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[A]0

where k is the rate constant for this reaction.

Figure

1 of 2The figure shows the concentration of A as a function of time for a zero-order reaction. The curve is a straight line which goes from the top left to bottom right.

zero-order (A) time

Part A

The rate constant for a certain reaction is k = 5.60×10⁻³ s−1 . If the initial reactant concentration was 0.950 M, what will the concentration be after 19.0 minutes?

Express your answer with the appropriate units.

Part B

A zero-order reaction has a constant rate of 2.60×10⁻⁴ M/s. If after 75.0 seconds the concentration has dropped to 2.00×10⁻² M, what was the initial concentration?

Express your answer with the appropriate units.

Answer 1

For nth order reaction, rak[A]" Unit of ris mole / mote t'i to, units of a becomes , concentration of A is in mole mole meante mokeyhty If we put order of reaction n=1 (1st order) Units of k will be st kc 5.60*103 5 ( reaction is 1st order) PART A: r= R [A reactant Cencentration. -d(4) K CA]' at CA] 7 & CA] =/ -k.dt (Integrating both the sides) CA]! J CA 2 Inita concentration 7 CAD In [n] - ln [A] = -kt en/CA1 - - Rt =[A] - (A). e-kt. so, [A) = 0.950m, t= 19 min = 198608 . In [A] - ln(0.950) = -5.60 x16 ? x19860 In CA] - (-0,0512) = -6.384 en C --6-384 +070512 - MUUT -6.4352 CADA PELA -6.4352 - -1.604 X703M. after E

PART B Giren Zero - order reaction. 8=2068109M t = 750 [4], - e xiorem \ final concentration of reactant) so, r = R[A]° z R = 2.6X10" M/A dt LA t pocef-adt (Integrating both the sides) 2. [A] - [A] = - kb (A) - [A] = kt [A]. - kt - CA] CABO - CA] + kt = 2x10 + 2,6x16"x 75 final concentration of reactant A final concent Initial concentration of ..0.02 +0.0195 = 0.0395 M A reactant