How to Find Instantaneous Rate Using a Tangent in HSC Chemistry

A reactant has a concentration of \(0.90\ \mathrm{mol\,L^{-1}}\) at the start of an experiment and \(0.47\ \mathrm{mol\,L^{-1}}\) after 30 seconds. You can calculate its average rate of disappearance over those 30 seconds. But what if an exam asks for its rate of disappearance at exactly 30 seconds?

The answer comes from the gradient of a tangent to the concentration-time graph at \(t=30\ \mathrm{s}\).

To find instantaneous reaction rate using a tangent, draw a straight line that matches the curve’s direction at the required time, then calculate its gradient using two points on that tangent. For a reactant being consumed, the gradient is negative, so reverse its sign to report a positive rate of disappearance.

The important part is choosing the correct line. Using two experimental points on the curve gives an average rate, not necessarily an instantaneous one.

01How concentration-time data are collected

Before calculating a tangent, it helps to understand where the concentration-time graph comes from.

Imagine a reaction in which a coloured reactant, A, is gradually converted into colourless products. We want to measure how quickly the concentration of A changes.

One possible approach uses a colorimeter.

Light passes through the cuvette containing the reaction mixture. A detector measures how much light emerges from the other side.

As the coloured reactant is consumed, the solution becomes less strongly coloured. Its absorbance decreases.

Using a calibration curve, the measured absorbance can be converted into the concentration of A. A data logger can record these measurements at regular time intervals.

For this experiment, we would predict that the concentration of A decreases as the reaction proceeds. If the reaction slows down, the concentration should decrease rapidly at first and then more gradually.

The following is an illustrative dataset, not a record of an actual experiment.

The measurements show a decreasing concentration. However, the amount of concentration lost in each 10-second interval also becomes smaller.

This suggests that the reaction is slowing down.

There is an experimental detail worth noticing. If a student removes samples manually and measures them later, the reaction might continue while each sample is being transferred. The recorded concentration would then correspond to a slightly later time than intended.

Continuous measurement in the cuvette reduces this timing problem, provided that mixing and the start of data collection are properly synchronised.

The data still have limitations. A colorimeter has finite resolution, and its readings depend on suitable calibration, consistent temperature, and a fixed optical path.

These measurements give us a series of concentration values. To estimate the rate at a particular instant, we need to turn those values into a concentration-time curve.

02Why a tangent gives instantaneous rate

A concentration-time graph has:

• Time on the horizontal axis.
• Concentration on the vertical axis.

The gradient tells us how quickly concentration changes with time.

Consider two different questions:

• What is the average rate between 10 and 40 seconds?
• What is the instantaneous rate at 30 seconds?

They sound similar, but they require different lines on the graph.

Average rate: the gradient of a secant

A secant is a straight line joining two points on a curve.

Its gradient is:

\[
\text{Average rate of change}
=
\frac{\Delta[\mathrm{A}]}{\Delta t}
\]

Here, \(\Delta[\mathrm{A}]\) is the change in concentration, and \(\Delta t\) is the time interval.

The two points are taken from the experimental curve.

This calculation gives the average rate over the entire interval. It does not tell us the rate at every individual moment within that interval.

Instantaneous rate: the gradient of a tangent

Now imagine shrinking the time interval around the instant we care about.

The secant gradually changes its direction as the two points move closer together. It approaches the tangent at the chosen point.

The gradient of that tangent represents the instantaneous rate of concentration change.

A useful mental picture is a winding road. From a distance, the road clearly curves. If you examine a tiny section, its direction is almost straight.

A tangent represents that local direction.

The graph does not actually become straight. The tangent simply describes the gradient at one particular point.

You do not need to calculate calculus limits to use this method in HSC Chemistry. You need to draw the tangent accurately and calculate its gradient.

The six-step tangent method

When an exam gives you a concentration-time graph, follow this sequence.

1. Identify the required time. Find the exact instant specified in the question.
2. Locate the point on the curve. Mark the corresponding concentration.
3. Draw a tangent. Use a ruler to draw a straight line that matches the curve’s direction at that point.
4. Choose two points on the tangent. Select convenient coordinates that are reasonably far apart.
5. Calculate the gradient. Divide the change in concentration by the change in time.
6. Interpret the result. Check whether the graph represents a reactant or product, then use the correct sign and units.

The most important detail is step four.

Use two points on the tangent, not two experimental points on the curve.

The tangent coordinates do not have to be actual experimental measurements. They are simply convenient points along the straight tangent line.

