A)  The reactant concentration in a zero-order reaction was 0.100 mol L−1 after 140 s and 1.00×10−2 mol L−1 after 345 s . What is the rate constant for this reaction? Express your answer with the appropriate units.     B) What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units.   C)  The reactant concentration in a first-order reaction was 8.70×10−2 mol L−1 after 30.0 s and 1.80×10−3 mol L−1 after 85.0 s . What is the rate constant for this reaction? Express your answer with the appropriate units.

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section: Chapter Questions
Problem 121QRT: If you know some calculus, derive the integrated first-order rate law for the reaction by following...
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A)  The reactant concentration in a zero-order reaction was 0.100 mol L−1 after 140 s and 1.00×10−2 mol L−1 after 345 s . What is the rate constant for this reaction?
Express your answer with the appropriate units.
 
 
B) What was the initial reactant concentration for the reaction described in Part A?
Express your answer with the appropriate units.
 
C)  The reactant concentration in a first-order reaction was 8.70×10−2 mol L−1 after 30.0 s and 1.80×10−3 mol L−1 after 85.0 s . What is the rate constant for this reaction?
Express your answer with the appropriate units.
 
D)  The reactant concentration in a second-order reaction was 0.180 mol L−1 after 240 s and 3.80×10−2 mol L−1 after 805 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.
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
[A]t = - kt + [A]o
[A]t vs. t
-k
In[A]; = -kt + In[A]o Im[A]t vs. t -k
1
1
kt +
[A],
1
ㅠ vs. t
[A]:
2
[A]:
1,
Transcribed Image Text: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 [A]t = - kt + [A]o [A]t vs. t -k In[A]; = -kt + In[A]o Im[A]t vs. t -k 1 1 kt + [A], 1 ㅠ vs. t [A]: 2 [A]: 1,
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