SCES 3163 PHYSICAL CHEMISTRY (PRACTICAL REPORT)

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SCES 3163 Science Intake June 2019

Physical Chemistry

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TABLE OF CONTENTS

1. PRACTICAL REPORT 1
Determining the enthalpy change of neutralization between strong acid
and strong base

1.1. Example of task (Ang Sian Ling)
1.2. Example of task (Eason Ting Sin Bing)

2. PRACTICAL REPORT 2
Determining the enthalpy change of reaction in conversation of sodium
hydrogen carbonate to sodium carbonate reaction using Hess Law

2.1. Example of task (Muhammad Zamir Syahmi bin S. Fadhirllah al-Jupry)
2.2. Example of task (Ho Jia Heng)

3. PRACTICAL REPORT 3
Investigating the effect of concentration effect on E.M.F cell

3.1. Example of task (Muhammad Zulhilmi bin Zulkifli)
3.2. Example of task (Ying Sii Mei)

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1. PRACTICAL REPORT 1

Determining the enthalpy change of neutralization between
strong acid and strong base

1.1. Example of task (Ang Sian Ling)

1.0 AIM: Determining the enthalpy change of neutralization between strong acid and strong
base.

2.0 INTRODUCTION:

According to the Arrhenius theory, an acid is anything that contains hydrogen and ionizes (H+)
into hydrogen ions in aqueous solution, while a base is anything that contains the hydroxyl
group (OH) (De & Pilot, 2001). An acid is a proton giver while a base is a proton acceptor, as
stated by Lowry-Bronsted (Manjula, 2016). Besides, according to the Lewis (1983), bases are
molecules or ions that have unshared electron pairs accessible for sharing with acids, while
acids are defined as molecules or ions capable of coordinating with unshared electron pairs.
A molecule needs to be electron deficient in order to be acidic in the Lewis sense. This is the
broadest definition of acid and base. Lewis acids, which include all Lowery Bronstead acids,
also include many additional reagents like boron trifluoride, aluminium chloride, etc.

Neutralization is a reaction where H+ ions from acid and OH- ions from base react to from
water and salt (Zakrewsky etl.al., 2014). A pH of 7 results from the neutralization of a potent
acid and potent base. A strong base neutralizing a weak acid will produce a pH result that is
more than 7, while the neutralization of a strong acid and weak base will produce a pH result
that is less than 7. A solution is neutralized when equal amounts of acid and base combine to
create salts. One mole of protons (H+) is equivalent to one mole of base, and one mole of acid
is equivalent to one mole of protons (H+) (OH-). N parts of acid will always neutralize N parts
of base because salts are created by neutralization reactions with equivalent weight
concentrations of acids and bases.

Enthalpy is defined as a thermodynamic property of a system linked to internal energy. The
capital H is used to represented the enthalpy. The energy of the bonds formed and broken
during a chemical reaction is related to enthalpy. On the other hand, change in enthalpy is
represented by the symbol ∆H. The formula to find the change in enthalpy (∆H) is enthalpy of
the products – enthalpy of the reactants. Change in enthalpy (∆H) for an exothermic reaction
is negative while for an endothermic reaction the change in enthalpy (∆H) is positive. The

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different between enthalpy and enthalpy is enthalpy cannot be measured and calculated but
enthalpy changes can be measured and calculated (Alyn, Teresa & Nora, 2015). The formula
to calculate enthalpy changes is:

∆H=Q (no. of mol of water)
∆H = mcθ

Where,
Q= change in heat energy in joules
m = mass in grams of the substance to which the temperature change occurs (usually water
(for combustion) or a solution)
c = specific heat capacity (energy required to raise the temperature of 1g of
a substance by 1ºC)
θ = temperature change in ºC or Kelvin (K)
Change in enthalpy (∆H) is affected by three factors which are quantity of acid and base,
basicity of the acid and base and strength of acid and base.

In term of quantity of acid and base:
Quantity of acid and base used is directly proportional to quantity of heat released in the
neutralization reaction. The example below shows how quantity if acid and base affecting the
change in enthalpy (∆H).
1 mole of HCl + 1 mole of NaOH Heat Released = 57.0kJ
2 mole of HCl + 2 mole of NaOH Heat released = 57.0 kJ x 2 = 114kJ
3 mole of HCl + 3 mole of NaOH Heat released = 57.0 kJ x 23= 171kJ

In term of basicity of acid and base:
Diprotic acid neutralization generates greater heat energy than monoprotic acid neutralization.
When monoprotic acid for example HCl react with base NaOH, the equation is (HCl + NaOH
→ H2O + NaCl) 1 mole of HCl will react with 1 mole of NaOH to produce 57.0kJ of heat.
Meanwhile, when diprotic acid for example H2SO4 react with base NaOH, the equation is
(H2SO4 + 2NaOH→Na2SO4 + 2H2O). The equation shows that 1 mole of H2SO4 will react with
2 moles of NaOH to produce 114 kJ of heat.

In term of strength of acid and base:

According to Ajit & Nameeta (2022), when a strong acid, HA, react with a strong base,

BOH, the reactions that occurs is

H+ + OH-  H2O.

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As neither the cation B+ nor the anion A are involved in the neutralization reaction, and the
acid and base are completely separated. At 25 °C, this reaction's enthalpy change is -57.62
kJ/mol.

On the other hands, for weak acids or bases, the heat of neutralization is pH-dependent
(Soumen & Asok, 2002). For complete dissociation to occur in the absence of any additional
mineral acid or base, heat is needed. There will be less overall heat produced during
neutralization. The following is an example for the heat of neutralization between reaction of
Hydrogen Cyanide (HCN) and Sodium Hydroxide (NaOH).

HCN+NaOH  NaCN+H2O; ∆H = -12kJ/mol
The experiment is conducted by using Nitric Acid HNO3, Sulfric Acid H2SO4, Hydrochloric
Acid HCl as acid and Sodium Hydroxide NaOH as base and we aim to determining the
enthalpy change of neutralization between strong acid and strong base.

3.0 HYPOTHESIS:

The enthalpy of neutralization for a mixture sulfuric acid and sodium hydroxide is higher
compared to the mixture of nitric acid with sodium hydroxide and a mixture of hydrochloric
acid.

4.0 VARIABLE

4.1 Manipulated Variable : Type of acid
4.2 Response Variable : Enthalpy of neutralization
4.3 Control Variable : Molarity of bas and acid

5.0 MATERIAL:

Sodium Hydroxide NaOH, Nitric Acid HNO3, Sulfric Acid H2SO4, Hydrochloric Acid HCl,
distilled water.
5.1 APPARATUS:
Plastic cup, thermometer, Measuring cylinder, filter paper

6.0 SAFETY PRECAUTIONS:
1. Lab coat must be worn throughout the experiment. The purpose of wearing lab coat is
to reduce harm to body as protection against spills, explosions and other unexpected
events that may occur during the experiment.
2. The doors and windows should always be open throughout the experiment to allow
good ventilation.

