Calculate the value of AG° (in k) for the combustion of 1 mole of propane (CaHg) with molecular oxygen to form carbon dioxide and gaseous water, using the values of AGf° (in kl/mol) given below. AGP (C3H9(g) - -20. AGP (CO2) -390. AG (H20g) - 238. Enter value as an integer (value 2).
Calculate the value of AG° (in k) for the combustion of 1 mole of propane (CaHg) with molecular oxygen to form carbon dioxide and gaseous water, using the values of AGf° (in kl/mol) given below. AGP (C3H9(g) - -20. AGP (CO2) -390. AG (H20g) - 238. Enter value as an integer (value 2).
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter8: Thermochemistry
Section: Chapter Questions
Problem 11QAP: When one mol of KOH is neutralized by sulfuric acid, q=56 kJ. (This is called the heat of...
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![Calculate the value of AG° (in k]) for the combustion of 1 mole of propane (C3HR) with molecular oxygen to form carbon dioxide and gaseous water, using the values of AGF° (in kJ/mol) given below.
AG (C3Hg(g) --20.
AG (CO2(g)) =-390.
AG (H20(g) --238.
Enter value as an integer (value 2).
QUESTION S](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff364b906-934b-4f43-b7b4-bcc25ebe3486%2F45d30d88-7419-426a-9289-81c013d75ee0%2Fji3fq4e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Calculate the value of AG° (in k]) for the combustion of 1 mole of propane (C3HR) with molecular oxygen to form carbon dioxide and gaseous water, using the values of AGF° (in kJ/mol) given below.
AG (C3Hg(g) --20.
AG (CO2(g)) =-390.
AG (H20(g) --238.
Enter value as an integer (value 2).
QUESTION S
Expert Solution
Step 1
The combustion reaction of propane can be written as,
=> C3H8 (g) + O2 (g) -------> CO2 (g) + H2O (g)
Balancing : Since we have 3 C in LHS. Hence making it 3 in RHS also.
=> C3H8 (g) + O2 (g) -------> 3 CO2 (g) + H2O (g)
Since we have 8 H in LHS. Hence making it 8 in RHS also.
=> C3H8 (g) + O2 (g) -------> 3 CO2 (g) + 4 H2O (g)
Now the number of O in RHS is 10. Hence making it 10 in LHS also.
=> C3H8 (g) + 5 O2 (g) -------> 3 CO2 (g) + 4 H2O (g)
Since all the elements are in equal number in both side of the reaction now. Hence the reaction is now balanced.
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