Paper 1

1. Potassium chloride is a soluble salt that can be formed in a variety of chemical reactions.

a) Potassium chloride contains the ions K+ and Cl-. Give the formula of potassium chloride.

Answer: The formula is KCl.

b) Give the name of two reactants that can be used to make a sample of potassium chloride in a neutralisation reaction.

Answer: hydrochloric acid and any one of potassium hydroxide, potassium carbonate, or potassium oxide

c) Potassium has two common isotopes. Their masses numbers are 39 and 41. The percentage abundance of each isotope is 93.3% of 39K and 6.7% 41K. Calculate the relative atomic mass (Ar) of potassium.

Answer: Ar = [(39 × 93.3) + (41 × 6.7)] ÷ 100 = 39.134

d) Describe and explain the properties of potassium chloride. Answer in terms of structure and bonding.

Answer: Potassium chloride is an ionic compound with strong forces of attraction between the oppositely charged ions in the lattice structure. So it has high melting and boiling points as a lot of energy is needed break the strong forces of attraction. It conducts electricity when molten (or dissolved) as the ions are free to move and carry the charge.

2. Early forms of the periodic table arranged the elements in order of their atomic weights. Modern periodic tables arrange the elements in order of their atomic numbers.

a) Based on atomic weight, argon should be placed after potassium and not before it. Explain why Mendeleev placed these two elements in their correct positions in his periodic table.

Answer: Mendeleev positioned the elements based on their properties. By placing argon before potassium, the properties of argon matched those of the other elements in the same column (group). If argon was placed after potassium, its properties did not match those of the elements in the same column (group).

b) Explain why lithium, sodium and potassium are in Group 1 of the periodic table.

Answer: They all have one electron in their outer shells.

c) The table below shows the observations made when three Group 1 metals react with water. Describe the trend in the reactivity of the Group 1 metals.

Answer: Reactivity increases down Group 1.

d) Name the gas produced during the reaction between lithium and water.

Answer: hydrogen

e) Suggest a value for the pH of the solution formed during the reaction between potassium and water.

Answer: Any pH value greater than 7 as the solution will contain hydroxide ions, making it alkaline.

f) During the reaction between potassium and water, potassium atoms react to form potassium ions (K+ ). Explain why this is an example of oxidation.

Answer: The potassium atoms lose one electron to form K+ ions. Oxidation is the loss of electrons.

g) Give the half equation for the oxidation of potassium atoms.

Answer: K  →  K+ + e-  or  K - e-  →  K+

h) The halogens make up Group 7 of the periodic table. Describe the trend in the reactivity of the Group 7 elements.

Answer: Reactivity decreases down the Group 7. When halogens react they gain one electron to fill their outer shell of electrons. Going down the Group the number of electron shells increases, so the gained electron is further from the nucleus. So the electron is less strongly attracted to the halogen nucleus, and it is harder for halogens further down the Group to gain an electron.

3. Hydrazine (N2H4 ) is a highly reactive compound and can be made in the following reaction: 2NH3 + NaClO  →  N2H4 + NaCl + H2O

a) Calculate the percentage atom economy for this reaction to produce hydrazine. Relative formula masses (Mr): NH3 = 17 , NaClO = 74.5 , N2H4 = 32 , NaCl = 58.5 , H2O = 18

Answer: The total Mr of reactants is 108.5 . The percentage atom economy for this reaction is (32 ÷ 108.5) × 100 = 29.49%

b) When 1.0 kg of ammonia (NH3) reacts the maximum theoretical mass of hydrazine is 0.94 kg. The percentage yield of the reaction is 71%. Calculate the actual mass of hydrazine formed in the reaction.

Answer: actual mass = (71 ÷ 100) × 0.94 kg = 0.667 kg

c) The reaction between nitrogen and hydrogen to form hydrazine is N2 + 2H2  →  N2H4 . This reaction is endothermic. Draw a fully labelled reaction profile for this reaction between nitrogen and hydrogen. The reaction profile shows how the energy changes during the progress of the reaction.


4. Methane (CH4 ) is a flammable gas that is used in combustion reactions to provide heat for cooking and heating.

a) State whether the combustion of methane is exothermic or endothermic.