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3. The solution used in the experiment should be disposed of in a proper and safe way.
4. Pour the acid into the base. This is because when acid added into base the heat

generated heat that may causes mixture splash out and causes burns.

7.0 PROCEDURE
1. 30cm3 of sodium hydroxide was measured by using measuring cylinder.
2. Initial temperature of the sodium hydroxide was measured and recorded.
3. 30cm3 of hydrochloric acid was measured by using measuring cylinder.
4. Iinitial temperature of the hydrochloric acid was measured and recorded.
5. Sodium hydroxide was transferred into plastic cup. Hydrochloric acid was poured into
sodium hydroxide.
6. Stir well the mixture. The temperature of the mixture was measured with a thermometer,
observe the temperature change. The highest temperature of the mixture was recorded.
7. Calculate the heat release out (Q) and the enthalpy change during neutralization (∆H).
8. Repeat the experiment by replacing hydrochloric acid with nitric acid and sulfuric acid.

8.0 OBSERVATION AND RESULT:

8.1 Reaction between Nitric Acid and Sodium Hydroxide

Volume of HNO3 30cm3

Mole of HNO3 1.0 mol
Volume of NaOH 30cm3

Mole of NaOH 1.0 mol
Initial Temperature t1 ⁰C 29.5⁰C
Highest temperature t2 ⁰C 34.5⁰C

Change in temperature 5

Equation:
HNO3 (aq) + NaOH (aq) -- NaNO3(aq) + H2O(l)
1 mole of H+ from HCl react with 1 mole OH- from NaOH to form 1 mole of H2O. Therefore,
0.030 mole of H2SO4 react with 0.03 mole of NaOH to form 0.030 mole of H2O

Heat Given Out (Q) Enthalpy of neutralization
∆H=Q (no. of mol of water)
From the equation =(mcθ) (no. of mole of water)
Density = = (1260) J (0.03 mol)
= 37.80 kJ/mol
= - 37.80 kJ/mol

Mass of the mixture, m =
(volume of solution) (density of solution)

= (30+30)cm3 x 1.0 g cm3

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= 60g
∆H = mcθ
= (60) (4.2) (5)
= 1260J

8.2 Reaction between Hydrochloric Acid and Sodium Hydroxide

Volume of HCl 30cm3

Mole of HCl 1.0 mol

Volume of NaOH 30cm3

Mole of NaOH 1.0 mol
Initial Temperature t1 ⁰C 29.5⁰C
Highest temperature t2 ⁰C 35.0⁰C

Change in temperature 5.5

Equation:
HCl (aq) + NaOH (aq) -- NaCl(aq)+H2O (l)
1 mole of HCl react with 1 mole on NaOH to form 1 mole of H2O. Therefore, 0.030 mole of
HCl react with 0.03 mole of NaOH to form 0.030 mole of H20

Heat Given Out (Q) Enthalpy of neutralization
∆H=Q (no. of mol of water)
From the equation =(mcθ) (no. of mole of water)
Density = = (1386) J (0.03 mol)
= 26.26 kJ/mol
= - 41.58 kJ/mol

Mass of the mixture, m =
(volume of solution) (density of solution)

= (30+30) cm3 x 1.0 g cm3

= 60g
∆H = mcθ

= (60) (4.2) (5.5)

= 1386J

8.3 Reaction between Sulfuric Acid and Sodium Hydroxide

Volume of H2SO4 15cm3

Mole of H2SO4 1.0 mol
Volume of NaOH 30cm3

Mole of NaOH 1.0 mol

Initial Temperature t1 ⁰C 29.5⁰C

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Highest temperature t2 ⁰C 37.5⁰C

Change in temperature 8

Equation:
H2SO4 (aq) + NaOH (aq) -- Na2SO4 (aq) + H2O(l)
1 mole of H+ from HCl react with 1 mole OH- from NaOH to form 1 mole of H2O. Therefore,

0.030 mole of H2SO4 react with 0.03 mole of NaOH to form 0.030 mole of H2O

Heat Given Out (Q) Enthalpy of neutralization
∆H=Q (no. of mol of water)

From the equation =(mcθ) (no. of mole of water)
Density = = (1512) J (0.03 mol)
= 45.36 kJ/mol
= -45.36 kJ/mol

Mass of the mixture, m =
(volume of solution) (density of solution)

= (15+30) cm3 x 1.0 g cm3

= 45g
∆H = mcθ

= (45) (4.2) (8)

= 1512J

9.0 DISCUSSION:
While carry out this experiment, it was found that the reaction occurs is an exothermic reaction
that is the release of energy the environment. When comparing the heat of neutralization, the
negative mark should be ignored because it only represents a symbol of liberation energy to
the environment. The heat of neutralization will be affected by the acid base. Monobasic acids
such as hydrochloric acid, HCl and nitric acid, HNO3 will produce 1 mole of acid monobase
that will react with 1 mole of OH- ions (base) for produce 1 mole of water as the follow equation.

Reaction between Nitric Acid and Sodium Hydroxide:
HNO3 (aq) + NaOH (aq) -- NaNO3(aq) + H2O(l)
Reaction between Hydrochloric Acid and Sodium Hydroxide:
HCl (aq) + NaOH (aq) -- NaCl(aq)+H2O (l)
If the acid used is dibasic acid such as sulfuric acid (H2SO4), it will be neutralized completely
by 2 moles of NaOH and produce 2 moles of water as the follow equation:
Reaction between Sulfuric Acid and Sodium Hydroxide:
H2SO4 (aq) + NaOH (aq) -- Na2SO4 (aq) + H2O(l)

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In this experiment, produce the enthalpy of neutralization for reaction between HNO3 and
NaOH is -37.80kJ/mol. This result is not the same as the result stated by Ted & Janet (2020)
who stated the enthalpy of neutralization for reaction between HNO3 and NaOH is 53.7kJ/mol.

On the other hand, the findings of enthalpy of neutralization for HCl and NaOH is - 41.58
kJ/mol. However, William & Clifford (2010) state that enthalpy of neutralization for HCl and
NaOH is state that the -55.84kJ/mol.

Besides, this experiment show reaction between H2SO4 and NaOH has produce the
enthalpy of neutralization of -45.36 kJ/mol. However, Lawrie & Roger (2014) state enthalpy of
neutralization for H2SO4 and NaOH is state that the -57.3kJ/mol.

All this result error is due to the current error factor implement the process of implementing
the procedure for the mixture sulfuric acid is not poured as quickly as possible and as carefully
as possible can so that heat is not released to the environment. Besides, we only use regular
plastic cups but we didn’t use wide polythene bottles. This is also one of the causes of rapid
heat release. In addition, the rapid air flow in the laboratory also causes rapid heat loss.

As an improvement, we should use wide polythene bottles during the experiment.
Additionally, a lid can be made of insulating material and cover the cup to reduce the speed
of heat loss. Furthermore, the method of pouring the acid must be correct. For example, make
sure all the acid is poured at a fast rate. Moreover, the temperature should be recorded every
5 seconds to ensure that the highest temperature is not overlooked. Video can also be taken
throughout the experiment as it is possible for the operator to overlook the temperature reading
during the experiment.

