Required practical 3 - Temperature and Reaction Rate

Required practical 3 - Temperature and Reaction Rate

In this required practical, you do not track concentration continuously with an instrument. Instead, you time how long it takes for a fixed amount of sulfur to form as sodium thiosulfate reacts with hydrochloric acid. By keeping everything except temperature the same, you can use that time to compare rates and explain, in particle terms, why warmer mixtures react faster.

What the disappearing cross measures

The reaction used in this practical is:

NaX2SX2OX3(aq)+2 HCl(aq)→2 NaCl(aq)+SOX2(g)+S(s)+HX2O(l)\ce{Na2S2O3(aq) + 2HCl(aq) -> 2NaCl(aq) + SO2(g) + S(s) + H2O(l)}

The key observation is the formation of solid sulfur. As sulfur particles appear, the mixture turns cloudy. If the reaction vessel is placed over a black cross, there comes a point when the cloudiness is great enough that the cross can no longer be seen from above. That visible end-point gives us a practical way to compare how quickly sulfur is being produced.

Initial rate

The initial rate is the rate of reaction measured right at the start, when reactant concentrations have changed very little from their starting values.

Strictly speaking, this practical does not measure the true initial rate directly. What it does is compare the time taken to reach the same visible end-point each time. Because the same depth of solution and the same visual standard are used, a shorter time means sulfur is being produced more quickly, so the reaction rate is greater.

Temperature is the independent variable. When temperature increases, particles in both solutions have more kinetic energy. That means collisions happen with greater energy, and a larger fraction of collisions have energy greater than or equal to the activation energy. The result is more successful collisions per second, so sulfur appears more quickly and the cross disappears sooner.

The diagram below shows how the disappearing-cross end-point is judged and why a shorter time means a faster reaction.

[DIAGRAM: asset_name: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 1; asset_slug: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 1; recommended_method: deterministic_code; description: Top views compare the visible external cross with the just-obscured endpoint through the same depth of reacting solution.]
Diagram

Running the method fairly

To make this a fair test, you prepare the same reaction mixture each time and change only the temperature. A typical setup uses a measured volume of sodium thiosulfate solution in a tube placed over the cross, and a measured volume of hydrochloric acid in a second tube. Both tubes are placed in a warm water bath so they reach the chosen starting temperature before they are mixed.

Using a water bath matters because it heats the solutions more gently and evenly than a direct flame. It also matches the apparatus-and-technique requirement to use suitable heating equipment while still keeping the practical safe and controlled.

Control variable

A control variable is a factor kept the same so that any change in the result can be linked confidently to the independent variable.

In this experiment, the important control variables are the concentrations and volumes of both reactants, the total depth of liquid above the cross, the same viewing position, and the same end-point for "cross disappears". If one run used a deeper solution or a different volume, the optical path length and endpoint amount of sulfur would change even if the chemistry had not changed, so the comparison would be unfair.

A careful sequence is:

  1. Measure the sodium thiosulfate solution into the tube above the cross.
  2. Measure the hydrochloric acid into a separate tube.
  3. Place both tubes in the water bath and allow them to reach the chosen temperature.
  4. Record the starting temperature.
  5. Mix the reactants, start the timer immediately, and look straight down from above.
  6. Stop the timer when the cross just disappears.
  7. Record the final temperature and use the mean of the initial and final readings as the best estimate of the reaction temperature.
  8. Repeat at several temperatures and repeat runs when needed.

The diagram below shows a suitable water-bath setup for carrying out the method fairly and recording the timing correctly.

[DIAGRAM: asset_name: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 2; asset_slug: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 2; recommended_method: deterministic_code; description: A water bath covers the initial solution levels; mixing and timing then occur at an endpoint station with an external cross beneath the flask.]
Diagram
Measuring both initial and final temperature improves the method because the reaction mixture may warm or cool slightly during mixing and reaction. The mean temperature is only an estimate when drift is small. If the initial and final temperatures differ substantially, improve thermal control and repeat; a mean cannot turn a changing-temperature reaction into an isothermal one.

Safety is part of the chemistry, not an afterthought. Hydrochloric acid is irritant or corrosive depending on concentration, the reaction releases toxic sulfur dioxide, and hot water can cause burns. Wear goggles and use small quantities in good ventilation or a fume cupboard under the laboratory risk assessment. Do not seal a gas-producing mixture. Dispose of the mixture promptly using the teacher-provided alkaline stop bath or specified waste route to limit continued sulfur dioxide release.

Turning time into rate data

The clever part of this practical is the data handling. Each run is stopped when the same visible event happens: enough sulfur has formed to hide the cross. That means the amount of sulfur at the end-point is treated as fixed, so the rate is proportional to:

Rate proxy for the disappearing cross method

rate∝1t\text{rate} \propto \frac{1}{t}

Here, tt is the time in seconds for the cross to disappear. A smaller tt gives a larger value of 1/t1/t, which represents a faster reaction.

