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Investigating the Effect of Concentration on the Temperature Rise, Heat Evolved and Heat of Neutralization for the Reaction Between HCl and NaOH

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Investigating the Effect of Concentration on the Temperature Rise, Heat Evolved and Heat of Neutralization for the Reaction Between HCl and NaOH

Neutralization is the special name given for the reaction between an
acid and an alkali leading to the formation of water molecules and a
salt. The reactions between basic oxides, or hydroxides, and acids are
very important and are called neutralizations. Since the metallic ions
and anions from the acid do not change, the essential reaction of
neutralization is always the formation of non-ionized molecules of
water from the hydroxide and hydrogen ions.

H+ (aq) + OH-(aq) [IMAGE] H2o (l)

The following are examples of neutralizations. The metallic ions and
the negative ions from the acids remain to produce the salt.

½ H2+So4- (aq) + ½ Na+OH- (aq) → ½ Na+So4- (aq) + H2o (l)

Sulphuric acid Sodium hydroxide Sodium sulphate water

H+No3- (aq) + Na+OH- (aq) → Na+No3- (aq) + H2o (l)

Nitric acid Sodium hydroxide Sodium nitrate water

H+cl- (aq) + Na+OH- (aq) → Na+Cl- (aq) + H2o (l)

Hydrochloric acid Sodium hydroxide Sodium chloride water

Since all these reactions reduce to H+ + OH-→ H2o, the energy change
accompanying both should be the same. Because, in all the three
reactions only 1 mole of water is produced, and the metallic ions and
the negative ions in the acid are rather spectator ions and do not
participate in the reaction so much as to give an energy change. If
aqueous acidic solutions are made up containing one mole of ½ H2So4,
HNo3, and HCl in 25 cm3, and alkaline solutions containing one mole of
NaOH in 25 cm3 (strong acids and alkalis), any neutralization between
these solutions produces 1 mole of water and liberates the same amount
of heat energy and this heat is called Heat of neutralization which is
usually - 57.3 kJ. If a bond is broken, energy is needed and the
reaction is endothermic. However if a reaction is exothermic, it is a
recombination, which is bond making. Through this, energy is produced.
In all these three reactions a hydrogen ion bond with a hydroxide ion
to produce water, giving out some energy and this is the only bonding
taking place in the reaction. Neutralisation is an exothermic reaction
which means it gives out energy to the surroundings in the form of
heat. This obviously means there will be a rise in temperature in the
reacting container as the bonds are made.

H+ (aq) + OH-(aq)→ H2o (l)

Δ H neutralization = - 57.3 kJ

The standard enthalpy of neutralisation of an acid is the enthalpy
change (heat of reaction per mole) under standard condition when the
acid is neutralized by the base and I mole of water is produced.

This is clearly shown in the energy diagram next page.

[IMAGE]

Factors affecting temperature change and heat evolved

1. Concentration - when the concentration of the acid and alkali
increases the temperature change and the heat evolved increases while
the heat of neutralisation will remain constant. It is because, when
the concentration of the acid and alkali is increased there will be
more number of acid or alkali particles in the same volume, so more
heat energy will be required for the reaction of these acid and alkali
particles. In fact when the concentration is doubled, double the
amount of energy will be needed for the reaction to take place. Thus,
when the energy required increases the heat evolved will also increase
obviously along with the temperature change. But the heat of
neutralisation remains same because it is the heat evolved per mole,
and when the concentration increases the mole also increases in the
same proportion so as the heat evolved, so heat of neutralisation will
not change by concentration change.

[IMAGE][IMAGE]

2. Volume - when the volume increases the heat evolved will increase
but the temperature change and heat of neutralisation will remain
constant. It is because. When the volume of the reactants is increased
there will be more acids and alkalis present in aqueous medium. So
more heat will be produced. Thus the heat evolved will increase as
volume increases. There will not be any changes in the heat of
neutralisation for the reason stated in the previous point.

[IMAGE]

I will make several calculations based on which I will do my
prediction.

