Half-life and pseudo first order reactions

The half-life of a first order reaction

We met the half-life of a zero order reaction in the first lesson. Now take a first order reaction. Put $[R] = [R]_0/2$ and $t = t_{1/2}$ into $\ln\dfrac{[R]_0}{[R]} = kt$:

We want $t_{1/2}$, so $t_{1/2} = \dfrac{0.693}{k}$. Look closely at this. $[R]_0$ is not in it. The half-life of a first order reaction is the same whatever the start concentration. It is also the same at every stage: the time to go from $0.64\ \text{mol/L}$ to $0.32\ \text{mol/L}$ equals the time to go from $0.04\ \text{mol/L}$ to $0.02\ \text{mol/L}$.

If you know the half-life, you also know $k$: $k = \dfrac{0.693}{t_{1/2}}$. The unit of $k$ is one over time, so a half-life in minutes gives $k$ in $\text{min}^{-1}$.

The left graph shows this. Four equal gaps of time halve the concentration four times. The right graph is a zero order reaction. Its first half takes $24\ \text{min}$, the next takes $12\ \text{min}$ and the next $6\ \text{min}$, because each half-life is measured from the smaller start.

How much is left after n half-lives

After one half-life, half is left. After two, half of that is left, and so on. After $n$ half-lives the fraction left is $\left(\tfrac{1}{2}\right)^n$, where $n = \dfrac{t}{t_{1/2}}$. The number $n$ need not be a whole number.

• After 1 half-life: $\tfrac{1}{2}$ is left, which is $50\ \%$.
• After 2 half-lives: $\tfrac{1}{4}$, which is $25\ \%$.
• After 3 half-lives: $\tfrac{1}{8}$, which is $12.5\ \%$.
• After 4 half-lives: $\tfrac{1}{16}$, which is $6.25\ \%$.
• After 5 half-lives: $\tfrac{1}{32}$, which is about $3.1\ \%$.

A worked example

A first order reaction has a half-life of $12\ \text{min}$ and starts at $0.64\ \text{mol/L}$. How much is left after $36\ \text{min}$? How long does it take to fall to $0.01\ \text{mol/L}$?

• After 36 min. $n = \dfrac{36\ \text{min}}{12\ \text{min}} = 3$ half-lives. $[R] = 0.64\ \text{mol/L} \times \tfrac{1}{8} = 0.08\ \text{mol/L}$.
• To 0.01 mol/L. The ratio is $\dfrac{0.64}{0.01} = 64 = 2^6$. That is 6 half-lives, so $6 \times 12\ \text{min} = 72\ \text{min}$.
• Check with the equation. $k = \dfrac{0.693}{12\ \text{min}} = 0.0578\ \text{min}^{-1}$, and $t = \dfrac{\ln 64}{k} = \dfrac{4.159}{0.0578\ \text{min}^{-1}} = 72\ \text{min}$.

The time for any fraction

Half-life is one special point. For any fraction $f$ that has reacted, a first order reaction needs:

• 75 % reacted. The ratio is 4, so $t = \dfrac{1.386}{k}$. That is $2t_{1/2}$.
• 99 % reacted. The ratio is 100, so $t = \dfrac{4.606}{k}$. That is about $6.6t_{1/2}$.
• 99.9 % reacted. The ratio is 1000, so $t$ is about $10t_{1/2}$. A first order reaction is practically over in ten half-lives.

For a zero order reaction the times are different, because the line is straight. For 75 % to react, $t = \dfrac{0.75[R]_0}{k} = 1.5t_{1/2}$, and the reaction ends at $2t_{1/2}$.

Finding the order from the half-life

Run the reaction twice, with different start concentrations, and measure the half-life each time. The way it changes tells you the order:

• Zero order: the half-life is proportional to $[R]_0$. Half the start gives half the half-life.
• First order: the half-life does not change.
• Second order: the half-life is proportional to $\dfrac{1}{[R]_0}$. Half the start gives twice the half-life. For a reaction with one reactant, $t_{1/2} = \dfrac{1}{k[R]_0}$.

An example: a reaction has $t_{1/2} = 18\ \text{min}$ at $0.40\ \text{mol/L}$ and $t_{1/2} = 9\ \text{min}$ at $0.20\ \text{mol/L}$. Half the start gave half the half-life, so the reaction is zero order. Then $k = \dfrac{[R]_0}{2t_{1/2}} = \dfrac{0.40\ \text{mol/L}}{2 \times 18\ \text{min}} = 0.0111\ \text{mol L}^{-1}\,\text{min}^{-1}$.

