Employing the following data (at 25°C): 1/2H, (g)+1/20, (g)→ HO (g) AH=10.06 kcal mol" H, (g)+1/20, (g)→H,O (g) A,H=-57.80 kcal mol AH=104.178 kcal mol" AH=118.318 kcal mol- ' H, (g)→ 2H (g) O, (g) → 20 (g) assuming ideal behaviour, calculate the AH and AU of the following ctions: HO (g) → H (g)+ O (g) H,0 (g) → 2H (g) +O (g) H,0 (g) →H (g)+ HO (g)
Employing the following data (at 25°C): 1/2H, (g)+1/20, (g)→ HO (g) AH=10.06 kcal mol" H, (g)+1/20, (g)→H,O (g) A,H=-57.80 kcal mol AH=104.178 kcal mol" AH=118.318 kcal mol- ' H, (g)→ 2H (g) O, (g) → 20 (g) assuming ideal behaviour, calculate the AH and AU of the following ctions: HO (g) → H (g)+ O (g) H,0 (g) → 2H (g) +O (g) H,0 (g) →H (g)+ HO (g)
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...
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Question
Employing the following data (at 25ºC):
and assuming ideal behaviour, calculate the ΔH and ΔU of the following
reactions:
![Employing the following data (at 25°C):
1/2H, (g)+1/20, (g)→ HO (g) AH=10.06 kcal mol"1
A,H=-57.80 kcal mol"
H, (g)+1/20, (g)→H,O (g)
H, (g)→ 2H (g)
AH=104.178 kcal mol-
0, (g) →20 (g)
AH=118.318 kcal mol-
assuming ideal behaviour, calculate the AH and AU of the following
ctions:
HO (g) →H (g)+O (g)
H,0 (g) → 2H (g) +O (g)
H,0 (g) →H (g)+HO (g)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F960fefbe-e928-4d40-bfca-5d59ca2a178b%2Fe238af2a-7cbc-4498-83f7-20171eef6805%2Fnyi92ii_processed.png&w=3840&q=75)
Transcribed Image Text:Employing the following data (at 25°C):
1/2H, (g)+1/20, (g)→ HO (g) AH=10.06 kcal mol"1
A,H=-57.80 kcal mol"
H, (g)+1/20, (g)→H,O (g)
H, (g)→ 2H (g)
AH=104.178 kcal mol-
0, (g) →20 (g)
AH=118.318 kcal mol-
assuming ideal behaviour, calculate the AH and AU of the following
ctions:
HO (g) →H (g)+O (g)
H,0 (g) → 2H (g) +O (g)
H,0 (g) →H (g)+HO (g)
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