04The sign of the gradient depends on what you measure

A common source of mistakes is assuming that every concentration-time graph has a positive gradient.

It doesn’t.

Consider a reaction in which reactant A forms product C.

For a reactant:

\[
\text{Rate of disappearance}
=
-\text{Gradient}
\]

For a product:

\[
\text{Rate of formation}
=
+\text{Gradient}
\]

The sign is not a mistake. It tells us whether the concentration is increasing or decreasing.

What if the question asks for the rate of reaction?

There is one more distinction to understand.

The rate of disappearance of a particular reactant is not always numerically equal to the rate of reaction.

Consider:

\[
\ce{2A(aq) + B(aq) -> 3C(aq)}
\]

The balanced equation tells us that two moles of A disappear for every three moles of C produced.

The reaction rate accounts for these stoichiometric coefficients:

\[
r=-\frac{1}{2}g_A
=\frac{1}{3}g_C
\]

Here:

• \(r\) is the rate of reaction.
• \(g_A\) is the tangent gradient of the concentration of A against time.
• \(g_C\) is the tangent gradient of the concentration of C against time.

If the question asks for the rate of disappearance of A, use the full magnitude of its negative gradient.

If it asks for the rate of reaction, divide that disappearance rate by the coefficient of A.

05Common mistakes when drawing tangents

Before moving into practice, check these common errors.

How accurate is a tangent estimate?

A tangent drawn by hand is an estimate, not an exact measurement.

Experimental data are collected at separate time intervals. There are no infinitely many measurements between the recorded points.

Drawing a smooth curve requires some interpretation, and two students may draw slightly different tangents.

To improve your estimate:

• Plot all the available measurements accurately.
• Draw a smooth curve that follows the overall experimental trend.
• Avoid forcing the curve through an obvious anomalous reading.
• Match the tangent to the local direction of the curve.
• Use two well-separated coordinates on the tangent.
• Report an answer with sensible significant figures.

There is also an important assumption behind concentration-time calculations: the solution volume is constant.

If water is added, for example, the concentration can decrease because of dilution even if no reactant has been consumed. A concentration gradient would then reflect more than just the chemical reaction.

For gas-volume-time graphs, the same tangent principle applies, but the vertical axis and units are different. See How to Calculate Reaction Rate from a Gas-Volume-Time Graph for the gas-syringe version.

The underlying mathematical idea is also shared with Average and Instantaneous Velocity in HSC Physics. The difference is that a chemistry concentration-time graph describes changes in chemical concentration rather than displacement.

06Questions and solutions

The following questions progress from straightforward tangent calculations to more demanding problems involving stoichiometry, experimental uncertainty, and reaction pathways.

Question 1

A student investigates a reaction in which reactant R forms product P:

\[
\ce{R(aq) -> P(aq)}
\]

At \(t=40\ \mathrm{s}\), a tangent to the product concentration-time graph passes through:

• Point A: \((20\ \mathrm{s}, 0.16\ \mathrm{mol\,L^{-1}})\)
• Point B: \((60\ \mathrm{s}, 0.40\ \mathrm{mol\,L^{-1}})\)

Calculate the instantaneous rate of formation of P at 40 seconds.

Give your answer in \(\mathrm{mol\,L^{-1}\,s^{-1}}\).

Solution 1

The instantaneous rate of formation of P is \(0.0060\ \mathrm{mol\,L^{-1}\,s^{-1}}\).

Because P is a product, its concentration increases, so the tangent has a positive gradient.

Step 1: Identify the tangent coordinates.

\[
\begin{aligned}
t_1&=20\ \mathrm{s}\\
t_2&=60\ \mathrm{s}\\
[P]_1&=0.16\ \mathrm{mol\,L^{-1}}\\
[P]_2&=0.40\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Step 2: Calculate the gradient.

\[
\begin{aligned}
\text{Gradient}
&=\frac{0.40-0.16}{60-20}\\
&=\frac{0.24}{40}\\
&=0.0060
\end{aligned}
\]

Step 3: Interpret the sign and units.

The gradient is positive, so it directly represents the rate of product formation.

\[
\boxed{
0.0060\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

The reaction is forming P at an instantaneous rate of \(0.0060\ \mathrm{mol\,L^{-1}}\) each second at \(t=40\ \mathrm{s}\).

