Energy change will rise in proportion to the concentration as well. However, energy per mole will not rise. One mole of a solution is varied in different solutions, consequently the concentration will affect the amount of solution involved in making a mole of water, but not the energy in a mole.
Apparatus:
- 500ml beaker
- Two 50cm³ measuring cylinders
- Two conical flasks
- Thermometer
- Two pipettes
- Distilled water
- Polystyrene cup, with lid
- Labels, so that chemicals do not get mixed
- Acid- strong Hydrochloric, weak Ethanoic and Propanoic
- Alkalis- strong Sodium hydroxide, weak Ammonia solution
Fair Test:
To ensure that my experiment is as fair as possible I will first of all make sure that all of the variables I have not planned to change are kept constant. Therefore I will use the same concentration (1M) of everything and I will always use equal volumes of acid and alkali (25cm³). Every time that I repeat the experiment I will ensure it is carried out as the previous one was, and I will repeat each experiment three times so as to avoid anomalous results. I will make sure that I always hold the thermometer at the top so my body heat does not interfere with the readings. I will not take a reading off the thermometer before it has reached the highest temperature of that experiment, so I will wait for it to be constant for some time. I will also make sure that I do not stir some experiments more than others.
Safety:
To ensure that my experiment is safe I will follow these rules:
- Wear goggles throughout the experiment.
- Ensure my hair is tied back.
- Wear an overall.
- Wash hands before and after handling the chemicals.
- Be careful with the thermometer as it contains mercury.
- Make sure work area is tidy.
Due to the corrosive nature of the acids I must be very careful when handling them. If they come into contact with skin, the area must be washed.
Experiment order:
In my experiment I will carry it out in the following order-
- NaOH (Sodium Hydroxide) and HCl (Hydrochloric acid)
-
NaOH (Sodium Hydroxide) and CH3COOH (Ethanoic acid)
-
NaOH (Sodium Hydroxide) and CH3CH2COOH (Propanoic acid)
-
NH4OH (ammonia solution) and HCl (Hydrochloric acid)
-
NH4OH (ammonia solution) and CH3COOH (Ethanoic acid)
-
NH4OH (ammonia solution) and CH3CH2COOH (Propanoic acid)
Method:
- Set up apparatus as shown below:
- Fill two beakers, one with an acid and one an alkali (follow list above) and label them accordingly. The first one will be NaOH and HCl.
-
Using the measuring cylinder and pipette for accuracy measure 25cm3 of NaOH into the polystyrene cup.
- Record the temperature of the alkali.
-
Measure out 25cm3 of HCl using the second measuring cylinder and pipette.
- Record the temperature of the acid.
- Pour the acid into the cup along with the NaOH and very quickly place on the lid.
- Stir the solution five times with the thermometer.
- Wait until the temperature has reached its highest point, i.e. when there are no signs it will increase, and then record it.
- Wash out the measuring cylinders, thermometer and cup then repeat the experiment twice more.
- Repeat the whole experiment to obtain results for the different combinations of acids and alkalis shown above.
When completing the results table below, the initial temperature is the average of the initial temperature of acid and that of the alkali.
Results:
Finding the heat of neutralisation:
Key: ∆H= change in heat energy
m = mass (always 50g)
C = specific heat capacity (always 4.2 J/g/˚C)
θ = Temperature rise
Formula: ∆H= m x C x θ
For my results, in calculating the heat of neutralisation I rounded the temperature change to the nearest whole number and then found the average.
