Sulfur Worked ExamplesIGCSE Chemistry 0620

(a) Fossil fuels such as coal contain sulfur compounds. When they burn, the sulfur is oxidised to sulfur dioxide, $\text{SO}_2$. Sulfur dioxide dissolves in rainwater and makes it acidic — acid rain.

(b) The other acid-rain gas in the list is nitrogen dioxide (an oxide of nitrogen, formed in car engines). Nitrogen itself is unreactive and is not a pollutant — it makes up about 78% of clean air.

(c) The syllabus strategies for reducing sulfur dioxide are: use low-sulfur fuels (remove sulfur compounds from the fuel before it is burned), and flue gas desulfurisation — passing the waste gases from power stations through calcium oxide, a base that reacts with the acidic sulfur dioxide.

(d) Bubble the gas into acidified aqueous potassium manganate(VII). Sulfur dioxide turns it from purple to colourless.

(a) Look for a cation that gives a white precipitate soluble in excess with both reagents. Aluminium ions dissolve in excess sodium hydroxide but not in excess ammonia; calcium ions give little or no precipitate with ammonia. Only zinc, $\text{Zn}^{2+}$, fits both results.

(b) A white precipitate with acidified aqueous barium nitrate shows the sulfate ion, $\text{SO}_4^{\,2-}$. The precipitate is barium sulfate, which is insoluble (sulfates are soluble except those of barium, calcium and lead).

(c) Acidified silver nitrate tests for halide ions. No precipitate means W contains no chloride (or bromide or iodide) ions. A negative result is still information.

(d) Combining the two ions: W is zinc sulfate, $\text{ZnSO}_4$.

(a) The syllabus gives two sources of sulfur dioxide: burning sulfur, and roasting sulfide ores (heating a metal sulfide such as zinc sulfide in air).

(b) The oxygen comes from air.

(c) The catalyst is vanadium(V) oxide, $\text{V}_2\text{O}_5$. Typical conditions are a temperature of 450 °C and a pressure of 200 kPa (2 atm).

(d) Sulfur dioxide and oxygen form sulfur trioxide (gas Z) in a reversible reaction:

(a) Read up from each temperature to the curve and across: 96% at 450 °C and 75% at 600 °C.

(b) The percentage converted decreases as the temperature increases, so the position of equilibrium moves to the left on heating. Increasing the temperature favours the endothermic direction, so the reverse reaction is endothermic and the forward reaction is exothermic.

(c) At 350 °C the equilibrium yield would be slightly higher, but the rate of reaction would be too slow — particles have less energy, so fewer collisions have energy equal to or greater than the activation energy. 450 °C is a compromise: a fast enough rate (helped by the catalyst) with a still-high yield.

(d) $M_r$ of $\text{SO}_2 = 32 + 2(16) = 64$ and $M_r$ of $\text{SO}_3 = 32 + 3(16) = 80$. Convert 64.0 kg to grams and find the moles:

The mole ratio $\text{SO}_2 : \text{SO}_3$ is $2 : 2 = 1 : 1$. Only 96% is converted:

There are fewer moles of gas on the right ($3 \rightarrow 2$), so a higher pressure would increase the yield — but the conversion is already very high at 2 atm, so the extra cost and safety risk of high-pressure plant are not worth it. C is correct.

A describes the Haber process (iron catalyst, 200 atm). B is wrong because the forward reaction is exothermic: raising the temperature lowers the yield.

D is the tempting wrong answer. A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster but its position does not change.

(a) Fe and S appear once on the left. $\text{Fe}_2\text{O}_3$ has 2 Fe, so start with 2 $\text{Fe}_2\text{O}_3$ → 4 FeS → 4 $\text{SO}_2$. Now count oxygen on the right: $2 \times 3 + 4 \times 2 = 14$ atoms, so 7 $\text{O}_2$.

Check: Fe 4 = 4; S 4 = 4; O 14 = 6 + 8. Balanced.

(b)(i) $M_r$ of ZnS $= 65 + 32 = 97$.

Mole ratio ZnS : $\text{SO}_2$ = 2 : 2 = 1 : 1, so 0.200 mol of $\text{SO}_2$. One mole of gas occupies 24 $\text{dm}^3$ at r.t.p.

(ii) Ratio ZnS : $\text{O}_2$ = 2 : 3.

(iii) Ratio ZnS : ZnO = 1 : 1; $M_r$ of ZnO $= 65 + 16 = 81$.

(a) Purple to colourless — this is the syllabus test for sulfur dioxide.

(b) Oxygen is −2. In a compound the oxidation numbers add up to zero; in an ion they add up to the charge on the ion.

(c) In $\text{MnO}_4^{\,-}$ the total is −1:

(d) Sulfur's oxidation number increases (+4 → +6), so sulfur dioxide is oxidised. Manganese's decreases (+7 → +2), so the manganate(VII) ion is reduced. An oxidising agent oxidises another substance and is itself reduced, so the manganate(VII) ion, $\text{MnO}_4^{\,-}$, is the oxidising agent.

(a) Excess base makes sure all the acid reacts, so the salt solution contains no leftover sulfuric acid. It is in excess when black solid remains (it no longer disappears) after stirring and warming.

(b) Filter to remove the excess copper(II) oxide. Heat the filtrate to evaporate some water until it is saturated (crystals start to form on a cold glass rod), then leave to cool so crystals form. Filter off the crystals, and dry them between filter papers (or in a warm place). Do not heat to dryness — that removes the water of crystallisation.

(c) Moles of acid: convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000. The ratio acid : salt is 1 : 1.

$M_r$ of $\text{CuSO}_4\cdot 5\text{H}_2\text{O} = 64 + 32 + 4(16) + 5(18) = 250$.

(d)

(a) Burette scales read downwards, each small division is $0.1 \text{ cm}^3$, and readings are given to the nearest $0.05 \text{ cm}^3$. Start: $1.20 \text{ cm}^3$. End: the meniscus is halfway between 22.4 and 22.5, so $22.45 \text{ cm}^3$.

(b) Moles of sulfuric acid:

The ratio NaOH : $\text{H}_2\text{SO}_4$ is 2 : 1, so double it:

Divide by the volume of alkali in $\text{dm}^3$:

(c) $M_r$ of NaOH $= 23 + 16 + 1 = 40$.

(a) Sulfur dioxide is an acidic oxide (a non-metal oxide). Calcium oxide is a basic oxide (a metal oxide). A base reacts with an acidic oxide — it neutralises it — forming a salt, so the sulfur dioxide is removed instead of escaping to form acid rain.

(b)

(c) Mass of sulfur in 1.00 tonne of coal:

S : $\text{SO}_2$ is 1 : 1 and $\text{SO}_2$ : CaO is 1 : 1, so 500 mol of CaO is needed. $M_r$ of CaO $= 40 + 16 = 56$.

(a) Both lines become horizontal at 8 minutes. At equilibrium in a closed system: the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products no longer change.

(b) $\text{SO}_2$ falls from 1.0 to $0.2 \text{ mol/dm}^3$:

(c)(i) There are 3 moles of gas on the left and 2 on the right. Increasing the pressure moves the position of equilibrium to the side with fewer moles of gas — to the right — so more $\text{SO}_3$ forms.

(ii) The forward reaction is exothermic. Increasing the temperature favours the endothermic (reverse) reaction, so the position moves to the left — less $\text{SO}_3$.

(d) A catalyst increases the rate but does not change the position of equilibrium: the line is steeper at the start and becomes flat earlier, at the same final concentration of $0.8 \text{ mol/dm}^3$.

(a) Purple to colourless, caused by sulfur dioxide, formed when sulfur compounds in the fuel burn.

(b) P — it decolourised the solution in the shortest time, so it produced sulfur dioxide fastest, meaning the most sulfur per gram. R gave a negative result: it contains no sulfur (or too little to detect).

(c) The pump draws the combustion gases through the solution (they would otherwise rise away from the funnel). A fume cupboard is needed because sulfur dioxide (and any carbon monoxide) is toxic.

(d) Any one: keep the pump rate the same; make sure the fuels burn at the same rate (same size of flame / same wick); use a white tile behind the tube to judge the colour change; repeat each fuel and average. The same volume and concentration of manganate(VII) is already used, so saying that does not score.