Chemistry Required Practical: Enthalpy of Neutralisation

Understanding the energy changes in chemical reactions is fundamental to chemistry, from industrial process design to biological systems. The enthalpy of neutralisation—the heat change when one mole of water is formed from the reaction of an acid and a base—is a key example. This required practical teaches you the essential calorimetry techniques to measure this energy change accurately, developing your skills in precise measurement, data manipulation, and critical evaluation of experimental limitations.

The Theory Behind the Measurement

The enthalpy of neutralisation ($\Delta H_{neut}$) is defined as the enthalpy change when one mole of water is produced from the reaction between an acid and a base under standard conditions. For strong acids and strong bases, the reaction is essentially the same: $H_{(aq)}^{+}+O{H}_{(aq)}^{-}\to H_2{O}_{(l)}$. The accepted value for this reaction is approximately -57.1 kJ mol${}^{-1}$, a useful benchmark for evaluating your experimental results.

To measure this, we use a simple calorimeter—typically a polystyrene cup fitted with a lid and a thermometer. Polystyrene is an excellent insulator, minimising heat exchange with the surroundings, which is vital for accuracy. The core principle is that the heat energy released by the exothermic neutralisation reaction ($q$) is absorbed by the solution itself. We assume the specific heat capacity of the final solution is the same as that of water (4.18 J g${}^{-1}$ K${}^{-1}$). The basic calculation uses the formula $q=mc\Delta T$, where $m$ is the mass of the solution (in grams), $c$ is the specific heat capacity, and $\Delta T$ is the temperature change.

Apparatus Setup and Correct Technique

A meticulous setup is the first defence against systematic error. You will need a polystyrene cup placed inside a beaker for stability, a lid with holes for a thermometer and pipette, a thermometer readable to 0.1°C (often a digital temperature probe is better), and volumetric pipettes or burettes for measuring precise volumes of acid and alkali.

The choice of reagents is important. A typical experiment uses 25.0 cm${}^3$ of 1.0 mol dm${}^{-3}$ hydrochloric acid and 25.0 cm${}^3$ of 1.0 mol dm${}^{-3}$ sodium hydroxide solution. Using equal volumes and concentrations ensures neither reactant is limiting, allowing the calculation to be based on the moles of water formed. Before mixing, you must measure the initial temperatures of both the acid and the alkali separately. A critical step is to calculate the mean initial temperature of these two readings. This is because the solutions may be at slightly different temperatures, and the reaction begins the moment they are mixed.

The Procedure and Temperature Monitoring

The procedure must be performed methodically to obtain reliable data. First, place the measured acid into the polystyrene cup and record its initial temperature. Then, measure the temperature of the alkali. Quickly but carefully, add the alkali to the acid, fit the lid, and start a stopwatch. Gently swirl the cup to ensure uniform mixing.

You must record the temperature of the mixture at fixed time intervals—for example, every 30 seconds for 5 minutes. It is crucial to continue stirring gently before each reading. You will observe the temperature rise rapidly to a maximum and then gradually fall as heat is lost to the surroundings. The highest temperature you record is not the true maximum temperature rise because heat loss begins immediately. Therefore, you must extrapolate the cooling curve back to the time of mixing to find the corrected maximum temperature.

Data Processing and Calculation

This is where your analytical skills come into play. Begin by plotting a graph of temperature (y-axis) against time (x-axis). Your graph should show a steep increase, a brief plateau at the peak, and then a gradual downward slope. Draw two best-fit lines: one through the initial steep rising portion and another through the later, linear cooling section. Extrapolate both lines back to the time at which the reactants were mixed (time = 0). The difference between the two extrapolated temperatures at this point gives you the corrected temperature change, $\Delta T$.

