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Why more particles per litre react faster, reading a clock reaction's rate as one over a time, and keeping rate apart from total amount.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to explain, with the collision model, why raising a reactant's concentration or a gas's pressure increases the rate, and why it does not change the share of collisions that succeed. You will be able to turn clock-reaction times into rates, predict a new time when the rate is proportional to the concentration, and say whether a change to a reactant alters the rate, the total amount of product, both or neither.
Concentration is amount divided by volume, in mol/L, from Chemistry 1: 0.2 mol dissolved to make 500 mL is 0.4 mol/L. Last lesson gave the collision model — a rate depends on how often particles collide and on the share of collisions that succeed. Concentration is the first of four changes to put through that model.
Concentration is the amount of a substance per litre of solution. For a gas, partial pressure plays the same part: at a fixed temperature, doubling a gas's pressure doubles the number of its molecules in each litre. A clock reaction is one set up so that something sudden — a colour change, a cross disappearing — happens when a fixed amount of product has formed, so the time taken measures the rate. Proportional means that doubling one quantity doubles the other.
Put twice as many acid particles into every litre and each patch of a marble chip's surface is struck twice as often. In a solution where two dissolved substances react, doubling one of them doubles the number of times the other kind of particle meets it each second.
What concentration does not change is the other factor in the model. The share of collisions that succeed depends on the temperature, which sets the spread of particle energies, and on the activation energy, which belongs to the reaction. Neither has anything to do with how many particles are present.
So concentration is a frequency lever:
| collisions per second | share that succeed | rate | |
|---|---|---|---|
| higher concentration | up | unchanged | up |
| lower concentration | down | unchanged | down |
For many reactions the rate turns out to be proportional to a reactant's concentration: double it and the rate doubles. That is not guaranteed — some rates depend on the square of a concentration, some do not depend on a particular reactant at all — and finding out which is the job of the rate-equation lessons later in this unit. Every item in this lesson says when the rate is proportional.
For gases, pressure is concentration. Squeeze a gas into half the volume at the same temperature and there are twice as many molecules in each litre: the pressure has doubled and so has the concentration. Industrial gas reactions are run at high pressure partly for this reason.
Another way: picture
A crowded corridor between lessons. With twice as many people in the same corridor, you bump into someone twice as often — but each bump is no harder than before. Concentration changes how often, not how hard.
Another way: steps
To predict a concentration change:
In the thiosulfate clock reaction, sodium thiosulfate and hydrochloric acid slowly make a cloud of sulfur, and the flask stands on a paper cross. The cross disappears when a fixed amount of sulfur has formed. A faster reaction reaches that amount sooner, so
$$\text{rate} \propto \frac{1}{\text{time}}.$$
The thiosulfate solution is diluted with different volumes of water, always to the same total volume, so the thiosulfate volume is a direct measure of its concentration. Typical results:
| thiosulfate, mL | water, mL | time, s | 1/time, 1/s |
|---|---|---|---|
| 10 | 40 | 200 | 0.005 |
| 20 | 30 | 100 | 0.010 |
| 40 | 10 | 50 | 0.020 |
Each doubling of concentration halves the time and doubles the rate: the rate is proportional to the thiosulfate concentration. A time that goes down is a rate that goes up, which is the one thing to hold onto when reading clock data.
Magnesium ribbon in excess hydrochloric acid gives a certain volume of hydrogen, fixed by how much magnesium there is: the magnesium is the limiting reactant. Change the acid and ask two questions.
Twice the concentration — the rate goes up, because acid particles strike the metal twice as often. The total volume of hydrogen does not change: the magnesium still runs out, and it makes the same amount of hydrogen whatever acid it meets.
Twice the volume of the same acid — nothing changes at all. The concentration, and so the collision frequency at the metal's surface, is exactly what it was; the extra acid simply sits in the flask unused.
A more concentrated acid produces more gas only when the acid itself is the limiting reactant. That is a question about amounts, answered with moles, and it is a different question from how fast.
Concentration lowers the activation energy. The activation energy is a property of the reaction's route; the number of particles present has no effect on it. Concentration changes how often particles meet.
Particles move faster in a concentrated solution. Their speeds are set by the temperature. A concentrated solution at 20 °C has particles moving exactly as fast as a dilute one at 20 °C.
