How to Use a Cooling Correction in HSC Chemistry Calorimetry
Learn how to extrapolate cooling data to find a corrected maximum or minimum temperature, then use it to calculate molar enthalpy. Includes worked examples, sign conventions, and challenging HSC-style practice questions with solutions.
A calorimetry experiment gives you a temperature-time table. The solution reaches a recorded maximum of \(27.4^\circ\mathrm{C}\), but the first reading was taken one minute after the reactants were mixed. Has the reaction’s true temperature rise already been underestimated?
Yes. By the time the thermometer records \(27.4^\circ\mathrm{C}\), the solution has already started losing heat to the surroundings. Using the highest recorded temperature would therefore underestimate the energy released by the reaction.
The cooling correction fixes this problem by extending the post-reaction cooling trend backwards to the time the reactants were mixed. The temperature at that point is the corrected maximum temperature.
The key is knowing which data points to use, where to extend the trend, and how to use the corrected temperature in your enthalpy calculation.
01What is a cooling correction?
Imagine leaving a hot cup of tea on a desk. If you measure its temperature every minute, the readings gradually decrease as heat transfers to the surrounding air.
If you missed the first reading, you could use the later measurements to estimate the earlier temperature by extending the cooling trend backwards.
A calorimetry experiment has a similar problem, except that the reacting chemicals initially generate heat themselves.
When an exothermic reaction occurs in solution:
- The reaction releases energy.
- The solution absorbs some of that energy, so its temperature rises.
- Heat gradually escapes from the calorimeter to the surrounding environment.
- By the time the highest temperature is recorded, some energy has already escaped.
The cooling correction estimates the temperature that would have been reached at the time of mixing if the subsequent cooling had not reduced the measured temperature.
The tea analogy has one important limitation: tea does not generate additional heat, but a reacting mixture does. The correction therefore uses the cooling data collected after the reaction has finished, not the entire temperature-time record indiscriminately.
02Understand the calorimeter before interpreting its data
A typical solution calorimetry experiment uses an insulated polystyrene cup supported inside a beaker.
The reactants are mixed inside the cup. A lid reduces heat exchange with the air, while a thermometer or temperature probe passes through the lid into the solution. A stirrer helps distribute heat evenly.
The temperature probe measures the solution’s temperature, not the energy of the chemical reaction directly.
System and surroundings
Before calculating anything, identify what is transferring energy to what.
- System: The chemical reaction itself.
- Immediate surroundings: The solution and, where relevant, the calorimeter.
- External environment: The room surrounding the calorimeter.
For an exothermic reaction, the chemical system releases energy. The solution and calorimeter absorb that energy, while some may subsequently escape into the room.
For an endothermic reaction, the chemical system absorbs energy from its surroundings. The solution and calorimeter lose energy, so their temperatures fall.
This gives us the sign convention:
| Quantity | Meaning |
|---|---|
| \(q_{\text{reaction}}>0\) | Reaction absorbs heat; endothermic |
| \(q_{\text{reaction}}<0\) | Reaction releases heat; exothermic |
| \(q_{\text{solution}}>0\) | Solution gains heat and warms |
| \(q_{\text{solution}}<0\) | Solution loses heat and cools |
For an idealised calorimeter in which the calorimeter’s own heat capacity is negligible:
\[ q_{\text{reaction}}=-q_{\text{solution}} \]
The negative sign is important. If the solution gains energy, the reaction must have lost that energy.
The cooling correction improves your estimate of the solution’s temperature change. It does not automatically account for the calorimeter’s own heat capacity. If a calorimeter heat capacity is supplied, you must include that separately.
03How to read supplied temperature-time data
The procedure is slightly different depending on whether the reaction is exothermic or endothermic.
Case 1: An exothermic reaction
The solution warms up during the reaction, reaches a maximum, and then cools towards room temperature.
The cooling correction follows this sequence:
- Identify the time at which the reactants were mixed.
- Identify the temperature readings taken after the reaction has finished and the solution has started cooling.
- Select the approximately straight section of the cooling data.
- Draw a best-fit straight line through those cooling points.
- Extend the line backwards to the mixing time.
- Read the temperature at the intersection.
That intersection is the corrected maximum temperature.
Notice that it should be higher than the highest recorded temperature.
Case 2: An endothermic reaction
An endothermic process absorbs energy from the solution, causing its temperature to fall.
After the process finishes, the cooler solution gradually absorbs heat from the surrounding room. Its temperature begins rising again.
The same extrapolation idea applies, but the direction is reversed:
- Identify the time of mixing.
