Basic Chemistry — Final Exam 1

Name:________________________________

Student number:_______________________

Autumn Semester

Basic Chemistry for Engineering/ Fundamental Inorganic Chemistry

Final Exam

18 February 2013

Time: 60 minutes

Instructions: Answer any 4 questions out of 5 possible questions. You may answer more than 4 questions if you wish. Your marks will be taken from your top 4 questions. Show all relevant working. Write all answers and final working in pen. Give your answers to an appropriate number of significant figures. 9 pages including: 1 top page, 1 data page, 1 periodic table, 5 question pages, 1 page for rough working Hand in all pages at the end of the exam. Permissible materials: Calculator. English or English bilingual Dictionary.

Basic Chemistry — Final Exam 2

Equations and Data

Avagadro number 6.022 x 1023

pV = nRT where p = pressure, V = volume, n= no. of moles, R = ideal gas constant T = temperature R = 8.3144 J K–1mol–1 or R = 0.082057 L atm K–1mol–1 ∆Hrxn = ∑ ∆Hf products – ∑∆Hf reactants Density of water: 1.0 g ml–1 Enthalpies of formation CH4(g): –74.9 kJ mol–1 HCCH(g): 226.8 kJ mol–1 H2O(l): –285.83 kJ mol–1 H2O2(l): –187.8 kJ mol–1

H3PO4(l): –1288 kJ mol–1 PCl3(l): –320 kJ mol–1

Latent heat of fusion (heat of forming a liquid from a solid) of NaO2CCH3: 22.9 kJ mol–1

Basic Chemistry — Final Exam 3

Basic Chemistry — Final Exam 4

Question 1 A) An inventor asks a bank to invest in a new kind of heater that uses water as a fuel. He claims the

engine burns water to generate heat using oxygen and a secret catalyst. He says the reaction occurs

according to the following equation:

2H2O(l) + O2(g) ——> 2H2O2(l)

i) Use data from the data sheet to calculate the enthalpy change, per mole of H2O, for this reaction.

∆Hf = 2 mol (–187.8 kJ mol–1) – (2 mol (–285.83 kJ mol–1) + 0 kJ mol–1) = 196.06 kJ for 2 moles of H2O. => 98.03 kJ per mole H2O (4 s.f.)

[2]

iii) Which of the following words best describes the above reaction? Circle the answer.

exothermic concentrated dilute endothermic spontaneous

[1]

iii) Do you predict the entropy change for this reaction would be positive or negative? Why?

I expect it to be negative because two moles of liquid and one mole of gas become only two moles of liquid.

[2]

iv) Should the bank invest in this heater? Why?

No. This reaction is endothermic and will not give off heat. [2]

B) A pocket hand warmer pouch contains a supersaturated solution of 170 g sodium acetate

(NaO2CCH3) dissolved in 100 ml water. When the pouch is activated, 134 g of NaO2CCH3

crystallises out of solution.

i) Using data from the data sheet, calculate the enthalpy change of this crystallisation process.

Mr NaO2CCH3 = 22.9898 g mol–1 + 2(15.9994 g mol–1) + 2(12.0107 g mol–1) + 3(1.0079 g mol–1) = 82.0337 g mol–1 => No. moles crystallised NaO2CCH3 = 134 g ÷ 82.0337 g mol–1 = 1.63348 (6 s.f.) mol => enthalpy change = 1.63348 mol x –22.9 kJ mol–1(*) = –37.4 kJ (3.s.f) (*Note: The heat of fusion given in the data sheet is stated as the enthalpy change of forming a liquid from a solid. This reaction is forming a solid from a liquid and so the enthalpy change will have a negative sign.)

