Rate Determination Practical

CIE A-Level Chemistry 9701 · Thiosulfate Clock Reaction

In this experiment, you investigate how the rate of reaction between sodium thiosulfate and hydrochloric acid depends on the concentration of thiosulfate ions. The reaction produces a fine suspension of sulfur, which gradually turns the solution cloudy. By timing how long it takes for a cross beneath the flask to disappear, you obtain a measure of the reaction rate (1/t) at each concentration. You then plot log(1/t) against log(volume) to find the order of reaction from the gradient.

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01 — the chemistry
The reaction
S₂O₃²⁻(aq) + 2H⁺(aq) → S(s) + SO₂(g) + H₂O(l)
Thiosulfate ions decompose in acid to form a fine suspension of sulfur. The sulfur makes the solution cloudy — the reaction is timed until the cross beneath the flask is just no longer visible.
Why use 1/t as a measure of rate?
A faster reaction produces the fixed amount of sulfur in less time. So rate ∝ 1/t — the shorter the time, the faster the rate.
Key pointYou are not measuring the actual rate in mol dm⁻³ s⁻¹ — you are using 1/t as a proportional measure. This is valid as long as the endpoint (same amount of sulfur) is consistent across all experiments.
Why use volume as a measure of concentration?
Concentration ∝ volume of solution used, provided the total volume is kept constant in every experiment. This is why you add water to make the total up to 55 cm³ each time.
CriticalTotal volume must be exactly the same in every experiment. Only ONE variable (e.g. volume of FA 1) should change at a time. All other volumes must be adjusted accordingly.
Why plot log(1/t) vs log(V)?
The rate equation is rate = k[S₂O₃²⁻]ⁿ, which gives 1/t = k·Vⁿ. Taking logs of both sides:
log(1/t) = n·log(V) + log(k)
This is y = mx + c — a straight line where the gradient = n = order of reaction. Plotting 1/t vs V directly gives a curve, which is much harder to interpret.
Plot Graph shape Can find order?
1/t vs V Curve ✕ Not easily
log(1/t) vs log(V) Straight line ✓ Gradient = order
Investigating H⁺ concentration
To test the effect of acid concentration, keep the volume of FA 1 constant (same thiosulfate concentration) and change the volume of FA 2 (acid) only.
FA 1 volumeconstant — same as Expt 2
FA 2 volumechanges between experiments
Water top-up55 − V(FA1) − V(FA2)
Expected resultLittle change in reaction time as acid concentration is doubled — rate is approximately zero order with respect to H⁺.
02 — results table (headings & layout)
✕ Wrong headings
Vol FA1 Water T 1/T log V
50 5 35 0.029 1.70
10 45 180 0.006 1.00
✕
No units — every heading needs a unit. "Vol FA1" should be "Volume of FA 1 / cm³".
✕
"T" for time — must use lowercase t. Capital T conventionally means temperature.
✕
1/T has no unit — should be "1/t / s⁻¹" or "log(1/t)".
✕
Time not to nearest second — 35 and 180 are fine here, but values like 35.4 suggest false precision from a stopwatch.
✓ Correct headings
Volume of FA 1 / cm³ Volume of water / cm³ t / s 1/t / s⁻¹ log(V / cm³) log(1/t / s⁻¹)
50 5 35 0.0286 1.699 −1.544
10 45 180 0.00556 1.000 −2.255
✓
Full column names with /unit notation. Lowercase t for time.
✓
1/t given as s⁻¹. Log values given to 3 dp.
✓
Table drawn in advance with all columns before experiments begin.
Choosing your concentration range
You need a minimum of 5 experiments total (Expts 1 & 2 plus at least 3 additional). Volumes of FA 1 should be spread evenly across the full range.
✕ Bunched volumes — loses graph mark
Experiment Volume of FA 1 / cm³
1 50
2 10
3 40
4 42
5 46
✕
Experiments 3–5 are bunched between 40 and 46. Points on the graph will be too close together — the line spans less than 20×20 small squares and the gradient mark is lost.
✓ Evenly spread volumes
Experiment Volume of FA 1 / cm³
1 50
2 10
3 20
4 30
5 40
✓
Volumes spread evenly across 10–50 cm³. Points on the log–log graph will span the full axis range, giving a reliable gradient.
The Vt accuracy check
If the reaction is first order, then concentration × time = constant. Since volume ∝ concentration, V × t should be approximately constant across experiments. The examiner checks Vt for Expt 1 and your two largest-volume additional experiments.
Vt difference Marks awarded
Both within 10% of the larger of the closest pair 2 marks ✓
Difference between 10% and 20% 1 mark
Difference greater than 20% 0 marks ✕
Worked example
V = 50, t = 35 sVt = 1750
V = 40, t = 55 sVt = 2200
V = 30, t = 69 sVt = 2070
Closest pair: 2200 and 2070. Difference = 130. 10% of 2200 = 220. Since 130 < 220 → full marks ✓
03 — graph & gradient
✕ Wrong
Wrong graph: axes swapped, bunched points, dot-to-dot, forced through origin Log-log graph with common mistakes: axes swapped, points bunched in top-right, dots connected directly, and line forced through origin log(1/t) log(V) 1.5 1.0 0.5 0 bunched! axes swapped forced through 0