- Select the approximately straight section after the temperature minimum, where the solution is warming.
- Draw a best-fit straight line through those points.
- Extend it backwards to the mixing time.
- Read the corrected minimum temperature.
The corrected minimum should be lower than the lowest recorded temperature.
| Reaction | Temperature after mixing | Subsequent trend | Corrected temperature |
|---|---|---|---|
| Exothermic | Rises | Cools | Higher than recorded maximum |
| Endothermic | Falls | Warms | Lower than recorded minimum |
What if the mixing time is not zero?
This is an easy detail to overlook.
Suppose the experiment starts recording at 0 minutes, but the reactants are not mixed until 2 minutes.
You must extrapolate to \(t=2\), not \(t=0\).
The corrected temperature always corresponds to the actual mixing time. The clock reading itself is not important.
Mathematically, a straight-line temperature-time relationship can be written as:
\[ T(t)=st+b \]
where:
- \(T(t)\) is the temperature at time \(t\), in degrees Celsius.
- \(s\) is the gradient of the cooling or warming line, in \(\mathrm{^\circ C\,min^{-1}}\).
- \(b\) is the temperature intercept.
- \(t\) is time in minutes.
If mixing occurs at \(t=0\), the corrected temperature is simply \(b\). Otherwise, substitute the actual mixing time into the equation.
When the data are slightly scattered, use a best-fit line rather than forcing a line through every individual point. Only use the section representing cooling or warming after the reaction is complete.
04Connect the corrected temperature to enthalpy
Once you have the corrected temperature, the rest of the calculation follows the usual calorimetry method.
However, there are three things to establish before substituting numbers.
Step 1: Calculate the corrected temperature change
For an exothermic reaction:
\[ \Delta T=T_{\text{corrected}}-T_{\text{initial}} \]
For an endothermic process, use the same expression. The result will be negative because the corrected minimum is below the initial temperature.
Temperature differences in degrees Celsius have the same numerical value as temperature differences in kelvin.
Step 2: Calculate the heat transferred by the solution
Use:
\[ q_{\text{solution}}=mc\Delta T \]
where:
- \(m\) is the mass of the solution, in grams.
- \(c\) is its specific heat capacity, in \(\mathrm{J\,g^{-1}\,K^{-1}}\).
- \(\Delta T\) is the corrected temperature change, in kelvin.
For dilute aqueous solutions, questions commonly allow you to assume:
\[ c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}} \]
and a density of:
\[ \rho=1.00\ \mathrm{g\,mL^{-1}} \]
Only use these assumptions when permitted by the question.
Remember that the solution mass is generally based on the total solution volume, not just the volume of one reactant. For dissolution experiments, include the dissolved solid in the solution mass when appropriate.
If the calorimeter has a supplied heat capacity, calculate its energy change as well:
\[ q_{\text{cal}}=C_{\text{cal}}\Delta T \]
Here, \(C_{\text{cal}}\) is the calorimeter’s heat capacity in \(\mathrm{J\,K^{-1}}\).
The reaction heat is then:
\[ q_{\text{reaction}} =-(q_{\text{solution}}+q_{\text{cal}}) \]
The cooling correction and calorimeter heat-capacity correction solve different problems. Do not confuse them.
Step 3: Determine the amount of reaction
You must divide the reaction heat by the correct number of moles.
Start with the balanced chemical equation and calculate the initial moles of each reactant.
For solutions:
\[ n=cV \]
where \(n\) is the amount in moles, \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\), and \(V\) is volume in litres.
Identify the limiting reagent using the balanced equation. Assuming the reaction goes to completion, this determines the amount of reaction.
For example:
\[ \ce{HCl(aq)+NaOH(aq)->NaCl(aq)+H2O(l)} \]
The mole ratio is \(1:1\). Therefore, the amount of water formed equals the amount of the limiting reactant consumed.
For a reaction with different coefficients, account for those coefficients. If the balanced equation is \(aA+bB\rightarrow\text{products}\), the amount of reaction is determined by the smaller of \(n(A)/a\) and \(n(B)/b\).
Do not simply add the moles of the reactants together.
Step 4: Calculate the molar enthalpy change
At approximately constant atmospheric pressure, the reaction heat provides an experimental estimate of the enthalpy change.
\[ \Delta H=\frac{q_{\text{reaction}}}{n_{\text{reaction}}} \]
Convert the reaction heat into kilojoules before dividing if you want the final answer in \(\mathrm{kJ\,mol^{-1}}\).
A negative answer indicates an exothermic reaction. A positive answer indicates an endothermic reaction.
