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University College - Chemistry Chem 402 Lecture Notes - Lecture 27: Reaction Rate Constant, Reagent, Rate Equation

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School
WUSTL
Department
University College - Chemistry
Course
University College - Chemistry Chem 402
Professor
Barnes Alexander

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Document Summary

L27: 2nd order rate laws and kinetic parameters. 31 march 2017: 2nd order rate laws, from chem 112, 2a products, a + b products, differential rate law: (cid:1853)(cid:1857)= (cid:2869)(cid:2870)[(cid:3002)] =(cid:1863)[(cid:1827)](cid:2870) [(cid:3002)](cid:3116: integrated rate law: (cid:2869)[(cid:3002)]= (cid:2869)[(cid:3002)](cid:3116)+(cid:884)(cid:1863, half life: (cid:3117)(cid:3118)= (cid:2869)(cid:2870)[(cid:3002)](cid:3116, differential rate law: [(cid:3002)] =(cid:1863)[(cid:1827)][(cid:1828)]=(cid:1863)[(cid:1827)](cid:2870) (if [a] = [b]) [(cid:3002)](cid:3116: integrated rate law: (cid:2869)[(cid:3002)]= (cid:2869)[(cid:3002)](cid:3116)+(cid:1863, half life: (cid:3117)(cid:3118)= (cid:2869)[(cid:3002)](cid:3116, only works if [a] = [b] during reaction, plot of 1/[a] vs. t for 2nd order reactions. 1/[a: linear slope of k or 2k, y-intercept of 1/[a]0. 1/[a]0 t: deriving the 2nd order rate law, if [a] [b], [(cid:3002)] =(cid:1863)[(cid:1827)][(cid:1828)] =(cid:1863) (cid:2868) (cid:4666)[(cid:3002)](cid:3116) (cid:4667)(cid:4666)[(cid:3003)](cid:3116) (cid:4667) (cid:4666)[(cid:3002)](cid:3116) (cid:4667)+ (cid:3014)(cid:4666)[(cid:3003)](cid:3116) (cid:4667) (cid:4666)[(cid:3002)](cid:3116) (cid:4667)(cid:4666)[(cid:3003)](cid:3116) (cid:4667)= (cid:2869) (cid:3013: (cid:883)=(cid:1838)(cid:4666)[(cid:1828)](cid:2868) (cid:4667)+(cid:1839)(cid:4666)[(cid:1827)](cid:2868) (cid:4667, x0 1=[(cid:1828)](cid:2868)(cid:1838)+[(cid:1827)](cid:2868)(cid:1839, x1 (cid:882)= (cid:1838) (cid:1839) [(cid:3003)](cid:3116) : using the standard integral result: (cid:3028)+(cid:3029)=(cid:2869)(cid:3028)ln|(cid:1853)+(cid:1854)|+(cid:1855) [(cid:3002)](cid:3116) [(cid:3003)](cid:3116)(cid:4672)ln(cid:4666)[(cid:3002)](cid:3116) (cid:4667) (cid:4666)[(cid:3003)](cid:3116) (cid:4667)[(cid:3002)](cid:3116)(cid:4667) (cid:2869) (cid:2869: since (cid:1827)=(cid:4666)[(cid:1827)](cid:2868) (cid:4667) and (cid:1828)=(cid:4666)[(cid:1828)](cid:2868) (cid:4667)

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Related Questions

1/ derive the integrated form of the following rate equation using the SHILLADY equation.

dx/dt = k [A][B]

show easy-to-follow steps in your derivation. get your integrated equation in a form that matches Table 9 equation.

Show transcribed image text

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 = 3.50×10−3 s−1 . 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 5.00×10−4M/s. If after 75.0 seconds the concentration has dropped to 2.50×10−2M, what was the initial concentration?

Express your answer with the appropriate units.