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CHE 1302 Lecture Notes - Lecture 6: Rate Equation, Common Application, Reaction Rate Constant

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lilaceel7
13 Jul 2017
School
Baylor University
Department
Chemistry
Course
CHE 1302
Professor
Mc Grath

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Document Summary

Baker | che 1302 | summer 2017 | lecture 6 | page 1. The integrated rate law: dependence of concentration on time. For a first order reaction, the overall order is one: (cid:1853)(cid:1872)(cid:1857)=[(cid:1827)](cid:2871)(cid:2870) [(cid:1828)] (cid:2869)(cid:2870) (cid:2871)(cid:2870) (cid:2869)(cid:2870)=(cid:883) A simple first order rate law is (cid:1853)(cid:1872)(cid:1857)=(cid:1863)[(cid:1827)] An integrated first order rate law is ln[] [a]o is not necessarily (cid:1872)=(cid:882), just keep in mind [a]t is with respect to [a]0: now we have a time variable. If we know [a]0 we can determine [a] at time = t and it has the form of the equation for a straight line: ln[(cid:1827)]= (cid:1863)(cid:1872)+ln[(cid:1827)](cid:2868) (cid:1877)=(cid:1865)(cid:1876)+(cid:1854) A plot of ln[a]t against t should give a straight line if the reaction is first order: the slope of the line is -k ln[a]0 is the y-intercept of the line. Baker | che 1302 | summer 2017 | lecture 6 | page 2.

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Related Questions

order inter. rate law graph slope

0 [A]=-kt+[A]_0 [A] vs t -k

1 ln[A]=-kt+ln[A]_0 ln[A] vs t-k

2 1/A=kt + 1/[A]_0 1/[A] vs t k

<?xml:namespace prefix = o ns ="urn:schemas-microsoft-com:office:office" /?> Using Integrated Rates Laws

The integrated rate laws for zero-,first-, and second-order reaction may be arranged such that theyresemble the equation for a straight line,.


A. The reactant concentration in a zero-order reaction was9.00

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.

1.) The reactant concentration in a zero-order reaction was 6.00×10−2M after 165 s and 3.50×10−2Mafter 385 s . What is the rate constant for this reaction?

2.)What was the initial reactant concentration for the reaction described in Part A?

3.)The reactant concentration in a first-order reaction was 6.70×10−2

M after 40.0 s and 2.50×10−3Mafter 95.0 s . What is the rate constant for this reaction?

4.)The reactant concentration in a second-order reaction was 0.850

M after 255 s and 3.40×10−2M after 860 s . What is the rate constant for this reaction?

Order Integrated Rate Law Graph Slope
0 [A]=−kt+[A]0 [A] vs. t −k
1 ln[A]=−kt+ln[A]0 ln[A] vs. t −k
2 1[A]= kt+1[A]0 1[A] vs. t k

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.

Order Integrated Rate Law Graph Slope
0 [A]=?kt+[A]0 [A] vs. t ?k
1 ln[A]=?kt+ln[A]0 ln[A] vs. t ?k
2 1[A]= kt+1[A]0 1[A] vs. t k

Part A

The reactant concentration in a zero-order reaction was 5.00×10?2M after 175 s and 4.00×10?2M after 400 s . What is the rate constant for this reaction?

Part B

What was the initial reactant concentration for the reaction described in Part A?

Part C

The reactant concentration in a first-order reaction was 6.40×10?2M after 50.0 s and 4.60×10?3M after 60.0 s . What is the rate constant for this reaction?

Part D

The reactant concentration in a second-order reaction was 0.730 M after 210 s and 6.40×10?2M after 705 s . What is the rate constant for this reaction?