How to Calculate Reaction Rate from a Gas-Volume-Time Graph

Add a strip of magnesium to excess hydrochloric acid and collect the hydrogen in a gas syringe. At first, the syringe fills quickly. Then the gas arrives more slowly, until the volume stops increasing.

Suppose the syringe contains 44.2 mL of gas after 20 seconds. Is the reaction producing hydrogen at 44.2 ÷ 20 = 2.21 mL/s at that exact moment?

Not necessarily. That calculation gives the average rate over the first 20 seconds, not the rate at exactly 20 seconds.

To calculate reaction rate from a gas-volume-time graph, find its gradient:

• Average rate: calculate the gradient between two points on the curve.
• Instantaneous rate: draw a tangent at the required time and calculate the gradient of that tangent.

The distinction is important because reactions usually slow down as their reactants are consumed. The gradient changes throughout the experiment.

01What does the gradient of a gas-volume-time graph represent?

A gas-volume-time graph shows how the volume of gas collected changes as a reaction proceeds.

• The horizontal axis represents time, usually in seconds (s).
• The vertical axis represents the volume of gas collected, usually in millilitres (mL) or cubic centimetres (cm³).

Consider the reaction between magnesium and hydrochloric acid:

\[ \ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \]

Hydrogen gas is produced as the reaction proceeds. The gas syringe measures the total volume collected up to each point in time.

The key relationship is:

\[ \boxed{\text{Rate}=\text{gradient of the graph}} \]

A steep gradient means gas is being produced quickly. A shallow gradient means gas is being produced slowly.

When the graph becomes horizontal, its gradient is zero. No additional gas is being collected at a measurable rate.

Notice that the graph can be horizontal at 65 mL, not just at zero. The volume represents the total gas already collected, whereas the gradient represents how quickly that volume is changing.

Why does the graph become less steep?

At the beginning of the reaction, there are many available reactant particles. They collide frequently, so the reaction initially produces gas quickly.

As reactants are consumed, their concentrations decrease. There are fewer successful collisions per second, so the reaction slows down.

Eventually, the limiting reactant is consumed, and gas production effectively stops.

This explains the typical shape of a gas-volume-time graph:

There are a few experimental assumptions behind this interpretation. The apparatus must be airtight, the gas must be collected effectively, and its temperature and pressure must remain constant. Otherwise, a change in measured volume might not represent a change in the amount of gas produced.

02Calculating average reaction rate

Suppose an experiment records the following gas volumes:

To calculate the average rate between two times, choose the corresponding points on the graph and calculate the gradient of the straight line joining them.

This line is called a secant.

The formula is:

\[ \boxed{ \text{Average rate} = \frac{\Delta V}{\Delta t} } \]

where:

• \(\Delta V\) is the change in gas volume.
• \(\Delta t\) is the time interval.

In full:

\[ \text{Average rate} = \frac{V_2-V_1}{t_2-t_1} \]

The units are determined by the axes. If volume is measured in mL and time in seconds, the rate is in \(\mathrm{mL\,s^{-1}}\).

A common mistake: Using the final volume divided by the final time

Suppose a student calculates:

\[ \frac{54.4}{30}=1.81\ \mathrm{mL\,s^{-1}} \]

This is the average rate from 0 to 30 seconds, not from 10 to 30 seconds.

The mistake is using the wrong starting point.

Always check the time interval specified in the question before selecting your values.

03Calculating instantaneous reaction rate

Now suppose an HSC question asks:

Calculate the rate of reaction at exactly 20 seconds.

You cannot simply use the average gradient between 0 and 20 seconds. The reaction is slowing down, so its rate at 20 seconds is different from its average rate over the earlier interval.

Instead, calculate the gradient of a tangent.

A tangent is a straight line that touches the curve at the particular point where you want to measure its instantaneous gradient.

The calculation is:

\[ \boxed{ \text{Instantaneous rate} = \text{gradient of tangent} } \]

Therefore:

\[ \text{Instantaneous rate} = \frac{\Delta V_{\text{tangent}}} {\Delta t_{\text{tangent}}} \]

There is one important detail: use two convenient points on the tangent, not necessarily two experimental data points on the curve.

Choose points that are reasonably far apart. This reduces the effect of small reading errors when calculating the gradient.

Average rate versus instantaneous rate

A useful exam habit is to circle the time wording before calculating. The words between and at usually tell you which method to use.