9.1 COMPARISON BETWEEN ENTHAPLY OF NEUTRALIZATION AMONG THREE

MIXTURES

NAME OF TYPE OF ACID AND EQUATION

MIXTURE BASE

MIX A Reaction between HNO3 (aq) + NaOH (aq) -- NaNO3(aq) + H2O(l)
MIX B Nitric Acid and
MIX C Sodium Hydroxide
Reaction between HCl (aq) + NaOH (aq) -- NaCl(aq)+H2O (l)
Hydrochloric Acid and
Sodium Hydroxide
Reaction between H2SO4 (aq) + NaOH (aq) -- Na2SO4 (aq) + H2O(l)
Sulfuric Acid and
Sodium Hydroxide

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Through the comparison, we found that the reaction that produces highest enthalpy of
neutralization is reaction between H2SO4 and NaOH (∆H: -45.36 kj/mol) follow by the reaction
between HCl and NaOH (∆H: - 41.58 kj/mol) and lastly reaction between HNO3 and NaOH
(∆H: - 37.80 kJ/mol).
Comparison between Mixture A and B

Heat of neutralization for reaction between HCl and NaOH is (∆H: -41.58 kj/mol) which is
higher than heat of neutralization for reaction between HNO3 and NaOH is (∆H: - 37.80 kJ/mol).
Actually, heat of neutralization of a mixture of HCl and NaOH should be the same as that of a
mixture of HNO3 and NaOH because both of them are mixture of a monobasic acid with a
strong base. However, However, by refer at the list of relative strengths of acids, it was found
that the strength of HCl is stronger than HNO3. So, it is rational that the heat of neutralization
for reaction between HCl and NaOH is higher than the heat of neutralization for reaction
between HNO3 and NaOH.

Comparison between Mixture A, B with mixture C
The heat of neutralization will be affected by the acid base. Monobasic acids such as

hydrochloric acid, HCl and nitric acid, HNO3 will produce 1 mole of acid monobase that will
react with 1 mole of OH- ions (base) for produce 1 mole of water. If the acid used is dibasic
acid such as sulfuric acid (H2SO4), it will be neutralized completely by 2 moles of NaOH and
produce 2 moles of water as the follow equation. The heat of neutralization of a mixture of
H2SO4 with NaOH is higher than a mixture of HCl and NaOH and a mixture of HNO3 with
NaOH. It is because the mixture of H2SO4 with NaOH is the result of neutralization between
dibasic acid with strong base.

H2SO4 (aq) + NaOH (aq) -- Na2SO4 (aq) + H2O(l)
The value of the heat of neutralization is doubled when sulfuric acid is neutralized sodium
hydroxide to produce water and salt. This is caused by sulfuric acid produces 2 moles of
hydrogen ions indicating that it is a diprotic acid.

H2SO4 → 2H++ SO4-
Therefore, 2 moles of hydrogen ions react with base to produce 2 moles of water:

2H++ 2OH- → 2H2O

10.0 QUESTION
1.Why is the temperature of the initial solution of acid and base measured after 5
minutes?
Temperature of the initial solution of acid and base measured after 5 minutes because heat
transfer will occur from a high temperature to a low temperature taking about 5 minutes. After

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5 minutes, thermal balance is achieved by acid and base with the environment and at this time
the temperature of the initial can be measured accurately.

2.How to improve the practical procedure to obtain a more accurate value of the heat
of neutralization.
First of all, to improve the practical, we should use a proper apparatus which is wide polythene
bottles instead of using a plastic cup. In addition, the fan should be closed so that the
ventilation in the laboratory does not too fast and increase the rate of heat loss of the mixture.
Additionally, we should take temperature readings every 30 seconds and continue to do so
until the temperature reaches a maximum and hasn't changed much over the course of at
least ten observations.

3.Compare the value of the heat of neutralization between a strong acid and a strong
base obtained with the actual value from theory (reference). Give an explanation

In this experiment, produce the enthalpy of neutralization for reaction between HNO3 and
NaOH is -37.80kJ/mol. This result is not the same as the result stated by Ted & Janet (2020)
who stated the enthalpy of neutralization for reaction between HNO3 and NaOH is 53.7kJ/mol.

On the other hand, the findings of enthalpy of neutralization for HCl and NaOH is - 41.58
kJ/mol. However, William & Clifford (2010) state that enthalpy of neutralization for HCl and
NaOH is state that the -55.84kJ/mol.

Besides, this experiment show reaction between H2SO4 and NaOH has produce the
enthalpy of neutralization of -45.36 kJ/mol. However, Lawrie & Roger (2014) state enthalpy of
neutralization for H2SO4 and NaOH is state that the -57.3kJ/mol.

All this result error is due to the current error factor implement the process of implementing
the procedure for the mixture sulfuric acid is not poured as quickly as possible and as carefully
as possible can so that heat is not released to the environment. Besides, we only use regular
plastic cups but we didn’t use wide polythene bottles. This is also one of the causes of rapid
heat release. In addition, the rapid air flow in the laboratory also causes rapid heat loss.

As an improvement, we should use wide polythene bottles during the experiment.
Additionally, a lid can be made of insulating material and cover the cup to reduce the speed
of heat loss. Furthermore, the method of pouring the acid must be correct. For example, make
sure all the acid is poured at a fast rate. Moreover, the temperature should be recorded every
5 seconds to ensure that the highest temperature is not overlooked. Video can also be taken
throughout the experiment as it is possible for the operator to overlook the temperature reading
during the experiment.

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4.Give inferen of the value of heat of neutralization between ethanoic acid and sodium
hydroxide. Give justification.
Ethanoic acid is a carboxylic acid. A weak acid-strong base reaction occurs when ethanoic
acid and aqueous NaOH are combined. Water and sodium ethanoate (salt) are provided as
products. The pH level changes while acetic acid and NaOH are titrated. The neutralising
reaction between a weak acid and a strong base is also demonstrated by this reaction. The
following is the equation of reaction between ethanoic acid and sodium hydroxide:
CH3COOH(aq) + NaOH(aq) → CH3COO-Na+(aq) + H2O(aq)

Ethanoic acid Sodium Hydroxide Sodium ethanoate Water
Sodium ethanoate and water are given as products by the reaction of ethanoic acid and
aqueous NaOH and this is an exothermic reaction because heat is given out. Based on the
equation, 1 mol of ethanoic acid react with 1 mol of NaOH to from 1 mol of H2O. In this reaction
the heat of neutralization is -56.1kJ/mol (Gary, 2018).

11.0 CONSLUSION
In conclusion, the hypothesis is accepted. The heat of neutralization for the mixture of

sulfuric acid and sodium hydroxide is the highest which is -45.36 kJ/mol compared to the
mixture of hydrochloric acid with sodium hydroxide which is - 41.58 kJ/mol and the mixture of
nitric acid with sodium hydroxide which is -37.80kJ/mol. All three of these reactions are
exothermic reactions because they release heat to their surroundings. The complete
neutralization of a dibasic acid, ie sulfuric acid, releases a higher heat than the complete
neutralization of a monobasic acid because the enthalpy change in this reaction is the sum of
the heat of neutralization and also the heat of dilution of sulfuric acid. Sulfuric acid, which is a
dibasic acid, will also cause more water molecules to form and more heat to be released into
the environment.

