# The Question

Joseph Priestley (right) discovered sweet-smelling, colourless laughing gas, nitrous oxide (N2O), in 1763. N2O has been widely used as an anesthetic for medical and dental applications. Nitrogen dioxide (NO2) is a toxic reddish brown gas that is the major component of smog. Nitrogen oxide (NO) is a colourless, non-toxic gas that readily oxidizes to nitrogen dioxide in the presence of oxygen.

Using the reactions:

4 NO(g) → 2 N2O(g) + O2(g)             ΔG° = – 139.56 kJ

2 NO(g) + O2(g) → 2 NO2(g)             ΔG° = – 69.70 kJ

calculate the Gibbs free energy released or required when 2 moles of nitrous oxide combine with 3 moles of oxygen to produce 4 moles of nitrogen dioxide.

1. – 209.26 kJ
2. – 139.86 kJ
3. – 69.86 kJ
4. + 0.16 kJ
5. +139.72 kJ

## Solution

To solve this problem, first write the chemical equation from the word equation provided in the final sentence:

2 N2O (g) + 3O2 (g) → 4 NO2 (g)       ΔG° =?

Rearrange the given equations to solve for the value of the Gibbs free energy

Eq. 1                       4 NO(g) → 2 N2O(g) + O2(g)             ΔG° = – 139.56 kJ

Eq. 2                       2 NO(g) + O2(g) → 2 NO2(g)             ΔG° = – 69.70 kJ

Reverse Eq. 1   2 N2O(g) + O2(g) → 4 NO(g)   ΔG° = 139.56 kJ

Double Eq. 2   4 NO(g) + 2 O2(g) → 4 NO2(g) ΔG° = – 139.40 kJ

Add Eq. 1 & 2      2 N2O(g) + 3O2(g) → 4 NO2(g) ΔG° =+ 0.16 kJ

## Analysis

Only 45% of the students who took the exam answered this question correctly. Most of the students who answered correctly did well on the exam overall and very few of the students who scored poorly overall got this question right.

Gibbs free energy is not covered in depth in several provincial chemistry curricula in Canada. However, this question is analogous to a Hess’s law question in which determining enthalpy would parallel the logic used in this calculation. Students had to use logic, problem solving and several related steps to solve the problem. Little difference between provinces was shown in the differentiating indices, demonstrating strong students in all provinces tended to do well on this question.

## Acknowledgements

A team of educators from across Canada collaborate in the creation of the multiple choice questions for Part A of the CCC.  In 2015, contributors included Barbara Skrela, Danny Hickie, Jennifer Howell, Andy Dicks, Jenny Pitt-Lainsbury and Ken Hoffman.

## References and Notes

1. http://www.cheminst.ca/outreach/canadian-chemistry-contest.
2. D. DiBattista and L. Kurzawa, Examination of the Quality of Multiple-choice. The Canadian Journal for the Scholarship of Teaching and Learning2011, pages 1-23.

### Editor’s note

Information for the U.S. National Chemistry Olympiad can be found at the American Chemical Society website.

Many countries take part in the International Chemistry Olympiad. The Chemistry Olympiad 2016 will be held in Pakistan. Go to your country’s website for more details.

## Questions from 2015 (#2 and # 21)

2.     What is the correctly balanced form of the chemical reaction depicted in the figure below?

1. 4 A + 6 B → A4B6
2. A4 + B6 →  A4B6
3. A4 + 3 B2 → 2 A2 + 2 B3
4. *  A4 + 3 B2 → 2 A2B3
5. 4 A + 3 B2 → 2 A2B3

This question was too easy with 90% of students getting the correct answer. The discriminating index was only 0.2.

21.  Which of the following statements about the electrolysis of a solution of magnesium iodide, as depicted in the diagram, is false?

1. *  If the negative and positive terminals of the cell are reversed, magnesium would begin to plate on electrode B.
2. The reaction is non-spontaneous without an applied power source.
3. An acid-base indicator could be used to detect a product formed at electrode A.
4. I2 forms at the anode.
5. A gas is formed at the cathode.

I2(s) + 2 e → 2 I(aq)                             E° = + 0.54 V

Mg2+(aq) + 2 e → Mg(s)                       E° = − 2.37 V

O2(g) + 4 H+(aq) + 4 e → 2 H2O(l)     E° = + 1.23 V

2 H2O(l) + 2 e → H2(g) + 2 OH(aq)   E° = − 0.83 V

This question was too hard. Only 10% of students got this question right and the discriminating index was only 0.1.