If you need a broader explanation of the temperature-rise calculation itself, see How to Calculate Enthalpy Change from a Temperature Rise.
Assumptions behind the correction
The standard straight-line cooling correction relies on several assumptions:
- The reaction is effectively complete before the selected cooling section begins.
- The cooling or warming trend is approximately linear over the selected interval.
- The time of mixing is known.
- The temperature probe responds quickly enough to provide useful readings.
- The reaction is complete, and the supplied limiting-reagent calculation represents the amount that reacted.
- Any significant calorimeter heat capacity is included separately.
- The mass and specific heat capacity used in the calculation reasonably represent the solution.
The method is an estimate, not a perfect reconstruction of an ideal experiment.
If a reaction takes several minutes to finish, the simple correction may not be valid. During that time, the reaction can continue generating or absorbing energy while heat is also being exchanged with the room. We will return to this limitation in Question 5.
05Worked examples
Worked example: correcting the temperature rise in a neutralisation reaction
50.0 mL of \(1.00\ \mathrm{mol\,L^{-1}}\) hydrochloric acid is mixed with 50.0 mL of \(1.00\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. Both solutions initially have a temperature of \(21.0^\circ\mathrm{C}\).
The following illustrative temperature-time data are recorded:
| Time after mixing (min) | Temperature (\(^\circ\mathrm{C}\)) |
|---|---|
| 0 (immediately before mixing) | 21.0 |
| 1 | 27.4 |
| 2 | 27.0 |
| 3 | 26.6 |
| 4 | 26.2 |
| 5 | 25.8 |
| 6 | 25.4 |
The cooling data follow an approximately straight line.
Assume the solution density is \(1.00\ \mathrm{g\,mL^{-1}}\), its specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\), the calorimeter’s heat capacity is negligible, and the reaction is complete by the first post-mixing reading.
Calculate the corrected temperature change and the molar enthalpy of neutralisation per mole of water formed.
The balanced equation is:
\[ \ce{HCl(aq)+NaOH(aq)->NaCl(aq)+H2O(l)} \]
Step 1
Use the first and last cooling readings:
\[ \begin{aligned} s&=\frac{25.4-27.4}{6-1}\\ &=\frac{-2.0}{5}\\ &=-0.40\ \mathrm{^\circ C\,min^{-1}} \end{aligned} \]
The negative gradient confirms that the solution is cooling.
Step 2
At \(t=1\) minute, the temperature is \(27.4^\circ\mathrm{C}\).
The temperature would have been \(0.40^\circ\mathrm{C}\) higher at \(t=0\).
Therefore:
\[ T_{\text{corrected}}=27.4+0.40 =27.8^\circ\mathrm{C} \]
The corrected maximum is \(27.8^\circ\mathrm{C}\), not the highest recorded value of \(27.4^\circ\mathrm{C}\).
Step 3
\[ \begin{aligned} \Delta T &=T_{\text{corrected}}-T_{\text{initial}}\\ &=27.8-21.0\\ &=6.8\ \mathrm{K} \end{aligned} \]
Step 4
The total volume is:
\[ V=50.0+50.0=100.0\ \mathrm{mL} \]
Using the assumed density:
\[ \begin{aligned} m&=\rho V\\ &=1.00\times100.0\\ &=100.0\ \mathrm{g} \end{aligned} \]
Step 5
\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=(100.0)(4.18)(6.8)\\ &=2842.4\ \mathrm{J}\\ &=+2.84\ \mathrm{kJ} \end{aligned} \]
The solution gains \(2.84\ \mathrm{kJ}\).
Therefore, the reaction releases the same amount of energy:
\[ q_{\text{reaction}}=-2.84\ \mathrm{kJ} \]
Step 6
For hydrochloric acid:
\[ \begin{aligned} n(\ce{HCl}) &=cV\\ &=(1.00)(0.0500)\\ &=0.0500\ \mathrm{mol} \end{aligned} \]
The sodium hydroxide also has \(0.0500\ \mathrm{mol}\).
Because the reaction has a \(1:1\) mole ratio, both reactants are completely consumed and:
\[ n(\ce{H2O})=0.0500\ \mathrm{mol} \]
Step 7
\[ \begin{aligned} \Delta H &=\frac{q_{\text{reaction}}}{n(\ce{H2O})}\\ &=\frac{-2.84}{0.0500}\\ &=-56.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
The experimental molar enthalpy of neutralisation is:
\[ \boxed{\Delta H=-56.8\ \mathrm{kJ\,mol^{-1}}} \]
The negative sign confirms that neutralisation is exothermic.