Question 2

A reaction occurs in a constant-volume solution:

\[
\ce{2A(aq) + B(aq) -> C(aq)}
\]

At \(t=30\ \mathrm{s}\), a tangent to the concentration-time graph for A passes through:

• Point A: \((10\ \mathrm{s}, 0.84\ \mathrm{mol\,L^{-1}})\)
• Point B: \((50\ \mathrm{s}, 0.52\ \mathrm{mol\,L^{-1}})\)

Additional experimental measurements show that:

• At \(t=0\ \mathrm{s}\), \([A]=1.10\ \mathrm{mol\,L^{-1}}\).
• At \(t=30\ \mathrm{s}\), \([A]=0.68\ \mathrm{mol\,L^{-1}}\).

Calculate:

a) The instantaneous rate of disappearance of A at 30 seconds.

b) The instantaneous rate of reaction at 30 seconds.

c) The average rate of reaction between 0 and 30 seconds.

Explain why the instantaneous and average rates are different.

Solution 2

The instantaneous rate of disappearance of A is \(0.0080\ \mathrm{mol\,L^{-1}\,s^{-1}}\), and the instantaneous rate of reaction is \(0.0040\ \mathrm{mol\,L^{-1}\,s^{-1}}\). The average rate of reaction over the first 30 seconds is \(0.0070\ \mathrm{mol\,L^{-1}\,s^{-1}}\).

The difference comes from the changing gradient of the concentration-time curve.

Step 1: Calculate the tangent gradient.

\[
\begin{aligned}
\text{Gradient}
&=\frac{0.52-0.84}{50-10}\\
&=\frac{-0.32}{40}\\
&=-0.0080
\end{aligned}
\]

Step 2: Calculate the instantaneous rate of disappearance of A.

A is a reactant, so reverse the negative sign.

\[
\boxed{
0.0080\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 3: Calculate the instantaneous rate of reaction.

The coefficient of A is 2.

\[
\begin{aligned}
r&=\frac{0.0080}{2}\\
&=0.0040
\end{aligned}
\]

Therefore:

\[
\boxed{
r=0.0040\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 4: Calculate the average rate of reaction between 0 and 30 seconds.

First, calculate the average rate of disappearance of A:

\[
\begin{aligned}
\text{Average disappearance rate}
&=\frac{1.10-0.68}{30}\\
&=\frac{0.42}{30}\\
&=0.0140
\end{aligned}
\]

Now account for the coefficient of A:

\[
\begin{aligned}
r_{\text{average}}
&=\frac{0.0140}{2}\\
&=0.0070
\end{aligned}
\]

Therefore:

\[
\boxed{
r_{\text{average}}
=0.0070\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

The average rate is greater than the instantaneous rate at 30 seconds.

The average includes the early part of the reaction, when A was disappearing more quickly. The tangent measures the lower rate at the particular instant of 30 seconds.

Question 3

A student uses a colorimeter to investigate the disappearance of reactant A.

The following illustrative data are obtained:

Three repeat measurements at 40 seconds give concentrations of \(0.390\), \(0.392\), and \(0.389\ \mathrm{mol\,L^{-1}}\). The table shows their mean rounded to three decimal places.

A smooth best-fit curve is drawn. A tangent at 30 seconds passes through:

• Point A: \((10\ \mathrm{s}, 0.640\ \mathrm{mol\,L^{-1}})\)
• Point B: \((50\ \mathrm{s}, 0.310\ \mathrm{mol\,L^{-1}})\)

A student argues that the instantaneous rate at 30 seconds should instead be calculated using the experimental points at 20 and 40 seconds because they are closer to the required time.

a) Calculate the instantaneous rate of disappearance of A at 30 seconds using the supplied tangent.

b) Calculate the average rate of disappearance between 20 and 40 seconds.

c) Explain why the student’s method does not give the exact instantaneous rate at 30 seconds.

d) Explain what the repeated measurements tell us about the reliability of the experimental data.

Solution 3

The estimated instantaneous rate of disappearance is \(0.00825\ \mathrm{mol\,L^{-1}\,s^{-1}}\), while the average rate between 20 and 40 seconds is \(0.0100\ \mathrm{mol\,L^{-1}\,s^{-1}}\).

The student’s approach calculates a secant gradient rather than a tangent gradient.

Step 1: Calculate the instantaneous rate using the tangent.

\[
\begin{aligned}
\text{Gradient}_{\text{tangent}}
&=\frac{0.310-0.640}{50-10}\\
&=\frac{-0.330}{40}\\
&=-0.00825
\end{aligned}
\]

Since A is a reactant:

\[
\boxed{
\text{Instantaneous disappearance rate}
=0.00825\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 2: Calculate the average rate between 20 and 40 seconds.

