Challenge competing concentration–time laws

Try: one halving does not identify the order

The original synthetic packet starts with two exact anchors: 0.100 mol/L at time zero and 0.050 mol/L at 10 s. All three candidate laws match. Before opening a later packet, predict which concentration at 20 s would distinguish them. The two anchors calibrate each law; the later public observations are checks and do not refit its constant.

Understand three different continuations

Write C for concentration in mol/L, C₀ for its initial value, t for elapsed seconds and k for a constant under the stated conditions. Here A → P has coefficient one, so disappearance rate and reaction rate coincide. The candidates assume −dC/dt = kCⁿ with order n equal to 0, 1 or 2.

• Zero order removes the same amount per second: C = C₀ − kt. The anchors give k = 0.005 mol L⁻¹ s⁻¹. At 20 s this law reaches zero; its negative continuation is physically invalid.
• First order removes the same fraction in equal times: C = C₀ exp(−kt). The anchors give k = ln(2)/10 ≈ 0.0693147 s⁻¹. At 20 s, C = 0.025 mol/L.
• Second order has 1/C = 1/C₀ + kt. The anchors give k = 1 L mol⁻¹ s⁻¹. At 20 s, C = 1/30 ≈ 0.0333333 mol/L.

These k values have different units and cannot be ranked as comparable speeds merely by their numbers. The graph uses common concentration and time axes to compare the actual predictions.

Experiment: observations and their bounds

The equal-fraction packet selects first order from this family; the reciprocal and equal-amount packets discriminate differently. A broad concentration interval from 0.024 to 0.034 mol/L at 20 s still permits first and second order. Reporting one precise-looking best curve would conceal that ambiguity. The contradictory packet permits none, and the changed-temperature record cannot test the same k without additional temperature evidence.

All checks are original public synthetic records. Reopening them is practice, not a new independent observation. Candidate compatibility does not rule out other functional forms or identify a mechanism.

Connect the transformed graph

For zero order plot C−C₀, for first order plot ln(C/C₀), and for second order plot 1/C−1/C₀ against time. Each candidate predicts a line through zero. The natural logarithm ln reverses exponentiation; its argument C/C₀ is dimensionless. Reciprocals and logarithms are undefined at zero concentration. The table maps the actual interval endpoints, so transformed bounds can be asymmetric. Changing units or transformations changes graph scale; visual straightness alone is not a common error metric. Compare predictions with the original concentration bounds.

Apply: a fresh sample changes the test

All candidates take 10 s to halve the original 0.100 mol/L. But after that, zero order needs only 5 more seconds to halve the remainder, first order needs 10 and second order needs 20. A fresh 0.200 mol/L sample instead has first half-lives of 20, 10 and 5 s respectively at the same candidate k. Explain why the first half-life alone was insufficient evidence, then transfer repeated halving to the unfamiliar concentration and time in the practice question.

OpenStax Chemistry 2e §12.4 supplies the integrated-law relationships. Revisit the HS half-life explanation for a guided exponential refresher, or rate-window measurement for why estimated slopes require calibration and uncertainty. Actual empirical calibration, measurement dependence, volume changes, additional orders and independent practical review remain outside this bounded model.