Topic 11 · page 16 of 21

Activation energy by calculation

Subtract the Arrhenius equation at two temperatures and the constant A cancels. The result links two rate constants, two temperatures and the activation energy, and can be rearranged for any of them.

13 min read11.10

Read first: Activation energy from a graph

Using the equation directly 11.10

If you are given EaE_\text{a}, A and T, you can calculate k from the Arrhenius equation. The units are the trap: EaE_\text{a} must be in J mol⁻¹, so multiply kJ mol⁻¹ by 1000.

Worked exampleCalculating k

A reaction has EaE_\text{a} = 52.0 kJ mol⁻¹ and A = 2.0 × 10⁸ s⁻¹. Calculate k at 298 K. (R = 8.31 J mol⁻¹ K⁻¹)

  1. EaE_\text{a} = 52.0 × 1000 = 52 000 J mol⁻¹
  2. exponent = EaE_\text{a} ÷ RT = 52 000 ÷ (8.31 × 298) = 20.998
  3. e−20.998e^{-20.998} = 7.59 × 10⁻¹⁰
  4. k = 2.0 × 10⁸ × 7.59 × 10⁻¹⁰ = 0.15 s⁻¹

The two-temperature equation 11.10

You often have k at two temperatures and no value of A. Write the Arrhenius equation in log form at each temperature and subtract. The ln A terms cancel:

ln⁡k2−ln⁡k1=−EaR(1T2−1T1) \ln k_2 - \ln k_1 = -\frac{E_\text{a}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)
ln⁡k2k1=EaR(1T1−1T2) \ln\frac{k_2}{k_1} = \frac{E_\text{a}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

The second form is the easiest to use: the bracket is positive when T2T_2 is the higher temperature, and so is the left side when k rises.

MethodTwo-temperature calculations

  1. Convert temperatures to kelvin and label them T1T_1, T2T_2 with k1k_1, k2k_2.
  2. Work out ln(k2k_2 ÷ k1k_1) and (1/T1T_1 − 1/T2T_2). Keep extra figures, because the bracket is a small difference between two large numbers.
  3. Rearrange for the unknown and substitute.
  4. Give EaE_\text{a} in J mol⁻¹ and convert to kJ mol⁻¹ at the end.

Worked exampleThe rate constant doubles for a 10 K rise

The rate constant of a reaction doubles when the temperature rises from 298 K to 308 K. Calculate EaE_\text{a}.

  1. ln 2 = 0.6931
  2. 1/298 − 1/308 = 3.3557 × 10⁻³ − 3.2468 × 10⁻³ = 1.0895 × 10⁻⁴ K⁻¹
  3. EaE_\text{a} = R × ln 2 ÷ 1.0895 × 10⁻⁴ = 8.31 × 0.6931 ÷ 1.0895 × 10⁻⁴ = 5.29 × 10⁴ J mol⁻¹
  4. EaE_\text{a} = 52.9 kJ mol⁻¹

This shows where the rule of thumb "the rate roughly doubles for every 10 K rise" comes from. It is true for EaE_\text{a} of about 50 kJ mol⁻¹ near room temperature.

Worked exampleTwo measured rate constants, in °C

A reaction has k = 3.20 × 10⁻⁴ s⁻¹ at 25 °C and 1.90 × 10⁻³ s⁻¹ at 45 °C. Calculate EaE_\text{a}.

  1. T1T_1 = 25 + 273 = 298 K and T2T_2 = 45 + 273 = 318 K
  2. ln(k2k_2 ÷ k1k_1) = ln(1.90 × 10⁻³ ÷ 3.20 × 10⁻⁴) = ln 5.9375 = 1.7813
  3. 1/298 − 1/318 = 2.1105 × 10⁻⁴ K⁻¹
  4. EaE_\text{a} = 8.31 × 1.7813 ÷ 2.1105 × 10⁻⁴ = 7.01 × 10⁴ J mol⁻¹
  5. EaE_\text{a} = 70.1 kJ mol⁻¹

Worked exampleFinding k at a new temperature

A reaction has EaE_\text{a} = 60.0 kJ mol⁻¹ and k = 0.0200 s⁻¹ at 298 K. Calculate k at 318 K.