Use the actual experimental measurements:

\[
\begin{aligned}
\text{Average rate}
&=-\frac{0.390-0.590}{40-20}\\
&=\frac{0.200}{20}\\
&=0.0100
\end{aligned}
\]

Therefore:

\[
\boxed{
0.0100\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 3: Explain the difference between the methods.

The student’s calculation uses two experimental points on the curve.

The resulting line is a secant, so its gradient represents the average rate over the interval from 20 to 40 seconds.

It does not necessarily equal the instantaneous rate at 30 seconds because the concentration-time curve is changing direction gradually as the reaction slows.

The tangent provides the best estimate of the local gradient at the specified instant.

Step 4: Evaluate the repeated measurements.

The three measurements at 40 seconds are close together:

\[
0.389,\quad 0.390,\quad 0.392
\]

This suggests good repeatability at that measurement time.

However, repeatability does not guarantee that the entire experiment is free from systematic errors. Calibration errors, temperature changes, and delays in measurement could still affect the results.

The tangent itself is also an estimate based on the shape of the fitted curve.

Question 4

A student investigates the reaction:

\[
\ce{A(aq) -> P(aq)}
\]

The reaction occurs in a constant-volume solution. At \(t=50\ \mathrm{s}\), a catalyst is added. Assume the catalyst addition is effectively instantaneous and its added volume is negligible.

The concentration of A is \(0.30\ \mathrm{mol\,L^{-1}}\) at the moment the catalyst is added.

The following additional measurements are recorded:

The tangent immediately before catalyst addition passes through:

• \((30\ \mathrm{s}, 0.38\ \mathrm{mol\,L^{-1}})\)
• \((70\ \mathrm{s}, 0.22\ \mathrm{mol\,L^{-1}})\)

The tangent immediately after catalyst addition passes through:

• \((30\ \mathrm{s}, 0.50\ \mathrm{mol\,L^{-1}})\)
• \((70\ \mathrm{s}, 0.10\ \mathrm{mol\,L^{-1}})\)

a) Calculate the instantaneous rate of disappearance of A immediately before the catalyst is added.

b) Calculate the instantaneous rate of disappearance of A immediately after the catalyst is added.

c) Calculate the average rate of disappearance between 40 and 60 seconds.

d) Explain why there is no single tangent gradient at the exact instant of catalyst addition in this idealised model.

e) Does the catalyst cause an instantaneous decrease in the concentration of A? Explain.

Solution 4

The instantaneous disappearance rate increases from \(0.0040\) to \(0.0100\ \mathrm{mol\,L^{-1}\,s^{-1}}\) after the catalyst is added. The concentration remains continuous at the moment of addition, but the gradient changes.

Step 1: Calculate the rate immediately before catalyst addition.

\[
\begin{aligned}
\text{Gradient}_{\text{before}}
&=\frac{0.22-0.38}{70-30}\\
&=\frac{-0.16}{40}\\
&=-0.0040
\end{aligned}
\]

Therefore:

\[
\boxed{
0.0040\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 2: Calculate the rate immediately after catalyst addition.

\[
\begin{aligned}
\text{Gradient}_{\text{after}}
&=\frac{0.10-0.50}{70-30}\\
&=\frac{-0.40}{40}\\
&=-0.0100
\end{aligned}
\]

Therefore:

\[
\boxed{
0.0100\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

The rate of disappearance has increased by a factor of:

\[
\frac{0.0100}{0.0040}=2.5
\]

Step 3: Calculate the average rate between 40 and 60 seconds.

\[
\begin{aligned}
\text{Average disappearance rate}
&=\frac{0.35-0.22}{60-40}\\
&=\frac{0.13}{20}\\
&=0.0065
\end{aligned}
\]

Therefore:

\[
\boxed{
0.0065\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

This average lies between the two instantaneous rates.

It does not have to equal their arithmetic mean because the reaction rate may also change during the intervals before and after catalyst addition.

Step 4: Explain the tangent at the moment of catalyst addition.

Immediately before addition, the curve has a gradient of \(-0.0040\).

Immediately after addition, it has a gradient of \(-0.0100\).

The two one-sided tangents have different gradients.

Therefore, in this idealised model, there is no single tangent gradient at exactly 50 seconds.