Answer: Exothermic as heat energy is released.

b) Explain why methane is a gas at room temperature. Answer in terms of structure and bonding.

Answer: Methane is made up of small molecules (is simple molecule). It has weak intermolecular forces that require little energy to overcome, so at room temperature all the molecules have enough energy to spread out and become a gas.

c) The figure below shows the displayed formulae for the combustion reaction between methane and oxygen, while the table shows the bond energies in the reaction. Calculate the overall energy change for the combustion of methane.

Answer: bonds broken = (4 × 413) + (2 × 496) = 2644 kJ ; bonds made = (2 × 743) + (4 × 463) = 3338 kJ ; energy change = bonds broken - bonds made = 2644 - 3338 = -694 kJ

5. Titanium dioxide (TiO2 ) nanoparticles are used in sunscreen.

a) Which property of TiO2 nanoparticles makes them suitable for use as a sunscreen.

Answer: They filter out ultraviolet light.

b) Cubic nanoparticles used in sunscreen have a width of 50 nm. Calculate the surface area to volume ratio of these nanoparticles.

Answer: surface area = 50 × 50 × 6 = 15 000 nm2 ; volume = 50 × 50 × 50 nm = 125000 nm3 ; SA : V ratio = 15000 : 125000 = 0.12

c) Give the surface area to volume ratio of a cubic nanoparticle with a width of 5 nm.

Answer: 1.2 as ratio increases by a factor of 10 when width decreases by a factor or 10.

d) Explain why some people are concerned about the possible health effects of using TiO2 nanoparticles in sunscreens.

Answer: Nanoparticles have different properties to those of the bulk material and it is not safe to assume that they are harmless. TiO2 nanoparticles may be able to enter into cells and cause damage, unlike bulk TiO2 .

e) Titanium is a transition metal. Give some differences between the properties of titanium and potassium.

Answer: Titanium has a higher melting (boiling point), is denser, is harder, is stronger, is less reactive, forms coloured compounds, can act as a catalyst, can form ions with more than one charge.

f) The most common isotope of titanium has a mass number of 48 and an atomic number of 22. Give the number of protons, neutrons and electrons in this isotope of titanium.

Answer: 22 protons, 22 electrons and 26 neutrons

6. Fuel cells oxidise fuels electrochemically to produce electrical energy. Hydrogen fuel cells use hydrogen as the fuel. Direct methanol fuel cells use methanol as the fuel.

a) Give one advantage of fuel cells over rechargeable batteries.

Answer: Fuel cells can run continually (as long as the fuel is supplied). Rechargeable batteries will go flat and need to be recharged. Rechargeable batteries may release toxic chemicals when disposed of. The only waste product from the fuel cell is water.

b) Give the overall balanced equation for the reaction that occurs in a hydrogen fuel cell.

Answer: 2H2 + O2  →  2H2O

c) In a direct methanol fuel cell, 5.12 g of methanol (CH3OH) reacted with 10.24 g of oxygen to form 7.04 g of carbon dioxide and 5.76 g of water. Use this information to write the balanced symbol equation for the reaction between methanol and oxygen. Relative formula masses (Mr): CH3OH = 32 , O2 = 32 , CO2 = 44 , H2O = 18

Answer: mol CH3OH = mass ÷ Mr = 5.12 ÷ 32 = 0.16 mol ; mol O2 = 10.24 ÷ 32 = 0.32 mol ; mol CO2 = 7.04 ÷ 44 = 0.16 mol ; mol H2O = 5.76 ÷ 18 = 0.32 mol. Balanced equation is CH3OH + 2O2  →  CO2 + 2H2O

d) The table below gives information about hydrogen and direct methanol fuel cells. Evaluate the use of hydrogen fuel cells compared with direct methanol fuel cells.

Answer: Hydrogen fuel cells are more efficient (than direct methanol fuel cells). Hydrogen can be made using renewable energy sources. Methanol can be made sustainably from fermentation. Electrolysis requires a source of energy, whereas fermentation does not (or, at least, very little). Methanol can use existing infrastructure for refuelling, whereas hydrogen cannot. Methanol does not need high pressure storage tanks to be used. Hydrogen cells do not produce greenhouse gases (they are less polluting).

