Worked problem

Arrhenius Equation: How Temperature Changes Reaction Rate

Calculate how much faster a 75 kJ/mol reaction runs at 35 °C than at 25 °C with the Arrhenius equation, and the error of using Celsius.

Chemical & Process · Last reviewed

Given

A reaction has an activation energy Ea = 75 kJ/mol. The rate constant is k1 at 25 °C and k2 at 35 °C. Gas constant R = 8.314 J/mol·K.

Find

The ratio k2 ÷ k1, the temperature rise that would double the rate, and the result of the common error of using Celsius.

Solution

k2 ÷ k1 = exp[ (Ea ÷ R) × (1 ÷ T1 − 1 ÷ T2) ] (T in kelvin)
  1. Convert to kelvin: T1 = 25 + 273.15 = 298.15 K and T2 = 35 + 273.15 = 308.15 K.
  2. Ea ÷ R = 75,000 ÷ 8.314 = 9,021 K.
  3. 1 ÷ T1 − 1 ÷ T2 = 0.0033540 − 0.0032452 = 1.0884 × 10⁻⁴ K⁻¹.
  4. Exponent = 9,021 × 1.0884 × 10⁻⁴ = 0.982, so k2 ÷ k1 = e^0.982 = 2.67.
  5. For the rate to double (ratio 2) starting at 25 °C, T2 = 305.1 K (about 32 °C), a rise of 7 K.

Answer

k2 ÷ k1 = 2.67 | rate doubles for a rise of about 7 K at this activation energy

The Celsius error

The Arrhenius equation uses absolute temperature. Entering 25 and 35 instead of 298.15 and 308.15 gives an exponent of 9,021 × (1 ÷ 25 − 1 ÷ 35) = 103, so a ratio of about 6 × 10⁴⁴, which is obviously impossible. A result that large should always prompt a units check.

Activation energyk2 ÷ k1 (25 °C to 35 °C)
50 kJ/mol1.92
75 kJ/mol (this problem)2.67

Common mistakes

  • Using °C instead of K.
  • Mixing units of Ea (kJ/mol) and R (J/mol·K), which is a factor of 1,000 error.
  • Applying the rule that the rate doubles for every 10 K rise. It is only an approximation that fits some reactions near room temperature, and it does not fit this one.
  • Extrapolating far outside the measured temperature range, or past a temperature where the mechanism changes.

Related tools and labs

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Questions

What does a high activation energy mean?

The rate is more sensitive to temperature. A 75 kJ/mol reaction speeds up by a factor of 2.7 over 10 K, while a 50 kJ/mol one speeds up by 1.9.

Why does the rate rise with temperature?

A larger fraction of molecular collisions has enough energy to overcome the activation barrier. The Arrhenius equation captures this fraction.

Does this work for every reaction?

It describes many elementary and simple overall reactions over a limited range. Complex mechanisms and enzyme reactions can deviate.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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