Calculation for prediction

I am taking 1 M, 2 M, 3 M, 4 M, and 5 M HCl and NaOH to find out the
effect of concentration on the temperature rise, heat evolved, and
heat of neutralization for the reaction between HCl and NaOH. I will
keep the volume of acid, alkali as constant at 25 cm3 (considering the
direct mixing method). Hence the total mass of the neutral product
should be 50 cm3. I will use the following formulae for calculation of
my prediction:

Moles (mol) = concentration (M) X volume (cm3)

Heat evolved (J) = mass of neutral product (g) X specific heat
capacity of water (J/°C) X

Change in temperature (°C)

Heat evolved = Q

Change in temperature = ΔT

Heat of neutralisation = ΔH

Mass of neutral product = m

Specific heat capacity of water = c = 4.2 (J/°C)

ΔT = Q / (m x c)

ΔH = Q / moles

I will use direct proportion rule to predict the amount of heat that
should be evolved for each reaction after finding the number of moles
of NaOH used in the reaction by using the formula I stated before.

In each experiment both HCl and NaOH has same concentration. So I will
use NaOH in calculations. 1 M of NaOH gives 57 kJ of heat energy. In
exothermic reactions, ΔH is negative. The products are at a lower
energy than the reactants. In exothermic reactions, the energy
released in bond formation is greater than the energy used in breaking
old bonds. To work out ΔH, the equation that should be used is ΔH = m
x s x ΔT where m = mass, s = specific heat capacity of water (4.2 J/
°C), ΔT = change in temperature (initial temperature - final
temperature).

1. For the reaction of 25 cm3 of 1 M HCl and 1 M NaOH

Moles of NaOH = c x v = 1 M x 0.025 dm3 = 0.025 moles

Heat evolved = moles x 57 kJ = 0.025 x 57 kJ = 1425 J

Temperature change = Q / (m x c) = 1425 / (50 x 4.2) = 6.786 °C

2. For the reaction of 25 cm3 of 2 M HCl and 2 M NaOH

Moles of NaOH = c x v = 2 M x 0.025 dm3 = 0.05 moles

Heat evolved = moles x 57 kJ = 0.05 x 57 kJ = 2850 J

Temperature change = Q / (m x c) = 2850 / (50 x 4.2) = 13.571 °C

3. 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

4. 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

5. 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

Concentration of both acid and alkali (M)

Temperature change (°C)

Heat evolved (J)

Heat of neutralization (kJ)

1

[IMAGE]6.786

[IMAGE]1425

-57.3

2

13.571

2850

-57.3

3

20.357

4275

-57.3

4

27.143

5700

-57.3

5

33.929

7125

-57.3

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.

[IMAGE]

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

1. Polystyrene cup - for the reaction to take place without much heat
loss.

2. 25 ml measuring cylinders - to measure the acid and alkali.

3. Thermometer - to measure the temperature.

4. Small beaker - for keeping the 25 ml NaOH.

Materials

1. 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

Concentration of both acid and alkali (M)

Initial Temperature (°C)

Final

Temperature (°C)

Temperature change (°C)

1

2

3

4

5

Diagram

[IMAGE]

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

1. Thermometer - to measure the temperature.

2. Burette - to measure and keep the NaOH.

3. Pipette - to measure HCl.

4. Polystyrene cup - for the reaction to take place without much heat
loss.

5. Burette stand - to hold the burette.

Materials

1. 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

Volume (cm3)

Exp 1 - Final Temperature (°C)

Exp 2 - Final Temperature (°C)

Average Temperature (°C)

0

[IMAGE]5

10

15

For the accuracy of my experiment I will do my experiment twice and
take the average reacing as my original readings.

Diagram

[IMAGE]

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.