12.0 REFERENCE
Alyn G. McFarland, Teresa Quigg & Nora Henry (2015). AQA A-Level Chemistry For

A-level Year 1 and AS. United Kingdom: Hodder Education
De Vos, W., & Pilot, A. (2001). Acids and Bases in Layers: The Stratal Structure of an

Ancient Topic. Journal of Chemical Education, 78(4), 494. doi:10.1021/ed078p494
Gary Horner. (2018). MYP Chemistry Years 4 & 5: A concept-based approach. United

Kingdom: Oxford University Press.
Lawrie Ryan & Roger Norris. Cambridge International AS and A Level Chemistry

Coursebook with CD-ROM. United Kingdom: Cambridge University Press.
Lewis, G. N. (1938). Acids and bases. Journal of the Franklin Institute, 226(3), 293–

313. doi:10.1016/s0016-0032(38)91691-6

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Manjula Devi A.S & M. Pharm. (2016). Hypothesis are accepted. Effect Of Ph and
Polyvalent Cations o the Y-Site Incompatibility of Continuous Intravenous Infusions

of Selected Critical Care Drugs in The Medical Intensive Care Unit. Coimbatore:
Department of Pharmacy Practice, College of Pharmacy, SRIPMS,

Soumen Ghosh, Asok Banerjee. (2002). Biomacromolecules. 3(1): 9-16.
Ted Lister & Janet. (2020). Cambridge International AS & A Level Complete Chemistry.

United Kingdom: Oxford University Press.
William M. Davis & Clifford E. Dykstra. Physical Chemistry: A Modern Introduction,

Second Edition. United State: Taylor & Francis Group.
Zakrewsky, M., Lovejoy, K. S., Kern, T. L., Miller, T. E., Le, V., Nagy, A., Mitragotri,

S. (2014). Ionic liquids as a class of materials for transdermal delivery and
pathogen neutralization. Proceedings of the National Academy of Sciences,
111(37), 13313–13318. doi:10.1073/pnas.1403995111

13.0 ATTACHMENT

NaOH 30ml HNO3 30ml

HCl + NaOH H2SO4 + NaOH

13

Initial temperature HNO3 + NaOH

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1.2. Example of task (Eason Ting Sin Bing)

1.0 TITLE:

To determine the enthalpy change of neutralization between strong acid and strong base.

2.0 AIM:
To determine the enthalpy change of neutralization between:

i. Hydrochloric acid (HCl) and sodium hydroxide (NaOH)
ii. Nitric acid (HNO3) and sodium hydroxide (NaOH)
iii. Sulfuric acid (H2SO4) and sodium hydroxide (NaOH)

3.0 INTRODUCTION:

Neutralization reaction is a reaction that occurs between an acid and a base to produce
salt and water as the only products. Heat of neutralization is also known as enthalpy of
neutralization that is the heat evolved when one mole of hydrogen ions are neutralized
completely by one mole of hydroxide ions in dilute solution (Saxena, 2004). During this
neutralization, heat of formation of water is due to the combination of hydrogen, H+ and
hydroxide ions, OH- . Thus,

H+(aq) + OH−(aq) → H2O(lq), ∆H = negatif, unit = kJ mo −1

The heat of neutralization of an acid is known as the amount of heat released when
one equivalent of an acid and one equivalent of a base react with each other to form water
and a salt. However, the heat of neutralisation of a base is the amount of heat released when
one gram equivalent of the base is completely neutralised by a strong acid in a dilute solution.
Heat of neutralization is an exothermic reaction as heat is released during the reaction that
cause it is always in negative value and denoted by a symbol ∆H with unit equals to kJ/mol
(Ademba, Mwangi & Muturi, 2005).

Exothermic reaction is a chemical reaction in which heat is evolved to the
surroundings that cause the rising in surrounding temperature. As stated, the negative value
of heat of neutralization is due to the total energy absorbed to break bonds is less than the
energy liberated when bonds are formed. The more negative value of heat of neutralization
indicates the more stable the system. The change of energy in an exothermic reaction is from
chemical energy to heat energy.

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In this experiment, the heat of neutralization between acid and base is determined
through the reaction between strong acid and strong alkali. The volume and concentration of
strong acid and base together with the rise in temperature is measured and noted down in
order to calculate the heat of neutralization. Normally, the heat of neutralization between
strong acid and strong base is always in the rate of -57.3 kJmol-1 (Clark, 2020) .

There are three factors that affect the amount of heat released during the neutralization
reaction which are the quantity of acid and base, the basicity of the acid and base and strength
of acid. In this experiment, the amount of heat released in the neutralization reaction is directly
proportional to the quantity of acid and base used.

4.0 VARIABLES:

Manipulated Variable: Type of strong acid used
Responding variable: Temperature change
Constant variable: Type of strong base used

5.0 APPARATUS AND MATERIALS:

5.1 Apparatus:
25 cm3 pipette, thermometer (0 oC – 110 oC), 100 cm3 measuring cylinder, a plastic
cup, and a bottle of distilled water.

5.2 Materials:
1.0 mol dm-3 hydrochloric acid (HCl), 1.0 mol dm-3 sulphuric acid (H2SO4), 1.0 mol dm-
3 nitric acid (HNO3), 1.0 mol dm-3 sodium hydroxide (NaOH), plastic cups, a bottle of
distilled water.

6.0 PROCEDURE:

1. 50 cm³ of 1.0 mol dm-3 hydrochloric acid (HCl) is measured by using a
measuring cylinder.

2. The hydrochloric acid measured is then is poured into a plastic cup.
3. The initial temperature of the 1.0 mol dm-3 hydrochloric acid (HCl) is then

measured and recorded by using thermometer.

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4. Another measuring cylinder is used to measure 50 cm³ of 1.0 mol dm-3 sodium
hydroxide solution.

5. The thermometer is washed thoroughly. The initial temperature of the 1.0 mol
dm-3 sodium hydroxide solution is measured and recorded.

6. 1.0 mol dm-3 sodium hydroxide solution is then poured into the plastic cup
containing 1.0 mol dm-3 hydrochloric acid.

