Rate of a reaction

Fast and slow

A match burns in a second. Milk turns sour in a day in a hot kitchen and in a week in a fridge. An iron nail needs months to rust, and a diamond takes far longer than that to turn into graphite. Reactions run at very different speeds.

A chemist wants a number for the speed. A bus that covers $60\ \text{km}$ in one hour has a speed of $60\ \text{km/h}$: the distance divided by the time. A reaction has a speed in the same way. We ask how much reactant is used up, or how much product forms, in one unit of time. This number is the rate of the reaction.

Writing the rate

We measure the amount of a substance as a concentration, in mol/L. A square bracket stands for concentration: $[\text{R}]$ is the concentration of R.

Take the simplest reaction, $\text{R} \rightarrow \text{P}$. As time goes on, $[\text{R}]$ falls and $[\text{P}]$ rises. The rate can be written in two ways:

The symbol $\Delta$ means "change in". The change $\Delta[\text{R}]$ is negative because R is used up. A rate must be positive, so we put a minus sign in front. This gives the rate of disappearance of R. The change $\Delta[\text{P}]$ is already positive. This gives the rate of appearance of P.

The unit of rate is concentration divided by time: $\text{mol L}^{-1}\ \text{s}^{-1}$. For a slow reaction we may use $\text{mol L}^{-1}\ \text{min}^{-1}$ or $\text{mol L}^{-1}\ \text{h}^{-1}$. For a gas we can use pressure instead of concentration, and then the unit is $\text{bar s}^{-1}$.

A reaction with more than one substance

Most reactions have several reactants and products, and they do not change at the same speed. Look at the equation:

Four molecules of ammonia are used for every five of oxygen. So oxygen disappears faster than ammonia, and water appears faster than nitric oxide. Which of these speeds is "the rate of the reaction"?

We divide each rate by its coefficient in the balanced equation. All the results are then the same number, and that number is the rate of the reaction:

Suppose ammonia disappears at $0.0240\ \text{mol L}^{-1}\ \text{s}^{-1}$. Then:

• The rate of the reaction is $\dfrac{0.0240\ \text{mol L}^{-1}\ \text{s}^{-1}}{4} = 0.00600\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Oxygen disappears at $5 \times 0.00600 = 0.0300\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Nitric oxide appears at $4 \times 0.00600 = 0.0240\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Water appears at $6 \times 0.00600 = 0.0360\ \text{mol L}^{-1}\ \text{s}^{-1}$.

Check the ratios. Ammonia, oxygen, nitric oxide and water change in the ratio $4 : 5 : 4 : 6$, as the equation says.

For the general reaction $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$, the rate is $-\dfrac{1}{a}\dfrac{\Delta[\text{A}]}{\Delta t} = +\dfrac{1}{c}\dfrac{\Delta[\text{C}]}{\Delta t}$, and the same for B and D.

Average rate

Here are readings for a reactant A in solution. A coloured compound is breaking down, and a colour meter gives the concentration.

The average rate over an interval is the fall in concentration divided by the time taken. Take each 100 s in turn:

• $0$ to $100\ \text{s}$: $\dfrac{0.900\ \text{mol/L} - 0.675\ \text{mol/L}}{100\ \text{s}} = 2.25 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• $100$ to $200\ \text{s}$: $\dfrac{0.675\ \text{mol/L} - 0.506\ \text{mol/L}}{100\ \text{s}} = 1.69 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• $200$ to $300\ \text{s}$: $1.26 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• $300$ to $400\ \text{s}$: $0.95 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• $400$ to $500\ \text{s}$: $0.71 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• $500$ to $600\ \text{s}$: $0.54 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$.

The rate keeps falling. Why? As A is used up, there are fewer molecules of A in each litre. They meet other molecules less often, so fewer of them react in each second. Most reactions slow down as they go on.

The rate at one moment

An average rate hides the changes inside the interval. The reaction was faster at the start of the interval than at the end. What is the rate at exactly $200\ \text{s}$?

