Why does a small rise in temperature make reactions go so much faster, and why does a catalyst help? The answers come from the way molecules must collide to react, captured quantitatively by the Arrhenius equation.
Temperature dependence
It is found experimentally that the rate of most reactions roughly doubles for every $10\ ^\circ\text{C}$ rise in temperature; the ratio $\frac{k_{T+10}}{k_T}$ is the temperature coefficient (typically $2$ to $3$). This large effect cannot be due to more frequent collisions alone, because the collision frequency rises only slightly with temperature. The real reason is that more molecules acquire enough energy to react.
The Arrhenius equation
Arrhenius gave the temperature dependence of the rate constant as
$$k = Ae^{-E_a/RT}$$
where $A$ is the frequency (pre-exponential) factor (related to the collision frequency and orientation), $E_a$ is the activation energy, $R$ is the gas constant and $T$ the absolute temperature. The factor $e^{-E_a/RT}$ is the fraction of molecules with energy equal to or greater than $E_a$. Taking logarithms, $\ln k = \ln A - \frac{E_a}{RT}$, so a plot of $\ln k$ (or $\log k$) against $\frac{1}{T}$ is a straight line of slope $-\frac{E_a}{R}$ (or $-\frac{E_a}{2.303R}$).
Comparing the rate constant at two temperatures eliminates $A$:
$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{E_a}{2.303R}\cdot\frac{T_2 - T_1}{T_1 T_2}$$
This is the most-used working form for finding $E_a$ from two $(k, T)$ data points.
Activation energy
The activation energy $E_a$ is the minimum extra energy that reactant molecules must possess above their average energy to form the high-energy activated complex (transition state) at the top of the energy barrier. The difference between the threshold energy and the average energy of reactants is $E_a$. A reaction with a small $E_a$ proceeds fast; a large $E_a$ makes it slow.
Collision theory
According to collision theory, reactant molecules must collide for a reaction to occur, but not every collision is fruitful. Only collisions that are both sufficiently energetic (energy $\geq$ threshold energy) and correctly oriented lead to products. The rate is therefore $\text{Rate} = PZ_{AB}\,e^{-E_a/RT}$, where $Z_{AB}$ is the collision frequency and $P$ the steric (orientation) factor. This explains why $A$ in the Arrhenius equation depends on both collision frequency and proper orientation.
Effect of a catalyst
A catalyst speeds a reaction by providing an alternative pathway with a lower activation energy $E_a$. Because $e^{-E_a/RT}$ is very sensitive to $E_a$, even a modest lowering of the barrier raises $k$ dramatically. A catalyst does not change the enthalpy of reaction $\Delta H$, the position of equilibrium, or the energies of reactants and products — it only lowers the hump in between, and lowers $E_a$ equally for the forward and reverse reactions.
The rate constant of a reaction doubles when the temperature rises from $300\ \text{K}$ to $310\ \text{K}$. Calculate the activation energy. (Take $\log 2 = 0.301$, $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.)
Solution- $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$ with $\frac{k_2}{k_1} = 2$.
- $\log 2 = \frac{E_a}{2.303 \times 8.314}\left(\frac{310-300}{300 \times 310}\right)$.
- $0.301 = \frac{E_a}{19.15}\left(\frac{10}{93000}\right) = \frac{E_a}{19.15}(1.0753\times10^{-4})$.
- $E_a = \frac{0.301 \times 19.15}{1.0753\times10^{-4}} = \frac{5.764}{1.0753\times10^{-4}} = 5.36\times10^{4}\ \text{J mol}^{-1}$.
Answer: $E_a \approx 53.6\ \text{kJ mol}^{-1}$.
For a reaction $E_a = 100\ \text{kJ mol}^{-1}$ and $A = 6 \times 10^{14}\ \text{s}^{-1}$. Calculate the rate constant at $T = 300\ \text{K}$. (Take $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.)
Solution- $\log k = \log A - \frac{E_a}{2.303RT}$.
- $\frac{E_a}{2.303RT} = \frac{100000}{2.303 \times 8.314 \times 300} = \frac{100000}{5744.7} = 17.41$.
- $\log A = \log(6\times10^{14}) = 14.778$.
- $\log k = 14.778 - 17.41 = -2.63$, so $k = 10^{-2.63} = 2.34\times10^{-3}\ \text{s}^{-1}$.
Answer: $k \approx 2.34 \times 10^{-3}\ \text{s}^{-1}$.
The activation energy of a reaction is $209.5\ \text{kJ mol}^{-1}$ at $581\ \text{K}$. Find the fraction of molecules having energy equal to or greater than $E_a$. (Take $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.)
Solution- The fraction is $f = e^{-E_a/RT}$, so $\ln f = -\frac{E_a}{RT}$.
- $\frac{E_a}{RT} = \frac{209500}{8.314 \times 581} = \frac{209500}{4830.4} = 43.37$.
- $\log f = -\frac{43.37}{2.303} = -18.83$, so $f = 10^{-18.83} = 1.47\times10^{-19}$.
Answer: $f = e^{-E_a/RT} \approx 1.47 \times 10^{-19}$ (a very small fraction).
The rate constant of a first-order reaction is $1.0\times10^{-4}\ \text{s}^{-1}$ at $300\ \text{K}$ and $2.0\times10^{-3}\ \text{s}^{-1}$ at $320\ \text{K}$. Calculate $E_a$. (Take $\log 20 = 1.301$, $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.)
Solution- $\frac{k_2}{k_1} = \frac{2.0\times10^{-3}}{1.0\times10^{-4}} = 20$, so $\log\frac{k_2}{k_1} = \log 20 = 1.301$.
- $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1 T_2}\right) = \frac{E_a}{19.15}\left(\frac{20}{300\times320}\right) = \frac{E_a}{19.15}(2.083\times10^{-4})$.
- $E_a = \frac{1.301 \times 19.15}{2.083\times10^{-4}} = \frac{24.91}{2.083\times10^{-4}} = 1.196\times10^{5}\ \text{J mol}^{-1}$.
Answer: $E_a \approx 119.6\ \text{kJ mol}^{-1}$.
Explain, using an energy profile, how a catalyst increases the rate of a reaction. Does it change $\Delta H$ of the reaction?
Solution- A catalyst provides an alternative reaction path that passes over a lower energy barrier, i.e. a smaller activation energy $E_a$.
- Since $k = Ae^{-E_a/RT}$, lowering $E_a$ raises $e^{-E_a/RT}$ and so increases $k$ and the rate; more molecules now have energy above the lowered barrier.
- The catalyst lowers $E_a$ for both forward and reverse steps equally and leaves the energies of reactants and products unchanged, so $\Delta H$ is unchanged.
Answer: A catalyst lowers $E_a$ (alternative path), increasing the rate, but $\Delta H$ stays the same.
For a reaction, a plot of $\log k$ against $\frac{1}{T}$ is a straight line with slope $-5000\ \text{K}$. Calculate the activation energy. (Take $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.)
Solution- From $\log k = \log A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$, slope $= -\frac{E_a}{2.303R}$.
- So $\frac{E_a}{2.303R} = 5000\ \text{K}$, giving $E_a = 5000 \times 2.303 \times 8.314$.
- $E_a = 5000 \times 19.15 = 95750\ \text{J mol}^{-1}$.
Answer: $E_a \approx 95.75\ \text{kJ mol}^{-1}$.