enthalpy change in solution heading.

Enthalpy changes involving solutions

A coffee cup calorimeter

A diagram to shown a coffee cup calorimeter.

Many reactions in we study in chemistry take place in solution, for example neutralisation of an acid using an alkali is a chemical reaction that takes place in solution. In this reaction the reactants and the product are the system. The water and the calorimeter are part of the surroundings. If the reactants undergo an exothermic reaction then the heat produced will heat the water and as long as the calorimeter is well insulated we can use the temperature change to calculate the enthalpy change for the neutralisation reaction taking place. A simple but effective calorimeter is shown opposite.

Here a polystyrene coffee cup is used to hold the two solutions that are reacting. Polystyrene is an excellent insulator but this simple calorimeter can be improved if we simply place the polystyrene cup inside another cup to further insulate the reaction and prevent heat from leaving or entering the solution. The cork lid will also prevent heat loss by evaporation if the reaction is exothermic and also prevent heat from the surrounding entering the solution if the reaction is endothermic.

Calculating enthalpy of neutralisation

Neutralisation reactions are exothermic. When an acid and an alkali react they form a salt and water.

Acid(aq) + alkali(aq) salt(aq) + water(l)

A definition you should learn is:

The standard enthalpy of neutralisation is defined as: The enthalpy change when an acid is neutralised by an alkali or base to form one mole of water under standard conditions (298K, 100 kPa).

Consider the neutralisation reaction between the strong acid; hydrochloric and the strong alkali sodium hydroxide. An equation for this neutralisation reaction is given below:

HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l)
Here the acid and alkali react in a molar ratio of 1:1. We can use the simple coffee cup calorimeter to measure the enthalpy of neutralisation here. A coffee cup calorimeter being used to calculate the enthalpy of neutralisation, hydrochloric acid and sodium hydroxide. Method.

Calculating the enthalpy of neutralisation

Graph using extraploation method to calculate the maximum temperature rise during a neutralisation reaction using a coffee cup calorimeter.

Neutralisation reactions are very exothermic and can release large amounts of heat energy. We are also recording temperature changes over a fairly long period of time and despite our best efforts with insulation and using double polystyrene cups some heat loss to the surrounding will take place, especially if the reaction is very exothermic and releases large amounts of heat energy. However it is possible to compensate for SOME of this heat loss by using an extrapolation method as shown in the graph opposite.

Here you can see a set of results obtained by a student, they have plotted their results on a graph and extrapolated to find the maximum temperature reached to try and compensate for heat loss to the surroundings.

Example 1:

Let's use the example above to calculate the enthalpy of neutralisation when 25ml of 1M hydrochloric acid is neutralised by a 25ml of a 1M sodium hydroxide solution. Let's assume that the maximum temperature rise obtained from the graph in the example above was 70C.

To calculate the molar enthalpy of neutralisation we simply need to scale up to work out the enthalpy change when 1 mole of water is formed. In the neutralisation equation above 1 mole of water is formed from 1 mole of acid, so if can work out the enthalpy change for 1 mole of acid and this will equate to the formation of 1 mole of water.
First calculate the number of moles of acid used in the experiment:

Moles of acid used = concentration of acid x volume
= 1 x 25/1000
= 0.025 mol
So 0.025 moles of acid gave a temperature rise of 70C. To scale this up to molar quantities simply divide the enthalpy change obtained from the experiment by the number of moles used to get the molar enthalpy of neutralisation:
ΔH = q/number of moles
=1.47kJ/0.025 mol
= -58.8 kJ mol-1
(The answer is negative since it is an exothermic reaction).

Example 2 - Displacement reactions.

Apparatus diagram to show the displacement reaction between a copper sulfate solution and zinc metal.

Displacement reactions occur when a more reactive metal removes or displaces a less reactive metal from a compound or solution. Displacement reactions can be very exothermic, especially if the two metals involved are far apart in the reactivity series. For example zinc will displace copper from a copper sulfate solution according to the equation below|:

Zn(s + CuSO4(aq) → ZnSO4(aq) + Cu(s)
1mol    1mol         1mol         1mol


The method used is very similar to that for the neutralisation reaction discussed above. 50 ml of 0.75M copper sulfate solution was pipetted into a coffee cup calorimeter and the temperature was recorded every 30 seconds for 4 minutes. 5g of zinc powdered was then added to the copper sulfate solution and stirred continuously, the temperature was recorded every 30 seconds until a maximum temperature was reached. Then the temperature was recorded for a further 5 minutes. The initial temperature of the solution was 24oC and the final temperature was 540C. This reaction is slow and the student obtained the final temperature by extrapolation as in the previous example above.
Moles of copper sulfate present Moles of zinc present
Number of moles = concentration x volume
= 0.75 mol dm-3 x 50/1000
= 0.0375 mol
Number of moles = mass/Ar
= 0.077 mol
The zinc is in excess, which is what is required. This will ensure that the reaction goes to completion and all the copper sulfate will react. In order to calculate the enthalpy change for this displacement reaction we only take into account the mass of the solution as this is what we are measuring the temperature of.
So enthalpy change is calculated from:
q = m x c x ΔT
= 50 x 4.2 x 30
= 6300J or 6.3kJ
To calculate the enthalpy change per mole of copper sulfate we simply scale up and use the formula below:
ΔH = q/number of moles of copper sulfate
=6.3kJ/0.0375 mol
= -168kJmol-1

Key Points

Practice questions

Check your understanding - Questions on enthalpy changes in solution.