CHM 2045C Module Seven Sample Exam Answers     

 

Module Seven Part H: Standard Enthalpies of Formation   5 points

 

Calculate  H for the reaction:

 

  2 Al (s)  + 1 Cr2O3 (s)  à 1  Al2O3 (s)  + 2 Cr (s)

 

Given:

     

f (Al2O3 (s) ) = -1676 kJ/mol    f (Cr2O3 (s) ) = - 1128 kJ/mol

 

From equation 6.6 page 267:

 

rxn  = ∑[∆f (products)] -  ∑[∆f (reactants)]

 

rxn  = [1∆H˚f (Al2O3 (s) ) + 2 ∆f (Cr(s) )] - [1∆H˚f (Cr2O3 (s) ) + 2∆H˚f (Al (s) )]

 

Remember that the heat of formation of elements is defined as 0 at standard state:

 

Also apply coefficients from the balanced equation to cancel the moles:

 

rxn  = 1(-1676)kJ + 2(0)kJ – 1(-1128 kJ) – 2(0) kJ

 

rxn  = -1676 kJ + 1128 kJ 

 

rxn  = - 548 kJ

 

See page 267 for a worked example in the discussion. On page 268-9 Example 6.9 shows you step by step.

 

On page 275 at the end of the chapter Problems #47-55 provide more examples of Part H problems for the exam.

 

 

 

 

 

 

 

 

 

Answer:    -548 kJ