Question
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they...
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 |
| 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 |
------------
Part A
The reactant concentration in a zero-order reaction was 5.00×10−2M after 110 s and 4.00×10−2M after 375 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 9.60×10−2M after 30.0 s and 5.20×10−3M after 60.0 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 0.650 M after 175 s and 5.60×10−2M after 815 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 multiplication, for example write a Newton-meter as N*m.
Homework Answers
Answer #1
A. rate constant, k = - slope
= - d[A]/dt
= - [0.04-0.05] M /[375-110]s
= 3.77*10-5 M.s-1
B. Initial reactant concentration = [A]0
= intercept in y axis
look at the graph(the graph is qualitative). Draw your own graph in a graph paper to determine the initial concentration.
C. rate constant, k = - slope
= - dln[A]/dt
= - [ln 0.0052 M - ln 0.096 M]/(60-30)s
= - ln (0.0052/0.096)/30 s
= 0.097 s-1
D. rate constant, k = slope
= d(1/[A])/dt
= (1/0.056 - 1/0.65) M-1 /(815-175)s
= 0.0255 M-1*s-1
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