Arrhenius Equation

Multiply the pre-exponential factor A by the Boltzmann factor $e^{(-\frac{E_{A}}{(R\cdot{}T)})}$: $k = A\cdot{}e^{(-\frac{E_{A}}{(R\cdot{}T)})}$. Insert the activation energy E_A and the gas constant $R =\frac{ 8.314 J}{(mol\cdot{}K)}$ in matching units and the temperature T in kelvin. The exponent is dimensionless, so E_A and $R\cdot{}T$ must have the same energy unit. A common mistake is to leave E_A in $\frac{kJ}{mol}$ while R is in $\frac{J}{(mol\cdot{}K)}$; then convert E_A to $\frac{J}{mol}$, that is with a factor of 1000. The pre-exponential factor A has the same unit as k and describes the collision frequency with suitable orientation.

Because the temperature sits in the exponent of the Arrhenius equation and the exponential term reacts very sensitively. The Boltzmann factor $e^{(-\frac{E_{A}}{(R\cdot{}T)})}$ gives the fraction of particles that have enough energy to overcome the activation barrier. Even a small temperature increase shifts the energy distribution so that markedly more particles lie above the threshold. As a rule of thumb the reaction rate often doubles to triples per 10 kelvin. Example: from 300 K to 310 K, k rises by about a factor of 1.9 at $E_{A} =\frac{ 50 kJ}{mol}$. This strong temperature dependence explains why many reactions proceed dramatically faster when heated.

Take the logarithm of the Arrhenius equation, then you get the linear form $\ln{} k = \ln{} A - (\frac{E_{A}}{R})\cdot{}(\frac{1}{T})$. Plotting ln k against $\frac{1}{T}$ gives a straight line with slope $-\frac{E_{A}}{R}$. From the measured slope m it follows that $E_{A} = -m\cdot{}R$. The intercept yields ln A and thus the pre-exponential factor. You therefore need at least two measured values of k at different temperatures; more points make the evaluation more accurate. Make sure to insert the temperature as the reciprocal $\frac{1}{T}$ in kelvin. This linearization is the standard method to determine activation energies experimentally.

The activation energy E_A is the energy barrier the reactants must overcome to reach the transition state and then the products. It determines how fast a reaction proceeds, that is the kinetics. The reaction enthalpy ΔH, by contrast, is the energy difference between products and reactants and determines whether the reaction releases or absorbs energy, that is the thermodynamics. E_A affects the rate, not the position of equilibrium; ΔH affects the energy balance, not the rate. A catalyst lowers E_A and thus speeds up the reaction, but leaves ΔH unchanged. Both quantities are independent and describe different aspects of a reaction.

A catalyst lowers the activation energy E_A by providing an alternative reaction path with a lower energy barrier. In the Arrhenius equation $k = A\cdot{}e^{(-\frac{E_{A}}{(R\cdot{}T)})}$ the E_A sits negatively in the exponent, so a smaller barrier raises the Boltzmann factor and thus the rate constant k noticeably. Because E_A acts in the exponent, even a moderate lowering leads to a large acceleration. The catalyst changes neither the position of equilibrium nor the reaction enthalpy ΔH; it speeds up the forward and reverse reactions equally. After the reaction it is left unconsumed. This is exactly why catalysts are so important in industry and biochemistry.