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)
∆H˚f (Al2O3 (s) ) = -1676 kJ/mol ∆H˚f (Cr2O3 (s) ) = - 1128 kJ/mol
From equation 6.6 page 267:
∆H˚rxn = ∑[∆H˚f (products)] - ∑[∆H˚f (reactants)]
∆H˚rxn = [1∆H˚f (Al2O3 (s) ) + 2 ∆H˚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:
∆H˚rxn = 1(-1676)kJ + 2(0)kJ – 1(-1128 kJ) – 2(0) kJ
∆H˚rxn = -1676 kJ + 1128 kJ
∆H˚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