Effect of temperature: the Arrhenius equation

A warm kitchen and a cold one

A bowl of dough rises in an hour on a warm day. On a cold day it may need three hours. A pressure cooker cooks lentils in a short time, because the water in it is hotter than ordinary boiling water.

Heat speeds up almost every reaction. How much does it help? Chemists measured this long ago. For many reactions near room temperature, a rise of $10\ \text{K}$ makes the rate constant two or three times larger. The ratio $\dfrac{k_{T+10}}{k_T}$ is called the temperature coefficient. It is usually between 2 and 3.

Here is a puzzle. A rise from $300\ \text{K}$ to $310\ \text{K}$ makes the molecules move only about $1.7$ percent faster. Their average kinetic energy grows by only $3$ percent. Yet the rate doubles. How can a small push give such a large change?

The Swedish chemist Svante Arrhenius answered this in 1889. We follow his idea.

The Arrhenius equation

Arrhenius found that the rate constant follows this equation:

Let us read it one symbol at a time.

• $k$ is the rate constant.
• $E_a$ is the activation energy. It is the minimum extra energy that the colliding molecules must have before they can react. We write it in joules per mole, or in kilojoules per mole.
• $R$ is the gas constant, $8.314\ \text{J/(mol K)}$.
• $T$ is the temperature in kelvin. Never use degrees Celsius here.
• $A$ is the pre-exponential factor. It is also called the frequency factor. It has the same unit as $k$. It depends on how often the molecules collide and how they are oriented. We study it in the next lesson.

The part $e^{-E_a/RT}$ is a number between 0 and 1. It is the share of collisions that have enough energy. A large $E_a$ or a low $T$ makes it small. When $T$ becomes very large, this number goes towards 1 and $k$ goes towards $A$.

Look at the size of $\dfrac{E_a}{RT}$. For $E_a = 72\ \text{kJ/mol}$ at $310\ \text{K}$:

This has no unit, as an exponent must not. Then $e^{-27.9} = 7.4 \times 10^{-13}$. With $A = 5.0 \times 10^{11}\ \text{s}^{-1}$ we get $k = 0.37\ \text{s}^{-1}$.

Now go to $330\ \text{K}$. The exponent becomes $26.2$ and $k = 2.0\ \text{s}^{-1}$. A rise of only $20\ \text{K}$ made $k$ about $5.4$ times larger.

The left graph shows these two points on the curve. The curve is nearly flat when it is cold and climbs more and more steeply as it gets warmer. The exponent in the equation is the reason.

The straight line of the Arrhenius equation

Take the natural logarithm of both sides:

Compare this with $y = c + mx$. A graph of $\ln k$ against $\dfrac{1}{T}$ is a straight line. Its slope is $-\dfrac{E_a}{R}$ and its intercept is $\ln A$. The right graph above shows it.

This is how $E_a$ is found in a laboratory. Measure $k$ at several temperatures. Plot $\ln k$ against $1/T$. Draw the best line. Then $E_a = -R \times \text{slope}$.

With logarithms to base 10, the same equation reads:

The slope of $\log k$ against $1/T$ is now $-\dfrac{E_a}{2.303\,R}$.

Two temperatures

Often you know $k$ at only two temperatures. Write the equation for each and subtract. The term $\ln A$ cancels:

In base 10 logarithms:

This works when $E_a$ and $A$ stay the same over the range, which is true for a small range of temperature. We want $k_2$, or $E_a$, so we rearrange the equation in that step.

Example 1: a new rate constant

A reaction has $E_a = 72.0\ \text{kJ/mol}$ and $k = 0.37\ \text{s}^{-1}$ at $310\ \text{K}$. Find $k$ at $330\ \text{K}$.

Step 1. The exponent. $\dfrac{E_a}{R} = \dfrac{72\,000\ \text{J/mol}}{8.314\ \text{J/(mol K)}} = 8660\ \text{K}$.

Step 2. The temperature part. $\dfrac{1}{310\ \text{K}} - \dfrac{1}{330\ \text{K}} = 1.955 \times 10^{-4}\ \text{K}^{-1}$.

Step 3. The ratio. $\ln\dfrac{k_2}{k_1} = 8660\ \text{K} \times 1.955 \times 10^{-4}\ \text{K}^{-1} = 1.693$. So $\dfrac{k_2}{k_1} = e^{1.693} = 5.4$.

Step 4. The answer. $k_2 = 0.37\ \text{s}^{-1} \times 5.4 = 2.0\ \text{s}^{-1}$.