04Converting gas volume rate into molar reaction rate

So far, our answers have been in mL/s.

Sometimes, however, a question asks for the rate in mol/s or \(\mathrm{mol\,L^{-1}\,s^{-1}}\).

To do this, we need to connect gas volume to the amount of gas in moles.

At constant temperature and pressure, the volume of a gas is directly proportional to its number of moles.

This means that if you know the molar volume of the gas, you can convert a volume production rate into a molar production rate.

The relationship is:

\[ V=nV_m \]

where:

• \(V\) is gas volume in litres.
• \(n\) is the amount of gas in moles.
• \(V_m\) is molar volume in \(\mathrm{L\,mol^{-1}}\).

At 25°C and 100 kPa, the molar volume of an ideal gas is approximately:

\[ V_m=24.79\ \mathrm{L\,mol^{-1}} \]

Therefore:

\[ \text{Molar gas production rate} = \frac{\text{Gas volume rate}}{V_m} \]

Make sure the gas volume rate is in L/s before dividing.

If the temperature and pressure are different, use the ideal gas equation, \(PV=nRT\), with the appropriate conditions.

For a more detailed explanation of the mole-to-volume conversion, see Gas Volumes in Chemical Reactions for HSC Chemistry.

05What if the graph shows concentration or mass instead?

Gas volume is not the only way to monitor reaction rate.

HSC Chemistry questions can also provide concentration-time or mass-time data. The gradient method is still the same, but you need to interpret the sign and units correctly.

Concentration-time graphs

Consider a reaction in which reactant A forms product B:

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

As the reaction proceeds:

• The concentration of A decreases.
• The concentration of B increases.

For a reactant, the gradient is negative because its concentration decreases.

For a product, the gradient is positive because its concentration increases.

However, reaction rates are conventionally reported as positive values.

For a general reaction:

\[ \ce{aA + bB -> cC} \]

the average reaction rate is:

\[ r= -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = \frac{1}{c}\frac{\Delta[C]}{\Delta t} \]

The coefficients account for the different amounts of each substance consumed or produced.

For example, suppose the concentration of A decreases from 0.80 mol/L to 0.50 mol/L over 30 seconds.

The average disappearance rate of A is:

\[ \begin{aligned} -\frac{\Delta[A]}{\Delta t} &=-\frac{0.50-0.80}{30}\\[4pt] &=0.010\ \mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

But because the coefficient of A is 2, the average rate of reaction is:

\[ r=\frac{0.010}{2} \]

\[ \boxed{ r=0.0050\ \mathrm{mol\,L^{-1}\,s^{-1}} } \]

This is also the average rate of formation of B, because its coefficient is 1.

Do not forget the stoichiometric coefficients when the question asks for the rate of reaction rather than the rate of disappearance or formation of one particular substance.

Mass-time graphs

Another common experimental method involves measuring the mass of a reaction mixture while a gas escapes.

Imagine calcium carbonate reacting with hydrochloric acid in a flask:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)} \]

The flask sits on an electronic balance. A loose cotton-wool plug prevents liquid droplets from escaping while allowing carbon dioxide to leave.

As carbon dioxide escapes, the mass recorded by the balance decreases.

The gradient of the mass-time graph is negative.

To report a positive mass-loss rate, use:

\[ \boxed{ \text{Mass-loss rate} = -\frac{\Delta m}{\Delta t} } \]

where \(\Delta m\) is the change in mass in grams and \(\Delta t\) is the time interval in seconds.

If the question asks for the molar rate of gas production, divide the mass-loss rate by the molar mass of the gas:

\[ \text{Molar gas rate} = \frac{\text{Mass-loss rate}}{\text{Molar mass}} \]

For carbon dioxide:

\[ M(\ce{CO2})=44.0\ \mathrm{g\,mol^{-1}} \]

The calculation assumes that the measured mass loss comes entirely from carbon dioxide escaping. Evaporation or liquid splashing out of the flask would introduce additional mass loss and reduce the accuracy of the result.

Comparing the three graph types

The underlying mathematical idea never changes: calculate the gradient using the correct axes and interpret what that gradient represents.

06A reliable exam method

When you encounter a reaction-rate graph, use this sequence.

Step 1: Read the axes.

Identify what is changing and the units used. Is the graph showing gas volume, concentration, or mass?

Step 2: Identify what the question is asking for.

