Chemical Kinetics · Lesson 8 of 10
Arrhenius Calculations and Experimental Data
“Derive logarithmic and two-temperature Arrhenius relationships, extract parameters from data and make conditional kinetic predictions.”
• Derive the natural-log and base-10 Arrhenius forms. • Use two rate constants to determine activation energy and the prefactor. • Fit an Arrhenius line using all measured data and interpret scatter. • Combine conversion-time or half-life observations with temperature dependence.
Turn an exponential dependence into a straight line
The Arrhenius equation contains a product Aexp(−E_a/RT). Taking natural logarithms turns that product into a sum: ln k=ln A−E_a/(RT). For a fixed unit chosen for k, this is a straight-line relationship between the logarithm of its numerical value and reciprocal absolute temperature. The intercept gives the prefactor in the same rate-constant unit.
A negative slope means a positive activation energy: moving to the right increases 1/T, which means lowering temperature, so k becomes smaller. The reciprocal-temperature axis reverses the intuitive direction of warming. If the horizontal axis is scaled as 1000/T, the slope becomes −E_a/(1000R), so the scale factor must be restored before calculating E_a.
Eliminate A between two temperatures
Write ln k₁=ln A−E_a/(RT₁) and ln k₂=ln A−E_a/(RT₂). Subtract the first from the second. If A and E_a can be treated as constant over the interval, ln A cancels. Combine the logarithms into ln(k₂/k₁) and collect reciprocal temperatures. Both rate constants must use the same units and the same definition of measured rate.
Problem
A first-order process has k₁=0.0200 s⁻¹ at 500 K and k₂=0.0700 s⁻¹ at 700 K. Determine E_a and A.
- 1.Ratio k₂/k₁=3.5 and reciprocal-temperature difference=1/500−1/700. Substitution gives E_a=18227.1 J mol⁻¹≈18.23 kJ mol⁻¹.
- 2.A=0.0200exp[E_a/(8.314×500)]=1.60424 s⁻¹.
- 3.Check at 700 K: Aexp[−E_a/(8.314×700)] reproduces 0.0700 s⁻¹. This tests both the sign and the units.
- 4.The numerical value of A does not become a concentration or an energy; its inverse-time unit follows from the first-order k used here.
Problem
For C₂H₅I(g)→C₂H₄(g)+HI(g), k₁=1.60×10⁻⁵ s⁻¹ at 600 K and E_a=209 kJ mol⁻¹. Predict k₂ at 700 K.
- 1.Convert the barrier to 209000 J mol⁻¹. Then ln(k₂/k₁)=(209000/8.314)(1/600−1/700).
- 2.Exponentiating gives k₂=1.60×10⁻⁵×exp[(209000/8.314)(1/600−1/700)]=0.00636076 s⁻¹.
- 3.The result exceeds k₁ because the temperature is higher and the activation energy positive. Using Celsius or reversing the temperature difference would give an unphysical result for this model.
Problem
If k doubles from 298 K to 308 K, estimate E_a in the constant-parameter Arrhenius model.
- 1.Use the ratio 2 and the actual temperature pair: E_a=8.314ln2/(1/298−1/308)=52893.5 J mol⁻¹≈52.89 kJ mol⁻¹.
- 2.The result is specific to the temperature interval. A doubling over a different interval can imply a different barrier.
- 3.No concentration values are required because the observation concerns rate constants, rather than a rate change caused by composition.
Use all the data when a graph is requested
| Temperature / °C | Temperature / K | Measured k / s⁻¹ |
|---|---|---|
| 0 | 273.15 | 7.87×10⁻⁷ |
| 20 | 293.15 | 1.70×10⁻⁵ |
| 40 | 313.15 | 2.57×10⁻⁴ |
| 60 | 333.15 | 1.78×10⁻³ |
| 80 | 353.15 | 2.140×10⁻² |
Problem
Fit ln k against 1/T for the table and predict k at 30 °C and 50 °C.
- 1.Convert every Celsius temperature to kelvin and use all five measured rate constants. Fit a straight line by minimising the sum of squared vertical logarithmic residuals; do not assume that rounded experimental points lie exactly on one line.
- 2.The fitted line is ln(k/[1 s⁻¹])≈30.3124−12115.2 K/T. Thus E_a=100.726 kJ mol⁻¹ and A=1.46046e+13 s⁻¹.
- 3.At 30 °C=303.15 K, fitted k≈6.42972e-05 s⁻¹. At 50 °C=323.15 K, fitted k≈0.000762758 s⁻¹.
