First order reactions

A medicine in the blood

You take a tablet. The body starts to clear the medicine from the blood. In the first hour, say a fifth of it goes. In the second hour, a fifth of what is left goes. When there is a lot in the blood, a lot is cleared. When there is little, little is cleared.

The speed follows the amount present. This is a first order reaction, and it is the most common kind you will meet.

The rate law

For a first order reaction the rate is proportional to the concentration of one reactant:

The unit of $k$ is the unit of the rate divided by the unit of the concentration. That gives $\dfrac{\text{mol L}^{-1}\,\text{min}^{-1}}{\text{mol L}^{-1}} = \text{min}^{-1}$. So $k$ has the unit of one over time, such as $\text{s}^{-1}$. It does not depend on how the concentration is measured.

The integrated equation

We want $[R]$ at any time. First put every $[R]$ on one side:

Now integrate both sides. The integral of $1/[R]$ is $\ln[R]$:

At $t = 0$ we have $[R] = [R]_0$, so $I = \ln[R]_0$. The integrated rate equation of a first order reaction is:

It can be written in three other ways, and you should be able to move between them:

• $\ln\dfrac{[R]_0}{[R]} = kt$
• $[R] = [R]_0\,e^{-kt}$
• $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$, which uses logarithms to base 10, because $\ln x = 2.303\log x$.

We use the last form when we want $k$, and we use the second form when we want $[R]$.

Two graphs of one reaction

On the left, $[R]$ against $t$ is a curve. It falls fast at first and then more and more slowly. It only approaches zero and never quite reaches it. Look at the red dots. The concentration halves, then halves again, then again, and each step takes the same time.

On the right, $\ln[R]$ against $t$ is a straight line. Its slope is $-k$ and its intercept is $\ln[R]_0$. This is the test for a first order reaction: if $\ln[R]$ against $t$ is a straight line, the reaction is first order. A graph of $\log[R]$ is also straight, with slope $-k/2.303$.

A worked example

A first order reaction has $k = 0.035\ \text{min}^{-1}$ and starts at $[R]_0 = 0.60\ \text{mol/L}$. What is the concentration after $20\ \text{min}$?

• The exponent. $kt = 0.035\ \text{min}^{-1} \times 20\ \text{min} = 0.70$. It has no unit.
• The concentration. $[R] = [R]_0\,e^{-kt} = 0.60\ \text{mol/L} \times e^{-0.70} = 0.60\ \text{mol/L} \times 0.497 = 0.298\ \text{mol/L}$.

Now the other way round. In a different run, $[R]$ is $0.60\ \text{mol/L}$ at the start and $0.24\ \text{mol/L}$ after $40\ \text{min}$. We want $k$, so:

You do not have to start at $t = 0$. Any two readings work: $k = \dfrac{2.303}{t_2 - t_1}\log\dfrac{[R]_1}{[R]_2}$.

Practise with new numbers

The straight line in ln [R]

The straight line is the reason chemists plot $\ln[R]$. A rate constant is the hardest thing to read from a curve and the easiest to read from the slope of a line.

A gas reaction followed by pressure

A gas that is breaking up is not easy to sample. The total pressure in a closed vessel is easy to read. At a fixed volume and temperature the pressure of a gas is proportional to its concentration, so pressure can take the place of concentration.

Take $\text{A}(g) \rightarrow \text{B}(g) + \text{C}(g)$ in a closed vessel that starts with pure A at a pressure $p_i$. Suppose the pressure of A has fallen by $x$. Then B and C have each risen by $x$. The total pressure is:

So $x = p_t - p_i$, and the pressure of A is $p_i - x = 2p_i - p_t$. For a first order reaction:

Watch out. This form belongs to this equation, one gas giving two. For another equation, work out $x$ again in the same way.

Here is an example. A gas A starts at $0.36\ \text{bar}$. After $15\ \text{min}$ the total pressure is $0.47\ \text{bar}$. The pressure of A is $2 \times 0.36\ \text{bar} - 0.47\ \text{bar} = 0.25\ \text{bar}$. Then $k = \dfrac{2.303}{15\ \text{min}}\log\dfrac{0.36}{0.25} = 0.0243\ \text{min}^{-1}$.

Try it in Python

The program fits a straight line through $\ln[R]$ for five readings. Then it uses the slope to look ahead.

Common mistakes

• Using $\log$ and forgetting 2.303. Either use $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$ or use $\ln$ without the 2.303.
• Putting the readings upside down. It is $[R]_0$ over $[R]$. The ratio is more than 1, so the logarithm is positive.
• Giving $k$ the unit $\text{mol L}^{-1}\,\text{s}^{-1}$. For first order $k$ is $\text{s}^{-1}$.
• Taking the total pressure as the pressure of A. Work out $2p_i - p_t$ first.
• Thinking $k$ changes as the reaction goes on. The rate falls, but $k$ stays the same at a fixed temperature.

Remember these

• First order: rate $= k[R]$, and $k$ has the unit $\text{time}^{-1}$.
• $\ln[R] = \ln[R]_0 - kt$, or $[R] = [R]_0e^{-kt}$, or $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$.
• A graph of $\ln[R]$ against $t$ is a straight line with slope $-k$.
• For $\text{A}(g) \rightarrow \text{B}(g) + \text{C}(g)$, the pressure of A is $2p_i - p_t$.

Check yourself

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

1. The concentration of a reactant falls from $0.50\ \text{mol/L}$ to $0.40\ \text{mol/L}$ in $5\ \text{min}$. If the reaction is first order, what is $k$?

2. A graph of $\ln[R]$ against time is a straight line with a slope of $-0.020\ \text{min}^{-1}$. What do you learn from it?

3. The concentration of a reactant in a first order reaction is doubled. Does $k$ change? Does the rate change?

4. Why does a first order reaction never reach zero concentration in the equation, while a zero order reaction does?

5. A gas A breaks up as $\text{A}(g) \rightarrow \text{B}(g) + \text{C}(g)$ at a fixed volume. It starts at $0.30\ \text{bar}$, and later the total pressure is $0.40\ \text{bar}$. What is the pressure of A at that time?