Look for the distinction between:

• Average rate over a stated interval.
• Instantaneous rate at a particular time.

Step 3: Choose the correct points.

For an average rate, use two points on the curve.

For an instantaneous rate, draw a tangent and use two points on that tangent.

Step 4: Calculate the gradient.

Use:

\[ \text{Gradient}=\frac{\Delta y}{\Delta x} \]

Make sure the subtraction order is consistent for both axes.

Step 5: Check the sign and units.

A decreasing reactant concentration or mass produces a negative gradient. Report the positive disappearance or mass-loss rate when appropriate.

Step 6: Convert units only when necessary.

If the question asks for mol/s, convert the measured gas-volume or mass rate into moles per second.

If it asks for mol L⁻¹ s⁻¹, account for the reaction stoichiometry and the volume of the reaction mixture.

Step 7: Check the experimental conditions.

A gas-volume gradient directly represents a gas production rate only when temperature and pressure are constant, or when the measurements have been corrected to common conditions.

For more on why gas volume changes with temperature and pressure, see How to Read Pressure-Volume and Temperature Gas Graphs.

07Questions and solutions

The following questions progress from a straightforward gradient calculation to more demanding problems involving tangents, stoichiometry, experimental measurements, and changing gas conditions.

Question 1

A student reacts excess hydrochloric acid with calcium carbonate and collects the carbon dioxide produced in a gas syringe.

The following results are recorded:

Calculate the average rate of carbon dioxide production between 15 and 45 seconds. Give your answer in mL/s.

Solution 1

The average rate of carbon dioxide production is 0.733 mL/s.

The average rate is calculated using the gradient between the two specified points.

Step 1: Identify the starting and ending values.

At 15 seconds:

\[ V_1=18\ \mathrm{mL} \]

At 45 seconds:

\[ V_2=40\ \mathrm{mL} \]

Step 2: Calculate the changes in volume and time.

\[ \Delta V=40-18=22\ \mathrm{mL} \]

\[ \Delta t=45-15=30\ \mathrm{s} \]

Step 3: Calculate the average rate.

\[ \begin{aligned} \text{Average rate} &=\frac{\Delta V}{\Delta t}\\[4pt] &=\frac{22}{30}\\[4pt] &=0.733\ \mathrm{mL\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{0.733\ \mathrm{mL\,s^{-1}}} \]

The reaction produced carbon dioxide at an average rate of 0.733 mL/s over the specified interval.

Notice that we did not divide 40 mL by 45 seconds. That would calculate the average rate from the beginning of the experiment, which is not what the question asks.

Question 2

A reaction produces hydrogen gas, and the following measurements are obtained:

A tangent drawn to the gas-volume-time curve at 20 seconds passes through the points:

\[ A=(8,\ 25) \]

\[ B=(32,\ 47) \]

The time coordinates are in seconds, and the volume coordinates are in millilitres.

Calculate:

a) The instantaneous rate of hydrogen production at 20 seconds.

b) The average rate of hydrogen production between 0 and 20 seconds.

c) Explain why these two rates are different.

Solution 2

The instantaneous rate is 0.917 mL/s, while the average rate over the first 20 seconds is 1.85 mL/s.

The difference occurs because the reaction slows down as reactants are consumed.

Step 1: Calculate the instantaneous rate using the tangent.

The two coordinates are points on the tangent, not necessarily experimental measurements.

\[ \text{Instantaneous rate} = \frac{47-25}{32-8} \]

\[ =\frac{22}{24} \]

\[ \boxed{0.917\ \mathrm{mL\,s^{-1}}} \]

Step 2: Calculate the average rate between 0 and 20 seconds.

The volume increases from 0 mL to 37 mL.

\[ \begin{aligned} \text{Average rate} &=\frac{37-0}{20-0}\\[4pt] &=\frac{37}{20}\\[4pt] &=1.85\ \mathrm{mL\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{1.85\ \mathrm{mL\,s^{-1}}} \]

Step 3: Explain the difference.

The average rate includes the faster reaction occurring at the beginning of the experiment.

By 20 seconds, some reactants have already been consumed, reducing the frequency of successful collisions. The instantaneous rate is therefore lower than the average rate over the first 20 seconds.

A common mistake is to use the gradient between the measured points at 10 and 30 seconds to calculate the instantaneous rate at 20 seconds. That gives an average rate over that interval, not the exact instantaneous rate.

