Question

# Learning Goal: To understand how to use integrated rate laws to solve for concentration. A car...

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 = 5.90×10−3 s−1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 12.0 minutes?

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Part B

A zero-order reaction has a constant rate of 1.90×10−4M/s. If after 65.0 seconds the concentration has dropped to 2.00×10−2M, what was the initial concentration?

part-A

k = 5.90×10−3 s−1

[A0]   = 0.4M

[A] =

t     = 12min   = 12*60sec   = 720sec

K = 2.303/t log[A0]/[A]

5.9*10^-3   = 2.303/720 log0.4/[A]

log0.4/[A]    = 5.9*10^-3*720/2.303

log0.4/[A]    = 1.845

0.4/[A]        = 10^1.845

0.4/[A]        = 69.98

[A]              = 0.4/69.98   = 0.0057M

The concentration = 0.0057M

part-B

t   = 65sec

K = 1.90×10−4M/s

K t = [A0]- [A]

1.9*10^-4 *65 = [A0]-0.02

0.01235+0.02 = [A0]

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