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) 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