7. The mixture is stirred immediately using a thermometer. The maximum
temperature of the mixture is recorded.

8. The experiment is repeated from steps 1 to 7 by replacing 50 cm³ of 1.0 mol
dm-3 hydrochloric acid (HCl) to 25 cm³ of 1.0 mol dm-3 sulphuric acid (H2SO4)
and 50 cm³ of 1.0 mol dm-3 nitric acid (HNO3)

7.0 RESULT:

TYPE OF ACID INITIAL FINAL CHANGE IN
TEMPERATURE, TEMPERATURE, TEMPERATURE, ∆
50 cm³ of 1.0
mol dm-3 (oC) (oC) (oC)
29 32 3
hydrochloric
acid (HCl) 29 38 9
25 cm³ of 1.0
mol dm-3 29 34 5
sulphuric acid Table 1: The result of the experiment.
(H2SO4)
50 cm³ of 1.0
mol dm-3 nitric
acid (HNO3)

7.1 CALCULATION OF HEAT OF NEUTRALIZATION FOR REACTION BETWEEN
HYDROCHLORIC ACID (HCl) AND SODIUM HYDROXIDE (NaOH)

Chemical equation ∶ + → + 2
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Ionic equation ∶ + + − → 2

Change in temperature, ∆θ = final temperature − intial temperature = 32 ° −
29 ° = 3 °

Specific heat capacity, = θ = (50 c 3 + 50 c 3) × 4.2 × 3 ° = 1260

No. of mole of HCl = = 1.0 mol d −3× 50 c 3 = 0.05
1000 1000

No. of mole of NaOH = = 1.0 mol d −3× 50 c 3 = 0.05
1000 1000

Based on the chemical equation above, 1 mole of hydrogen ion, H+ from HCl reacts
with 1 mole of hydroxide ion, OH- from NaOH to produce 1 mole of water, H2O. In this
experiment, the number of moles for acid and base are the same in which there are
0.05 mole for both hydrogen ion, H+ from HCl and hydroxide ion, OH- from NaOH.
Therefore, 0.05 mole of hydrogen ion, H+ reacts with 0.05 mole of hydroxide ion, OH-
to produce 0.05 mole of water, H2O. The formation of 0.05 mole of water has evolved
1260 J of heat to the environment.

Thus, 1 mole of water, H2O has released the heat of neutralization as calculated
below:

0.05 mol 1.0 mol
1260 =

Therefore, = 1260 J = 25200 −1

0.05

∴ , ∆ = −25.2 −1

7.2 CALCULATION OF HEAT OF NEUTRALIZATION FOR REACTION BETWEEN
SULPHURIC ACID (H2SO4) AND SODIUM HYDROXIDE (NaOH)

Chemical equation ∶ 2 4 + 2 → 2 4 + 2 2
Ionic equation ∶ + + 2 − → 2 2

18

Change in temperature, ∆θ = final temperature − intial temperature = 38 ° −
29 ° = 9 °

Specific heat capacity, = θ = (25 c 3 + 50 c 3) × 4.2 × 9 ° = 2835

No. of mole of H2SO4 = = 1.0 mol d −3× 25 c 3 = 0.025
1000
1000

No. of mole of NaOH = = 1.0 mol d −3× 50 c 3 = 0.05
1000 1000

Based on the chemical equation above, 1 mole of hydrogen ion, H+ from H2SO4 reacts
with 2 moles of hydroxide ion, OH- from NaOH to produce 2 moles of water, H2O. In
this experiment, the number of moles for acid and base are different in which 0.025
mole of hydrogen ion, H+ from H2SO4 and 0.05 mole of hydroxide ion, OH- from NaOH.
Therefore, 0.025 mole of hydrogen ion, H+ reacts with 0.025 mole of hydroxide ion,
OH- to produce 0.05 mole of water, H2O. The formation of 0.05 mole of water has
evolved 2835 J of heat to the environment.

Thus, 1 mole of water, H2O has released the heat of neutralization as calculated
below:

0.05 mol 1.0 mol
2835 =

Therefore, = 2835 J = 56700 −1

0.05

∴ , ∆ = −56.7 −1

7.3 CALCULATION OF HEAT OF NEUTRALIZATION FOR REACTION BETWEEN
NITRIC ACID (HNO3) AND SODIUM HYDROXIDE (NaOH)

Chemical equation ∶ 3 + → 3 + 2
Ionic equation ∶ + + − → 2

Change in temperature, ∆θ = final temperature − intial temperature = 34 ° −
29 ° = 5 °

19

Specific heat capacity, = θ = (50 c 3 + 50 c 3) × 4.2 × 5 ° = 2100

No. of mole of HNO3 = = 1.0 mol d −3× 50 c 3 = 0.05
1000
1000

No. of mole of NaOH = = 1.0 mol d −3× 50 c 3 = 0.05
1000 1000

Based on the chemical equation above, 1 mole of hydrogen ion, H+ from HNO3 reacts
with 1 mole of hydroxide ion, OH- from NaOH to produce 1 mole of water, H2O. In this
experiment, the number of moles for acid and base are the same in which there are
0.05 mole for both hydrogen ion, H+ from HCl and hydroxide ion, OH- from NaOH.
Therefore, 0.05 mole of hydrogen ion, H+ reacts with 0.05 mole of hydroxide ion, OH-
to produce 0.05 mole of water, H2O. The formation of 0.05 mole of water has evolved
2100 J of heat to the environment.

Thus, 1 mole of water, H2O has released the heat of neutralization as calculated
below:

0.05 mol 1.0 mol
2100 =

Therefore, = 2100 J = 42000 −1

0.05

∴ , ∆ = −42 −1

SOLUTION HEAT OF NEUTRALIZATION,
∆ ( − )
HCl + NaOH 25.2
H2SO4 + NaOH 56.7
HNO3 + NaOH 42

TABLE 2: Shows the heat of neutralization for each mixture, ∆ ( −1)

20

8.0 DISCUSSION:

Heat of neutralization is defined as the amount of heat that has been released or
enthalpy change invloved when 1 mole of H+ from acid is neutralized by 1 mole of OH- from
base to produce 1 mole of water. The ionic equation for this reaction is + + − → 2 . The
heat of neutralization is ∆ = -57.3 kJ mol-1 (Khattar, Dudeja, Arora, 2012).

According to the result obtained in this experiment, the difference between initial and
final temperature indicates that there are rising in temperature among the three mixtures which
shows that heat has been evolved and detected by the thermometer. Thus, this proven again
that the reaction takes place is an exothermic reaction. An exothermic reaction is a type of
reaction that release heat energy to the surroundings. Therefore, the negative sign of the value
of heat of neutralization represents the release of heat energy to the surroundings (Yeap &
Chien, 2021). Based on the camparison of heat of neutralization among the reaction of sodium
hydroxide that is strong alkali with the three strong acids, the heat of neutralization for the
reaction between hydrochloric acid and sodium hydroxide solution is -25.2 kJ mol-1 while the
heat of neutralization between sulfuric acid and sodium hydroxide solution is -56.7 kJ mol-1
and the heat of neutralization between nitric acid and sodium hydroxide solution is -42 kJ mol-
1. From the results obtained, it found that the value of heat of neutralization for the reaction
between sulphuric acid and sodium hydroxide solution is the highest followed by the reaction
between nitric acid and sodium hydroxide solution and the reaction between hydrochloric acid
and sodium hydroxide solution.

