🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
Chemistry
1
answer
0
watching
218
views
10 Nov 2019

The integrated rate laws for zero-, first-, and second-orderreaction may be arranged such that they resemble the equation for astraight 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

The reactant concentration in a zero-order reaction was 6.00

Watch
1
answer
0
watching
218
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Sixta Kovacek
Sixta KovacekLv2
13 Feb 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share
discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Basic Chemistry
5 Edition,
Timberlake
ISBN: 9780134138046
Principles of Chemistry Molecular Approach
4th Edition,
Tro
ISBN: 9780134112831
Principles of Chemistry Molecular Approach
3rd Edition, 2014
Tro
ISBN: 9780321971944
Chemistry: Structure and Properties
2nd Edition,
Tro
ISBN: 9780134293936
Chemistry: A Molecular Approach
3rd Edition,
Tro
ISBN: 9780321809247
Chemistry: A Molecular Approach
5th Edition,
Tro
ISBN: 9780134874371
Principles of Chemistry: A Molecular Approach
4th Edition,
Tro
ISBN: 9780134895741
Chemistry: The Central Science
14th Edition, 2017
Brown
ISBN: 9780134414232

Related questions

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?

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 0.100 M after 195 s and 1.50×10−2 M after 330 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 6.90×10−2 M after 10.0 s and 8.10×10−3 M after 100 s . What is the rate constant for this reaction?

Part D The reactant concentration in a second-order reaction was 0.130 M after 215 s and 6.10×10−2 M after 845 s . What is the rate constant for this reaction? Please show your work.

Weekly leaderboard

Home
Homework Help 3,900,000
Chemistry 830,000
Start filling in the gaps now
Log in