A zero-order reaction has a constant rate of 3.90×10¬4 M/s. If after 55.0 seconds the concentration has dropped to 7.50x10-2 M , what was the initial concentration? Express your answer with the appropriate units. • View Available Hint(s)
A zero-order reaction has a constant rate of 3.90×10¬4 M/s. If after 55.0 seconds the concentration has dropped to 7.50x10-2 M , what was the initial concentration? Express your answer with the appropriate units. • View Available Hint(s)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please answer the question
![A zero-order reaction has
constant rate of 3.90x10-4 M/s. If after 55.0 seconds the concentration has dropped to 7.50×10-2 M, what was the initial concentration?
Express your answer with the appropriate units.
• View Available Hint(s)
Templates Symbols undo redo Teset keyboard shortcuts help
[A]o =
Value
Units](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06a07530-2782-4976-b3ec-15951dfb63a9%2F95526584-2a8d-4fc0-ba0d-06363ffe07f3%2Fqvo4rog_processed.png&w=3840&q=75)
Transcribed Image Text:A zero-order reaction has
constant rate of 3.90x10-4 M/s. If after 55.0 seconds the concentration has dropped to 7.50×10-2 M, what was the initial concentration?
Express your answer with the appropriate units.
• View Available Hint(s)
Templates Symbols undo redo Teset keyboard shortcuts help
[A]o =
Value
Units
![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 x 2 hr = 110 miles traveled
milemarker 145 – 110 miles = milemarker 35
If we were to write a formula for this calculation, we might express it as follows:
milemarker = milemarker, – (speed × time)
where milemarker is the current milemarker and milemarker, is the initial milemarker.
Similarly, the integrated rate law for a zero-order reaction is expressed as follows:
[A] = [A]o – rate x time
or
[A] = [A]o – kt
since
rate = k[A]o = 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]ge¬kt
where k is the rate constant for this reaction.
The integrated rate law for a second-order reaction is expressed as follows:
1
JA = kt +
where k is the rate constant for this reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06a07530-2782-4976-b3ec-15951dfb63a9%2F95526584-2a8d-4fc0-ba0d-06363ffe07f3%2Fe6n4p7_processed.png&w=3840&q=75)
Transcribed Image Text: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 x 2 hr = 110 miles traveled
milemarker 145 – 110 miles = milemarker 35
If we were to write a formula for this calculation, we might express it as follows:
milemarker = milemarker, – (speed × time)
where milemarker is the current milemarker and milemarker, is the initial milemarker.
Similarly, the integrated rate law for a zero-order reaction is expressed as follows:
[A] = [A]o – rate x time
or
[A] = [A]o – kt
since
rate = k[A]o = 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]ge¬kt
where k is the rate constant for this reaction.
The integrated rate law for a second-order reaction is expressed as follows:
1
JA = kt +
where k is the rate constant for this reaction.
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