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.

k

The reactant concentration in a zero-order reaction was 0.100 M after 195 s and 2.00×102M after 365 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

Part B

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

Express your answer with the appropriate units.

Part C

The reactant concentration in a first-order reaction was 6.90×102M after 40.0 s and 8.20×103Mafter 80.0 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

Part D

The reactant concentration in a second-order reaction was 0.620 M after 295 s and 6.30×102M after 830 s . What is the rate constant for this reaction?

Express your answer with the appropriate units. Include an asterisk to indicate a compound unit with mulitplication, for example write a Newton-meter as N*m.

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

Part A. Given,

[A] = [A]o - kt

let us calculate slope by given values,

slope = [A]t at 365 s - [A]t at 195 s/365-195

          = 2-0.1/365-195

          = 0.011

So, rate constant k = 0.011 M/s

Part B. Initial concentration [A]o

0.1 = [A]o - 195 x 0.011

[A]o = 2.245 M

Part C. Using the slope = -k

slope = -(ln 8.20 x 10^-3 - ln 6.90 x 10^-2/80 - 40)

          = 0.053 s-1 is the rate constant

Part D. Let us find slope,

slope = (1/6.3 x 10^-2 - 1/0.62)/(830 - 295)

          = 0.027 M-1*s-1 is the rate constant

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