[3]

Basic Chemistry — Final Exam 5

Question 2 A) i) What is the chemical formula of the following compounds? (Write CnHm followed by other

elements, e.g. C2H4O2)

C3H6O C6H12 C6H15N C3H7NO

[4]

ii) Identify pairs of isomers. If no isomer exists, circle 'No isomer'.

a) b) c) d) e)

Isomer with: e) ______ ______ ______ a)

or: No isomer No isomer No isomer No isomer No isomer

[4]

B) Choose your answers from the molecules bearing the following functional groups:

a) b) c) d) i) Which molecule can exist as geometrical isomers? c) ii) Which molecule bears the most easily-detachable functional group? d)

[2]

H3C

O

CH3

H3C

CH3

CH3N NH2

O

OH

O OOH

NH2OH

O Cl

Basic Chemistry — Final Exam 6

Question 3 A) Choose the most appropriate concentration measure from the following list to complete parts i) — iv). molarity, molality, percentage by weight, percentage by volume, parts per million i) Reporting the amount of flavour syrup in a sweet drink. percentage by volume ii) Measuring the effect of a solute on solvent vapour pressure across a range of temperatures. molality iii) Calculating the volume of solution needed for a chemical reaction. molarity iv) Reporting levels of trace pollutants in the atmosphere. parts per million

[4] B) Refer to the following chart of solubility: i) Demonstrate, using calculations, whether it is, or is not, possible to make a 12M aqueous solution of potassium chloride (KCl). Mr KCl = 74.5513 g mol–1 so a 12M solution is equivalent to (74.5513 g mol–1 x 12 mol ÷ 1 dm3) = 894.616 g dm–3 (6 s.f.). Density of water = 1 g ml–1 so 894.616 g dm–3

H2O ≡ 89.4616 g 100 g–1

H2O. The chart shows there is no temperature at which liquid water can dissolve this much KCl.

[3] You will need one or more of the following terms for part ii): concentrated dilute saturated dissolve precipitate ii) Na2SO4 is dissolved in water at 30 ˚C until no more will dissolve. At this point, the solution is saturated. The solution is heated to boiling which makes some Na2SO4 precipitate.

[2] iii) Will the Na2SO4 solution boil at exactly 100 ˚C? Why? No. Solutes increase the boiling point of a solution.

[1]

Basic Chemistry — Final Exam 7

Question 4 A) i) Why is tin used to protect the iron in 'tin' cans. Explain why.

Because tin is unreactive. [1]

ii) Explain one method of protecting metal against corrosion with one advantage and one

disadvantage.

Paint: It's cheap and prevents contact with air and water but it does not protect the metal if it gets scratched. or Metal plating: It gives better protection against air and water than paint but it doesn't protect the metal if scratches penetrate the plating. or Sacrificial anodes: They protect all areas of metal in electrical contact with the anode but they are expensive.

[3]

B)

A student makes three mistakes when designing a voltaic

cell. Identify the mistakes and briefly explain why they are a

problem.

i) The electrodes are in the wrong solutions. This arrangement will result in an electrolytic reaction directly between the (copper) electrode and the (AgNO3) solution. ii) The salt bridge has been replaced by a wire. A wire will not allow the counter ions to balance between the cells. iii) There is a battery instead of a voltmeter or similar device. Batteries are used in electrolytic cells.

[6]

wires

wire

battery

Cu(NO3)2

silver (Ag) electrode

copper (Cu) electrode

Ag(NO3)2

Basic Chemistry — Final Exam 8

Question 5

A) Explain what effect the following changes will make to the rate of a reaction and why:

i) heating the reaction: This will increase the rate of reaction because more molecules have enough

energy for reaction.

ii) diluting the reaction: This will decrease the rate of reaction because there will be fewer collisions

in any moment.

iii) using a catalyst: This will increase the rate of reaction by either changing the activation energy

of the reaction or by helping the molecules collide with the correct orientation for reaction (or

both).

[4]

B) i) Chemists rarely perform chemical reactions between two solids. Instead, they prefer to

dissolve the solids in solvents and run the reactions in solution. Explain why, in terms of collision

theory.

In a solution, more molecules can collide at any moment and it is more likely that molecules can collide with the correct orientation.

[2]

ii) Gunpowder burns charcoal (pure carbon) as a fine powder (~0.1 mm diameter), whilst barbeques

burn large pieces of charcoal (~ 50 — 150 mm length). Explain this difference.

Gunpowder needs to release all of its energy in an instant and smaller particles increase the rate of reaction. Barbeques need to burn for a long time and larger pieces take longer to completely react (burn).

[2]

C) According to thermodynamics calculations, wood should spontaneously react with oxygen at

room temperature. Explain why wooden furniture does not spontaneously catch fire at room

temperature.

The combustion of wood has a high activation energy. [2]

Basic Chemistry — Final Exam 9

Rough Working