Practise with new numbers

Radioactive decay

The nuclei of a radioactive atom decay at random. The number that decay in each second is proportional to the number present, so the decay follows a first order law, and the half-life is constant. The same equations work, with the number of nuclei in place of the concentration.

Carbon-14 has a half-life of $5730$ years. A living tree takes in carbon-14 from the air and holds a fixed share of it. After the tree is cut, no more is taken in, and the share falls. A wooden bowl from a dig has one quarter of the carbon-14 share of a living tree. One quarter is two half-lives, so the bowl is about $2 \times 5730\ \text{y} = 11\,460\ \text{y}$ old.

Pseudo first order reactions

Many reactions have two reactants. Take $\text{A} + \text{B} \rightarrow$ products, with rate $= k[\text{A}][\text{B}]$. The reaction is first order in each, so the overall order is two.

Now suppose B is in large excess, perhaps 100 times as much as A. When all of A has reacted, only a tiny part of B has been used, so $[\text{B}]$ hardly changes. Then $[\text{B}]$ is almost equal to $[\text{B}]_0$ for the whole run, and the rate is:

where $k' = k[\text{B}]_0$. The reaction looks first order in A. It is called a pseudo first order reaction, because the word pseudo means "seems to be". The unit of $k'$ is $\text{s}^{-1}$, while the true rate constant $k$ has the unit $\text{L mol}^{-1}\,\text{s}^{-1}$.

Two examples:

• Sucrose in dilute acid. $\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6$, giving glucose and fructose. The acid is a catalyst. Water is the solvent, about $55.5\ \text{mol/L}$, and almost none of it is used. So the rate behaves as $k'[\text{sucrose}]$.
• A dye in alkali. A coloured dye D fades as it reacts with hydroxide ions. With hydroxide in large excess, the colour fades by a first order law.

A worked example for the dye: the rate constant is $k = 0.45\ \text{L mol}^{-1}\,\text{s}^{-1}$, and hydroxide is at $0.10\ \text{mol/L}$. Then $k' = 0.45\ \text{L mol}^{-1}\,\text{s}^{-1} \times 0.10\ \text{mol/L} = 0.045\ \text{s}^{-1}$, and the half-life of the dye is $\dfrac{0.693}{0.045\ \text{s}^{-1}} = 15.4\ \text{s}$. With hydroxide at $0.20\ \text{mol/L}$, $k'$ doubles to $0.090\ \text{s}^{-1}$ and the half-life is halved, to $7.70\ \text{s}$.

That is the test for a pseudo first order reaction. Change the excess and $k'$ changes. A true first order reaction has a rate constant that does not depend on the concentration of anything.

Try it in Python

The program takes the half-life of the dye in three runs, each with a different excess of alkali. It tests whether $k'$ is proportional to the excess.

Common mistakes

• Using a different half-life formula for each stage. A first order half-life is the same at every stage.
• Turning the formula upside down. $t_{1/2} = 0.693/k$, so a bigger $k$ gives a shorter half-life.
• Counting half-lives as if they took away equal amounts. After 3 half-lives, $\tfrac{1}{8}$ is left. It is not zero and it is not $\tfrac{1}{6}$.
• Calling a pseudo first order reaction first order overall. The true order is two. It only looks first order.
• Using $k'$ as the true rate constant. $k = k'/[\text{B}]_0$.

Remember these

• First order: $t_{1/2} = \dfrac{0.693}{k}$, the same at every stage and for every start concentration.
• Zero order: $t_{1/2} = \dfrac{[R]_0}{2k}$, which is proportional to the start concentration.
• After $n$ half-lives, $\left(\tfrac{1}{2}\right)^n$ is left. A first order reaction is practically over in about ten half-lives.
• Pseudo first order: one reactant in large excess, $k' = k[\text{B}]_0$, and $k = k'/[\text{B}]_0$.

Check yourself

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

1. A first order reaction has $k = 0.0231\ \text{s}^{-1}$. What is its half-life?

2. What fraction of a first order reactant is left after 5 half-lives?

3. The half-life of a reaction falls from $24\ \text{min}$ to $12\ \text{min}$ when the start concentration is halved. What is the order?

4. Why does the fading of a dye in a large excess of alkali follow a first order law, although the reaction involves two reactants?

5. In a pseudo first order reaction the excess of B is doubled. What happens to $k'$, and what happens to the true rate constant $k$?

6. A first order reaction takes $20\ \text{min}$ for 50 % to react. Roughly how long until it is 99.9 % complete?