Rates of reaction — return questions key

1. UR1 · Return

Model answer. Row C: 0.50 g of carbon dioxide is a change a balance reading to 0.01 g follows easily. Row A: 50 cm³ of hydrogen weighs only about 0.004 g, so the balance hardly changes; collect the gas instead. Row B: 270 cm³ overfills a 100 cm³ syringe; use smaller amounts of reactants. Row D: a balance that reads to 1 g cannot show changes within 0.50 g.

A full answer includes

• I chose row C.
• I rejected row A: 0.004 g of hydrogen hardly changes a balance.
• I rejected row B: 270 cm³ of gas will not fit in a 100 cm³ syringe.
• I rejected row D: a balance that reads to 1 g cannot follow a change of 0.50 g.

Also accept

• Row C

Watch for

• “Row A.” — A balance can follow a gas only if its mass is big enough to show. 0.004 g of hydrogen hardly moves a balance that reads to 0.01 g.
• “Row B.” — A gas syringe is a good method, but this one holds only 100 cm³. 270 cm³ of gas would push the plunger to the end long before the reaction stops.
• “Row D.” — Loss of mass works for carbon dioxide, but a balance that reads to 1 g would show 0 g lost for most of the reaction. It needs to read to 0.01 g.

2. UR2 · Return

Model answer. (a) Volume = rate × time = 1.4 cm³/s × 35 s = 49 cm³. (b) Time = volume ÷ rate = 66 cm³ ÷ 1.2 cm³/s = 55 s.

Checked automatically

• (a) Volume in the first 35 s: 49 cm³ (within ±0.5 cm³)
  • Common wrong answer 25 : That is time ÷ rate. Volume = rate × time: 1.4 × 35.
  • Common wrong answer 0.04 : That is rate ÷ time. Multiply instead: 1.4 × 35.
• (b) Time taken: 55 s (within ±0.5 s)
  • Common wrong answer 79.2 : That is volume × rate. Time = volume ÷ rate: 66 ÷ 1.2.
  • Common wrong answer 0.018 : That is rate ÷ volume. Divide the volume by the rate: 66 ÷ 1.2.

A full answer includes

• (a) I worked out 1.4 cm³/s × 35 s.
• (a) I got 49 cm³.
• (b) I worked out 66 cm³ ÷ 1.2 cm³/s.
• (b) I got 55 s.
• (b) I did not multiply 66 by 1.2.

Also accept

• 49 cm³; 55 s

Watch for

• “Volume = time ÷ rate.” — Check the units: s ÷ cm³/s does not give cm³. Volume = rate × time.
• “Time = volume × rate.” — Check the units: cm³ × cm³/s does not give seconds. Time = volume ÷ rate.

3. UR3 · Return

Model answer. (a) A catalyst speeds up a reaction and is unchanged at the end. The platinum keeps working for years, so it is not used up. (b) The platinum lowers the activation energy, so more of the collisions between the gas particles have enough energy to react. (c) Hot gas particles move faster: they collide more often, and a much larger share of their collisions have at least the activation energy. The second reason matters more. (d) The same platinum spread thinly over the honeycomb has a much larger surface area, so the gas particles reach and collide with it more often.

A full answer includes

• (a) I wrote that the platinum speeds up the reactions.
• (a) I wrote that it is not used up: the same platinum still works after years.
• (b) I wrote that the platinum gives a route with a lower activation energy.
• (b) I wrote that a larger share of the collisions have at least Ea, so more succeed.
• (b) I did not write that the platinum gives the particles energy.
• (c) I wrote that the particles move faster, so they collide more often.
• (c) I wrote that a larger share of the collisions have at least Ea.
• (c) I did not give "more frequent collisions" as my only reason.
• (d) I wrote that the platinum has a much larger surface area.
• (d) I wrote that the gas particles collide with the platinum more often.

Also accept

• Not used up; a lower Ea; more frequent and more energetic collisions; a larger surface area

Watch for

• “A catalyst is used up in the reaction.” — A catalyst is unchanged at the end: the same mass of the same substance, ready to speed up the reaction again.
• “Heating only makes the particles collide more often.” — They do collide more often, but the bigger reason is energy: a larger share of the collisions have at least the activation energy.
• “The platinum gives the gas particles extra energy.” — A catalyst does not add energy. It lowers the activation energy, so more of the collisions that happen have enough.

4. UR4 · Return

Model answer. More concentrated acid has more acid particles in the same volume of rain. They collide with the stone more often, so there are more successful collisions each second. The particles are no faster and the activation energy is the same.

A full answer includes

• I chose: more acid particles in the same volume, so they collide with the stone more often.
• I did not choose a lower activation energy.

Also accept

• More particles in the same volume; more frequent collisions

Watch for

• “The acid particles in more concentrated acid move faster.” — Concentration changes how many particles are in the same volume, not how fast they move. Temperature changes their speed.
• “More concentrated acid lowers the activation energy.” — Ea is the same for the same reaction; only a catalyst gives a route with a lower one. Here there are more acid particles to collide.
• “The particles of a more concentrated acid are bigger.” — The acid particles are the same size. There are just more of them in each cm³ of rain.