Learning Goal: To understand how to use integrated rate laws to solve for concen

Question

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 145?110 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]0?rate×time

or

[A]=[A]0?kt

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]0e?kt

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 = 3.30×10?3 s?1 . If the initial reactant concentration was 0.300 M, what will the concentration be after 10.0 minutes?

Express your answer with the appropriate units.

Part B

A zero-order reaction has a constant rate of 4.50×10?4M/s. If after 50.0 seconds the concentration has dropped to 6.00×10?2M, what was the initial concentration?

Express your answer with the appropriate units.

Explanation / Answer

Part A)

rate constant k = 3.30 x 10^-3 s-1

initial concentration = 0.300 M

time = 10 min = 600 sec

k = 1/t ln (Ao / At)

3.30 x 10^-3 = 1/ 600 ln (0.300 / At)

At = 0.0414 M

concentration = 0.0414 M

Part B)

rate constant = Ao - At / t

4.50 x 10^-4 = Ao - 6.00 x 10^-2 / 50

Ao = 0.0825

initial concentration = 0.0825 M

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