In a real experiment, mixing and instrument response take a finite amount of time, so the transition may appear smoother.

Step 5: Explain what happens to concentration.

The concentration of A does not suddenly decrease when the catalyst is added.

It remains \(0.30\ \mathrm{mol\,L^{-1}}\) at the instant of addition.

The catalyst changes the reaction rate, not the amount of reactant instantaneously present.

This distinction matters when interpreting concentration-time graphs that change shape after an experimental intervention.

Question 5

A reaction occurs through two consecutive steps in a constant-volume solution:

\[
\ce{A -> B}
\]

\[
\ce{B -> C}
\]

Assume:

• Both reaction steps have a 1:1 stoichiometric relationship.
• B is produced only by the first reaction.
• B is consumed only by the second reaction.
• There are no competing reactions or side products.

At \(t=60\ \mathrm{s}\), the concentration-time graph for A has a tangent passing through:

• \((40\ \mathrm{s}, 0.28\ \mathrm{mol\,L^{-1}})\)
• \((80\ \mathrm{s}, 0.12\ \mathrm{mol\,L^{-1}})\)

At the same instant, the concentration-time graph for B is horizontal at \(0.20\ \mathrm{mol\,L^{-1}}\). In fact, its concentration remains approximately constant between 50 and 70 seconds.

A student concludes:

“The tangent to B’s graph is horizontal, so B is not reacting at 60 seconds. The reaction must have stopped producing and consuming B.”

Evaluate this conclusion.

Calculate:

a) The instantaneous rate of disappearance of A.

b) The instantaneous rate of formation of B from A.

c) The instantaneous rate of disappearance of B to form C.

d) The instantaneous rate of formation of C.

Explain why a constant concentration of B does not mean that the reactions involving B have stopped.

Solution 5

The student’s conclusion is incorrect. B is being formed and consumed simultaneously at equal instantaneous rates. Its concentration remains constant because the amount entering the B pool equals the amount leaving it.

Step 1: Calculate the instantaneous rate of disappearance of A.

Use the tangent coordinates:

\[
\begin{aligned}
\text{Gradient}_{A}
&=\frac{0.12-0.28}{80-40}\\
&=\frac{-0.16}{40}\\
&=-0.0040
\end{aligned}
\]

Since A is a reactant:

\[
\boxed{
\text{Rate of disappearance of A}
=0.0040\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 2: Calculate the rate of formation of B.

The first reaction is:

\[
\ce{A -> B}
\]

The stoichiometric relationship is 1:1.

Therefore, B forms at the same rate that A disappears:

\[
\boxed{
\text{Rate of formation of B}
=0.0040\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 3: Calculate the rate of disappearance of B.

The tangent to the B concentration-time graph is horizontal.

Therefore, the net rate of concentration change of B is zero.

However, this does not mean that its formation and consumption rates are both zero.

The concentration of B can only remain constant if the rate at which B is produced equals the rate at which it is consumed.

Therefore:

\[
\begin{aligned}
\text{Formation rate of B}
&=\text{Disappearance rate of B}\\
&=0.0040
\end{aligned}
\]

Hence:

\[
\boxed{
\text{Rate of disappearance of B}
=0.0040\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 4: Calculate the rate of formation of C.

The second reaction is:

\[
\ce{B -> C}
\]

Again, the stoichiometric relationship is 1:1.

Therefore, C forms at the same rate that B disappears:

\[
\boxed{
\text{Rate of formation of C}
=0.0040\ \mathrm{mol\,L^{-1}\,s^{-1}}
}
\]

Step 5: Explain why the horizontal tangent is not evidence that the reaction has stopped.

The key distinction is between the net change in concentration and the individual rates of formation and consumption.

For B:

\[
\begin{aligned}
\text{Net rate of change of B}
&=\text{Formation rate}\\
&\quad-\text{Disappearance rate}\\
&=0.0040-0.0040\\
&=0
\end{aligned}
\]

The horizontal tangent correctly shows that the concentration of B is not changing.

It does not show that individual B particles are inactive or that neither reaction is occurring.

This is a steady concentration of an intermediate, maintained by two equal and opposing flows. It is not necessarily chemical equilibrium.

The important lesson is that a tangent measures the net rate of concentration change of the species plotted. To understand what is happening throughout a reaction pathway, you must also consider the balanced equations and the other species involved.

That is the next step in interpreting more complex reaction profiles: moving from the gradient of one concentration-time curve to the relationships between the rates of several simultaneous or consecutive reactions.