The chemical equation for the reaction between sulphuric acid and sodium hydroxide
solution is 2 4 + 2 → 2 4 + 2 2 . The highest heat of neutralization in this
reaction is due to the basicity of the acid. According to Veerendra (2020), a complete
neutralization of a strong diprotic acid will produce double amount of heat energy than a
monoprotic acid. Therefore, sulphuric acid as a diprotic acid produces two moles of hydrogen
when it dissociates in water as shown in the ionic equation below;

+ + 2 − → 2 2
This means 1 mole of sulfuric acid contains 2 moles of hydrogen ion, H+ which also shows
that 1 mole of sulfuric acid requires 2 moles of NaOH to neutralize and produce 2 moles of
water. This is then further promoting the formation of more more water molecules and more
heat to be released to the surroundings.

21

However, for monoprotic acid such as hydrochloric acid and nitric acid, 1 mole of
hydrochloric acid and nitric acid will only react with 1 mole of hydroxide ion, OH- from the
sodium hydroxide solution to produce 1 mole of water. This is because 1 mole of
hydrochloric acid and 1 mole of nitric acid dissociates to produce 1 mole of H+ as the ionic
equation shown below;

+ + − → 2

Theoretically, the heat of neutralisation between a strong acid and a strong alkali
should always be same that is ∆ = −57.3 −1 (Chau, 2012). Through this experiment,
it found that the heat of neutralization for the reaction between nitric acid and sodium hydroxide
is higher than the heat of neutralization for the reaction between hydrochloric acid and sodium
hydroxide although both of them are categorized as monoprotic acids that are strong acids.
The value of heat of neutralization of hydrochloric acid and nitric acid is also lower than -57.3
kJ mol-1. Furthermore, for the heat of neutralization for diprotic acid which is sulfuric acid that
also supposed to be same with the theoretical value but it is slightly less which is -56.7 kJ mol-
1 only. This inaccuracy results obtained may due to several errors when conducting the
experiment such as the fans are opened that allow the heat to escape from the plastic cups
and the mixture didn’t stirred continuously and slowly until an even temperature is obtained.
The sodium hydroxide solution is also not poured quickly into the strong acids for reaction to
occur which may also cause the temperature to be released to the surrounding. Besides, part
of the heat energy has been lost and released to the surrounding and other heat energy
absorbed by the plastic cups are ignored when doing the calculation. Inaccuracy in measuring
the volume of acid and base will also cause the problem to occur. Then, the reading of the
thermometer is only taking once during the experiment that may also affect the accuracy of
the result obtained. Therefore, it is better to repeat the reading three times and obtain the
average temperature for a more accurate reading.

9.0 DISCUSSION QUESTION:

1. Why is the temperature of the initial solution of acid and alkali
measured after 5 minutes?
To ensure the temperature of both acid and alkali solution to rise to room
temperature. Based on Rastogi and Goyal (2020), in heat of neutralization
experiment, the temperature of both acid and alkali solution must be same
or wait until the solution obtain the same temperature.

22

2. How to improve the practical procedure to obtain a more accurate
value of the heat of neutralization?
 The fan should be closed to prevent heat to escape from the cup when
conducting the experiment.
 Hydrochloric acid, sulfuric acid and nitric acid should be mixed to the
sodium dioxide solution as quickly as possible so that the reaction can
be completed in a short time.
 The mixture should be stirred slowly and continuously to obtain an even
solution temperature.
 The thermometer reading should be observed at all times so that the
highest temperature reached by the reaction mixture can be recorded,
 The experiment should be repeated three times to get the average
temperature in order to obtain a more accurate reading.

3. Compare the value of the heat of neutralization between a strong acid
and a strong base obtained with the actual value from theory
(reference). Give an explanation.
The value of the heat of neutralization between a strong acid and a strong
base obtained is slightly less than the actual value, that is -57.3 kJ mol-1.
Theoretically, the heat of neutralisation between a strong acid and a strong
base is always constant as all strong acids and strong bases are completely
ionised in dilute solution (Byjus, 2022). This involves the combination of
hydrogen ions, H+ from acid and hydroxide ions, OH- from base to form
unionised water molecules with the heat releasion of -57.3 kJ mol-1. To
obtain the most accurate result, polystyrene cups should be used instead of
plastic cups as polystyrene cups are able to reduce the heat released to the
surrounding compared to plastic cups. In addition, the strong base should
be poured quickly into the strong acid to avoid the temperature being
released to the surrounding.

4. Give inferen of the value of heat of neutralization between ethanoic
acid and sodium hydroxide. Give justification.

23

Ethanoic acid is a weak acid while sodium hydroxide is a strong base (Clark,
2002). From here, we can know that the heat of neutralization between a
weak acid and a strong alkali is lower than -57.3 kJ mol-1. This is because
strong alkali ionize completely in water to produce hydroxide ions, OH-, but
weak acid can only ionize partially in water, where the low degree of
dissociation results in only a small number of hydrogen ions, H+ being
produced. Most of the weak acid molecules are existing as undissociated
molecules. Thus, some amount of heat of neutralization is used to ionise the
undissociated molecules. Therefore, when a weak acid reacts with a strong
alkali, lesser water molecules are formed which due to low in heat of
neutralization.

10.0 CONCLUSION:

In conclusion, the heat of neutralization between a strong acid and a strong base is ∆ =
−57.3 −1 . In this experiment, the heat of neutralization between sulphuric acid and
sodium hydroxide is the highest which is -56.7 kJ mol-1 followed by the heat of neutralization
between nitic acid and sodium hydroxide which is -42 kJ mol-1 and lastly the heat of
neutralization between hydrochloric acid and sodium hydroxide is the lowest which is -25.2 kJ
mol-1. All the three reactions show an exothermic reaction that the heat is released to the
surrounding. The complete neutralization of a diprotic acid that is sulphuric acid shows a
higher heat of neutralization than other monoprotic acids that are hydrochloric acid and nitric
acid which causes more water molecules to be formed and more heat to be released to the
surrounding.

11.0 REFERENCE:

Ademba, M., Mwangi, H. & Muturi, J. M. (2005). Certificate Chemistry Form 4. Kenya: East
African Educational Publishers Ltd.

Byjus. (2022). Enthalpy of Neutralization of Strong Acid and Strong Base. Retrieved from
https://byjus.com/chemistry/enthalpy-of-neutralization-of-strong-acid-and-strong-base/

Chau, K. K. (2012). Express Chemistry Form 5. Selangor: Pelangi ePublishing Sdn Bhd.

Clark, J. (2002). STRONG AND WEAK ACIDS. Retrieved from
https://www.chemguide.co.uk/physical/acidbaseeqia/acids.html

24

Clark, J. (2020). Enthalpy Change of Neutralization. Retrieved from
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_
Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/
Energies_and_Potentials/Enthalpy/Enthalpy_Change_of_Neutralization

Khattar, D., Dudeja, R. R. & Arora, K. K. (2012). The Pearson Complete Guide for the AIEEE.
India: Pearson.

Rastogi, S. C. & Goyal, M. E. (2020). Practical/Laboratory Manual Chemistry Class XII. Agra:
SBPD Publications.

Saxena, P. B. (2004). I.I.T. Chemistry-II. India: Krishna Prakashan Media (P) Ltd.
Veerendra. (2020). What is the enthalphy of neutralization? Retrieved from

https://www.aplustopper.com/enthalpy-heat-neutralization/
Yeap, T. K. & Chien, M. (2022). Xpress Kendiri: SPM Chemistry. Selangor: Sasbadi.
12.0 APPENDICES:

Figure 1: Apparatus and materials used in this experiment.

25

Figure 2: Type of acid used in this experiment.
26

Figure 3: Procedures in conducting the experiment.
27

2. PRACTICAL REPORT 2

Determining the enthalpy change of reaction in conversation of sodium
hydrogen carbonate to sodium carbonate reaction using Hess Law

2.1. Example of task (Muhammad Zamir Syahmi bin S.
Fadhirllah al-Jupry)

1.0 INTRODUCTION

When sodium hydrogen carbonate, NaHCO3 is heated, it decomposes to form sodium
carbonate, Na2CO3. The enthalpy change of this reaction is difficult to measure directly so
an indirect method is used. By determining the molar enthalpy change of reaction between
sodium carbonate and hydrochloric acid and that between sodium hydrogen carbonate and
hydrochloric acid, it is possible to obtain the enthalpy change for the decomposition of
sodium hydrogen carbonate (studylib, 2013). The enthalpy of a chemical system refers to
the “heat content” of the system. Enthalpy change refers to the amount of heat released or
absorbed when a chemical reaction occurs at constant pressure. Enthalpy change is given
the symbol, ∆H. An exothermic product of the reaction is a chemical reaction or physical
change in which heat is released (LibreText, 2021). ∆H for an exothermic reaction is
negative. In contrast, an endothermic product of the reaction is a chemical reaction or
physical change in which heat is absorbed. ∆H for an endothermic reaction is positive.
Calorimetry was the method used to collect the data required for the calculation of ‘Q’ of
the reactions. To calculate the temperature change of the reaction (of either experiment),
a styrofoam cup was used as the calorimeter. The main purpose of a calorimeter is to
maintain the heat energy within a system and keep the system closed, since heat energy
can easily disperse into the surrounding environment if not insulated properly. The heat
that is released during this reaction can be calculated using formula as shown below:

= ∆

Where:
m = mass of the solution
c = specific heat capacity of the solution
∆ = temperature change

28

2.0 AIM
To determine the enthalpy change of reaction in conversion of sodium hydrogen
carbonate to sodium carbonate reaction using Hess Law.

3.0 MATERIALS AND APPARATUS
1) EC1, 2.0 mol dm-3 hydrochloric acid, HCl
2) EC2, solid sodium carbonate, Na2CO3
3) EC3, solid sodium hydrogen carbonate, NaHCO3
4) 50 cm3 burette
5) Retort stand
6) Thermometer
7) Plastic cup
8) Electronic balance

4.0 PROCEDURES
1) 30 cm3 2.0 mol dm-3 HCl was measured using a 50 cm3 burette into a plastic cup.
2) The initial temperature of EC1 was measured using a thermometer.
3) 2.40 g of EC2 was measured by using an electronic balance.
4) 2.40 g of EC2 was poured into the plastic cup containing 30 cm3 2.0 mol dm-3 HCl.
5) The mixture was stirred until fully dissolves.
6) The maximum temperature was recorded in Table 1.
7) Steps 1 to 6 were repeated using a 3.0 g EC3 to replace 2.4 G EC2.

5.0 Data collection

Table 1: Observation in increase of temperature

Solution Initial temperature (oC) Final temperature (oC) Temperature
difference, ∆T (oC)
HCl 29 -
30 cm3 2.0 mol dm- 29 31 (Initial – Final)
3 HCl + 2.40 g of -
Na2CO3
-2.0

29

30 cm3 2.0 mol dm- 29 24 +5.0
3 HCl + 3.00 g of
NaHCO3

6.0 Data analysis

Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
1. Calculate moles of 2.40 g of Sodium Carbonate (Na2CO3)

Molar Mass of Sodium Carbonate, Na2CO3:
2 (22.990 g/mol) + 12.011 g/mol + 3 (15.999 g/mol) = 105.988 g/mol
Moles of Sodium Carbonate:
2.40 g ÷ 105.988 g/mol = 0.0226 mol of Na2CO3

2. Calculate the enthalpy change (∆H)
Q = mc∆T

= 30 g x 4.18 J g-1 C-1 x (-2.0) °C
= - 0.250 kJ
Enthalpy
∆H = Q ÷ mol
= - 0.250 kJ ÷ 0.0226 mol
= - 11.0619 kJ / mol (exothermic)

NaHCO3 + HCl ==> NaCl + H2O + CO2
1. Calculate moles of 3.00 g of Sodium Hydrogen Carbonate (NaHCO3)

Molar Mass of Sodium Hydrogen Carbonate, NaHCO3:
22.990 g/mol + 1.007 g/mol + 12.011 g/mol + 3(15.999 g/mol) = 84.005 g/mol
Moles of Sodium Carbonate:
3.00 g ÷ 84.005 g/mol = 0.0357 mol of NaHCO3

2. Calculate the enthalpy change (∆H)
Q = mc∆T

= 30 g x 4.18 J g-1 C-1 x (+5.0) °C
= + 0.627 kJ

30

Enthalpy
∆H = Q ÷ mol

= + 0.627 kJ ÷ 0.0357 mol
= + 17.5630 kJ / mol (endothermic)

∆H decomposition

2NaHCO3 Na2CO3 + CO2 + 2H2O

∆H 2 ∆H 1
+ 17.5630 kJ / mol - 11.0619 kJ / mol

2NaCl + CO2 + 2H2O

3. Calculating ∆H decomposition

∆H decomposition = ∆H 2 - ∆H 1
= (2 x + 17.5630 kJ / mol) – (- 11.0619 kJ / mol)
= (35.1260 kJ / mol) + (11.0619 kJ / mol)

Multiply by 2 = + 46.1879 kJ / mol
because
2NaHCO3

7.0 DISCUSSION
Hess's Law states that when two or more reactions can be added together to give a net
chemical reaction, the energy associated with each of the reactions can also be added
together to give the net enthalpy (ΔH) change. The law can be used to obtain thermodynamic
data that cannot be measured directly. Hess’ law is based on the state function character of
enthalpy and the first law of thermodynamics. Energy (enthalpy) of a system (molecule) is a
state function. So, enthalpy of reactant and product molecules is a constant and does not
change with origin and path of formation (Byju’s, 2022). According to the data collected and
processed, the final experimental enthalpy change value (∆H) for the decomposition of sodium
hydrogen carbonate into sodium carbonate is 46.2 kJ/mol. On the other hand, the literature
value of the reaction, is 91.3 kJ/mol (savvy-chemist, 2015). The enthalpy changes obtained in
the experiment is different than the literature value.

