DAT General Chemistry: The Arrhenius Equation

DAT High-Yield Concept

The rate constant $k$ increases exponentially with absolute temperature $T$ .
A linear Arrhenius plot of $\ln k$ versus $\frac{1}{T}$ yields a slope equal to $-\frac{E_a}{R}$ and a y-intercept of $\ln A$ .
For many standard reactions near room temperature, a $\pu{10 degC}$ rise in temperature roughly doubles the reaction rate because it significantly increases the fraction of molecules with energy exceeding the activation threshold.
Always convert temperatures to Kelvin ( $\pu{K}$ ) and ensure energy units match the gas constant $R = \pu{8.314 J/(mol K)}$ .

The Arrhenius Equation Expressions

The rate constant $k$ quantifies the intrinsic speed of a reaction. The Arrhenius equation relates $k$ to absolute temperature $T$ , the frequency factor $A$ , and the ideal gas constant $R = \pu{8.314 J/(mol K)}$ :

k = A e^{-\frac{E_a}{R T}}

Taking the natural logarithm yields the linearized form, which follows the standard slope-intercept line equation $y = mx + b$ :

\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A

Plotting $\ln k$ on the y-axis against $\frac{1}{T}$ on the x-axis produces a straight line with a negative slope equal to $-\frac{E_a}{R}$ .

To compare rate constants at two different temperatures without needing the pre-exponential factor $A$ , use the two-point form:

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

Equation Form	Mathematical Expression	Primary DAT Use Case
Exponential Form	$k = A e^{-\frac{E_a}{R T}}$	Qualitative analysis of temperature and rate constant relationships
Linear Form	$\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A$	Determining activation energy from graphical plot slopes
Two-Point Form	$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$	Calculating $k_2$ at temperature $T_2$ given $k_1$ at $T_1$

Temperature Dependence of the Rate Constant

As absolute temperature $T$ increases, the term $\frac{E_a}{RT}$ decreases, making the exponent $-\frac{E_a}{RT}$ less negative. Consequently, the exponential factor $e^{-\frac{E_a}{RT}}$ increases, driving an increase in the rate constant $k$ .

Key Distinction: Rate vs. Rate Constant

Increasing temperature increases the rate constant $k$ . Because reaction rate is directly proportional to $k$ (e.g., $\text{Rate} = k[\ce{A}]^x$ ), elevated temperature speeds up reactions regardless of whether the overall reaction is exothermic or endothermic.

Why a Small Temperature Rise Produces a Large Rate Increase

At the molecular level, temperature measures average kinetic energy. The energy distribution of reactant molecules follows the Maxwell-Boltzmann distribution curve.

Raising the temperature fattens the tail past the activation energy — At any temperature the collisions in a sample carry a whole spread of energies, and only those out past the activation energy can react. Warming the sample does not move every molecule up by the same amount; it stretches the distribution to the right, and because the reactive fraction lives in the thin tail, a modest temperature rise multiplies that fraction several times over. That is what the exponential in the Arrhenius equation is measuring, and it is why a rise of roughly ten degrees near room temperature is often enough to double the rate of a typical reaction even though the average energy has barely moved.

When the temperature is increased by a small increment (such as $\pu{10 degC}$ ), the average kinetic energy increases only slightly. However, this shifts and flattens the distribution curve, causing an exponential increase in the fraction of molecules residing in the high-energy tail past the minimum activation threshold. Because far more collisions possess energy equal to or greater than the activation energy, the reaction rate increases dramatically.

Mnemonic: Think KITE for Arrhenius graphs: Konstant rises nonlinearly with Increasing Temperature; Exponent controls the curve.

Worked Example & DAT Exam Traps

Problem: A first-order decomposition reaction, $\ce{2NO2 -> 2NO + O2}$ , has a rate constant $k_1 = \pu{2.00 e-3 s-1}$ at $T_1 = \pu{298 K}$ ( $\pu{25 degC}$ ). Calculate the new rate constant $k_2$ at $T_2 = \pu{308 K}$ ( $\pu{35 degC}$ ) if the reaction has an activation energy of $\pu{50.0 kJ/mol}$ .

Step-by-Step Solution:

Convert activation energy to $\pu{J/mol}$ : $\pu{50.0 kJ/mol} = \pu{50000 J/mol}$ .
Calculate inverse temperature differences: $\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{\pu{298 K}} - \frac{1}{\pu{308 K}} = \pu{0.0033557 K-1} - \pu{0.0032468 K-1} = \pu{1.0895 e-4 K-1}$
Substitute into the two-point Arrhenius equation using $R = \pu{8.314 J/(mol K)}$ : $\ln\left(\frac{k_2}{k_1}\right) = \left(\frac{\pu{50000 J/mol}}{\pu{8.314 J/(mol K)}}\right) \times (\pu{1.0895 e-4 K-1}) = (6013.95) \times (\pu{1.0895 e-4}) = 0.6552$
Solve for $k_2$ : $\frac{k_2}{k_1} = e^{0.6552} = 1.925$ $k_2 = 1.925 \times \pu{2.00 e-3 s-1} = \pu{3.85 e-3 s-1}$

Common DAT Exam Traps:

Forgetting Kelvin: Plugging in Celsius directly leads to calculation errors. Always convert to Kelvin ( $\pu{K} = \pu{degC} + 273.15$ ).
Energy Unit Mismatch: Using $\pu{kJ/mol}$ for $E_a$ alongside $R = \pu{8.314 J/(mol K)}$ creates a factor-of-1000 error. Always convert $E_a$ to $\pu{J/mol}$ .
Inverting Temperatures: Note the term $\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ when writing $\ln\left(\frac{k_2}{k_1}\right)$ . Switching the order flips the sign.

Critical Thinking Questions

If two different reactions have identical frequency factors $A$ at $\pu{298 K}$ , but Reaction 1 has a larger activation energy than Reaction 2, which reaction will experience a larger percentage increase in rate when both are heated to $\pu{318 K}$ ?
How does an Arrhenius plot of $\ln k$ versus $\frac{1}{T}$ visually demonstrate that a reaction with a higher activation energy is more sensitive to temperature changes than one with a lower activation energy?
Why does adding a catalyst alter the rate constant $k$ without changing the temperature $T$ of the system?

References

Sources returned by Google Search grounding during lesson verification.

14.4 Collision Theory and the Arrhenius Equation - Chad's Prep — chadsprep.com
6.2.3.1: Arrhenius Equation - Chemistry LibreTexts — chem.libretexts.org

The Arrhenius Equation

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