More of the same solution reacts faster. Pouring in a larger volume at the same concentration changes neither the particles per litre nor the collision frequency, so the rate is unchanged.
A longer clock time is a faster reaction. The time is how long the reaction took to reach a fixed point, so a longer time is a slower rate. The rate is one over the time.
With 15 mL of thiosulfate made up to 50 mL, the cross vanishes in 120 s. The rate is proportional to the thiosulfate concentration. Predict the time with 45 mL made up to 50 mL.
Both flasks have the same total volume, so the thiosulfate volume stands for its concentration.
$45 \div 15 = 3$, so the concentration is three times larger and the rate three times faster.
Proportional: the same factor for rate as for concentration.
Three times the rate reaches the same amount of sulfur in a third of the time: $120 \div 3 = 40$ s.
A faster rate means a shorter time, so the time is divided.
A mixture of gases reacts at 200 kPa and is compressed to 600 kPa at the same temperature.
Start with what happened to the gas: the same molecules in a third of the volume.
Three times the pressure at the same temperature is three times as many molecules per litre, so three times as many collisions per second.
For a gas, pressure stands in for concentration.
The temperature and activation energy are unchanged, so the share that succeed is the same, and the rate rises — by a factor of three if it is proportional to that concentration.
Frequency up, share unchanged, rate up.
$0.1$ is a third of $0.3$, so the rate is a third as fast.
Same factor for rate as for concentration.
A third of the rate takes three times as long: $90 \times 3 = \ldots$
…$270$ s, because a more dilute solution reacts more slowly.
Sodium thiosulfate reacts with hydrochloric acid to give a cloud of sulfur, and the time for a cross under the flask to vanish is recorded. Each flask holds the stated volume of thiosulfate solution, topped up with water to 50 mL, plus the same acid. Fill in the water added and the rate, taken as one divided by the time, in 1/s.
| thiosulfate solution, in mL | water added, in mL | time for the cross to vanish, in s | rate as one over the time, in 1/s | |
|---|---|---|---|---|
| flask A | 11 | 400 | ||
| flask B | 22 | 200 | ||
| flask C | 44 | 100 |
For a reaction whose rate is proportional to the concentration of thiosulfate, the cross vanishes in $100$ s. The thiosulfate is made $2$ times more concentrated and nothing else changes. How long will the cross take to vanish now? Give the time with its unit.
Answer: unit: h / s / min
Hydrochloric acid at $0.1$ mol/L reacts with a marble chip. The same chip in acid at $0.2$ mol/L, at the same temperature, fizzes faster. What is the reason, according to the collision model?
The initial rate of a reaction is measured at four concentrations. At 1, 2, 3 and 4 units of concentration (each unit is 0.1 mol/L) the initial rates are $2$, $4$, $6$ and $8$ units (each unit is 0.001 mol/(L s)). Plot the four measurements, concentration across and rate up.
Plot your answer on the grid:
A $2$ cm strip of magnesium ribbon reacts with $40$ mL of hydrochloric acid. The acid is in excess, so the magnesium is the limiting reactant. For each change, say what happens to the initial rate and to the total volume of hydrogen collected.
| initial rate | total volume of hydrogen | |
|---|---|---|
| the acid is twice as concentrated | ||
| twice the volume of the same acid | ||
| the acid is diluted to half its concentration |
Four solutions of the same acid react with identical marble chips at the same temperature. Put them in order of initial rate, fastest first.
Number the steps in order (write the number in the box):
In an industrial reactor, a gas-phase reaction runs at a fixed temperature at $400$ kPa. Engineers compress the gas to $2000$ kPa. For this reaction the rate is proportional to the concentration of the gas. By what factor does the rate increase?
Answer: times
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In a clock reaction whose rate is proportional to the concentration of iodate, a solution at $0.25$ mol/L changes colour after $160$ s. What concentration of iodate would make it change colour after $80$ s, with everything else the same? Answer in mol/L.
Answer: mol/L
You can use concentration as a collision-frequency lever and read a clock reaction's rate as one over its time. Say out loud why pouring in twice as much of the same acid does not make magnesium react faster. Next: temperature, the lever that changes the share of collisions that succeed.
10. Your turn: a reaction whose rate is proportional to concentration takes 90 s at 0.3 mol/L. How long at 0.1 mol/L?, step 3