SCH 4U - Heat of Formation

Standard Enthalpy of Formation (ΔH_f^°)

The standard enthalpy of formation relates to the formation of 1 mole of a compound from its constituent elements at 25°C and 101.325 kPa.

ex. 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)   ΔH_f^° = -277.7 kJ/mol

Use the most stable form of the element at 25°C (H₂(g), O₂(g), N₂(g), F₂(g), Cl₂(g), Br₂(l), I₂(s)); on the table it is the one with ΔH_f^° = 0 kJ/mol.

⮞⮞⮞The values are found in the tables under the "Data Tables" tab at the top of the blog.⮜⮜⮜

 

Using the Standard Enthalpies of Formation

ex. What is the ΔH_c for propane?  C₃H₈(g) + 5 O₂(g) → 3CO₂(g) + 4H₂O(g)

ΔH_rxn = ∑ΔH_f^°(products) - ∑ΔH_f^°(reactants)

ΔH_c^° = (3(-393.5)+4(-241.8)) - (1(-103.85)+5(0))

         = (-1180.5 + -967.2) - (-103.85 + 0)

         = (-2147.7) - (-103.85)

         = -2043.9 kJ/mol

* Notice that each enthalpy value is multiplied by the appropriate coefficient from the balanced equation.  This occurs because the enthalpy of formation is per mol of substance.  So, if there are 4 mol of substance, we multiply the enthalpy value by 4.

 

Multi-step Energy Calculations

• Chemical engineers need to do calculations involving the heat produced by an internal combustion engine; these calculations involve the skills we have learned to date.

• Determine the enthlapy of combustion for octane:
  

        C₈H₁₈(l)    +   25/2O₂(g)  →  8CO₂(g)    +    9H₂O(g)

ΔH_f^°  -250.1 kJ/mol      0 kJ/mol      -393.5 kJ/mol  -241.8 kJ/mol

ΔH_c^° = (8(-393.5kJ/mol)+9(-241.8kJ/mol))–(1(-250.1kJ/mol)+25/2(0kJ/mol)) 

ΔH_oct^° = -5074.1 kJ/mol

• Determine the total amount of heat absorbed by the ethylene glycol:

m_eg = 20.0 kg          q_eg = m ΔT c

T₁ = 10.0°C                = (20.0 x 10³ g)(70.0°C – 10.0°C)(3.5 J/g°C)

T₂ = 70.0°C                = 4.2 x 10³ kJ

c_eg = 3.5 J/g°C

• We can assume the amount of heat absorbed by the ethylene glycol is equivalent to the amount of heat lost by the octane, so q_oct = -4.2 x 10³ kJ.

• Determine the number of moles of octane burned:
  

                              ΔH_oct = q_oct/n_oct                                                     

           -5074.1 kJ/mol = -4.2 x 10³ kJ/ n                  

                                 n = -4.2 x 10³ kJ/-5074.1 kJ/mol            

                                    = 0.83 mol

 

•   Determine the mass of octane burned:
  

                              n_oct = m_oct/M_oct

                     0.83 mol = m_oct/114.231 g/mol

                             m_oct = 94.8 g

 ∴ the mass of octane required is 95 g.

 

Homework # 16-20