Worked example: suppose the cross disappears in 80 s80\,\mathrm{s} at 25∘C25 ^\circ\mathrm{C} and in 40 s40\,\mathrm{s} at 35∘C35 ^\circ\mathrm{C}.

  • at 25∘C25 ^\circ\mathrm{C}, 1/t=1/80=0.0125 s−11/t = 1/80 = 0.0125\,\mathrm{s}^{-1}
  • at 35∘C35 ^\circ\mathrm{C}, 1/t=1/40=0.0250 s−11/t = 1/40 = 0.0250\,\mathrm{s}^{-1}

The relative percentage increase is 100[(1/t2)/(1/t1)−1]100[(1/t_2)/(1/t_1)-1]; here it is 100%. The second value is twice as large, so the reaction is proceeding about twice as fast at the higher temperature under those conditions.

This is why tables and graphs for this practical often use either time or 1/t1/t. If you plot time against temperature, the graph usually slopes downward because time gets shorter. If you plot 1/t1/t against temperature, the graph rises because the rate proxy increases.

The diagram below compares the two common ways of plotting the results from this practical.

[DIAGRAM: asset_name: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 3; asset_slug: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 3; recommended_method: deterministic_code; description: Illustrative measured endpoint times and their reciprocals plotted against temperature; no unsupported extrapolation to zero rate at zero Celsius.]
Diagram
One subtle but important point is that 1/t1/t is an approximation, not a perfect rate measurement. The disappearing cross method assumes that the cross disappears when a fixed amount of sulfur has formed and that this threshold is judged consistently. That is good enough for school kinetics work, but it also explains why repeat results and careful technique matter.

Reliability, errors, and higher-level analysis

The largest uncertainty in this practical is usually the end-point. Different students may decide the cross has disappeared at slightly different moments, and even the same student may not judge it identically every time. There is also a short delay between mixing the reactants and starting the timer, and the temperature may drift if the tubes are not left in the water bath for long enough before mixing.

Simply saying "human error" does not tell you much. The real value comes from identifying the source of uncertainty and explaining how to reduce it. Using the same observer throughout improves consistency. Viewing from directly above keeps the path length through the liquid the same. Repeating each temperature and calculating a mean reduces the effect of random error. A thermostatically controlled water bath is better than estimating temperature with hot tap water because it holds the temperature steady. Using a light sensor or colorimeter would reduce subjective judgement even further, although the classic required practical uses the visible cross.

Record repeated times in a table with units in headings. Inspect an anomalous result for a specific cause, repeat it where possible and explain any exclusion; do not silently discard a time just because it weakens the trend. If the endpoint timing uncertainty is ±1 s, it contributes about 2% at 50 s but 10% at 10 s. Very fast runs therefore make mixing and judgement delays a larger fraction of the measured time.

Activation energy

Activation energy is the minimum energy that colliding particles must have for a reaction to occur.

This definition explains the temperature effect at particle level. A higher temperature does not lower the activation energy. Instead, it increases the fraction of particles with energy greater than or equal to that activation energy, so there are more successful collisions each second.

An A-level extension links these results to the Arrhenius equation taught in section 3.1.9. If 1/t1/t is treated as proportional to the rate constant kk, then plotting ln⁡(1/t)\ln(1/t) against 1/T1/T in kelvin gives a straight line with gradient −Ea/R-E_a/R, where R=8.31 J mol−1 K−1R = 8.31\,\mathrm{J}\,\mathrm{mol}^{-1}\, \mathrm{K}^{-1}. Calculate T=θ+273.15T=\theta+273.15 before taking the reciprocal. A gradient of −5000 K gives Ea=−R×gradient=41550 J mol−1=41.6 kJ mol−1E_a=-R\times\text{gradient}=41550\,\mathrm{J\,mol^{-1}}=41.6\,\mathrm{kJ\,mol^{-1}}. Use two well-separated points on the best-fit line to find its gradient, not two neighbouring raw data points. This estimate assumes that the fixed endpoint and the other concentrations make 1/t proportional to k across the temperature range.

The diagram below shows the straight-line plot used in the higher-level analysis of this practical.

[DIAGRAM: asset_name: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 4; asset_slug: RP 03 - Investigation of How the Rate of a Reaction Changes with Temperature - Diagram 4; recommended_method: deterministic_code; description: A natural-log reciprocal-time plot against reciprocal kelvin temperature has gradient −5000 K and gives an activation energy of 41.6 kJ per mole.]
Diagram
The core practical conclusion is simpler than that extension: when temperature increases, the time for the cross to disappear decreases, so the rate increases. A sound interpretation connects that observation to a fair method, clear control of variables, sensible safety measures, and data handling that uses 1/t1/t carefully rather than blindly.

Keep a dated results record with units, apparatus precision, observations and any justified change to the method. Present processed results in a clearly labelled table or graph, explain the uncertainty and cite the method or reference data used. These records support practical assessment; reading a method alone does not demonstrate competent practical performance.