Notice how using the recorded maximum of \(27.4^\circ\mathrm{C}\) instead would have produced a smaller temperature change and underestimated the magnitude of the enthalpy change.
Worked example: correcting an endothermic dissolution
A student dissolves \(4.00\ \mathrm{g}\) of ammonium nitrate, \(\ce{NH4NO3}\), in \(80.0\ \mathrm{g}\) of water.
The initial water temperature is \(24.0^\circ\mathrm{C}\). The following illustrative temperature-time data are recorded:
| Time after mixing (min) | Temperature (\(^\circ\mathrm{C}\)) |
|---|---|
| 0 (immediately before mixing) | 24.0 |
| 1 | 20.6 |
| 2 | 20.9 |
| 3 | 21.2 |
| 4 | 21.5 |
| 5 | 21.8 |
After the temperature minimum, the solution gradually warms.
Assume the solution has a specific heat capacity of \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\), the calorimeter heat capacity is negligible, and the reaction is complete by the first post-mixing reading.
The molar mass of ammonium nitrate is \(80.04\ \mathrm{g\,mol^{-1}}\).
Calculate the corrected temperature change and molar enthalpy of dissolution.
Step 1
\[ \begin{aligned} s&=\frac{21.8-20.6}{5-1}\\ &=\frac{1.2}{4}\\ &=+0.30\ \mathrm{^\circ C\,min^{-1}} \end{aligned} \]
The positive gradient shows that the solution is warming as it absorbs heat from the room.
Step 2
The first reading is \(20.6^\circ\mathrm{C}\), taken one minute after mixing.
The extrapolated temperature at \(t=0\) is:
\[ T_{\text{corrected}}=20.6-0.30 =20.3^\circ\mathrm{C} \]
This is lower than the lowest recorded temperature.
Step 3
\[ \begin{aligned} \Delta T &=T_{\text{corrected}}-T_{\text{initial}}\\ &=20.3-24.0\\ &=-3.7\ \mathrm{K} \end{aligned} \]
The negative temperature change indicates that the solution has lost thermal energy.
Step 4
The solution contains both the water and the dissolved ammonium nitrate:
\[ m=80.0+4.00=84.0\ \mathrm{g} \]
Step 5
\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=(84.0)(4.18)(-3.7)\\ &=-1299\ \mathrm{J}\\ &=-1.30\ \mathrm{kJ} \end{aligned} \]
The solution loses \(1.30\ \mathrm{kJ}\).
Therefore:
\[ q_{\text{reaction}}=+1.30\ \mathrm{kJ} \]
The positive sign tells us that the dissolution absorbs energy.
Step 6
\[ \begin{aligned} n(\ce{NH4NO3}) &=\frac{m}{M}\\ &=\frac{4.00}{80.04}\\ &=0.0500\ \mathrm{mol} \end{aligned} \]
Step 7
\[ \begin{aligned} \Delta H &=\frac{q_{\text{reaction}}}{n}\\ &=\frac{+1.30}{0.0500}\\ &=+26.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
The molar enthalpy of dissolution is:
\[ \boxed{\Delta H=+26.0\ \mathrm{kJ\,mol^{-1}}} \]
The positive sign confirms that the dissolution is endothermic.
This example exposes an important sign trap. The solution’s temperature decreases, so \(q_{\text{solution}}\) is negative. The reaction’s heat change is positive because it absorbs the energy lost by the solution.
06Common mistakes to avoid
| Mistake | Why it is wrong | What to do instead |
|---|---|---|
| Using the highest recorded temperature | Some heat may already have escaped | Extrapolate the cooling trend backwards |
| Extrapolating the rising section | The reaction may still be generating heat | Use the post-reaction cooling section |
| Extrapolating to the start of the experiment rather than mixing | The reaction may have started later | Use the actual mixing time |
| Making \(q_{\text{reaction}}\) positive when the solution warms | The reaction and solution have opposite heat changes | Reverse the sign |
| Dividing by the total moles of both reactants | Not all reactants necessarily react | Determine the limiting reagent and reaction amount |
| Ignoring the calorimeter heat capacity | The calorimeter may absorb significant energy | Include \(C_{\text{cal}}\Delta T\) when supplied |
| Using only the mass of water in a dissolution | The dissolved solid contributes to solution mass | Include the solute mass when appropriate |
| Treating every cooling curve as perfectly linear | Heat loss and temperature difference can change over time | Use an appropriate straight section and state the limitation |
One particularly important distinction is that the cooling correction does not fix every experimental error.
It estimates heat lost during the measurement delay under a particular model. It does not correct incomplete reactions, inaccurate concentrations, poor mixing, or an omitted calorimeter heat capacity.
If acid and base volumes differ, you must still calculate the limiting reagent correctly. The correction does not change the stoichiometry. See How to Calculate Enthalpy of Neutralisation When Volumes Differ for that separate calculation.
07Questions and solutions
Question 1
50.0 mL of \(0.800\ \mathrm{mol\,L^{-1}}\) hydrochloric acid is mixed with 50.0 mL of \(0.800\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. Both solutions initially have a temperature of \(22.0^\circ\mathrm{C}\).
The following illustrative data are recorded:
| Time after mixing (min) | Temperature (\(^\circ\mathrm{C}\)) |
|---|---|
| 0 | 22.0 |
| 1 | 26.6 |
| 2 | 26.3 |
| 3 | 26.0 |
| 4 | 25.7 |
| 5 | 25.4 |
Assume the cooling trend is linear, the solution density is \(1.00\ \mathrm{g\,mL^{-1}}\), the specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\), and the calorimeter heat capacity is negligible.
Calculate:
a. The corrected maximum temperature.
b. The corrected temperature change.
c. The molar enthalpy of neutralisation per mole of water formed.
Solution 1
The corrected maximum temperature is \(26.9^\circ\mathrm{C}\), and the molar enthalpy of neutralisation is approximately \(-51.2\ \mathrm{kJ\,mol^{-1}}\).
Step 1: Calculate the cooling gradient.
\[ \begin{aligned} s&=\frac{25.4-26.6}{5-1}\\ &=\frac{-1.2}{4}\\ &=-0.30\ \mathrm{^\circ C\,min^{-1}} \end{aligned} \]
Step 2: Extrapolate to the mixing time.
The first reading is \(26.6^\circ\mathrm{C}\) at one minute.
\[ T_{\text{corrected}}=26.6+0.30 =26.9^\circ\mathrm{C} \]
Step 3: Calculate the corrected temperature change.
\[ \Delta T=26.9-22.0=4.9\ \mathrm{K} \]
Step 4: Calculate the solution mass.
\[ m=(100.0\ \mathrm{mL})(1.00\ \mathrm{g\,mL^{-1}}) =100.0\ \mathrm{g} \]
Step 5: Calculate the reaction heat.
\[ \begin{aligned} q_{\text{solution}} &=(100.0)(4.18)(4.9)\\ &=2048.2\ \mathrm{J}\\ &=+2.05\ \mathrm{kJ} \end{aligned} \]
Therefore:
\[ q_{\text{reaction}}=-2.05\ \mathrm{kJ} \]
Step 6: Determine the amount of water formed.
\[ \begin{aligned} n(\ce{HCl})&=(0.800)(0.0500)=0.0400\ \mathrm{mol}\\ n(\ce{NaOH})&=(0.800)(0.0500)=0.0400\ \mathrm{mol} \end{aligned} \]
The reaction ratio is \(1:1\), so:
\[ n(\ce{H2O})=0.0400\ \mathrm{mol} \]
Step 7: Calculate the molar enthalpy.
\[ \begin{aligned} \Delta H &=\frac{-2.05}{0.0400}\\ &=-51.2\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
The reaction is exothermic. Using the recorded maximum instead of the corrected maximum would underestimate the magnitude of the energy released.
Question 2
A student dissolves \(3.20\ \mathrm{g}\) of ammonium nitrate in \(60.0\ \mathrm{g}\) of water initially at \(25.0^\circ\mathrm{C}\).
The following illustrative data are obtained:
| Time after mixing (min) | Temperature (\(^\circ\mathrm{C}\)) |
|---|---|
| 0 | 25.0 |
| 1 | 21.4 |
| 2 | 21.7 |
| 3 | 22.0 |
| 4 | 22.3 |
| 5 | 22.6 |
The molar mass of ammonium nitrate is \(80.04\ \mathrm{g\,mol^{-1}}\).
Assume the specific heat capacity of the solution is \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\), and the calorimeter heat capacity is negligible.
Calculate the molar enthalpy of dissolution. Explain why using the lowest recorded temperature would underestimate the result.
Solution 2
The corrected molar enthalpy of dissolution is approximately \(+25.8\ \mathrm{kJ\,mol^{-1}}\).
