Introduction to Integrated Rate Laws Learning Goal: To understand how to use int
ID: 945116 • Letter: I
Question
Introduction to Integrated Rate Laws
Learning Goal:
To understand how to use integrated rate laws to solve for concentration.
This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145.
55 mi/hr×2 hr=110 miles traveled
milemarker 145110 miles=milemarker 35
If we were to write a formula for this calculation, we might express it as follows:
milemarker=milemarker0(speed×time)
where milemarker is the current milemarker and milemarker0 is the initial milemarker.
Similarly, the integrated rate law for a zero-order reaction is expressed as follows:
[A]=[A]0rate×time
or
[A]=[A]0kt
since
rate=k[A]0=k
Because the rate of a first-order reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zero-order reaction.
The integrated rate law for a first-order reaction is expressed as follows:
[A]=[A]0ekt
where k is the rate constant for this reaction.
The integrated rate law for a second-order reaction is expressed as follows:
1[A]=kt+1[A0]
where k is the rate constant for this reaction.
Part A
The rate constant for a certain reaction is k = 5.20×103 s1 . If the initial reactant concentration was 0.200 M, what will the concentration be after 4.00 minutes?
Express your answer with the appropriate units.
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Part B
A zero-order reaction has a constant rate of 5.00×104M/s. If after 55.0 seconds the concentration has dropped to 8.00×102M, what was the initial concentration?
Express your answer with the appropriate units.
Please spare no detail. Thank you.
Introduction to Integrated Rate Laws
Learning Goal:
To understand how to use integrated rate laws to solve for concentration.
A car starts at mile marker 145 on a highway and drives at 55 mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours?This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145.
55 mi/hr×2 hr=110 miles traveled
milemarker 145110 miles=milemarker 35
If we were to write a formula for this calculation, we might express it as follows:
milemarker=milemarker0(speed×time)
where milemarker is the current milemarker and milemarker0 is the initial milemarker.
Similarly, the integrated rate law for a zero-order reaction is expressed as follows:
[A]=[A]0rate×time
or
[A]=[A]0kt
since
rate=k[A]0=k
A zero-order reaction (Figure 1) proceeds uniformly over time. In other words, the rate does not change as the reactant concentration changes. In contrast, first-order reaction rates(Figure 2) do change over time as the reactant concentration changes.Because the rate of a first-order reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zero-order reaction.
The integrated rate law for a first-order reaction is expressed as follows:
[A]=[A]0ekt
where k is the rate constant for this reaction.
The integrated rate law for a second-order reaction is expressed as follows:
1[A]=kt+1[A0]
where k is the rate constant for this reaction.
Part A
The rate constant for a certain reaction is k = 5.20×103 s1 . If the initial reactant concentration was 0.200 M, what will the concentration be after 4.00 minutes?
Express your answer with the appropriate units.
SubmitHintsMy AnswersGive UpReview Part
Part B
A zero-order reaction has a constant rate of 5.00×104M/s. If after 55.0 seconds the concentration has dropped to 8.00×102M, what was the initial concentration?
Express your answer with the appropriate units.
Please spare no detail. Thank you.
Explanation / Answer
Part A
k = 1/tlog a/x
5.20 X 10^-3 = 1/4 log(0.2/x)
20.8 X 10^-3 X 60 = -0.699 -logx
logx = -1.248 - 0.699
= -1.947
x = 10^-1.947
= 0.0113 M
Part B
k = a-(a-x)/t
5 X 10^-4 = a-(0.08)/55
0.0275 + 0.08 = a
a = 0.1075 M
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