Make the interval shorter and shorter. On the graph, the line joining two points on the curve is a secant. As the two points come together, the secant turns into the tangent, a line that touches the curve at one point. The slope of the tangent is the rate at that moment, the instantaneous rate:

To find the rate at $200\ \text{s}$ from the graph:

1. Draw the tangent to the curve at $200\ \text{s}$.
2. Pick two points far apart on the tangent. The two places where it meets the axes are convenient.
3. Divide the change in concentration by the change in time. Give the answer a positive sign.

The tangent in the picture meets the concentration axis at $0.798\ \text{mol/L}$ and the time axis at $548\ \text{s}$. So the rate at $200\ \text{s}$ is $\dfrac{0.798\ \text{mol/L}}{548\ \text{s}} = 1.46 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$.

The average rate from $100\ \text{s}$ to $300\ \text{s}$ is $\dfrac{0.675\ \text{mol/L} - 0.380\ \text{mol/L}}{200\ \text{s}} = 1.48 \times 10^{-3}\ \text{mol L}^{-1}\ \text{s}^{-1}$. This is close to the rate at the middle of the interval, but it is not the same number.

The rate at the very start, at $t = 0$, is the initial rate. We use it in the next lesson.

Slide the interval down to $20\ \text{s}$. The average rate and the rate at time $t$ become almost equal. A short interval is as good as a tangent.

Practise with new numbers

How do we follow a reaction?

We cannot count molecules one by one. We measure some property that changes as the reaction goes, and convert it to concentration:

• Colour. A colour meter gives the amount of a coloured substance.
• Gas. A reaction that makes or uses a gas changes the volume or the pressure of a closed vessel.
• Titration. Take a small sample at fixed times, stop the reaction by cooling or diluting it, and titrate.
• Electrical conductivity. Ions are made or used up, so the conductivity changes.

What changes the rate?

The rate of a reaction changes with the concentration of the reactants, with temperature, with a catalyst, with the surface area of a solid and with light. Powdered chalk reacts faster with acid than one lump of the same mass. This is why the surface area matters.

The effect of concentration is the main topic of this lesson and the next. Temperature and catalysts need a study of their own.

Try it in Python

The program takes the table above and finds the average rate in each interval. It ends with the rate over the whole run.

Common mistakes

• A negative rate. The change in a reactant is negative, so put a minus sign in front. The rate itself is positive.
• One rate for every substance. Different substances change at different speeds. Divide by the coefficient to get the rate of the reaction.
• Dividing by the coefficient twice, or not at all. To go from the rate of ammonia to the rate of oxygen, divide by 4 and multiply by 5.
• Mixing time units. If the concentration is in mol/L and the time is in minutes, the rate is in $\text{mol L}^{-1}\ \text{min}^{-1}$. Convert before you compare.
• Calling an average rate the rate at a moment. The average is a secant slope. The rate at a moment is a tangent slope.
• Expecting a constant rate. Most reactions slow down as the reactants are used up.

Remember these

• Rate $= -\dfrac{\Delta[\text{reactant}]}{\Delta t} = +\dfrac{\Delta[\text{product}]}{\Delta t}$, in $\text{mol L}^{-1}\ \text{s}^{-1}$.
• For $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$, divide each rate by its coefficient to get the rate of the reaction.
• Average rate: the slope of a secant. Instantaneous rate: the slope of a tangent.
• The initial rate is the rate at $t = 0$.
• Most reactions slow down as the reactants are used up.

Check yourself

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

1. In the reaction $2\text{HI}(g) \rightarrow \text{H}_2(g) + \text{I}_2(g)$, hydrogen iodide is used up at $0.0180\ \text{mol L}^{-1}\ \text{min}^{-1}$. At what rate does hydrogen form?

2. A rate is always written as a positive number, but the concentration of a reactant falls. How do the two fit together?

3. For $2\text{A} \rightarrow 3\text{B}$, the concentration of B rises at $0.045\ \text{mol L}^{-1}\ \text{s}^{-1}$. At what rate does A fall?

4. On a graph of concentration against time, the first minute has a steeper curve than the second minute. Which minute has the larger average rate, and why?

5. How would you find the rate at $250\ \text{s}$ from a graph of concentration against time?