So the problem is im solving this textbook problem and when im doing the math im getting 271K, what am i doing wrong. I prefer paper responses instead of computer so i can really understand. im horrible at math , plz help. This is the textbook problem: A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased from 22 C to a higher temperature, the rate constant increases by a factor of 7.00. Calculate the higher temperature.

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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
Question

So the problem is im solving this textbook problem and when im doing the math im getting 271K, what am i doing wrong. I prefer paper responses instead of computer so i can really understand. im horrible at math , plz help. This is the textbook problem: A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased from 22 C to a higher temperature, the rate constant increases by a factor of 7.00. Calculate the higher temperature. 

別
My work
In (K₁) = ( Ea ) ( ½/₂2 - 7 )
T2
In (7) = (540x1013/1) ( 1₂ -295)
T2
1.9.459 = 6995,0685 K ( 72
11.9459 = 6495,0685 K (7/7/₂2
1
6495,0685
2.995×10-4 =
1
72
64950685
+ tel:649560685
IZ
295
1
72
= 3₁6 88 5 x 10-3
-
1
3.6885×10-3
295
= 2.995× 10 -4+ 3,389 x 10-31
к
295
700
= 271K
Transcribed Image Text:別 My work In (K₁) = ( Ea ) ( ½/₂2 - 7 ) T2 In (7) = (540x1013/1) ( 1₂ -295) T2 1.9.459 = 6995,0685 K ( 72 11.9459 = 6495,0685 K (7/7/₂2 1 6495,0685 2.995×10-4 = 1 72 64950685 + tel:649560685 IZ 295 1 72 = 3₁6 88 5 x 10-3 - 1 3.6885×10-3 295 = 2.995× 10 -4+ 3,389 x 10-31 к 295 700 = 271K
#15.
(2) -
In
In(7.00) =
1
12
=
=
Е
1
RT₁ Т2
-
К2
k₁
textbook
=7.00, T1=295 K, Ea = 54.0 x 103 J/mol
5.4 x 104 J/mol
1
8.3145/K . mol 295K
=3.09×10-3, (Tz=324 к=51°C
T2
1
295K
-
1
Т2
=
3.00 × 10-4
Transcribed Image Text:#15. (2) - In In(7.00) = 1 12 = = Е 1 RT₁ Т2 - К2 k₁ textbook =7.00, T1=295 K, Ea = 54.0 x 103 J/mol 5.4 x 104 J/mol 1 8.3145/K . mol 295K =3.09×10-3, (Tz=324 к=51°C T2 1 295K - 1 Т2 = 3.00 × 10-4
Expert Solution
steps

Step by step

Solved in 3 steps with 18 images

Blurred answer
Knowledge Booster
Theories of Reaction Rates
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
  • SEE MORE 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:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY