Answers · Rates of reaction

The gas syringe race — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. UQ1 · Practice

Model answer. (a) The syringe held 0 cm³ at the start and 33 cm³ at 30 s: 33 cm³. (b) 54 − 33 = 21 cm³. (c) Less hydrogen came in the second 30 s (21 cm³) than in the first (33 cm³), so the reaction is slowing down as the reactants are used up.

Checked automatically

(a) Hydrogen in the first 30 s: 33 cm³ (exact value)
(b) Hydrogen from 30 s to 60 s: 21 cm³ (exact value)
- Common wrong answer 54 : That is the total at 60 s. Take away the 33 cm³ already there at 30 s.

A full answer includes

(a) I wrote 33 cm³.
(b) I worked out 54 − 33.
(b) I got 21 cm³.
(c) I wrote that less gas was made in the second 30 s than in the first.
(c) I wrote that the reaction is slowing down.
(c) I did not write that the rate goes up because the volume goes up.

Also accept

33 cm³; 21 cm³; slowing down

Watch for

“The rate goes up because the volume goes up.” — The volume keeps rising, but by less each time. The rate is how fast it rises, and that falls.

2. UQ2 · Practice

Model answer. (a) Curve A goes flat on the 60 cm³ gridline, so reaction A made 60 cm³ of gas in total. (b) At 30 s curve A is just below the 50 cm³ gridline: about 49 cm³. (c) Rate = volume ÷ time = 49 cm³ ÷ 30 s = 1.63 cm³/s, about 1.6 cm³/s (1.6 to 1.7 cm³/s, depending on your reading).

Checked automatically

(a) Total volume of gas in A: 60 cm³ (within ±1 cm³)
- Common wrong answer 100 : That is the top of the axis. Read the height where curve A goes flat.
- Common wrong answer 58 : That is curve B at 180 s. Read the flat part of curve A, the solid line.
(b) Volume in A at 30 s: 49 cm³ (within ±2 cm³)
- Common wrong answer 26 : That is curve B at 30 s. Read curve A, the solid line.
(c) Average rate over the first 30 s: 1.63 cm³/s (within ±5%)
- Common wrong answer 0.61 : That is time ÷ volume. Divide the volume by the time: 49 ÷ 30.

A full answer includes

(a) I read the height of the flat part of curve A: 60 cm³.
(a) I did not read the top of the axis, 100 cm³.
(b) I read about 49 cm³ (anything from 47 to 51 cm³).
(c) I divided my volume at 30 s by 30 s.
(c) I got about 1.6 cm³/s.
(c) I did not divide the time by the volume.

Also accept

60 cm³; about 49 cm³; about 1.6 cm³/s

Watch for

“Rate = time ÷ volume.” — That gives the seconds for each cm³. A rate is how much gas is made each second: volume ÷ time.
“The total volume is the top of the axis.” — The axis only gives the scale. The total is the height where the curve goes flat.

3. UQ3 · Practice

Options as numbered on the pupil sheet: 1 — The reaction slows down as a reactant is used up; 2 — The total volume of gas made; 3 — The reaction is fastest; 4 — The reaction has stopped.

Model answer. The gradient (steepness) shows the rate. It is steepest at the start, when there is the most reactant. It gets less steep as a reactant is used up, and becomes flat when one runs out. The height of the flat part is how much gas was made in total.

A full answer includes

I matched 1, steep at the start, to the reaction being fastest.
I matched 2, less steep, to the reaction slowing down as a reactant is used up.
I matched 3, flat, to the reaction having stopped.
I matched 4, the height of the flat part, to the total volume of gas made.
I did not match the flat part to a reaction going on steadily.

Also accept

Steep: fastest; less steep: slowing; flat: stopped; height of flat part: total gas

Watch for

“A flat line means the reaction is going steadily.” — A flat line means the volume is not changing: no more gas is being made.
“The height of the flat part shows how fast the reaction was.” — The height shows how much gas was made in total. How fast is shown by the steepness.
“A less steep curve means the reaction has stopped.” — While the curve still rises, gas is still being made, only more slowly. Stopped means flat.

4. UQ19 · Practice

Options as numbered on the pupil sheet: 1 — Too much gas for the syringe to hold; 2 — Gas escaped before the bung went in; 3 — Acid spray left the flask; 4 — The gas is too light to weigh; 5 — The plunger stuck, then jumped.

Model answer. 1: the volume stays at 22 cm³ for three readings while gas is still being made, then jumps to 41 cm³: the plunger stuck, then jumped. 2: the syringe is full at 100 cm³ while gas is still coming: the reaction makes too much gas, so smaller amounts of reactants are needed. 3: hydrogen is so light that the gas made weighs almost nothing, so the gas should be collected instead. 4: drops of acid left the flask and their mass counted as lost; a plug of cotton wool stops this. Gas lost before the bung goes in would make every syringe reading a little low, which none of these results shows.

A full answer includes

I matched 1 to the plunger sticking, then jumping.
I matched 2 to too much gas for the syringe to hold.
I matched 3 to a gas too light to weigh.
I matched 4 to acid spray leaving the flask.
I did not match 1 to gas escaping before the bung went in.

Also accept

1 plunger stuck; 2 too much gas for the syringe; 3 gas too light; 4 acid spray

Watch for

“Readings that stay the same mean the reaction stopped for a while.” — A reaction slows down smoothly; it does not stop and start again. A stuck plunger stays put, then jumps when the gas pushes it free.
“Gas lost before the bung goes in makes the readings jump.” — Gas lost at the start makes every reading a little low, but the curve stays smooth.
“A balance can follow any gas.” — Only a gas heavy enough to change the reading, such as carbon dioxide. A little hydrogen weighs almost nothing.
“Extra mass lost means more gas was made.” — The amount of gas is fixed by the reactants. Mass lost beyond that is something else leaving the flask: acid spray.

5. UQ4 · Practice

Model answer. The powder has a much larger surface area, so it reacts faster: its curve is steeper at the start. The acid is in excess, so the magnesium is the limiting reactant. There is 0.05 g of magnesium both times, so the same 49 cm³ of hydrogen is made; the powder gets there sooner.

A full answer includes

I chose: steeper at the start; it levels off at 49 cm³, sooner.
I used the larger surface area of the powder for the steeper start.
I used the same mass of magnesium, the limiting reactant, for the same 49 cm³.
I did not choose a higher final volume because the powder is faster.

Also accept

Steeper, levels off at the same 49 cm³, sooner

Watch for

“A faster reaction makes more product.” — A faster reaction makes the same product sooner. How much is made in the end depends on how much of the limiting reactant there is.
“A powder changes only how long the reaction lasts.” — It ends sooner because it is faster all the way, and that shows first at the start: the curve is steeper.
“A powder is slower, but makes more gas.” — A powder has more surface for the acid to reach, so it is faster. The mass of magnesium, and so the volume of gas, is the same.

6. UQ5 · Practice

Model answer. A is steeper at the start, so it is faster. A is flat from about 90 s at 60 cm³. B has not finished: it is still rising slowly at 180 s and is heading for the same 60 cm³. Being lower part-way through does not mean it will make less gas.

A full answer includes

I chose: A is faster at the start; B is still going and is heading for the same volume.
I used the steeper start of curve A to show that A is faster.
I saw that curve B is still rising at 180 s.
I did not choose that B makes less gas because its curve is lower.

Also accept

A faster; B heading for the same volume

Watch for

“B is faster, because it keeps going for longer.” — Taking longer to make the gas means slower, not faster. Compare the steepness at the start: A is steeper.
“A makes more gas in the end, because it is higher at 60 s.” — Being higher part-way shows that A is faster. How much gas is made in the end is the height where the curve goes flat.
“B makes less gas in the end, because its curve is lower.” — Look at where B is heading, not where it is part-way through. It is still rising at 180 s, towards the same 60 cm³.

7. UQ6 · Practice

Model answer. (a) 124.85 − 123.94 = 0.91 g. (b) 0.91 g ÷ 40 s = 0.02275 g/s, which is 0.023 g/s to 2 significant figures. (c) The reaction makes carbon dioxide gas, which escapes through the cotton wool, so the flask loses mass. The cotton wool lets the gas out but stops acid spray escaping too.

Checked automatically

(a) Mass lost in 40 s: 0.91 g (exact value)
- Common wrong answer 1.38 : That is the mass lost by the end. Use the reading at 40 s.
(b) Rate over 40 s: 0.023 g/s (within ±5%)
- Common wrong answer 44 : That is time ÷ mass. Divide the mass by the time: 0.91 ÷ 40.
- Common wrong answer 0.02 : Right method: 0.91 ÷ 40 = 0.02275. To 2 significant figures that is 0.023.
- Common wrong answer 0.0345 : That is the total mass lost ÷ 40 s. Use the mass lost in the first 40 s, 0.91 g.

A full answer includes

(a) I worked out 124.85 − 123.94.
(a) I got 0.91 g.
(b) I worked out 0.91 g ÷ 40 s = 0.02275 g/s.
(b) I gave 0.023 g/s, to 2 significant figures.
(b) I did not round to 0.02 g/s, which is only 1 significant figure.
(c) I wrote that the reaction makes carbon dioxide gas.
(c) I wrote that the gas escapes from the flask through the cotton wool.
(c) I did not write that the magnesium carbonate is destroyed.

Also accept

0.91 g; 0.023 g/s; the carbon dioxide escapes

Watch for

“The mass goes down because the magnesium carbonate is destroyed.” — No atoms are destroyed. Some of them leave the flask in the carbon dioxide gas.
“Rate = time ÷ mass.” — That gives seconds for each gram. Divide the mass lost by the time.

8. UQ7 · Practice

Model answer. Iron is in excess, so the acid is the limiting reactant. 15 cm³ of acid that is twice as concentrated holds the same amount of acid as 30 cm³ of the first acid, so the same volume of hydrogen is made in the end. The more concentrated acid reacts faster, so the curve is steeper at the start. Row A.

A full answer includes

I chose row A.
I wrote that the more concentrated acid makes the start steeper.
I wrote that half the volume at twice the concentration holds the same amount of acid.
I used the acid as the limiting reactant, so the final volume is the same.
I did not choose row B, twice as much gas.

Also accept

Row A

Watch for

“Row B.” — The acid is twice as concentrated, but there is only half the volume: the same amount of acid. The acid is the limiting reactant, so the volume of gas is the same.
“Row C.” — Half the volume of the same acid would make half the gas. Here each cm³ holds twice as much acid, so the total amount of acid is the same.
“Row D.” — The same amount of acid does not mean the same rate. Each cm³ holds more acid, so more of it reacts each second: the start is steeper.

9. UT1 · Apply

Model answer. (a) A gas syringe collects and measures the hydrogen; the readings are taken at set times from a stopclock started as the magnesium is added. (b) When the volume stays the same, no more gas is made: the curve is flat. (c) The same volume of hydrogen has a much smaller mass than carbon dioxide, so the balance reading hardly changes. (d) Gas that escapes before the bung is in, or through a leak, is never measured, and a sticking plunger moves less than it should.

A full answer includes

(a) I named a gas syringe joined to the flask by a delivery tube.
(a) I wrote that she starts a stopclock as the magnesium goes in.
(a) I wrote that she reads the volume every 10 s.
(b) I wrote that the curve becomes flat.
(b) I wrote that flat means the volume stops changing: no more gas is made.
(c) I wrote that hydrogen is very light.
(c) I wrote that the mass lost is too small for the balance to show.
(c) I did not write that hydrogen is made too slowly.
(d) I gave one: gas escaping before the bung goes in, a leak at the bung or tube, or a sticking plunger.

Also accept

Gas syringe and stopclock; flat curve; hydrogen too light; gas escapes or plunger sticks

Watch for

“Any reaction that makes a gas can be followed by loss of mass.” — Only if the gas is heavy enough for the balance to show the change, like carbon dioxide.