Topic 9 · page 1 of 15

Measuring rate with time

Rate is how fast an amount changes. When the same change is timed in every run, the rate is proportional to 1 ÷ time, and the thiosulfate and acid experiment gives you the numbers.

11 min read9.39.3(i)

What rate means 9.3

Some reactions are over in a flash and some take years. Rate puts a number on that. It says how quickly a reactant is used up or a product is made.

In symbols, for a reactant A that is being used up:

rate=−Δ[A]Δt \text{rate} = -\frac{\Delta[\text{A}]}{\Delta t}

The square brackets mean "concentration in mol dm⁻³". The minus sign is there because [A] falls, and a rate is always a positive number.

You rarely measure concentration directly. You measure something that changes as the reaction goes: the volume of gas, the mass of the flask, the colour, or the time for a fixed change to happen. The specification asks for two ways of getting a rate from such data:

Rate = 1 ÷ time 9.3(i)

Suppose you run the same reaction three times and time how long each takes to reach the same stage. The run with the shortest time is the fastest. Half the time means twice the speed. So the rate is proportional to 1 ÷ time:

rate=1timeunit: s−1 \text{rate} = \frac{1}{\text{time}} \qquad \text{unit: s}^{-1}

Two conditions must hold:

  1. The same amount of change must have happened when you stop the clock, in every run. Otherwise the times are not comparable.
  2. You are measuring the average rate from the start to the end point. If the end point comes early, before much reactant is used, this is close to the initial rate.

The result is a relative rate. Its unit is s⁻¹, not mol dm⁻³ s⁻¹, because you did not measure a concentration. It tells you how many times faster one run is than another, and that is usually all you need.

Worked exampleA rate from a time

A reaction reaches its end point after 25 s. Calculate the rate.

  1. rate = 1 ÷ time = 1 ÷ 25
  2. = 0.040 s⁻¹ (2 significant figures)

Worked exampleComparing two runs

The end point takes 40 s at 30 °C and 10 s at 50 °C. How many times faster is the reaction at 50 °C?

  1. Rate at 30 °C = 1 ÷ 40 = 0.025 s⁻¹
  2. Rate at 50 °C = 1 ÷ 10 = 0.100 s⁻¹
  3. 0.100 ÷ 0.025 = 4 times faster. You can also get this straight from the times: 40 ÷ 10 = 4.

The disappearing cross 9.3(i)

The classic experiment for this idea is the reaction of sodium thiosulfate with acid. A pale yellow solid, sulfur, slowly clouds the solution. You time how long it takes to hide a cross drawn under the flask.

SX2OX3X2−(aq)+2 HX+(aq)→S(s)+SOX2(g)+HX2O(l) \ce{S2O3^2-(aq) + 2H+(aq) -> S(s) + SO2(g) + H2O(l)}
The disappearing cross experiment. Left: side view of a conical flask holding sodium thiosulfate solution and acid, standing on paper with a cross on it, viewed from above. Right: three views from above as time passes. At the start the black cross is clear, part way it is faint behind a cloudy yellow precipitate of sulfur, and at the end point it is just hidden.
The end point is the same amount of sulfur every time: just enough to hide the cross.

PracticalDisappearing cross

  1. Draw a cross in black ink on a piece of paper and stand a conical flask on it.
  2. Measure a fixed volume of sodium thiosulfate solution into the flask (for example 50 cm³). Note its temperature.
  3. Add a small fixed volume of dilute hydrochloric acid (for example 5 cm³). Swirl once and start the clock.
  4. Look down through the solution from above. Stop the clock when the cross can no longer be seen.
  5. Repeat at different temperatures, warming the thiosulfate in a water bath first. Or repeat at different concentrations, diluting the thiosulfate with water so that the total volume stays the same.
  6. Work out rate = 1 ÷ time for each run.

Keep the same flask, the same cross, the same volumes, the same lighting and, if you can, the same observer. Sulfur dioxide is toxic and can trigger asthma: work in a ventilated room and pour the mixture away down a sink straight afterwards.

Illustrative results for a temperature series:

Temperature / °C Time / s Rate = 1 ÷ time / s⁻¹
20 80 0.0125
30 40 0.025
40 20 0.050
50 10 0.100

Each 10 °C rise halves the time and doubles the rate. This is typical of many reactions near room temperature, and page 7 explains why. These figures are illustrative.

For a concentration series, the same reasoning applies. Illustrative times of 25, 31, 42, 63 and 125 s for 50, 40, 30, 20 and 10 cm³ of thiosulfate (each made up to 50 cm³ with water) give rates of 0.040, 0.032, 0.024, 0.016 and 0.0080 s⁻¹. The rate is proportional to the concentration of thiosulfate in this experiment.

Check yourself

  1. Define the rate of a reaction.
  2. In a thiosulfate and acid experiment the cross disappears after 25 s. Calculate the rate, with its unit.
  3. At 20 °C the cross disappears after 80 s. At 40 °C it takes 20 s. How many times faster is the reaction at 40 °C?
  4. A student dilutes the thiosulfate by adding extra water to the flask, so the total volume is different in each run. Explain why 1 ÷ time is now unreliable.
  5. Give one limitation of the disappearing cross method and one way to improve it.
  6. Times for a concentration series, with total volume kept at 50 cm³, are 25, 31, 42, 63 and 125 s for 50, 40, 30, 20 and 10 cm³ of thiosulfate. Calculate each rate and state what they show.

Answers cover these until you have tried the questions

  1. The change in the concentration of a reactant or product in unit time, in mol dm⁻³ s⁻¹.
  2. rate = 1 ÷ 25 = 0.040 s⁻¹.
  3. Rates: 1 ÷ 80 = 0.0125 s⁻¹ and 1 ÷ 20 = 0.050 s⁻¹. 0.050 ÷ 0.0125 = 4 times faster (also 80 ÷ 20 = 4).
  4. The depth of the liquid above the cross differs, so the amount of sulfur needed to hide the cross is not the same each time. The end point no longer means the same amount of change, so the times cannot be compared.
  5. Limitation: the end point is judged by eye, so it is subjective (different observers or lighting give different times). Improvement: use a light sensor or colorimeter and stop the clock at a fixed reading.
  6. 1 ÷ 25 = 0.040; 1 ÷ 31 = 0.032; 1 ÷ 42 = 0.024; 1 ÷ 63 = 0.016; 1 ÷ 125 = 0.0080 s⁻¹. The rate is proportional to the concentration of thiosulfate: 50 cm³ gives 0.040 and 10 cm³ gives 0.0080, which is five times less for one fifth of the concentration.