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:

∆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