The Arrhenius Equation

This lesson covers the Arrhenius equation, its logarithmic form, how to determine the activation energy from experimental data, and the effect of temperature on the rate constant.

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The Arrhenius Equation

The rate constant k depends on temperature according to the Arrhenius equation:

k = Ae^(−Ea/RT)

where:

• k = rate constant (units depend on overall order)
• A = the pre-exponential factor (or frequency factor) — related to collision frequency and orientation
• Ea = activation energy (J mol⁻¹)
• R = gas constant = 8.31 J mol⁻¹ K⁻¹
• T = temperature in kelvin (K)
• e = Euler's number (≈ 2.718)

The term e^(−Ea/RT) represents the fraction of molecules with energy ≥ Ea. As T increases, −Ea/RT becomes less negative, so e^(−Ea/RT) increases, meaning k increases and the reaction is faster.

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The Logarithmic Form

Taking natural logarithms of both sides:

ln k = ln A − Ea/(RT)

This can be rearranged to:

ln k = −(Ea/R) × (1/T) + ln A

This has the form y = mx + c, where:

• y = ln k
• x = 1/T
• gradient (m) = −Ea/R
• y-intercept (c) = ln A

A graph of ln k (y-axis) against 1/T (x-axis) gives a straight line with:

• Gradient = −Ea/R
• y-intercept = ln A

Therefore: Ea = −gradient × R

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Worked Examples

Example 1: Determining Ea from a graph.

A plot of ln k against 1/T gives a straight line with gradient = −5200 K. Calculate Ea.

Ea = −gradient × R = −(−5200) × 8.31 = 43,212 J mol⁻¹ = 43.2 kJ mol⁻¹

Example 2: Using two data points.

At 300 K, k₁ = 1.4 × 10⁻³ s⁻¹. At 320 K, k₂ = 5.6 × 10⁻³ s⁻¹. Calculate Ea.

Using: ln(k₂/k₁) = (Ea/R) × (1/T₁ − 1/T₂)

ln(5.6 × 10⁻³ / 1.4 × 10⁻³) = (Ea/8.31) × (1/300 − 1/320)

ln(4.0) = (Ea/8.31) × (3.333 × 10⁻³ − 3.125 × 10⁻³)

1.386 = (Ea/8.31) × (2.083 × 10⁻⁴)

Ea = (1.386 × 8.31) / (2.083 × 10⁻⁴) = 11.52 / 2.083 × 10⁻⁴ = 55,300 J mol⁻¹ = 55.3 kJ mol⁻¹

Example 3: Calculating k at a new temperature.

A reaction has Ea = 50.0 kJ mol⁻¹ and k = 2.0 × 10⁻² s⁻¹ at 298 K. Calculate k at 308 K.