Now, perform the calculation step-by-step:

1. Calculate the heat evolved, $q$: The total mass of solution, $m$, is the sum of the masses of acid and alkali (assuming density is 1 g cm${}^{-3}$, 25 cm${}^3$ + 25 cm${}^3$ = 50.0 g). Use $q=mc\Delta T$, where $c=4.18$ J g${}^{-1}$ K${}^{-1}$. This gives $q$ in joules.
2. Find moles of water formed: The reaction equation is 1:1. Moles of $H^{+}$ = (concentration in mol dm${}^{-3}$) × (volume in dm${}^3$). For 25.0 cm${}^3$ of 1.0 M HCl, moles = 0.025 mol. Therefore, 0.025 mol of water is formed.
3. Calculate $\Delta H_{neut}$: $\Delta H_{neut}=q/\text{moles of water}$. As the reaction is exothermic, $q$ is negative (energy released), so $\Delta H_{neut}$ will be negative. Convert your answer from J mol${}^{-1}$ to kJ mol${}^{-1}$ by dividing by 1000.
4. Compare to the literature value: Calculate the percentage error: $\frac{|\text{experimental value}-\text{accepted value}|}{|\text{accepted value}|}\times 100\%$.

Evaluation of Accuracy and Limitations

No experimental value will perfectly match -57.1 kJ mol${}^{-1}$. Your evaluation is a test of your scientific reasoning. You must identify specific systematic errors that would consistently make your result less exothermic (less negative) than the accepted value. The most significant source is heat loss to the surroundings, despite the insulation. Even with extrapolation, some initial heat loss is unaccounted for, leading to a smaller observed $\Delta T$ and therefore a smaller magnitude of $\Delta H$.

Other errors include: assuming the specific heat capacity of the solution is exactly that of water; heat absorbed by the thermometer or the polystyrene cup itself; incomplete transfer or measurement of solutions; and not accounting for evaporation. You should also consider the precision of your instruments—a thermometer readable only to 1°C introduces more uncertainty than one readable to 0.1°C. A thorough evaluation links each identified limitation to its specific directional effect on the final result.

Common Pitfalls

Pitfall 1: Using the highest recorded temperature instead of the extrapolated value.

• Correction: Always plot the temperature-time graph and extrapolate the cooling curve back to the mixing time. The highest recorded point is always an underestimate due to continuous heat loss.

Pitfall 2: Using the initial temperature of only one reactant in the $\Delta T$ calculation.

• Correction: You must take the initial temperatures of both the acid and alkali and use the mean. Adding a warmer alkali to a cooler acid (or vice versa) means the starting energy of the system is an average of the two.

Pitfall 3: Confusing the sign of $\Delta H$ or forgetting unit conversions.

• Correction: Neutralisation is exothermic—$\Delta H$ is negative. Explicitly state this. Furthermore, remember that $q=mc\Delta T$ gives an answer in joules, but $\Delta H$ is typically quoted in kJ mol${}^{-1}$. Dividing by 1000 is a common but critical final step.

Pitfall 4: Vague evaluation blaming "human error" or "heat loss" without detail.

• Correction: Be specific. Instead of "heat loss," write: "Heat was lost to the air from the exposed surface of the solution and through the polystyrene cup, causing the measured temperature rise to be less than the theoretical maximum, resulting in a less exothermic $\Delta H$ value."

Summary

• The enthalpy of neutralisation is the heat change per mole of water formed in an acid-base reaction, with a theoretical value of approximately -57.1 kJ mol${}^{-1}$ for strong acid-strong base reactions.
• Accurate measurement requires a well-insulated calorimeter (polystyrene cup), precise temperature readings at intervals, and graphical extrapolation of the cooling curve to correct for heat loss and find the true $\Delta T$.
• The calculation follows a clear sequence: use $q=mc\Delta T$ to find the heat change, determine the moles of water formed from the limiting reagent (or the reagents if equimolar), and then find $\Delta H_{neut}=q/\text{moles of water}$.
• The main source of error is heat loss to the surroundings, leading to a measured $\Delta H$ that is less exothermic (closer to zero) than the literature value.
• A strong evaluation for this experiment identifies systematic errors like heat loss, assumptions about heat capacity, and instrumental limitations, and explains their specific directional effect on the final result.
• Mastery of this practical builds foundational skills in thermochemistry, data analysis, and critical experimental thinking, which are directly applicable to exam questions on energy changes.