Two thermostatted water baths flank a colorimeter on the kinetics bench, with a rack of matched cuvettes between them and a timer lying open on the notebook. A blue dye fades by a first-order reaction, and the fading is followed as a falling trace on the colorimeter, each absorbance reading converted back to a concentration on the calibration line. In the first run the timer starts at 25.0 °C the moment the dye is mixed in at 0.0400 mol/L, and the trace crosses the mark for 0.0250 mol/L at 6.00 min. The identical run in the warmer bath, at 45.0 °C, sweeps between the same two concentrations in only 1.25 min. Find the rate constant at each temperature, the activation energy the pair implies, the rate constant predicted for a 60.0 °C water bath, and the half-life of the dye at that temperature.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
[A]₀ = 0.04 mol/L — Dye at the start of each run
[A] = 0.025 mol/L — Dye when the timer stops
t₁ = 6 min — Run time at 25.0 °C
t₂ = 1.25 min — Run time at 45.0 °C
T₃ = 60 °C — Water bath to predict
Determine
(a)the rate constant at each temperature
(b)the activation energy the pair implies
(c)the rate constant predicted at 60.0 °C
(d)the half-life of the dye at that temperature
Step 1 of 5(a) · solve for First-order rate constant
A rate constant, not a rate. For first-order decay k = ln([A]₀/[A])/t, so only the ratio of the concentrations matters — which is why a colorimeter reading works as well as a titration here.
Rearranged for k
k=t1ln[A][A]0
Your values, in your units
k=(6min)1ln(0.025M)(0.04M)
Converted to base units
k=(360s)1ln(0.025M)(0.04M)
Answer
k=1.3056mHz
Carried onward at full precision, not this rounded figure.
The same equation at the warmer bath. Same concentrations, shorter clock: the rate constant is exactly 360/75 = 4.8 times larger, because the logarithm is identical and only the time changed.
Rearranged for k
k=t1ln[A][A]0
Your values, in your units
k=(1.25min)1ln(0.025M)(0.04M)
Converted to base units
k=(75s)1ln(0.025M)(0.04M)
Answer
k=6.2667mHz
Carried onward at full precision, not this rounded figure.
Two rate constants at two temperatures are enough to eliminate the pre-exponential factor entirely, leaving the activation energy — about 61.9 kJ/mol. This is why kineticists measure ratios, not absolutes.
Now run the same equation forwards with the barrier you just measured, to predict a run nobody has performed. This is the first step that draws on two earlier answers at once.
For a first-order process the half-life is ln2/k and nothing else — independent of how much dye you started with, which is exactly the property that makes first-order kinetics recognisable.
Rearranged for t1/2
t1/2=λln2
17.959 mHzcarried from step 4
Your values, in your units
t1/2=(0.0179595Hz)(0.693147)
Answer
t1/2=38.595s
Carried onward at full precision, not this rounded figure.
Therefore the dye fades with k = 1.31 × 10⁻³ s⁻¹ at 25.0 °C and 6.27 × 10⁻³ s⁻¹ at 45.0 °C — a factor of exactly 4.8 — the pair fixes the activation energy at 61.9 kJ/mol, the 60.0 °C bath should run at k = 1.80 × 10⁻² s⁻¹, and dye dropped into it is half gone in 38.6 s.
Why this order
This chain is the standard experimental route through kinetics, and each step exists because the one before it cannot be skipped. You cannot measure an activation energy directly; you measure concentrations against a clock, convert each run into a rate constant, and only then does the Arrhenius comparison have anything to compare. Steps 1 and 2 are deliberately the same equation run twice: the fading is first order, so k comes from a ratio of concentrations, and a student who reports Δ[A]/Δt instead has computed a rate — a quantity that changes every second of the run and cannot be compared between temperatures.
Step 3 is where kelvin becomes non-negotiable. The two-temperature form contains 1/T₁ − 1/T₂, and 25 °C to 45 °C is a 6.7% change in absolute temperature but an 80% change in the Celsius number; use Celsius and the activation energy is nonsense. Reversing the reciprocal difference is the other standard slip, and it announces itself as a negative Eₐ. Step 4 is genuinely predictive rather than descriptive, and the payoff is worth noticing: 35 degrees above the first run the dye fades with a half-life of about 38.6 s, against roughly 531 s at 25 °C — a factor of nearly 14 for a barrier of only 61.9 kJ/mol, which is what an exponential does to a modest change in temperature.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
