The data of the initial rate experiment of the following reaction are given below. The rate law expression of the reaction is Rate = k[S208 J*[ r}Y. Determine the rate constant k with 3 significant figures (key-in only your value; no units are needed). Note that your k value should be reported based on second (the time units here). Do not covert second to other time units. S208° (aq) + 3 I" (aq) → 2 so4²" (aq) + 13` (aq) (S20g Jo (M) [rJo (M) initial rate (M/sec) Trial 1 0.800 0.140 0.0288 Trial 2 2.400 0.140 0.2592 Trial 3 0.800 0.280 0.0288 Note that Trial 1 and Trial 3 have the same initial rate.

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter12: Chemical Kinetics
Section: Chapter Questions
Problem 4RQ: The initial rate for a reaction is equal to the slope of the tangent line at t 0 in a plot of [A]...
icon
Related questions
Question
100%
The data of the initial rate experiment of the following reaction are given below. The rate law expression of the reaction is Rate
= k[S208 *[ rY. Determine the rate constant k with 3 significant figures (key-in only your value; no units are needed). Note
that your k value should be reported based on second (the time units here). Do not covert second to other time units.
S20g (aq) + 31 (aq) → 2 SO,2 (aq) + 13 (aq)
[S208 lo (M)
[]o (M)
initial rate (M/sec)
Trial 1
0.800
0.140
0.0288
Trial 2
2.400
0.140
0.2592
Trial 3
0.800
0.280
0.0288
Note that Trial 1 and Trial 3 have the same initial rate.
Transcribed Image Text:The data of the initial rate experiment of the following reaction are given below. The rate law expression of the reaction is Rate = k[S208 *[ rY. Determine the rate constant k with 3 significant figures (key-in only your value; no units are needed). Note that your k value should be reported based on second (the time units here). Do not covert second to other time units. S20g (aq) + 31 (aq) → 2 SO,2 (aq) + 13 (aq) [S208 lo (M) []o (M) initial rate (M/sec) Trial 1 0.800 0.140 0.0288 Trial 2 2.400 0.140 0.2592 Trial 3 0.800 0.280 0.0288 Note that Trial 1 and Trial 3 have the same initial rate.
Expert Solution
steps

Step by step

Solved in 4 steps

Blurred answer
Knowledge Booster
Rate Laws
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781133611097
Author:
Steven S. Zumdahl
Publisher:
Cengage Learning
Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
Chemistry
ISBN:
9781305079243
Author:
Steven S. Zumdahl, Susan A. Zumdahl
Publisher:
Cengage Learning
Chemistry: Principles and Practice
Chemistry: Principles and Practice
Chemistry
ISBN:
9780534420123
Author:
Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:
Cengage Learning
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781133949640
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781337399074
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning