Temperature change = Q / (m x c) = 2850 / (50 x 4.2) = 13.571 °C
-
For the reaction of 25 cm3 of 3 M HCl and 3 M NaOH
Moles of NaOH = c x v = 3 M x 0.025 dm3 = 0.075 moles
Heat evolved = moles x 57 kJ = 0.075 x 57 kJ = 4275 J
Temperature change = Q / (m x c) = 4275 / (50 x 4.2) = 20.357 °C
-
For the reaction of 25 cm3 of 4 M HCl and 4 M NaOH
Moles of NaOH = c x v = 4 M x 0.025 dm3 = 0.025 moles
Heat evolved = moles x 57 kJ = 0.1 x 57 kJ = 5700 J
Temperature change = Q / (m x c) = 5700 / (50 x 4.2) = 27.143 °C
-
For the reaction of 25 cm3 of 5 M HCl and 5 M NaOH
Moles of NaOH = c x v = 5 M x 0.025 dm3 = 0.125 moles
Heat evolved = moles x 57 kJ = 0.125 x 57 kJ = 7125 J
Temperature change = Q / (m x c) = 7125 / (50 x 4.2) = 33.929 °C
Prediction
-
I predict that the concentration will be directly proportional to the temperature change when volume is kept constant. It is because when the concentration of the acid and alkali increases, the acid and alkali particles also increase, since the volume is kept constant. When there are more reacting particles (H+ ions and OH- ions) more heat will be liberated, thus temperature difference between the initial and final temperature, which is nothing but the temperature change, increases as the concentration increases.
-
My next prediction is that the concentration will be directly proportional to the heat evolved from the reaction. It is because when the concentration increases the reacting ions (H+ and OH-) in the solution increases. When there are more particles to react more collisions will take place and more reactions occur, giving off more heat. Thus it is clear that the heat evolved increases will concentration increase.
- I also predict that the heat of neutralization will remain constant at any concentration. It is because the heat of neutralization is nothing but the amount of heat evolved per mole. The concentration change will change the mole and as well as the heat evolved. Concentration is directly proportional to the heat evolved and it is also directly proportional to the mole, thus any increase in concentration will increase both the factors affecting heat of neutralization so that no difference can be noted in the ΔH whether the concentration is increased or decreased.
From this I can conclude that concentration will be directly proportional to temperature change and heat evolved, and heat of neutralization will be constant at any concentration. So when concentration is doubled temperature change and heat evolved will be doubled while the heat of neutralization remains constant. And when the concentration triples temperature change and heat evolved will be tripled while heat of neutralization will remain same.
Investigation of heat of neutralization
There are two main methods of doing my experiment:
-
Direct mixing method – This is a quicker method where we just add 25 cm3 of alkali to the 25 cm3 of acid and measure the temperature change. Because of its quickness this method is fairly inaccurate.
- Thermometric titration – This is a time taking method where a small amount of alkali is added to the acid consequently and the temperature is measured each time so that the point of neutralization and temperature change can be identified more accurately.
Direct mixing
Apparatus
- Polystyrene cup – for the reaction to take place without much heat loss.
- 25 ml measuring cylinders – to measure the acid and alkali.
- Thermometer – to measure the temperature.
- Small beaker – for keeping the 25 ml NaOH.
Materials
- 1 M, 2 M, 3 M, 4 M, and 5 M NaOH and HCl solutions
Factors to control
-
Volume – the volume of the acid and alkali should be exactly 25 cm3.
-
Temperature – the temperature of the atmosphere should be same throughout the experiments. Keeping it as 20 Co is good.
Precautions
All chemicals must be used carefully, and all experiments should be conducted with extreme care. Here are some points that must be followed.
- Good ventilation is required - for example, HCl has a very potent smells.
- Sodium hydroxide is corrosive. Wash hands carefully and thoroughly.
- Hydrochloric acid is irritants.
- Handle the thermometer with care - it contains mercury, which is poisonous.
- Handle all glassware with care - it may be hot after the reaction has taken place.
Procedure
Measure 25 ml of 1 M NaOH in one measuring cylinder. Measure 25 ml of 1 M HCl in another measuring cylinder. Pour the HCl into the polystyrene cup and measure its temperature with a thermometer. Measure the temperature of the NaOH with another thermometer. Pour the NaOH solution into the HCl solution. Immediately stir the mixture and note down the final temperature with one of the thermometer. Repeat the procedure with 2 M, 3 M, 4 M and 5 M concentrations.
Table of observation
Initial temperature of HCl – Ta
Initial temperature of NaOH – Tb
Initial temperature – T1
T1 = Ta + Tb / 2
Final temperature – T2
Rise in temperature or Temperature change = T2 – T1
Diagram
Thermometric titration
This is the method I am going to use to investigate my aim, mostly because this method is more practical and accurate than any other methods available for me to examine my aim.
Apparatus
- Thermometer – to measure the temperature.
- Burette – to measure and keep the NaOH.
- Pipette – to measure HCl.
- Polystyrene cup – for the reaction to take place without much heat loss.
- Burette stand – to hold the burette.
Materials
- 1 M, 2 M, 3 M, 4 M, and 5 M NaOH and HCl solutions
Factors to control
-
Volume – the volume of the acid in the polystyrene cup should be exactly 25 cm3 and the volume of the alkali in the burette should be exactly 50 cm3.
-
Temperature – the temperature of the atmosphere should be same throughout the experiments. Keeping it as 20 Co is good.
Precautions
- When noting the volume of the alkali in the burette, take the lower meniscus as the measure.
- Be careful while using pipette to take 25 ml of HCl, since it can be very dangerous if it is accidentally drunk.
- Handle the glassware with care.
- Handle the thermometer with care since mercury is poisonous.
- Do not hit the walls of the cup while stirring the mixture with the thermometer.
Procedure
Measure 50 ml of 1 M NaOH using a burette and fix it on a stand. Measure 25 ml of 1 M HCl in a pipette and pour completely into the polystyrene cup. Place this cup under the nozzle of the burette so that when alkali drops from the burette it doesn’t spill out. Measure the temperature of the acid in the polystyrene cup. Then add 5 ml of alkali into the cup, stir a bit with the thermometer and measure the temperature quickly and record it. Again add another 5 ml of alkali and follow the same procedures until all 50 ml of NaOH had been added. Repeat the same actions with 2 M, 3 M, 4 M & 5 M concentrations.
Table of observation
Concentration = ___M
For the accuracy of my experiment I will do my experiment twice and take the average reacing as my original readings.
Diagram
As I planned I did the experiment by thermometric method. I intended to do the experiment with five concentrations of acid and alkali for more accuracy but unfortunately I was only provided with three concentrations by the school management, so I was forced to do my experiment with the given three concentrations. I performed my experiment twice, for all the three provided concentrations, so that my results can be considered accurate enough to draw conclusions with. In both time I took 25 cm3 of acid in a polystyrene cup and 50 cm3 of alkali in the burette, and dropped 5 cm3 of acid at first, and then 10, then 15, so on, into the acid until I dropped all 50 cm3 and I noted the temperature of the reacting mixture at these intervals. I noted down both the experiments’ readings and found the average of those both readings. I repeated this procedure for all the three concentrations.
I was so conscious about the accuracy of my test, so that I took the following precautions.
Major Precautions
- Stir the solution constantly without hitting the walls of the polystyrene cup. Stirring helps to spread the heat evenly to all the parts of the solution. But when the thermometer, which acts as a stirrer, hits the walls of the cup, small amount of heat is produced. This extra heat may affect my experiment. So I stirred the solution constantly without striking the walls of the container.
- Measure the temperature quickly. Since the room temperature is vastly lower than the heat evolved during the reaction, there is a great chance of heat escaping to the atmosphere. So to get precise results I measured the temperature as quickly as I finished pouring the particular volumes of alkali.
- Thermometer readings have to be read properly. Since I was stirring with the thermometer which actually used to measure the temperature, I was extra careful about the measurement of temperature from the thermometer. I made the thermometer stand erect and saw the temperature mark in eye level.
These are the results I obtained for 1 M concentration of acid and alkali.
These are the results I obtained for 2 M concentration of acid and alkali.
These are the results I obtained for 3 M concentration of acid and alkali.
I have repeated my experiment twice for all the three concentrations of acid and alkali so that I will get an accurate enough average results of these two experiments.
I will utilize my graph to prove my prediction that temperature change and heat evolved are directly proportional to the concentration and heat of neutralization remains constant at varying concentrations.
Analyzing the graphs plotted in obtaining evidence
I have drawn six graphs. Three graphs are drawn of Volume of NaOH against temperature for the three concentrations, 1 M, 2 M & 3 M. One graph is of concentration against temperature change, one graph is of concentration against heat evolved and another graph is of concentration against heat of neutralization.
Although I can directly find the temperature change, heat evolved and heat of neutralization using the results I got. But I feel that this is not so accurate. So I decided to use the graph to find the point of neutralization, final temperature and alkali volume of neutralization. To do this, I first plotted the points from 0 cm3 of NaOH to 20 cm3 of NaOH and drew a line of best fit along those points, and I drew another line of best fit through the points from 30 cm3 of NaOH to 50 cm3 NaOH. I extended both the lines lavishly and marked the point at which they intersect and this point is called point of neutralization. The x-axis value of the point was the alkali volume of neutralization while the y-axis value of the point was the final temperature. Temperature change can be calculated by subtracting final temperature from the initial temperature. As I am taking the point of neutralization from the intersection of two lines of best fit, the readings from the graph would be exact than the results obtained during the experiment.
Anomaly: - I was forced to draw a line of best fit in all my graphs because the points were a bit scattered; some were above the line of best fit, like, in the 3 M graph of volume of NaOH against temperature points like (10, 35) and (5, 31) are like this, while some were below the line of best fit, like, in the same graph points like (40, 40) and (45, 39) are like this. I think the reasons why some points are above the line of best fit are that I might have poured slightly less volume of alkali or that I might have taken too long to read the readings. And the reasons of some points being below the line of best fit indicates that I might have added more volume of alkali or that I might have misread or that I might have hit the wall of the container with the thermometer too many times while I stirred the solution.
Analyzing the point of neutralization
The point at which all the H+ ions in the HCl solution and OH- ions in the NaOH solution have bonded together is known as point of neutralization. It is called such because at this point the acid is completely neutralized by the alkali. At this point temperature change would be the maximum because all ions have formed bonds and maximum heat is released. This is what I found in the previous section using my graph. I will now draw a table to show the volume of alkali, NaOH used to neutralize 25 cm3 of different concentrations of acid, HCl. I will find the volume of alkali by taking the x-axis value of the point of neutralization that I found for the three graphs with three concentrations.
In my experiments I had used 25 cm3 of HCl acid expecting that the point of neutralization would occur at 25 cm3 of alkali, since one mole of H+ ions bond with one mole of OH- ions to form one mole of water molecules, moreover I expected this because HCl and NaOH in a particular volume will give same amount of moles of H+ ions and OH- ions. So when the temperature reached maximum it is obvious that all the H+ ions and OH- ions have bonded, and no more H+ ions left in the mixture to react. Since the temperature is at its peak, the heat evolved will also be at its maximum, and the heat of neutralization can be calculated from the heat evolved and this ∆H should remain constant for all the concentrations.
The bar graph below shows the difference between the expected volume of alkali and experimental volume of alkali. I can clearly see that my investigational volume of alkali is less than what I calculated according to the logical theory.
Analyzing the relationship between concentration and temperature change
Each experiment I did twice to improve my accuracy on the results. I have predicted before that the relationship between concentration and temperature change is that concentration is directly proportional to the temperature change and that when concentration doubles temperature change doubles and when concentration triples temperature change also triples.
These are some of the magnificent pieces of results I ever obtained. The experimental temperature change of the 1 M concentration is the best result of all, but others too are close enough to be called accurate. One of the perfect trends I can se over here is the doubling and tripling of the temperature change. Although the doubled 2 M value is short of the exact expected doubled value, the tripled 3 M value is almost exactly the expected tripled value. This is because when the concentration is doubled the number of H+ and OH- ions doubles, and when this doubles the number of reactions taking place doubles, and when this doubles the energy released doubles and when this doubles the final temperature also increases in such a way that the temperature change is doubled. Even in tripling this is the case, because of tripled number of reactions producing tripled energy, the temperature change triples.
I go some anomalous results in the graph since it is impractical to get a straight line graph for this kind of experiments. The strange alikeness existing between all the three graphs of concentration against temperature is that the temperatures after adding 5 cm3 and 10 cm3 NaOH are anomalous for all the three graphs. Moreover, the points existing after 30cm3 volume are almost straight for all the three graphs except for the 3 M where the points are above the line of best fit; this maybe because of error in my measurement of the temperature. Overall I am getting less temperature change than I ought to get, and I think it is because I took less volume of HCl or NaOH or I took the temperature reading after too long or that I stirred the reacting solution less than I ought to.
Analyzing the relationship between concentration and heat evolved
During a neutralization reaction, heat is given out. Because neutralization reaction is basically the bond formation between the H+ and OH- ions, bond formation always gives out energy, so neutralization reaction gives out energy. Exothermic reaction is a reaction which gives out heat energy. Since neutralization gives out energy it is an exothermic reaction.
I have already predicted that concentration is directly proportional to the heat evolved.
Heat evolved can be calculated by:
Heat evolved = mass of neutral product (g) X specific heat capacity of water (J/°C) X change
in temperature (°C)
Mass of neutral product is the mass of acid and alkali reacted together to form and neutral product or the mass of acid and alkali of the point of neutralization.
Mass of neutral product = 25 g (acid weight) + weight of neutralization of alkali
Specific heat capacity of water = 4.2 J/°C
Temperature change = Temperature change found from the graph.
Calculation
Heat evolved in 1 M concentration: -
Heat evolved = (25 g + 23.5 g) X 4.2 J/°C X 6.5 °C
= 48.5 g X 4.2 J/°C X 6.5 °C
= 1324.05 J
Heat evolved in 2 M concentration: -
Heat evolved = (25 g + 23.75 g) X 4.2 J/°C X 12.5 °C
= 48.75 g X 4.2 J/°C X 12.5 °C
= 2559.38 J
Heat evolved in 3 M concentration: -
Heat evolved = (25 g + 23.5 g) X 4.2 J/°C X 19.54 °C
= 48.5 g X 4.2 J/°C X 19.54 °C
= 3980.30J
The table summarizes the results I got by calculating
I did my experiment two times and my result is based on the average of both the experiments, so I am convinced that my results are accurate as possible. The result for the 1 M concentration was the best of all and this justifies my act of comparing the 1 M result with the others with respect to doubling and tripling. The heat evolved for the 2 M concentration is only about 89 J different from double the heat evolved for the 1 M concentration. And the heat evolved for 3 M concentration is only about 8 J different from triple the heat evolved for 1 M concentration. This goes on to prove that the heat evolved directly varies along with concentration. This is because when the concentration increases the reacting particles increase and when this increase more reactions happen and when this happens more heat is given out. The overall results state that the heat evolved is less than the predicted heat evolved. This may be because I didn’t notice the temperature reading properly, or I might have added less acid or alkali, or I might have waited too long before I took the reading or I stirred the solution less.
Analyzing the relationship between concentration and heat of neutralization.
Heat of neutralization is the heat evolved per mole. It is the amount of heat produced when one mole of acid (H+ ions) react completely with one mole of alkali (OH- ions) to form one mole of neutral product.
Heat of neutralization = heat evolved / moles
Moles = concentration X volume
Calculation
Heat of neutralization of 1 M concentration = 1324.05 J / (1 M X 0.025 dm3)
= 1324.05 J / 0.025 mol
= -52,962 J or -52.962 KJ
Heat of neutralization of 2 M concentration = 2559.38 J / (1 M X 0.05 dm3)
= 2559.38 J / 0.05 mol
= -51,187.5 J or -51.1875 KJ
Heat of neutralization of 3 M concentration = 3980.30 J / (1 M X 0.075 dm3)
= 3980.30 J / 0.075 mol
= -53070.64 J or -53.07064 KJ
I have predicted that the heat of neutralization remains constant even if concentration differs. Although I did not get same ∆H for all the concentrations, I did get a very close range of results. The highest range between my highest ∆H and lowest ∆H is 1.88314 KJ which is very close. But at the same time my results are about 5 KJ short of the predicted value. I think it is may be because I did not take the readings properly or added less volume of acid or alkali or added less concentration of acid or alkali.
Conclusion
My experiment is very successful and I strongly believe that I have attained my aim of investigating the relationship between concentration and temperature change, concentration and heat evolved and concentration and heat of neutralization.
- Concentration is directly proportional to the temperature change, at constant volume.
- Concentration is directly proportional to the heat evolved, at constant volume.
- Concentration does not affect the heat of neutralization.
I think that the success of my experiment solely depended on the accuracy of the experiment and I managed well enough to perform my experiment with great accuracy. I did my experiment twice for each concentration and got the average readings as the final readings and this method of doing helped me in disguising the error I might have done in one experiment.
% Accuracy
% error = ((Predicted value – experimental value) / predicted value) X 100
% error for temperature change in 1 M concentration = ((6.79 – 6.5) / 6.79) X 100
= (0.29 / 6.79) X 100
= 4.21 %
% error for temperature change in 2 M concentration = ((13.57 – 12.5) / 13.57) X 100
= (1.07 / 13.57) X 100
= 7.89 %
% error for temperature change in 3 M concentration = ((20.36 – 19.54) / 20.36) X 100
= (0.82 / 20.36) X 100
= 4.01 %
% error for heat evolved in 1 M concentration = ((1425 – 1324.05) / 1425) X 100
= (100.95 / 1425) X 100
= 7.08 %
% error for heat evolved in 2 M concentration = ((2850 – 2559.38) / 2850) X 100
= (290.63 / 2850) X 100
= 10.2 %
% error for heat evolved in 3 M concentration = ((4275 – 3980.3) / 3980.3) X 100
= (294.7 / 3980.3) X 100
= 6.89 %
% error for heat of neutralization in 1 M concentration= ((57.3 – 52.96) / 57.3) X 100
= (4.34 / 57.3) X 100
= 7.57 %
% error for heat of neutralization in 2 M concentration= ((57.3 – 51.19) / 57.3) X 100
= (6.11 / 57.3) X 100
= 10.67 %
% error for heat of neutralization in 3 M concentration= ((57.3 – 53.07) / 57.3) X 100
= (4.23 / 57.3) X 100
= 7.38 %