All heat of neutralisations will be negative as they are exothermic reactions
Results:
1. NaOH (Sodium Hydroxide) and HCl (Hydrochloric acid):
25 x 1 moles = 50 x 4.2 x 7
1000 1000
0.025 moles = 1.47
1 mole = 1.47
0.025
= -58.8
When sodium hydroxide reacts with hydrochloric acid the heat of neutralisation is –
-58.8kj
2. NaOH (Sodium Hydroxide) and CH3COOH (Ethanoic acid)
25 x 1 moles = 50 x 4.2 x 6
1000 1000
0.025 moles = 1.26
1 mole = 1.26
0.025
= -50.4
When sodium hydroxide reacts with ethanoic acid the heat of neutralisation is -
-50.4kj
3. NaOH (Sodium Hydroxide) and CH3CH2COOH (Propanoic acid)
25 x 1 moles = 50 x 4.2 x 6
1000 1000
0.025 moles = 1.26
1 mole = 1.26
0.025
= -50.4
When sodium hydroxide reacts with ethanoic acid the heat of neutralisation is -
-50.4kj
4. NH4OH (ammonia solution) and HCl (Hydrochloric acid)
25 x 1 moles = 50 x 4.2 x 7
1000 1000
0.025 moles = 1.47
1 mole = 1.47
0.025
= -58.8
When ammonia solution reacts with hydrochloric acid the heat of neutralisation is -
-58.8kj
5. NH4OH (ammonia solution) and CH3COOH (Ethanoic acid)
25 x 1 moles = 50 x 4.2 x 6
1000 1000
0.025 moles = 1.26
1 mole = 1.26
0.025
= -50.4
When ammonia solution reacts with ethanoic acid the heat of neutralisation is -
-50.4kj
6. NH4OH (ammonia solution) and CH3CH2COOH (Propanoic acid)
25 x 1 moles = 50 x 4.2 x 6
1000 1000
0.025 moles = 1.26
1 mole = 1.26
0.025
= -50.4
When ammonia solution reacts with propanoic acid the heat of neutralisation is -
-50.4kj
Analysis:
When doing my calculations for the heat of neutralisation I assumed that 1cm³ weighs one gram. I also assumed that they had the same specific heat capacity as water and that no heat was lost to the surrounding because of the polystyrene, although I am sure that some was. As can be seen in my results, when a strong acid and alkali are reacted together a larger temperature change was given off then when a strong and a weak acid and alkali are reacted. This confirms what I said in my prediction was accurate. As predicted the temperatures in all my reactions increased because the reactions are exothermic. The heat of neutralisation is always smaller if both the acid and alkali are not strong because energy is needed to fully ionise them before they can react. The temperature rise and heat of neutralisation was always the same when reacting two strong acids and alkalis. Having looked in a data book the typical neutralisation reaction has an energy value of -57kj/mol, this value is very close to my results.
From my results it is evident that if the alkali and acid reacting together are weak then there is a low heat of neutralisation. This is because in the reaction it is the H+ ions and OH- ions that are reacting together. Therefore the weaker acids and alkalis require energy to fully ionize before they can neutralise. The breaking of the bonds uses the energy so less is given off as heat, so all of these reactions are exothermic, but some give off more heat.
Evaluation:
Although my experiment was successful, there were still some errors made. I managed to get the results that I had predicted and had no anomalous results. I also followed my safety instructions carefully and carried out my method as planned. The flaws in the experiment were not in the application of it but in the planning, and the restriction of equipment. Although heat was lost to the surroundings, it was lost in all my experiments equally.
If I were to redo the experiment there are a few improvements I would make.
I feel that I could have been more accurate in measuring my volumes of acid and alkali, but then flaws in my readings could be attributed to human error. Also, because of human error the readings on the thermometer may not have been accurate, however, using a thermometer connected to a computer and having the computer take the readings could solve this problem. Repeating my experiment another time for each of the reactions would be more accurate as I would have more data. If repeating the experiment I would insulate the lid of the cup and the measuring cylinders to prevent heat loss through them.
To extend the investigation I could also look at some of the variables that I mentioned above. If I were to investigate a wider range of acids and alkalis it would extend the investigation and enhance my results. Another extension would be to experiment with different concentrations and seeing how they affect the heat of neutralisation, however this was not possible due to time limitations.