The tangent is the correct method.

Question 3

A reaction occurs in aqueous solution according to:

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

The concentration of A is measured at two different times:

Assume the reaction occurs in a constant-volume solution.

Calculate:

a) The average rate of disappearance of A.

b) The average rate of reaction.

c) The average rate of formation of B.

Solution 3

The average rate of disappearance of A is 0.00700 mol L⁻¹ s⁻¹, while the average rate of reaction and formation of B are both 0.00350 mol L⁻¹ s⁻¹.

The important detail is the coefficient of A in the balanced equation.

Step 1: Calculate the change in concentration of A.

\[ \Delta[A]=0.560-0.840 \]

\[ \Delta[A]=-0.280\ \mathrm{mol\,L^{-1}} \]

The negative value indicates that A is being consumed.

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

\[ \begin{aligned} -\frac{\Delta[A]}{\Delta t} &=-\frac{-0.280}{40}\\[4pt] &=0.00700\ \mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{ 0.00700\ \mathrm{mol\,L^{-1}\,s^{-1}} } \]

Step 3: Calculate the average rate of reaction.

The balanced equation is:

\[ \ce{2A -> B} \]

Two moles of A disappear for every mole of B produced.

Therefore:

\[ r=-\frac{1}{2}\frac{\Delta[A]}{\Delta t} \]

Substituting:

\[ r=\frac{0.00700}{2} \]

\[ \boxed{ r=0.00350\ \mathrm{mol\,L^{-1}\,s^{-1}} } \]

Step 4: Calculate the rate of formation of B.

The coefficient of B is 1, so its formation rate equals the rate of reaction.

\[ \boxed{ \frac{\Delta[B]}{\Delta t} = 0.00350\ \mathrm{mol\,L^{-1}\,s^{-1}} } \]

The important distinction is that the disappearance rate of A is twice the rate of reaction. Reporting the disappearance rate as the reaction rate would ignore the balanced equation.

Question 4

Calcium carbonate reacts with excess hydrochloric acid:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)} \]

The reaction takes place in a conical flask fitted with a loose cotton-wool plug. The flask sits on an electronic balance, allowing carbon dioxide to escape while the remaining contents are weighed.

The following readings are obtained:

Calculate the average molar rate of carbon dioxide production between 20 and 50 seconds.

The molar mass of carbon dioxide is 44.0 g/mol.

Also explain why the gradient of the mass-time graph is negative, even though the rate of carbon dioxide production is positive.

Solution 4

The average molar rate of carbon dioxide production is 2.73 × 10⁻⁴ mol/s.

The mass decreases because carbon dioxide escapes from the flask. We therefore need to reverse the sign of the mass gradient before converting it into a positive gas production rate.

Step 1: Calculate the mass lost between 20 and 50 seconds.

\[ \Delta m=151.86-152.22 \]

\[ \Delta m=-0.36\ \mathrm{g} \]

The magnitude of the mass loss is:

\[ 0.36\ \mathrm{g} \]

Step 2: Calculate the average mass-loss rate.

\[ \begin{aligned} \text{Mass-loss rate} &=\frac{0.36}{50-20}\\[4pt] &=\frac{0.36}{30}\\[4pt] &=0.0120\ \mathrm{g\,s^{-1}} \end{aligned} \]

Step 3: Convert the mass-loss rate into a molar rate.

Using:

\[ n=\frac{m}{M} \]

the molar production rate is:

\[ \text{Molar rate} = \frac{\text{Mass-loss rate}}{M(\ce{CO2})} \]

Substitute:

\[ \begin{aligned} \text{Molar rate} &=\frac{0.0120}{44.0}\\[4pt] &=2.73\times10^{-4}\ \mathrm{mol\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{ 2.73\times10^{-4}\ \mathrm{mol\,s^{-1}} } \]

Step 4: Explain the sign.

The mass-time graph has a negative gradient because the mass remaining in the flask decreases as carbon dioxide escapes.

However, the amount of carbon dioxide produced is increasing. Its production rate is therefore positive.

The calculation also assumes that all mass lost is carbon dioxide. If water evaporates or liquid escapes from the flask, the measured mass loss will overestimate the amount of carbon dioxide produced.

Question 5

A student investigates the reaction between magnesium and excess hydrochloric acid:

\[ \ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \]

Hydrogen is collected in a gas syringe. During the experiment, the gas cools as it moves towards room temperature.