Check. The new temperature is higher, so $k$ must be larger. It is. The factor $5.4$ is a little more than $2$ for each $10\ \text{K}$, which fits the rule of thumb.

Example 2: the activation energy

A reaction has $k = 1.36 \times 10^{-3}\ \text{s}^{-1}$ at $315\ \text{K}$ and $k = 6.10 \times 10^{-3}\ \text{s}^{-1}$ at $335\ \text{K}$. Find $E_a$.

Step 1. The ratio. $\dfrac{k_2}{k_1} = \dfrac{6.10 \times 10^{-3}}{1.36 \times 10^{-3}} = 4.485$, and $\log 4.485 = 0.652$.

Step 2. The temperature part. $\dfrac{1}{315\ \text{K}} - \dfrac{1}{335\ \text{K}} = 1.895 \times 10^{-4}\ \text{K}^{-1}$.

Step 3. Solve. We want $E_a$, so $E_a = \dfrac{2.303 \times R \times \log(k_2/k_1)}{1/T_1 - 1/T_2}$. The top of the fraction is $2.303 \times 8.314\ \text{J/(mol K)} \times 0.652 = 12.48\ \text{J/(mol K)}$. So $E_a = \dfrac{12.48\ \text{J/(mol K)}}{1.895 \times 10^{-4}\ \text{K}^{-1}} = 65\,800\ \text{J/mol}$.

Answer: $E_a = 65.8\ \text{kJ/mol}$.

Why the rate doubles

Now we can answer the puzzle. Suppose the rate constant doubles between $300\ \text{K}$ and $310\ \text{K}$. Then $\ln 2 = \dfrac{E_a}{R}\left(\dfrac{1}{300} - \dfrac{1}{310}\right)$, which gives $E_a = 53.6\ \text{kJ/mol}$.

So a reaction with an activation energy near $54\ \text{kJ/mol}$ doubles its rate for every $10\ \text{K}$ near room temperature. Many reactions have barriers of this size. The rate grows much faster than the molecular speed because only the fastest molecules can react, and a warmer gas has many more of them. The next lesson shows this with a picture.

A reaction with a larger $E_a$ is more sensitive to temperature. A reaction with a smaller one is less sensitive. Try it in the simulation.

The solid and the dashed lines start at the same point at $300\ \text{K}$. The dashed line, with the higher barrier, climbs faster. A higher barrier means a stronger response to heat.

Practise with new numbers

Try it in Python

The program below finds $k$ for five temperatures from the Arrhenius equation. Then it works out the factor for each step of $10\ \text{K}$ and the activation energy from the two ends.

Common mistakes

• Using degrees Celsius. The equation needs kelvin. A change from $27\ \text{°C}$ to $37\ \text{°C}$ is from $300\ \text{K}$ to $310\ \text{K}$.
• Mixing the units of $E_a$ and $R$. $R$ has joules in it. Turn kilojoules into joules before you divide.
• Swapping $T_1$ and $T_2$. The bracket is $\dfrac{1}{T_1} - \dfrac{1}{T_2}$, and $k_2$ belongs to $T_2$. A wrong order gives a negative $E_a$.
• Forgetting 2.303 when you use $\log$. The natural logarithm needs no factor. The logarithm to base 10 does.
• Thinking $E_a$ changes with temperature. For one reaction over a small range it stays the same. The rate constant changes.

Remember these

• $k = Ae^{-E_a/RT}$. $T$ is in kelvin, $E_a$ in J/mol, $R = 8.314\ \text{J/(mol K)}$.
• A graph of $\ln k$ against $1/T$ is a straight line with slope $-E_a/R$.
• $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$.
• A higher activation energy gives a stronger response to temperature.
• Near room temperature, $E_a \approx 54\ \text{kJ/mol}$ doubles the rate for each $10\ \text{K}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A reaction goes three times faster when it is warmed by $10\ \text{K}$. Did its activation energy change?

2. You plot $\ln k$ against $1/T$ and get a line that falls from left to right. What does its slope equal?

3. Reaction X has $E_a = 40\ \text{kJ/mol}$ and reaction Y has $E_a = 120\ \text{kJ/mol}$. Which one is more sensitive to a rise in temperature?

4. Why must the temperature be in kelvin in the Arrhenius equation?

5. The rate constant of a reaction has the unit $\text{L mol}^{-1}\ \text{s}^{-1}$. What is the unit of $A$?