7. The figure below shows apparatus that can be used to electrolyse molten magnesium chloride.

a) Explain why magnesium chloride must be molten for electrolysis to occur.

Answer: When molten the ions are free to move, so can carry the charge.

b) Explain why graphite is a suitable material to use for the electrodes. Answer in terms of structure and bonding.

Answer: In graphite each carbon atom forms three bonds. One electron per carbon atom is delocalised. These electrons can move and carry the charge through the graphite.

c) Give the product formed at the negative electrode.

Answer: magnesium

d) Give the products formed at the positive electrode and negative electrode when a solution of magnesium chloride is electrolysed.

Answer: Hydrogen is formed at the negative electrode, while chlorine is formed at the positive electrode.

e) A student tested the hypothesis: "Hydrogen is always produced at the negative electrode when the metal in solution is more reactive than hydrogen". The table below shows the student’s results. Suggest what the student must do next to determine whether their hypothesis is correct or not.

Answer: Collect the gases produced and test to identify them.

8. A student investigated how the concentration of iron sulfate solution affected the temperature change of the displacement reaction between magnesium powder and iron sulfate.

a) Describe a method for this investigation.

Answer: Measure out 20 cm3 of iron sulfate solution using a measuring cylinder and pour into a beaker. Record the starting temperature of the solution using a thermometer. Add 2g of magnesium, stir with a glass rod and record the highest temperature reached. Repeat with a different concentration of iron sulfate solution. Use the same mass of magnesium and the same volume of iron sulfate solution in each experiment.

b) The student used 0.10 g of magnesium powder in each experiment. The figure below shows their results. Describe the trend shown in the student’s results.

Answer: Initially as the concentration of iron sulfate increases, the temperature rise also increases. After 4 g/dm3 the temperature rise is constant.

c) Explain why magnesium was the limiting reactant in the displacement reactions where the concentration of iron sulfate solution was greater than 4 g/dm3.

Answer: The limiting reactant is completely used up before the other reactant. After 4 g/dm3 increasing the concentration of iron sulfate does not cause the temperature to rise further. There must not be enough magnesium present to react with the extra iron sulfate to cause a further increase in temperature, so it has been used up.

d) Calculate the concentration of a 4.00 g/dm3 solution of iron sulfate in mol/dm3. Relative formula mass (Mr) FeSO4 = 152

Answer: concentration = 4.00 g/dm3 ÷ 152 = 0.0263 mol/dm3

9. This question is about a titration between sodium hydroxide (NaOH) and sulfuric acid (H2SO4 ).

a) Give the type of reaction that occurs between sodium hydroxide and sulfuric acid.

Answer: neutralisation

b) Sulfuric acid is a strong acid. Explain the difference between the strength of an acid and the concentration of an acid.

Answer: Strength is a measure of how completely the acid is ionised in solution. Concentration is a measure of the amount of acid dissolved in a given volume.

c) Describe a suitable method for carrying out a titration between sodium hydroxide solution and sulfuric acid, using phenolphthalein indicator.

Answer: Fill a burette with sulfuric acid solution. Using a glass pipette, measure out 25.0 cm3 of sodium hydroxide solution into a conical flask. Add 3 drops of the indicator to the sodium hydroxide (it will turn pink). Add the acid to the alkali and swirl until the indicator changes colour (goes colourless). Record the volume added. Repeat and find the mean volume of sulfuric acid added.

d) The reaction between sodium hydroxide and sulfuric acid is 2NaOH + H2SO4  →  Na2SO4 + 2H2O. A student titrated 25.0 cm3 portions of sodium hydroxide solution with a 0.215 mol/dm3 solution of sulfuric acid. The average titre was 20.65 cm3. Calculate the concentration of the sodium hydroxide solution in mol/dm3.

Answer: mol H2SO4 = concentration × volume = 0.215 × (20.65 ÷ 1000) = 0.00444 ; mol NaOH = 2 × mol H2SO4 = 0.00888 ; concentration NaOH = moles ÷ volume = 0.00888 ÷ (25 ÷ 1000) = 0.355 mol/dm3