Volume (cm3)

Exp 1 - Final Temperature (°C)

Exp 2 - Final Temperature (°C)

Average

Temperature (°C)

0.00

26.00

26.00

26.00

5.00

27.80

28.20

28.00

10.00

29.00

30.00

29.50

15.00

30.30

30.50

30.40

20.00

31.00

32.00

31.50

25.00

31.80

32.20

32.00

30.00

31.60

32.20

31.90

35.00

31.50

31.50

31.50

40.00

31.00

31.00

31.00

45.00

30.70

30.50

30.60

50.00

30.00

30.20

30.10

These are the results I obtained for 2 M concentration of acid and
alkali.

Volume (cm3)

Exp 1 - Final Temperature (°C)

Exp 2 - Final Temperature (°C)

Average Temperature (°C)

0.00

26.00

26.00

26.00

5.00

29.70

30.30

30.00

10.00

32.30

32.70

32.50

15.00

33.90

34.10

34.00

20.00

36.30

36.70

36.50

25.00

38.50

38.50

38.50

30.00

37.00

37.40

37.20

35.00

36.00

36.00

36.00

40.00

35.00

35.00

35.00

45.00

34.00

34.20

34.10

50.00

33.00

33.20

33.10

These are the results I obtained for 3 M concentration of acid and
alkali.

Volume (cm3)

Exp 1 - Final Temperature (°C)

Exp 2 - Final Temperature (°C)

Average Temperature (°C)

0.00

26.00

26.00

26.00

5.00

31.00

31.20

31.10

10.00

35.00

35.00

35.00

15.00

38.10

38.90

38.50

20.00

42.40

42.60

42.50

25.00

44.00

44.00

44.00

30.00

43.50

43.50

43.50

35.00

41.30

41.70

41.50

40.00

40.00

40.00

40.00

45.00

39.00

39.00

39.00

50.00

38.00

38.20

38.10

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.

Concentration of acid + alkali (M)

Volume of acid (cm3)

Volume of alkali (cm3)

Comment about volume of alkali

1

25

23.5

1.5 cm3 less than the calculated 25 cm3

2

25

23.75

1.25 cm3 less than the calculated 25 cm3

3

25

23

1.5 cm3 less than the calculated 25 cm3

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.

[IMAGE]

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.

Concentration (M)

Predicted ∆t (°C)

Experimental ∆t (°C)

Expected ∆t (°C)

Comment

1

6.786

6.5

Experimental ∆t is very close to the predicted ∆t; in fact they have a
difference of 0.286 °C which is excellent.

2

13.571

12.5

13

Experimental ∆t is 0.5 °C less than the expected ∆t and it is 1.071 °C
less than the predicted ∆t. This result is satisfactory.

3

20.357

19.54

19.5

Experimental ∆t is only 0.04 °C more than the expected temperature and
it is 0.817 °C less than the predicted ∆t. Another great result
compared to the first one.

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.

[IMAGE]

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

Concentration (M)

Predicted heat evolved (J)

Experimental heat evolved (J)

Expected heat evolved (J)

Comment

1

1425

1324.05

Experimental Q is 100.95 J less than the predicted Q and this margin
is quite good and the result is acceptable.

2

2850

2559.38

2648.10

Experimental Q is 290.8 J less than the predicted Q and 88.72 J less
than the expected value of Q which is fairly accurate enough.

3

4275

3980.30

3972.15

Experimental Q is 294.7 J less than the predicted Q and 8.15 J more
than the expected Q which is once again a very good result.

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.

[IMAGE]

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

Concentration

Predicted ∆H (KJ)

Experimental ∆H (KJ)

Comment

1

-57.3

-52.96200

Experimental ∆H is 4.338 less than the predicted value which is best
among the three.

2

-57.3

-51.18750

Experimental ∆H is 6.1125 less than the predicted value which is
agreeable.

3

-57.3

-53.07064

Experimental ∆H is 4.22936 less than the predicted value which is very
good.

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.

[IMAGE]

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 %

How to Cite this Page

MLA Citation:
"Investigating the Effect of Concentration on the Temperature Rise, Heat Evolved and Heat of Neutralization for the Reaction Between HCl and NaOH." 123HelpMe.com. 22 Aug 2014
    <http://www.123HelpMe.com/view.asp?id=147759>.




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