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
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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Transcribed Image Text:Calculate the value of ΔG° (in kJ) for the combustion of 1 mole of propane (C₃H₈) with molecular oxygen to form carbon dioxide and gaseous water, using the values of ΔG° (in kJ/mol) given below.
ΔG° (C₃H₈(g)) = -20
ΔG° (CO₂(g)) = -390
ΔG° (H₂O(ℓ)) = -238
Enter value as an integer (value ± 2).
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**QUESTION 8**
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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