Exercise 10.105 Part A Palmitic acid (C16 H322) is a dietary fat found in beef a
ID: 994990 • Letter: E
Question
Exercise 10.105 Part A Palmitic acid (C16 H322) is a dietary fat found in beef and butter. The caloric content of palmitic acid is typical of fats in general Write a balanced equation for the complete combustion of palmitic acid. Use H2O() n the balanced chemical equation because the metabolism of these compounds produces liquid water. Express your answer as a chemical equation. Identify all of the phases in your answer. Submit My Answers Give Up Part B Calculate the standard enthalpy of combustion. The standard enthalpy of formation of palmitic acid 208kJ/mol Express your answer using four significant figures. kJ Submit My Answers Give Up Part C What is the caloric content of palmitic acid in Cal/g?Explanation / Answer
A) Palmitic acid is C16H32O2. The complete combustion of palimitic acid gives CO2 (g) and H2O (l) as
C16H32O2 (s) + 23 O2 (g) ------> 16 CO2 (g) + 16 H2O (l)
B) The enthalpy of combustion can be found as
H0rxn = 16(H0f, CO2) + 16(H0f, H2O) – (H0f, C16H32O2) (the H0f terms denote enthalpy of formation; enthalpy of formation of diatomic molecules like O2 from its atoms is zero).
Now, we note the enthalpy of formation of CO2 (g), H2O(l) and palmitic acid as
H0f : CO2 (g) = -393.5 kJ/mol
H2O (l) = -285.8 kJ/mol
C16H32O2 (s) = -208 kJ/mol
Therefore, the enthalpy of combustion is
H0rxn = 16.(-393.5 kJ/mol) + 16.(-285.8 kJ/mol) – (-208 kJ/mol) = -6296 kJ/mol + (-4572.8 kJ/mol) + 208 kJ/mol = -10,660.8 kJ/mol.
C) We know that 1 cal = 4.18 J.
Hence the heat released by the combustion reaction is (-10,660.8 kJ/mol)/(4.18 J)* 1 cal = -2550430.62 cal/mol.
Molar mass of palmitic acid is (16*12 + 32*1 + 2*16) = 256 gm/mol
Hence caloric content of palmitic acid is (-2550430.62 cal/mol)/256 gm/mol = -9.96 kcal/gm (ans)
D) The balanced equation is
C12H22O11 (s) + 12 O2 (g) --------> 12 CO2 (g) + 11 H2O (l)
E) The standard enthalpy of combustion is
H0rxn = 12.(-393.5 kJ/mol) + 11.(-285.8 kJ/mol) – (-2226.1 kJ/mol) = (-4722 kJ/mol) + (-3143.8 kJ/mol) + 2226.1 kJ/mol = -5639.7 kJ/mol (ans)
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