  1. ln(k2k_2 ÷ k1k_1) = (60 000 ÷ 8.31) × (1/298 − 1/318) = 7220.2 × 2.1105 × 10⁻⁴ = 1.5238
  2. k2k_2 ÷ k1k_1 = e1.5238e^{1.5238} = 4.590
  3. k2k_2 = 0.0200 × 4.590 = 0.0918 s⁻¹

Check yourself

  1. Write the equation that links two rate constants, the two temperatures and EaE_\text{a}.
  2. The rate constant triples when the temperature rises from 300 K to 310 K. Calculate EaE_\text{a} in kJ mol⁻¹.
  3. k is 2.5 × 10⁻⁴ s⁻¹ at 25 °C and 3.0 × 10⁻³ s⁻¹ at 55 °C. Calculate EaE_\text{a} in kJ mol⁻¹.
  4. A reaction has EaE_\text{a} = 45.0 kJ mol⁻¹ and k = 1.20 × 10⁻³ s⁻¹ at 300 K. Calculate k at 320 K.
  5. A catalyst lowers EaE_\text{a} from 80.0 to 50.0 kJ mol⁻¹. Assuming A is unchanged, calculate by what factor k increases at 298 K.
  6. A student uses 25 and 35 in place of the temperatures in kelvin and gets a different answer from the correct one. Explain why this is wrong, and say whether the result would be too large or too small.

Answers cover these until you have tried the questions

  1. ln⁡(k2÷k1)=EaR(1T1−1T2)\ln(k_2 \div k_1) = \dfrac{E_\text{a}}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right), with T in kelvin and EaE_\text{a} in J mol⁻¹.
  2. ln 3 = 1.0986. 1/300 − 1/310 = 1.0753 × 10⁻⁴ K⁻¹. EaE_\text{a} = 8.31 × 1.0986 ÷ 1.0753 × 10⁻⁴ = 8.49 × 10⁴ J mol⁻¹ = 84.9 kJ mol⁻¹.
  3. T1T_1 = 298 K, T2T_2 = 328 K. k2k_2 ÷ k1k_1 = 3.0 × 10⁻³ ÷ 2.5 × 10⁻⁴ = 12, ln 12 = 2.4849. 1/298 − 1/328 = 3.0692 × 10⁻⁴. EaE_\text{a} = 8.31 × 2.4849 ÷ 3.0692 × 10⁻⁴ = 6.73 × 10⁴ J mol⁻¹ = 67.3 kJ mol⁻¹.
  4. ln(k2k_2 ÷ k1k_1) = (45 000 ÷ 8.31) × (1/300 − 1/320) = 5415.2 × 2.0833 × 10⁻⁴ = 1.1282. k2k_2 ÷ k1k_1 = e1.1282e^{1.1282} = 3.090. k2k_2 = 1.20 × 10⁻³ × 3.090 = 3.7 × 10⁻³ s⁻¹.
  5. The ratio of rate constants = e−50 000/RT÷e−80 000/RT=e30 000/RTe^{-50\,000/RT} \div e^{-80\,000/RT} = e^{30\,000/RT}. 30 000 ÷ (8.31 × 298) = 12.11, and e12.11e^{12.11} = 1.8 × 10⁵. k increases by a factor of about 1.8 × 10⁵.
  6. Kelvin is needed because the equation uses the absolute temperature, and 1/T with T = 25 or 35 is a very different number from 1/298 or 1/308. The bracket (1/25 − 1/35 = 0.0114) is about 100 times larger than the real bracket (1/298 − 1/308 = 1.09 × 10⁻⁴), so the calculated EaE_\text{a} would be far too small.