- 4.The points show scatter around the line, especially the 60 °C datum. These are model-based predictions from the chosen all-point fit, not exact values implied independently by every pair of measurements.
- 5.Using only two endpoints or rounding temperatures to integer kelvin yields somewhat different parameter estimates. State the method so the result is reproducible.
Conversion-time comparisons can reveal a rate-constant ratio
For a first-order process, time to leave fraction f is t=ln(1/f)/k. If two different conversion fractions are reached in equal times at two temperatures, equating those expressions determines k₂/k₁. The remaining fractions, rather than percentages consumed, go inside the logarithms. That ratio can then enter the two-temperature Arrhenius equation.
Problem
The time for 10% consumption at 298 K equals the time for 25% consumption at 308 K. If A=4.0×10¹⁰ s⁻¹, estimate E_a and k at 318 K.
- 1.The remaining fractions are 0.90 and 0.75. Equal times give ln(1/0.90)/k₂₉₈=ln(1/0.75)/k₃₀₈. Therefore k₃₀₈/k₂₉₈=ln(1/0.75)/ln(1/0.90).
- 2.Insert that ratio into the two-temperature equation: E_a=76.6502 kJ mol⁻¹.
- 3.At 318 K, k=Aexp[−E_a/(8.314×318)]=0.0102579 s⁻¹.
- 4.This combines a fractional-decay observation with a temperature model; using 0.10 and 0.25 as remaining fractions would reverse the meaning of the data.
The intercept is an extrapolation to 1/T=0, not a direct measurement at an experimentally attainable infinite temperature. Parameters can be correlated and sensitive to scatter. A useful line supports Arrhenius behaviour over the studied range, rather than guaranteeing it through a phase or mechanism change.
Quick Quiz
Quiz
On an ln k versus 1/T plot, what is the slope?
Explanation: Taking natural logarithms of Arrhenius gives slope −E_a/R against unscaled reciprocal temperature.
What must be held approximately common to eliminate A in the two-temperature equation?
Explanation: Subtracting cancels the same prefactor only when the constant-parameter approximation is adequate.
If k₂>k₁ for T₂>T₁, the calculated E_a in the simple model should be:
Explanation: Both ln(k₂/k₁) and 1/T₁−1/T₂ are positive, so their ratio gives positive E_a.
The prefactor A has which units?
Explanation: The exponential is dimensionless, leaving A with the unit of k.
Equal times for 10% and 25% consumption at different temperatures require which quantities inside first-order logarithms?
Explanation: The integrated decay law uses initial amount divided by remaining amount.
Practice Problems
- A hydrocarbon decomposition has k=2.418×10⁻⁵ s⁻¹ at 546 K and E_a=179.9 kJ mol⁻¹. Find A.
- A law is written k=(4.5×10¹¹ s⁻¹)exp(−28000 K/T). Find E_a.
- The numerical first-order constant in s⁻¹ obeys log₁₀(k/[1 s⁻¹])=14.34−1.25×10⁴ K/T. Find E_a and the temperature for a 256 min half-life.
- A process has k₁=4.5×10³ s⁻¹ at 10 °C and E_a=60 kJ mol⁻¹. Find the temperature for k₂=1.5×10⁴ s⁻¹.
- The rate quadruples from 293 to 313 K at unchanged composition and under one rate law. Estimate E_a.
- For the five-temperature dataset, why can different pairs give different E_a values, and what should be reported with an all-point fit?
- A decimal-log plot against 1000/T has slope −5.00. Calculate E_a with R=8.314.
Rearrange k=Aexp(−E_a/RT) to A=kexp(E_a/RT). Convert E_a to 179900 J mol⁻¹, giving A=2.418×10⁻⁵exp[179900/(8.314×546)]=3.93298e+12 s⁻¹. The positive sign in the rearranged exponent is essential. Substituting this prefactor back into the original negative-exponent expression must reproduce the measured rate constant.
Key Takeaways
• Taking logarithms turns the simple Arrhenius equation into a straight-line relation. • Natural- and decimal-log plots have slopes differing by ln10. • Check whether reciprocal temperature is plotted as 1/T or 1000/T. • The two-temperature equation removes A only under an adequate constant-parameter model. • Use kelvin, matching energy units and the same definition and units of k. • All-point fitting uses the complete dataset but still requires examining scatter. • Conversion-time and half-life information can supply k values or their ratios. • Predictions should be labelled with the fitted model and its experimental range.