Nitrogen dioxide, NO2, is an atmospheric pollutant and is formed in car engines.
(a) Explain how nitrogen dioxide is formed in car engines.
(b) Nitrogen dioxide causes respiratory problems.
State one other adverse effect of nitrogen dioxide.
(c) Nitrogen dioxide emissions can be reduced by adding an aqueous solution of urea, (NH2)2CO, to car exhaust gases.
The heat of the exhaust gases breaks down the urea into simpler substances.
(i) Name the type of reaction which occurs when a substance is heated and breaks down into simpler substances.
(ii) One molecule of urea breaks down to form one molecule of ammonia and one other molecule.
Complete the chemical equation to show the formula of the other molecule formed in this reaction.
(NH2)2CO → NH3 + ……..
(iii) State the test for ammonia.
(d) The ammonia formed reacts with nitrogen dioxide to form nitrogen and water.
(i) Balance the equation for this reaction.
NO2 + NH3 → N2 + 12H2O
(ii) State how the equation shows that the nitrogen in nitrogen dioxide is reduced.
(iii) This reaction is a redox reaction.
State the meaning of the term redox.
(e) 135 moles of urea, (NH2)2CO, is stored in the tank of a car.
Calculate the mass, in kg, of the stored (NH2)2CO.
(f) Another oxide of nitrogen formed in car engines is nitrogen monoxide, NO. A catalytic converter removes NO by reacting it with a gas formed by incomplete combustion of the fuel. Two non‑toxic gases are formed.
(i) Name the gas formed by incomplete combustion of the fuel.
(ii) Name the two non‑toxic gases formed.
▶️ Answer/Explanation
(a) Ans: Nitrogen (from air) and oxygen (from air) react due to high temperatures in the engine.
At high temperatures in car engines, nitrogen (\(N_2\)) and oxygen (\(O_2\)) from the air react to form nitrogen dioxide (\(NO_2\)).
(b) Ans: Acid rain.
Nitrogen dioxide dissolves in rainwater to form nitric acid (\(HNO_3\)), contributing to acid rain.
(c)(i) Ans: Thermal decomposition.
When a substance breaks down into simpler substances upon heating, it is called thermal decomposition.
(c)(ii) Ans: HCNO (isocyanic acid).
The balanced equation is: \((NH_2)_2CO \rightarrow NH_3 + HCNO\).
(c)(iii) Ans: Test: Damp red litmus paper. Observation: Turns blue.
Ammonia (\(NH_3\)) is alkaline and turns damp red litmus paper blue.
(d)(i) Ans: \(6NO_2 + 8NH_3 \rightarrow 7N_2 + 12H_2O\).
Balanced equation showing the reaction of \(NO_2\) with \(NH_3\) to form nitrogen and water.
(d)(ii) Ans: Nitrogen loses oxygen.
In \(NO_2\), nitrogen has an oxidation state of +4, which reduces to 0 in \(N_2\) (loss of oxygen).
(d)(iii) Ans: Reduction and oxidation occur simultaneously.
Redox reactions involve both reduction (gain of electrons) and oxidation (loss of electrons).
(e) Ans: 8.1 kg.
Molar mass of urea (\((NH_2)_2CO\)) = 60 g/mol. Mass = \(135 \times 60 = 8100\) g = 8.1 kg.
(f)(i) Ans: Carbon monoxide (CO).
Incomplete combustion of fuel produces carbon monoxide.
(f)(ii) Ans: Carbon dioxide (\(CO_2\)) and nitrogen (\(N_2\)).
In the catalytic converter, \(NO\) reacts with \(CO\) to form non-toxic \(CO_2\) and \(N_2\).
A student adds excess large pieces of magnesium carbonate, MgCO3, to dilute hydrochloric acid, HCl, and measures the volume of carbon dioxide gas, CO2, given off.
(a) Add the missing state symbols to the chemical equation for the reaction.
MgCO3 ….. + 2HCl….. → MgCl2(aq) + H2O ….. + CO2 …..
(b) Complete the dot-and-cross diagram to show the electron arrangement of the ions in magnesium chloride.
The inner shells have been drawn.
Give the charges on the ions.
(c) Complete the dot-and-cross diagram to show the electron arrangement in a molecule of carbon dioxide.
Show outer shell electrons only.
(d) The graph shows how the volume of carbon dioxide gas changes with time.
(i) Describe how the graph shows that the rate of this reaction decreases as time increases.
(ii) Explain, in terms of particles, why the rate of this reaction decreases as time increases.
(iii) The student repeats the experiment using powdered MgCO3 instead of large pieces.
All other conditions stay the same.
On the grid, draw the line expected when powdered MgCO3 is used instead of large pieces.
(e) Determine the volume of CO2 gas given off when excess MgCO3 is added to 25.0cm3 of 0.400mol/dm3 HCl at room temperature and pressure.
MgCO3 + 2HCl → MgCl2 + H2O + CO2
Use the following steps.
- Calculate the number of moles of HCl in 25.0cm3 of 0.400mol/dm3 of acid.
- Determine the number of moles of CO2 gas given off.
- Calculate the volume of CO2 gas given off in cm3.
▶️ Answer/Explanation
(a) Ans: MgCO3(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) + CO2(g)
Explanation: MgCO3 is solid, HCl is aqueous solution, H2O is liquid, and CO2 is gas.
(b) Ans:
• Mg2+ ion: 8 electrons in outer shell (2 lost)
• 2 Cl– ions: 8 electrons each (1 gained each)
• Charges: Mg2+ and Cl–
(c) Ans:
• Each oxygen shares 2 pairs of electrons with carbon (double bonds)
• 4 lone pairs (2 on each oxygen)
(d)(i) Ans: The gradient of the curve decreases with time.
Explanation: A decreasing slope indicates the reaction is slowing down.
(d)(ii) Ans:
1. HCl concentration decreases as it’s used up
2. Fewer successful collisions between particles per unit time
(d)(iii) Ans:
• Steeper initial slope (faster reaction)
• Reaches same final volume (same amount of HCl)
(e) Ans:
1. Moles HCl = 0.400 × (25/1000) = 0.0100 mol
2. Moles CO2 = 0.0100/2 = 0.00500 mol (1:2 ratio)
3. Volume CO2 = 0.00500 × 24000 = 120 cm3
Calculation: At RTP, 1 mole of gas occupies 24,000 cm3.