Zero order reactions

A queue at one counter

A ticket counter serves four people a minute. It does not matter whether the queue has twenty people or two hundred. The counter works at its own steady speed, so the queue gets shorter by four people every minute.

Some reactions behave like that counter. There is plenty of reactant waiting, but only a fixed amount can react in each minute. The speed does not depend on how much is waiting. We call this a zero order reaction.

The rate law

For a zero order reaction, the rate does not depend on the concentration:

Here $[R]$ is the concentration of the reactant in $\text{mol/L}$. The symbol $d[R]/dt$ is how fast it changes, and the minus sign is there because the reactant is falling. Anything to the power zero is 1, so the concentration drops out and the rate is just the constant $k$.

Because rate equals $k$, the unit of $k$ is the unit of rate: $\text{mol L}^{-1}\,\text{min}^{-1}$, or $\text{mol L}^{-1}\,\text{s}^{-1}$ if time is in seconds.

The integrated equation

We want the concentration at any time. The rate law says $d[R] = -k\,dt$. Add up all the small changes, which is to integrate both sides:

The letter $I$ is a constant. At the start, $t = 0$ and $[R] = [R]_0$. So $I = [R]_0$, and the integrated rate equation of a zero order reaction is:

In words: every minute the concentration falls by the same amount, $k$. This has the shape of a straight line, $y = c + mx$. A graph of $[R]$ against $t$ is a straight line with intercept $[R]_0$ and slope $-k$. The right-hand graph shows that the rate against concentration is a flat line.

A worked example

A zero order reaction starts at $0.45\ \text{mol/L}$ and has $k = 0.015\ \text{mol L}^{-1}\,\text{min}^{-1}$. What is the concentration after $12\ \text{min}$? When is the reactant all used up?

• After 12 min. $[R] = 0.45\ \text{mol/L} - 0.015\ \text{mol L}^{-1}\,\text{min}^{-1} \times 12\ \text{min} = 0.27\ \text{mol/L}$.
• All used up. Put $[R] = 0$ in the equation. We want $t$, so $t = \dfrac{[R]_0}{k} = \dfrac{0.45\ \text{mol/L}}{0.015\ \text{mol L}^{-1}\,\text{min}^{-1}} = 30\ \text{min}$.

Notice something new. A zero order reaction ends at a fixed time. After $30\ \text{min}$ the equation stops being true, because a concentration cannot be negative.

Half-life

The half-life, $t_{1/2}$, is the time for the concentration to fall to half of its starting value. Put $[R] = [R]_0/2$ and $t = t_{1/2}$ into the equation:

We want $t_{1/2}$, so $t_{1/2} = \dfrac{[R]_0}{2k}$.

For the example, $t_{1/2} = \dfrac{0.45\ \text{mol/L}}{2 \times 0.015\ \text{mol L}^{-1}\,\text{min}^{-1}} = 15\ \text{min}$. The reaction ends at $2t_{1/2} = 30\ \text{min}$, which agrees.

The half-life has $[R]_0$ in it. Start with twice as much and the half-life doubles. A reaction that begins with a longer queue takes longer to halve it.

Practise with new numbers

Where do zero order reactions happen?

In every case, something other than the reactant concentration sets the speed.

• A gas on a hot metal surface. The surface has a fixed number of places where molecules can sit and react. When there is enough gas to fill all of them, the surface is the ticket counter. More gas does not make it faster.
• An enzyme. The liver breaks down ethanol with an enzyme. When there is a lot of ethanol, every enzyme molecule is busy all the time. So the liver removes about the same mass of ethanol each hour.
• A reaction driven by steady light. If the light, not the reactant, limits the speed, the rate stays the same while the concentration changes.

A zero order reaction does not have a rate of zero. The rate is $k$. "Zero order" means only that the rate does not change when the concentration changes, over the range we study. If the gas on the metal surface falls to a very low pressure, the surface is no longer full. Then the rate does start to fall, and the order changes.

Try it in Python

The program fits a straight line through five readings of a zero order reaction. The slope gives $k$.

Common mistakes

• Thinking a zero order reaction has no rate. The rate is $k$. It is constant, not zero.
• Using the equation after the reactant is gone. $[R] = [R]_0 - kt$ holds only up to $t = [R]_0/k$.
• Giving $k$ the unit $\text{s}^{-1}$. For zero order, $k$ has the unit of a rate, $\text{mol L}^{-1}\,\text{s}^{-1}$.
• Calling the half-life constant. For zero order it is $[R]_0/2k$, so it changes with the start concentration.
• Reading the slope as $+k$. The line falls. The slope is $-k$, and $k$ itself is positive.

Remember these

• Zero order: rate $= k$, and the unit of $k$ is $\text{mol L}^{-1}\,\text{time}^{-1}$.
• $[R] = [R]_0 - kt$. A graph of $[R]$ against $t$ is a straight line with slope $-k$.
• $t_{1/2} = \dfrac{[R]_0}{2k}$. The reaction ends at $t = \dfrac{[R]_0}{k}$, which is $2t_{1/2}$.
• Zero order appears when a surface, an enzyme or the light limits the speed.

Check yourself

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

1. At $[R] = 0.50\ \text{mol/L}$ the rate of a zero order reaction is $4.0 \times 10^{-3}\ \text{mol L}^{-1}\,\text{s}^{-1}$. What is the rate when $[R] = 0.10\ \text{mol/L}$?

2. The concentration is in $\text{mol/L}$ and the time is in hours. What is the unit of $k$ for a zero order reaction?

3. The concentration in a zero order reaction falls from $0.50\ \text{mol/L}$ to $0.30\ \text{mol/L}$ in $10\ \text{min}$. Find $k$ and the time at which the reactant is all used up.

4. A gas breaks down on a hot metal wire by a zero order reaction. Why does the order change when the pressure of the gas becomes very low?

5. In one run a zero order reaction starts at $0.20\ \text{mol/L}$ and its half-life is $6\ \text{min}$. A second run starts at $0.40\ \text{mol/L}$. What is the half-life now?