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.

Appl Of Ms Excel In Analytical Chemistry
2nd Edition
ISBN:9781285686691
Author:Crouch
Publisher:Crouch
Chapter13: Kinetic Methods
Section: Chapter Questions
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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
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