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 zeroorder reaction is expressed as follows: [A]=[A]0−rate×time or [A]=[A]0−kt since rate=k[A]0=k A zeroorder reaction (Figure 1)proceeds uniformly over time. In other words, the rate does not change as the reactant concentration changes. In contrast, firstorder reaction rates (Figure 2) do change over time as the reactant concentration changes. Because the rate of a firstorder reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zeroorder reaction. The integrated rate law for a firstorder reaction is expressed as follows: [A]=[A]0e−kt where k is the rate constant for this reaction. The integrated rate law for a secondorder 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?
Express your answer with the appropriate units.
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Part B
A zeroorder reaction has a constant rate of 1.90×10^{−4}M/s. If after 65.0 seconds the concentration has dropped to 2.00×10^{−2}M, what was the initial concentration?
Express your answer with the appropriate units.
partA
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
partB
t = 65sec
K = 1.90×10^{−4}M/s
K t = [A0] [A]
1.9*10^4 *65 = [A0]0.02
0.01235+0.02 = [A0]
[A0] = 0.03235M >>>>answer
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