The method and the results are on the rate law lab. Do the analysis yourself first, then check the answers below.
Quick answer: A rate law, rate = k[A]^m[B]^n, shows how reaction rate depends on reactant concentrations, with orders found by experiment. These answers go with the AP Chemistry lab on rate law: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
The model answers use the sample table. The 1/t values rise in proportion to thiosulfate concentration, so the order is one. Percentage uncertainty is largest for fast runs, which affects which points you trust most in the graph. Remember to give each calculated value to a sensible number of significant figures.
Rate depends on the concentration of every reactant. If total volume or acid volume changed, the acid concentration would change too, so the effect of thiosulfate would be mixed with the effect of acid. Keeping them fixed means thiosulfate is the only variable.
Question 2 Mixed concentration
[S₂O₃²⁻] = V × 0.100/55. 50 cm³: 0.0909 M; 40: 0.0727 M; 30: 0.0545 M; 20: 0.0364 M; 10: 0.0182 M.
Question 3 Relative rate
1/t = 0.0417, 0.0333, 0.0244, 0.0161, 0.00800 s⁻¹. Each run forms the same small amount of sulfur (the same cloudiness hides the cross), so rate ∝ amount/time ∝ 1/t. 1/t ∝ initial rate.
Question 4 Graph
Plot 1/t on the y-axis and concentration on the x-axis. For first order the points lie on a straight line through the origin, because rate ∝ [S₂O₃²⁻]. A curve bending upwards would suggest order 2.
Question 5 Order from two points
Concentration ratio 0.0727/0.0364 = 2.0. Rate ratio (1/30)/(1/62) = 62/30 = 2.07. Rate ∝ [ ]n: 2.0n = 2.07, so n ≈ 1 (first order).
Question 6 Uncertainty
24 s: 1/24 = 4 %. 125 s: 1/125 = 0.8 %. The slow runs are measured with a smaller relative error; the 50 cm³ point has the largest uncertainty.
Question 7 Improvement
Use a light sensor or a standard viewing distance so the end-point does not depend on the observer; use a burette for volumes; repeat each run and average. At 10 °C higher the times would be about half, since the rate typically doubles for a 10 °C rise.
