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Two kinetics runs to an activation energy and a half-life

SCH4U Grade 12 Chemistry · Energy Changes and Rates of Reaction

A blue dye fades by a first-order reaction, followed in a colorimeter. At 25.0 °C the dye falls from 0.0400 mol/L to 0.0250 mol/L in 6.00 min; the identical run at 45.0 °C reaches the same concentration in 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.

Step 1 of 5 · 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=1tln⁡ ⁣[A]0[A]k = \frac{1}{t}\ln\!\frac{[\mathrm{A}]_0}{[\mathrm{A}]}
Your values, in your units
k=1(6 min)ln⁡ ⁣(0.04 M)(0.025 M)k = \frac{1}{\left( 6\ \text{min} \right)}\ln\!\frac{\left( 0.04\ \text{M} \right)}{\left( 0.025\ \text{M} \right)}
Converted to base units
k=1(360 s)ln⁡ ⁣(0.04 M)(0.025 M)k = \frac{1}{\left( 360\ \text{s} \right)}\ln\!\frac{\left( 0.04\ \text{M} \right)}{\left( 0.025\ \text{M} \right)}
Answer
k=113k = 113

Carried onward at full precision, not this rounded figure.

Open the First-Order Integrated Rate Law solver →

Step 2 of 5 · solve for First-order rate constant

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=1tln⁡ ⁣[A]0[A]k = \frac{1}{t}\ln\!\frac{[\mathrm{A}]_0}{[\mathrm{A}]}
Your values, in your units
k=1(1.25 min)ln⁡ ⁣(0.04 M)(0.025 M)k = \frac{1}{\left( 1.25\ \text{min} \right)}\ln\!\frac{\left( 0.04\ \text{M} \right)}{\left( 0.025\ \text{M} \right)}
Converted to base units
k=1(75 s)ln⁡ ⁣(0.04 M)(0.025 M)k = \frac{1}{\left( 75\ \text{s} \right)}\ln\!\frac{\left( 0.04\ \text{M} \right)}{\left( 0.025\ \text{M} \right)}
Answer
k=22.6k = 22.6

Carried onward at full precision, not this rounded figure.

Open the First-Order Integrated Rate Law solver →

Step 3 of 5 · solve for Activation energy

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.

Rearranged for Ea
Ea=Rln⁡(k2/k1)1T1−1T2E_a = \frac{R\ln(k_2/k_1)}{\dfrac{1}{T_1} - \dfrac{1}{T_2}}
113 1/dcarried from step 1
22.6 1/hcarried from step 2
Your values, in your units
Ea=Rln⁡((0.00626672 Hz)/(0.00130557 Hz))1(25 ∘C)−1(45 ∘C)E_a = \frac{R \ln\left(\left( 0.00626672\ \text{Hz} \right)/\left( 0.00130557\ \text{Hz} \right)\right)}{\dfrac{1}{\left( 25\ ^{\circ}\text{C} \right)} - \dfrac{1}{\left( 45\ ^{\circ}\text{C} \right)}}
Answer
Ea=61.9E_a = 61.9

Carried onward at full precision, not this rounded figure.

Open the Arrhenius Two-Temperature Form solver →

Step 4 of 5 · solve for Rate constant at T2

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.

Rearranged for k2
k2=k1exp⁡ ⁣[EaR(1T1−1T2)]k_2 = k_1 \exp\!\left[\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]
113 1/dcarried from step 1
61.9 kJ/molcarried from step 3
Your values, in your units
k2=(0.00130557 Hz)exp⁡ ⁣[(61,856.8 J/mol)R(1(25 ∘C)−1(60 ∘C))]k_2 = \left( 0.00130557\ \text{Hz} \right) \exp\!\left[\frac{\left( 61{,}856.8\ \text{J/mol} \right)}{R}\left(\frac{1}{\left( 25\ ^{\circ}\text{C} \right)} - \frac{1}{\left( 60\ ^{\circ}\text{C} \right)}\right)\right]
Converted to base units
k2=(0.00130557 Hz)exp⁡ ⁣[(61.8568 kJ/mol)R(1(25 ∘C)−1(60 ∘C))]k_2 = \left( 0.00130557\ \text{Hz} \right) \exp\!\left[\frac{\left( 61.8568\ \text{kJ/mol} \right)}{R}\left(\frac{1}{\left( 25\ ^{\circ}\text{C} \right)} - \frac{1}{\left( 60\ ^{\circ}\text{C} \right)}\right)\right]
Answer
k2=64.7k_2 = 64.7

Carried onward at full precision, not this rounded figure.

Open the Arrhenius Two-Temperature Form solver →

Step 5 of 5 · solve for Half-life

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=ln⁡2λt_{1/2} = \frac{\ln 2}{\lambda}
64.7 1/hcarried from step 4
Your values, in your units
t1/2=ln⁡2(0.0179595 Hz)t_{1/2} = \frac{\ln 2}{\left( 0.0179595\ \text{Hz} \right)}
Answer
t1/2=38.6t_{1/2} = 38.6

Carried onward at full precision, not this rounded figure.

Open the Half-Life and Decay Constant solver →

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.