How does temperature affect rate — explain Arrhenius equation

Question

How does temperature affect the rate of a chemical reaction? Explain with the Arrhenius equation. What is the significance of the activation energy and pre-exponential factor?

Solution — Step by Step

A useful rule of thumb: the rate of most reactions doubles for every 10°C rise in temperature (this is sometimes stated as temperature coefficient $Q_{10} = 2$ ). But this is just an approximation. The actual relationship between rate and temperature is exponential — and was quantified by Svante Arrhenius in 1889.

k = A \cdot e^{-E_a/RT}

Where:

$k$ = rate constant
$A$ = pre-exponential factor (also called frequency factor or Arrhenius factor)
$e$ = base of natural logarithm (≈ 2.718)
$E_a$ = activation energy (J/mol or kJ/mol)
$R$ = gas constant = 8.314 J/mol·K
$T$ = absolute temperature (Kelvin)

Taking natural log of both sides:

\ln k = \ln A - \frac{E_a}{RT}

In log base 10:

\log k = \log A - \frac{E_a}{2.303RT}

Pre-exponential factor (A): Represents the frequency of collisions between reactant molecules AND the fraction that have the correct orientation. Even if molecules have enough energy, they must collide in the right geometry. A is a measure of both collision frequency and steric factor.

Activation Energy ( $E_a$ ): The minimum energy that reacting molecules must have for a successful collision. It’s the “energy barrier” between reactants and products. Low $E_a$ → reaction proceeds easily even at low temperatures. High $E_a$ → reaction needs high temperature.

The exponential term $e^{-E_a/RT}$ : This is the fraction of molecules that have energy ≥ $E_a$ at temperature $T$ . As $T$ increases, this fraction increases exponentially — this is why rate increases dramatically with temperature.

If we know $k_1$ at $T_1$ and $k_2$ at $T_2$ :

\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Or:

\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)

Example: Rate doubles from 300 K to 310 K. Find $E_a$ .

\log 2 = \frac{E_a}{2.303 \times 8.314} \times \frac{10}{300 \times 310}

0.3010 = \frac{E_a}{19.147} \times 1.075 \times 10^{-4}

E_a = \frac{0.3010 \times 19.147}{1.075 \times 10^{-4}} = 53,600 \text{ J/mol} \approx 53.6 \text{ kJ/mol}

From $\ln k = \ln A - \frac{E_a}{RT}$ , plotting $\ln k$ vs $\frac{1}{T}$ gives a straight line:

Slope = $-\frac{E_a}{R}$
Intercept = $\ln A$

This is the standard method for experimentally determining $E_a$ — measure $k$ at several temperatures, plot $\ln k$ vs $1/T$ , find the slope.

Why This Works

The Arrhenius equation works because it captures the Maxwell-Boltzmann distribution of molecular energies. At any temperature, molecules have a distribution of kinetic energies — most near the average, few at high energies. Only molecules with energy above $E_a$ can react successfully. The fraction of such molecules follows a Boltzmann distribution: $f = e^{-E_a/RT}$ . Increasing $T$ shifts the entire energy distribution toward higher values, dramatically increasing the fraction above $E_a$ .

A numerically common JEE problem: “Find $E_a$ given that rate triples when temperature increases by 10°C from 300 K to 310 K.” Use the two-temperature Arrhenius formula. Memorise that $\log 2 = 0.3010$ and $\log 3 = 0.4771$ — these appear repeatedly in Arrhenius numericals.

Common Mistake

Students often write $E_a$ in kJ/mol but use $R = 8.314$ J/mol·K without converting — giving $E_a/R$ a value 1000 times too large. Always check units: if $E_a$ is in J/mol, use $R = 8.314$ J/mol·K. If $E_a$ is in kJ/mol, either convert to J/mol or use $R = 8.314 \times 10^{-3}$ kJ/mol·K. Dimensional consistency is everything in Arrhenius calculations.

Question

Solution — Step by Step

Why This Works

Common Mistake

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