✕
Axes swapped — log(V) must be on the x-axis, log(1/t) on the y-axis.
✕
Points bunched — cover less than half the axis range.
✕
Dot-to-dot — points joined directly instead of a single best-fit line.
✕
Forced through origin — line drawn to (0,0) regardless of data.
✓ Correct
Correct graph: log(1/t) on y-axis, log(V) on x-axis, evenly spread points, best-fit line, construction triangle A correctly drawn log-log graph with five evenly spread data points, a best-fit straight line, and a large construction triangle for the gradient log(V / cm³) log(1/t / s⁻¹) 1.6 1.4 1.2 1.0 -2.5 -2.0 -1.7 -1.5 -1.2 Δx Δy grad ≈ 1.0
✓
Correct axes — log(1/t) on y, log(V) on x, with units.
✓
Evenly spread — five points spanning the full axis range.
✓
Best-fit line — single straight line through the data.
✓
Construction triangle — large Δx and Δy for accurate gradient.
✕ Wrong gradient method
✕
Using two data points — always use two points on the best-fit line, not original data points.
✕
Small triangle — construction lines must span at least half the length of the line. A small triangle gives an inaccurate gradient.
✕
Answer given as integer — gradient must be given to at least 1 decimal place (e.g. 1.0, not just 1).
✓ How to draw construction lines
How to draw construction lines for gradient calculation on a log-log graph Step-by-step illustration showing two points chosen far apart on the best-fit line, horizontal and vertical construction lines drawn to form a right-angle triangle, and the gradient calculated as delta-y over delta-x log(V / cm³) log(1/t / s⁻¹) 1.0 1.2 1.4 1.6 1.8 1.0 1.2 1.4 1.6 1.8 2.0 Δx = 1.08 Δy = 1.08 ① pick point on line (1.02, −2.10) ② pick point on line (2.10, −1.02) gradient calculation Δx = 2.10 − 1.02 = 1.08 Δy = −1.02 − (−2.10) = 1.08 grad = Δy/Δx = 1.08/1.08 = 1.0 order of reaction = 1.0 ✓ ③ draw horizontal line from ① ④ draw vertical line from ② ⑤ read off Δx and Δy
1
Choose two points on the best-fit line (not data points), as far apart as possible — ideally spanning more than half the line length.
2
From the left point, draw a horizontal dashed line to the right. From the right point, draw a vertical dashed line downward. They meet at a right angle.
3
Read Δx and Δy from the axis scale — not by measuring the page in cm. Calculate gradient = Δy / Δx and give to at least 1 decimal place.
Worked example
Point 1 on linelog V = 1.02, log(1/t) = −2.10
Point 2 on linelog V = 2.10, log(1/t) = −1.02
Δx2.10 − 1.02 = 1.08
Δy−1.02 − (−2.10) = 1.08
Gradient = order1.08 / 1.08 = 1.0 → first order ✓
04 — uncertainty & percentage error
Source of uncertainty in this experiment
The endpoint — when the cross just disappears — is a judgment call. Different observers may stop the clock at slightly different moments, introducing a human reaction time uncertainty.
✕ Common mistake
✕
Same uncertainty for both experiments: Writing "uncertainty = ±3 s for all experiments." The uncertainty in seconds may be similar, but the percentage uncertainty is very different.
✕
Saying Expt 1 has larger uncertainty — Expt 1 (large volume, fast reaction) has a smaller percentage uncertainty, not larger.
✓ Correct analysis
Experiment Reaction speed Absolute uncertainty % uncertainty
Expt 1 (large V) Fast — clear endpoint ≤ 3 s Smaller
Expt 2 (small V) Slow — gradual endpoint ≥ 3 s Larger
% uncertainty = (uncertainty in seconds / reaction time) × 100
WhyExpt 2 has a longer reaction time (e.g. 180 s) so even ±5 s gives only ~2.8%. But if that same ±5 s applies to Expt 1 (t = 35 s), it would be ~14%. The key point is that Expt 2 has a larger percentage uncertainty because the change in cloudiness is more gradual and harder to judge.
05 — quick reference checklist
Before and during experiments
  • Total volume is the same in every experiment (e.g. 55 cm³)
  • Table drawn in advance with correct headings and units
  • Lowercase t used for time, not T
  • Times recorded to the nearest second
  • At least 3 extra experiments beyond Expts 1 & 2
  • Volumes of FA 1 spread evenly across the full range
Graph and gradient
  • log(1/t) on y-axis, log(V) on x-axis
  • Points cover more than half of each axis
  • All experiments plotted (including Expts 1 & 2)
  • Single best-fit straight line drawn, not dot-to-dot
  • Construction lines span at least half the line length
  • Gradient calculated to ≥ 1 decimal place
Uncertainty and H⁺ investigation
  • Uncertainty for Expt 2 stated as larger than for Expt 1
  • For H⁺ investigation: only acid volume changes, FA 1 volume constant
  • Water adjusted so total volume stays at 55 cm³
  • Expected conclusion: rate approximately zero order with respect to H⁺