Step 1: Calculate the gradient of the warming section.
\[ \begin{aligned} s&=\frac{22.6-21.4}{5-1}\\ &=+0.30\ \mathrm{^\circ C\,min^{-1}} \end{aligned} \]
Step 2: Find the corrected minimum temperature.
\[ T_{\text{corrected}}=21.4-0.30 =21.1^\circ\mathrm{C} \]
Step 3: Calculate the temperature change.
\[ \Delta T=21.1-25.0=-3.9\ \mathrm{K} \]
Step 4: Calculate the total solution mass.
\[ m=60.0+3.20=63.2\ \mathrm{g} \]
Step 5: Calculate the solution’s heat change.
\[ \begin{aligned} q_{\text{solution}} &=(63.2)(4.18)(-3.9)\\ &=-1030\ \mathrm{J}\\ &=-1.03\ \mathrm{kJ} \end{aligned} \]
Therefore:
\[ q_{\text{reaction}}=+1.03\ \mathrm{kJ} \]
Step 6: Calculate the moles of ammonium nitrate.
\[ n=\frac{3.20}{80.04} =0.0400\ \mathrm{mol} \]
Step 7: Calculate the molar enthalpy.
\[ \begin{aligned} \Delta H &=\frac{+1.03}{0.0400}\\ &=+25.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
The dissolution is endothermic.
The tempting alternative is to use \(21.4^\circ\mathrm{C}\), the lowest recorded temperature. However, the solution has already started warming by the time that reading is taken. This produces a smaller temperature decrease and therefore underestimates the magnitude of the energy absorbed.
Question 3
Magnesium reacts with hydrochloric acid according to:
\[ \ce{Mg(s)+2HCl(aq)->MgCl2(aq)+H2(g)} \]
A student reacts \(0.120\ \mathrm{g}\) of magnesium with \(40.0\ \mathrm{mL}\) of \(0.200\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
The initial temperature is \(20.0^\circ\mathrm{C}\). The following illustrative data are recorded:
| Time after mixing (min) | Temperature (\(^\circ\mathrm{C}\)) |
|---|---|
| 0 | 20.0 |
| 1 | 29.8 |
| 2 | 29.6 |
| 3 | 29.4 |
| 4 | 29.2 |
Assume:
- The corrected cooling trend is linear.
- The mass of the reacting liquid is \(40.1\ \mathrm{g}\).
- The solution has a specific heat capacity of \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).
- The calorimeter heat capacity is \(18.0\ \mathrm{J\,K^{-1}}\).
- The reaction is complete.
The molar mass of magnesium is \(24.3\ \mathrm{g\,mol^{-1}}\).
Calculate the molar enthalpy change per mole of reaction as written. Identify the limiting reagent.
Solution 3
Hydrochloric acid is the limiting reagent, and the molar enthalpy change is approximately \(-464\ \mathrm{kJ\,mol^{-1}}\).
Step 1: Calculate the corrected maximum temperature.
The cooling gradient is:
\[ \begin{aligned} s&=\frac{29.2-29.8}{4-1}\\ &=-0.20\ \mathrm{^\circ C\,min^{-1}} \end{aligned} \]
Extrapolating to \(t=0\):
\[ T_{\text{corrected}}=29.8+0.20 =30.0^\circ\mathrm{C} \]
Step 2: Calculate the corrected temperature change.
\[ \Delta T=30.0-20.0=10.0\ \mathrm{K} \]
Step 3: Calculate the initial moles of each reactant.
For magnesium:
\[ \begin{aligned} n(\ce{Mg}) &=\frac{0.120}{24.3}\\ &=0.00494\ \mathrm{mol} \end{aligned} \]
For hydrochloric acid:
\[ \begin{aligned} n(\ce{HCl}) &=cV\\ &=(0.200)(0.0400)\\ &=0.00800\ \mathrm{mol} \end{aligned} \]
Step 4: Identify the limiting reagent.
The equation requires two moles of HCl for every mole of Mg.
The amount of reaction possible according to magnesium is:
\[ \frac{0.00494}{1}=0.00494\ \mathrm{mol} \]
According to hydrochloric acid:
\[ \frac{0.00800}{2}=0.00400\ \mathrm{mol} \]
Hydrochloric acid is limiting.
The amount of reaction is therefore:
\[ n_{\text{reaction}}=0.00400\ \mathrm{mol} \]
Step 5: Calculate the heat absorbed by the solution.
\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=(40.1)(4.18)(10.0)\\ &=1676\ \mathrm{J} \end{aligned} \]
Step 6: Include the calorimeter heat capacity.
\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\ &=(18.0)(10.0)\\ &=180\ \mathrm{J} \end{aligned} \]
The total energy gained by the surroundings is:
\[ \begin{aligned} q_{\text{surroundings}} &=1676+180\\ &=1856\ \mathrm{J} \end{aligned} \]
Therefore:
\[ q_{\text{reaction}}=-1856\ \mathrm{J} =-1.856\ \mathrm{kJ} \]
Step 7: Calculate the molar enthalpy.
\[ \begin{aligned} \Delta H &=\frac{-1.856}{0.00400}\\ &=-464\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
The reaction is exothermic.
The important trap is assuming that magnesium is limiting simply because its mass is smaller. Limiting reagents must be compared using mole ratios from the balanced equation, not by comparing masses or raw mole amounts.
Question 4
A student mixes \(50.0\ \mathrm{mL}\) of hydrochloric acid at \(18.0^\circ\mathrm{C}\) with \(50.0\ \mathrm{mL}\) of sodium hydroxide at \(26.0^\circ\mathrm{C}\).
Both solutions have a concentration of \(0.500\ \mathrm{mol\,L^{-1}}\).
The following illustrative temperature-time data are recorded after mixing:
| Time after mixing (min) | Temperature (\(^\circ\mathrm{C}\)) |
|---|---|
| 0 | Mixing begins |
| 1 | 25.0 |
| 2 | 24.8 |
| 3 | 24.6 |
| 4 | 24.4 |
| 5 | 24.2 |
A student uses \(26.0^\circ\mathrm{C}\) as the initial temperature because it was the temperature of the warmer solution.
Assume the solutions have equal densities of \(1.00\ \mathrm{g\,mL^{-1}}\), equal specific heat capacities of \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\), and negligible calorimeter heat capacity.
Calculate the correct molar enthalpy of neutralisation per mole of water formed. Explain why the warmer solution’s initial temperature cannot be used as the baseline.
Solution 4
The correct molar enthalpy of neutralisation is approximately \(-53.5\ \mathrm{kJ\,mol^{-1}}\).
The initial temperature for the reaction calculation must account for the thermal energy already present in both solutions before they react.
Step 1: Calculate the baseline temperature before the reaction’s heat is included.
Both solutions have equal masses and equal specific heat capacities.
The temperature they would reach through thermal mixing alone is:
\[ \begin{aligned} T_{\text{initial}} &=\frac{(50.0)(18.0)+(50.0)(26.0)} {50.0+50.0}\\ &=\frac{900+1300}{100.0}\\ &=22.0^\circ\mathrm{C} \end{aligned} \]
This is the correct baseline.
Using \(26.0^\circ\mathrm{C}\) would ignore the thermal energy contributed by the cooler acid solution.
Step 2: Calculate the cooling gradient.
\[ \begin{aligned} s&=\frac{24.2-25.0}{5-1}\\ &=-0.20\ \mathrm{^\circ C\,min^{-1}} \end{aligned} \]
Step 3: Calculate the corrected maximum temperature.
The first reading is \(25.0^\circ\mathrm{C}\) at one minute.
\[ T_{\text{corrected}}=25.0+0.20 =25.2^\circ\mathrm{C} \]
Step 4: Calculate the corrected temperature change.
\[ \Delta T=25.2-22.0=3.2\ \mathrm{K} \]
Step 5: Calculate the heat gained by the solution.
The total mass is:
\[ m=50.0+50.0=100.0\ \mathrm{g} \]
Therefore:
\[ \begin{aligned} q_{\text{solution}} &=(100.0)(4.18)(3.2)\\ &=1337.6\ \mathrm{J}\\ &=+1.34\ \mathrm{kJ} \end{aligned} \]
The reaction heat is:
\[ q_{\text{reaction}}=-1.34\ \mathrm{kJ} \]
Step 6: Determine the amount of water formed.
For hydrochloric acid:
\[ n(\ce{HCl})=(0.500)(0.0500) =0.0250\ \mathrm{mol} \]
For sodium hydroxide:
\[ n(\ce{NaOH})=(0.500)(0.0500) =0.0250\ \mathrm{mol} \]
The reactants are present in a \(1:1\) ratio, so:
\[ n(\ce{H2O})=0.0250\ \mathrm{mol} \]
Step 7: Calculate the molar enthalpy.
\[ \begin{aligned} \Delta H &=\frac{-1.3376}{0.0250}\\ &=-53.5\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
The reaction is exothermic.
The tempting mistake is to use the warmer reactant’s temperature as the initial temperature. But mixing two solutions at different temperatures already involves energy transfer, even before considering the chemical reaction. You must account for that thermal mixing first.
