To understand how to use integrated rate laws to solve for concentration. A car

Question

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 = 3.50×103 s1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 5.00 minutes?

Express your answer with the appropriate units.

PART B

A zero-order reaction has a constant rate of 5.00×104M/s. If after 75.0 seconds the concentration has dropped to 2.50×102M, what was the initial concentration?

Express your answer with the appropriate units.

Explanation / Answer

A)

we know that

the unit of rate constant is M^(1-n) s-1

here

n is the order of the reaction

given

rate constant = 3.5 x 10-3 s-1

so

1-n = 0

n = 1

so

this reaction is 1st order

now

for 1st order reactions

[A] = [Ao] x e^(-kt)

ln A = ln Ao - kt

given

time (t) = 5 min = 5 x 60 = 300 s

so

ln A = ln 0.4 - ( 3.5 x 10-3 x 300)

ln A = -1.96629

A = 0.14

so

the concentration after 5 minutes is 0.14 M

2)

for zero order reaction

[A] = [Ao] - kt

using given values

2.5 x 10-2 = [Ao] - ( 5 x 10-4 x 75)

[Ao] = 0.0625

so

the initial concentration is 6.25 x 10-2 M

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