concentration in a zero-order reaction was 0.100 M after 195 s and 1.50×102 M after 330 s . What is the rate constant for this reaction?

Part B What was the initial reactant concentration for the reaction described in Part A?

Part C The reactant concentration in a first-order reaction was 6.90×102 M after 10.0 s and 8.10×103 M after 100 s . What is the rate constant for this reaction?

Part D The reactant concentration in a second-order reaction was 0.130 M after 215 s and 6.10×102 M after 845 s . What is the rate constant for this reaction? Please show your work.

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

Explanation / Answer

A) [A] = k*t + [A0]

Thus, k = (0.1-0.015)/(330-195) = 0.00063 M/s

B) [A] = k*t + [A0]

or, 0.1 = -0.00063*195 + [A0]

or, [A0] = 0.223 M

C) ln[A] = k*t + ln[A0]

k*(100-10) = ln(0.069/0.0081)

or, k = 0.024 s-1

D) 1/[A] = k*t + 1/[A0]

k*(845-215) = (1/0.061) - (1/0.13)

or, k = 0.0138 M-1s-1

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