Question 5
A student investigates a reaction in a calorimeter. The room temperature is \(20.0^\circ\mathrm{C}\), and the reacting solution initially has the same temperature.
Independent observations establish that the reaction continues for the first four minutes and is complete at \(t=4\) minutes.
The following illustrative data are recorded:
| Time (min) | Temperature (\(^\circ\mathrm{C}\)) |
|---|---|
| 0 | 20.0 |
| 1 | 24.0 |
| 2 | 27.0 |
| 3 | 28.5 |
| 4 | 29.0 |
| 5 | 28.8 |
| 6 | 28.6 |
| 7 | 28.4 |
| 8 | 28.2 |
The student selects the final four readings, calculates a cooling gradient of \(-0.20\ \mathrm{^\circ C\,min^{-1}}\), and extrapolates the straight line backwards to \(t=0\).
The student obtains \(29.8^\circ\mathrm{C}\) and claims this is the corrected maximum temperature.
Evaluate this claim.
In your response:
a. Calculate the extrapolated temperature at \(t=0\).
b. Explain why the usual cooling correction is not automatically valid in this experiment.
c. Explain what additional experimental information would be needed to obtain a defensible correction for the heat lost during the reaction.
Solution 5
The extrapolated value is \(29.8^\circ\mathrm{C}\), but the student has not established that this is a valid corrected maximum. The standard cooling correction cannot automatically be applied because the reaction continues generating heat while the temperature is rising.
Step 1: Calculate the gradient of the post-reaction cooling section.
Using the readings at five and eight minutes:
\[ \begin{aligned} s&=\frac{28.2-28.8}{8-5}\\ &=\frac{-0.6}{3}\\ &=-0.20\ \mathrm{^\circ C\,min^{-1}} \end{aligned} \]
Step 2: Calculate the student’s extrapolated value.
The temperature at five minutes is \(28.8^\circ\mathrm{C}\).
Extending the line backwards by five minutes gives:
\[ \begin{aligned} T(0)&=28.8+(0.20)(5)\\ &=29.8^\circ\mathrm{C} \end{aligned} \]
The arithmetic is correct.
The interpretation is the problem.
Step 3: Identify the assumption that has failed.
The usual correction assumes the reaction is effectively complete at the mixing time. After that point, the temperature changes because of heat exchange with the surroundings, rather than continued chemical heat generation.
Here, the measured temperature rises from \(20.0^\circ\mathrm{C}\) to \(29.0^\circ\mathrm{C}\) over four minutes.
The reaction is still generating heat during that period.
The later cooling line describes what happens after the reaction has finished. It does not automatically describe the heat-loss rate throughout the earlier reaction period.
Step 4: Consider why the room temperature matters.
The room and solution both begin at \(20.0^\circ\mathrm{C}\).
Initially, there is almost no temperature difference driving heat transfer between them. As the reaction warms the solution, that temperature difference grows.
Consequently, the heat-loss rate is unlikely to remain constant throughout the reaction.
Extending the final cooling gradient backwards across the entire four-minute reaction assumes a constant cooling rate that the experiment has not established.
The tempting interpretation is that \(29.8^\circ\mathrm{C}\) represents the temperature the solution would have reached without heat loss. That might be a model-based estimate under an additional assumption of constant heat-loss rate, but the supplied data do not justify that assumption.
Step 5: Identify what additional information is required.
A defensible correction would require an independent understanding of the calorimeter’s heat-loss behaviour during the reaction, such as a suitable calibration or a validated heat-transfer model using time-resolved temperature data.
Alternatively, the experiment could be redesigned so the reaction is sufficiently rapid that the standard extrapolation is appropriate.
The important conclusion is that neither the observed maximum nor the backwards extrapolated intercept necessarily gives the true no-heat-loss temperature in this situation.
A mathematically correct straight-line extrapolation is not automatically a physically valid correction. You must first check that the assumptions behind the line match what is happening in the experiment.
07Where to go next
The cooling correction helps you obtain a more realistic temperature change from experimental calorimetry data. Once that value is established, you can calculate the reaction heat and convert it into a molar enthalpy change.
The next useful distinction is between an experimentally measured enthalpy change and the enthalpy change of a reaction inferred from other reactions.
That leads to Hess’s law, where known enthalpy changes can be combined to calculate the enthalpy change of a reaction that is difficult to measure directly.
The quality of those calculations still depends on the quality of the original measurements. If heat loss was handled incorrectly in the calorimetry experiment, that error can carry through into every later calculation.