Temperature, Catalysts and Collision Theory: Quick Revision

A brief note on Temperature, Catalysts and Collision Theory. Read it before a test. Try each question at the end before you open its answer.

The Arrhenius equation

• $k = A\,e^{-E_a/RT}$. $T$ is in kelvin, $E_a$ in J/mol and $R = 8.314\ \text{J/(mol K)}$.
• $A$ is the pre-exponential factor. It has the unit of $k$.
• $e^{-E_a/RT}$ is the share of collisions with enough energy.
• $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$. A graph of $\ln k$ against $1/T$ is a straight line with slope $-E_a/R$.
• Two temperatures: $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$. With $\log$, divide $E_a$ by $2.303\,R$.
• Near room temperature, $E_a \approx 54\ \text{kJ/mol}$ doubles the rate for each $10\ \text{K}$.
• A higher $E_a$ means a stronger response to temperature.

Collisions and the energy barrier

• A reaction needs a collision, enough energy and the right orientation.
• $k = P\,Z\,e^{-E_a/RT}$, so $A = PZ$. $P$ is the probability factor and is below 1.
• A rise in temperature makes the tail of the energy distribution much larger. The collision frequency grows only as $\sqrt{T}$.
• Energy profile: $\Delta H = E_a - E_a'$, so $E_a' = E_a - \Delta H$.
• The activated complex is at the top of the barrier. An intermediate is in a dip.

Catalysts

• A catalyst gives a path with a lower activation energy and is not used up.
• It lowers the forward and the reverse barrier by the same amount. $\Delta H$ and the equilibrium constant do not change.
• $\dfrac{k_{cat}}{k} = e^{\Delta E_a/RT}$ when $A$ stays the same.
• It cannot make a reaction with a positive $\Delta G$ run.
• Solid catalysts work on active sites. A poison blocks them. A promoter helps them.
• Enzymes are specific protein catalysts. Their rate rises with temperature and then falls.

A worked example

A reaction has $E_a = 56.0\ \text{kJ/mol}$ and $k = 1.0 \times 10^{-3}\ \text{s}^{-1}$ at $298\ \text{K}$. Find $k$ at $308\ \text{K}$.

1. Temperature part. $\dfrac{1}{298\ \text{K}} - \dfrac{1}{308\ \text{K}} = 1.09 \times 10^{-4}\ \text{K}^{-1}$.
2. Exponent. $\ln\dfrac{k_2}{k_1} = \dfrac{56\,000}{8.314}\ \text{K} \times 1.09 \times 10^{-4}\ \text{K}^{-1} = 0.734$.
3. Ratio and answer. $\dfrac{k_2}{k_1} = e^{0.734} = 2.08$, so $k_2 = 2.08 \times 10^{-3}\ \text{s}^{-1}$.

Practice with new numbers

A program to try

The program works out how many times $k$ grows for each step of $10\ \text{K}$ for three activation energies.

Mistakes to avoid

• Degrees Celsius in the equation. Use kelvin.
• $E_a$ in kJ/mol with $R$ in J/(mol K). Change to joules first.
• Saying more collisions explain the rise in rate. The share with enough energy does.
• Reverse barrier as $E_a + \Delta H$. It is $E_a - \Delta H$.
• A catalyst changes $\Delta H$ or the equilibrium. It changes neither.

Check yourself

1. A reaction doubles its rate for a rise of $10\ \text{K}$. Does its activation energy change?

2. What does the slope of $\ln k$ against $1/T$ equal?

3. A reaction has $E_a = 90\ \text{kJ/mol}$ and $\Delta H = -20\ \text{kJ/mol}$. What is the reverse barrier?

4. Does a catalyst change the equilibrium constant?