Two measurements are recorded:

The absolute pressure of the gas is 100 kPa at both times.

A student calculates:

\[ \frac{57.0-60.0}{40-30} = -0.300\ \mathrm{mL\,s^{-1}} \]

They conclude that the reaction has stopped producing hydrogen and that the amount of hydrogen must have decreased.

Evaluate this conclusion. Calculate the actual average molar rate of hydrogen production between 30 and 40 seconds.

Assume:

• Hydrogen behaves as an ideal gas.
• No gas escapes from the apparatus.
• The gas has reached the stated temperature at each measurement.
• All measured gas is hydrogen.
• The gas pressure is constant at 100 kPa.

Use:

\[ PV=nRT \]

where:

\[ R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}} \]

Solution 5

The student’s conclusion is incorrect. The amount of hydrogen actually increases between 30 and 40 seconds, even though the measured gas volume decreases.

The reason is that the gas cools significantly during this interval. Its volume contracts as its temperature falls, partially hiding the additional hydrogen being produced.

This is a case where calculating the gradient of the raw volume-time data gives a misleading result because temperature is not constant.

Step 1: Convert both temperatures into kelvin.

At 30 seconds:

\[ T_1=40.0+273.15=313.15\ \mathrm{K} \]

At 40 seconds:

\[ T_2=20.0+273.15=293.15\ \mathrm{K} \]

Step 2: Convert both gas volumes into litres.

\[ V_1=0.0600\ \mathrm{L} \]

\[ V_2=0.0570\ \mathrm{L} \]

Step 3: Calculate the amount of hydrogen at 30 seconds.

Rearrange the ideal gas equation:

\[ n=\frac{PV}{RT} \]

Substitute:

\[ \begin{aligned} n_1 &=\frac{(100)(0.0600)} {(8.314)(313.15)}\\[4pt] &=2.30\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the amount of hydrogen at 40 seconds.

\[ \begin{aligned} n_2 &=\frac{(100)(0.0570)} {(8.314)(293.15)}\\[4pt] &=2.34\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Step 5: Calculate the change in the amount of hydrogen.

\[ \begin{aligned} \Delta n &=n_2-n_1\\[4pt] &=(2.34-2.30)\times10^{-3}\\[4pt] &=3.4\times10^{-5}\ \mathrm{mol} \end{aligned} \]

The amount of hydrogen has increased, not decreased.

Step 6: Calculate the average molar production rate.

\[ \begin{aligned} \text{Average molar rate} &=\frac{\Delta n}{\Delta t}\\[4pt] &=\frac{3.4\times10^{-5}}{40-30}\\[4pt] &=3.4\times10^{-6}\ \mathrm{mol\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{ 3.4\times10^{-6}\ \mathrm{mol\,s^{-1}} } \]

Because the coefficient of hydrogen in the balanced equation is 1, this is also the average reaction rate expressed in mol/s.

Step 7: Check why the original volume calculation was misleading.

If the amount of hydrogen had stayed constant while the gas cooled, its volume would have changed in proportion to its absolute temperature.

Using the combined gas relationship at constant pressure and amount:

\[ V_2=V_1\frac{T_2}{T_1} \]

\[ \begin{aligned} V_2 &=60.0\times\frac{293.15}{313.15}\\[4pt] &=56.2\ \mathrm{mL} \end{aligned} \]

So, if no additional hydrogen had formed, the expected volume would have been approximately 56.2 mL.

However, the actual measured volume is 57.0 mL.

That extra volume is evidence that additional hydrogen was produced while the gas cooled.

The student was right that the measured volume decreased, but wrong to assume that the amount of gas must also have decreased.

This is the important limitation of gas-volume-time graphs: their gradients directly represent gas production rates only when temperature and pressure are controlled.

08What to learn next

Once you can calculate average and instantaneous reaction rates, the next step is explaining why those rates change.

Collision theory helps explain why increasing concentration, temperature, or surface area can produce a steeper initial gradient, and why a catalyst can increase reaction rate without necessarily changing the final amount of product.

When comparing two reaction-rate experiments, look at the gradients to compare how quickly reactions proceed. Look at the final plateaus to compare how much gas was ultimately produced.

Those are two different measurements, and knowing which one an exam question is testing will help you avoid a surprisingly common mistake.