Using a plastic cup as the calorimeter posed a threat to the reliability of the results as it is not
the perfect insulator and therefore cannot maintain all the heat energy. If the heat energy
cannot be maintained, the experimental values will not be reliable because not all the energy
will have been accounted for. Through our qualitative observations, the change in temperature

31

could be felt from the outside of the cup, indicating heat escaping through the cup. Due to this
insulation problem with the plastic cup, a large amount of heat was lost in the reaction, which
resulted a large percentage discrepancy between the literature value and the experimental
value. For improvement in the future, we need to add a lid to the plastic cup to prevent more
heat loss to the environment. A small hole is placed in the polystyrene lid to allow a
thermometer to be pushed through. The fit must be snug enough to hold the thermometer in
place, suspended off the bottom of the cup and immersed in the reactant. It is assumed no
heat will be lost through the lid or the hole in the lid. Besides, if we have advanced equipment,
we can use of a bomb calorimeter when calculating the heat energy change in a reaction.
Unlike styrofoam cup calorimeters, bomb calorimeters are equipped with better insulation,
allowing for more reliable and accurate results. (Avinash Bharwaney, 2022).

 Figure 7.1: Styrofoam cup with lid and bomb calorimeter

8.0 CONCLUSION
The enthalpy changes of reaction in conversion of sodium hydrogen carbonate to sodium
carbonate reaction using Hess Law based on literature is +/- ∆H = + 91 kJ mol-1. But the results
obtained in this experiment are different than the literature value due to some weaknesses
that can be improve. For better results, we must follow the precautions to minimize the heat
lose during reaction occur.

9.0 REFERENCES
Avinash Bharwaney. (2022). Hess’s Law Lab from

https://pdfcoffee.com/hess-law-lab-equation-required-h-products-h-reactants-h-
reaction-pdf-free.html

32

Byju’s. (2022). What is Hess Law? from
https://byjus.com/jee/hess-law-of-constant-heat-summation/

LibreText. (Sep 23, 2021). Calorimetry and Hess’s Law (Experiment) from
https://chem.libretexts.org/Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Ex
periments/General_Chemistry_Labs/Online_Chemistry_Lab_Manual/Chem_11_Expe
riments/12%3A_Calorimetry_and_Hess's_Law_(Experiment)

Ssavvy-chemist. (December 29, 2015). Chemical Energetics (8) Using enthalpy changes of
formation to determine enthalpy change’s reaction from http://derekcarrsavvy-
chemist.blogspot.com/2015/12/

Studylib. (2013). Topic 6: Using Hess Law Sodium Hydrogen Carbonate from
https://studylib.net/doc/7374024/topic-6-using-hess-law-sodium-hydrogen-carbonate

ATTACHMENT

 Materials and apparatus used in the experiment

33

 Some procedures to investigate the enthalpy change the enthalpy change of reaction
in conversion of sodium hydrogen carbonate to sodium carbonate

34

2.2. Example of task (Ho Jia Heng)

Introduction
Heat of reaction is defined as difference in the enthalpy value of a chemical reaction under
constant pressure. It also known as enthalpy of reaction (Siddhesh, 2022). In other word, it is
a thermodynamic unit of measurement used to determine the total amount of energy produced
or released per mol for a material in a reaction. Heat of reaction is related closely to enthalpy.
According to Britannica, T. Editors of Encyclopaedia (2022), enthalpy is a state of function to
measure the transfer of heat energy in a thermodynamic system. Since enthalpy involved an
energy-like property, it is bound to the law of conservation of energy, where the change in
internal energy is equal to the heat transferred to, less the work done by, the system. If the
only work done is a change of volume at constant pressure, the enthalpy change is exactly
equal to the heat transferred to the system. In this experiment, Hess Law is involved. Hess
Law stated that heat that being absorbed or released during a reaction is determined by its
concentration of the materials instead of its path of reaction (Clark, 2010). Therefore, it can be
concluded that Hess Law is important to identify total of heat either getting absorbed or
released when changing from one form to another in a constant pressure.

Aim
To determine the enthalpy change of reaction in conversion of sodium hydrogen carbonate to
sodium carbonate reaction using Hess Law.

Materials and Apparatus  Anhydrous sodium carbonate, Na2CO3
 2.0 mol dm-3 hydrochloric acid, HCl  Thermometer
 Plastic cups  Sodium hydrogen carbonate solid,
 Burette
 Electronic balance NaHCO3
 Retort stand

Procedure
1. 30 ml of 2.0 M hydrochloric acid, HCl is measured by using a burette and poured into a

plastic cup labelled with “HCl”.
2. The initial temperature, θi of HCl is measured by using a thermometer and recorded in

Table 1.
3. 2.40 g anhydrous sodium carbonate solid, Na2CO3 is weighed accurately by using an

electronic balance.

35

4. All the sodium carbonate is put into the plastic cup containing hydrochloric acid. Make sure
all the solids are entirely put into the cup and there is no solution splashing out.

5. The mixture is stirred by using a thermometer and its maximum temperature, θf is
measured and recorded in Table 1.

6. Steps 1 to 5 are repeated by using 3.00 g sodium hydrogen carbonate solid, NaHCO3.

Data Collection
Table 1
Observation in enthalpy change

Solid Mass of Initial Final Temperature change,
Solid (g) Temperature Temperature ∆T
of Acid, θi (°C) of Mixture, θf
(Initial - Final
(°C) Temperature) (°C)

Sodium 2.4 30 -1
carbonate 3.0
29
solid,
Na2CO3 23 6
Sodium
hydrogen
carbonate

solid,
NaHCO3

Data Analysis
Formula of Quantity of Heat Released

= ∆
Where:
Q = Quantity of heat released(J)
m = mass of the reactor (g)
c = specific heat capacity (4.2 Jg-1 C-1)
∆ = temperature change (°C)

Formula of Reaction Heat
Where: ∆ =

Q = Quantity of heat released (J)

Table 2

Calculation of Reaction Heat

Reaction Calculation

Reaction equation:
Na2CO3 (s) + 2HCl (aq)  2NaCl (aq) + H2O (l) + CO2(g)
Mass of reactor, m:

0.03 kg + 0.0024 kg = 0.0324 kg

Specific heat capacity, c:
4.2 kJ kg-1 C-1
Temperature change, ∆T:

-1 °C

Quantity of heat released:
Q = -(mc∆T)

= -(0.0324) x (4.2) x (-1)

= 0.136 kJ

Number of mol of HCl:

N =
1000

Na2CO3 + HCl = 2 30
1000

= 0.06 mol

Number of mol of Na2CO3:

N =



= (23 2.4 3)
2)+(12)+(16

= 0.023 mol

Hence, it is inferred that 0.023 mol of Na2CO3 reacted with 0.06 mol of
HCl to create 0.023 mol of water molecule while released 0.136 kJ of

heat energy.

From this, we are able to calculate its reaction heat, whereby

∆H =


= 0.136

0.023

= 5.91 kJ mol-1

∴ The reaction heat between Na2CO3 and HCl is 5.91 kJ mol-1.

NaHCO3 + HCl Reaction equation:

NaHCO3 (s) + HCl (aq)  NaCl (aq) + H2O (l) + CO2(g)

Mass of reactor, m:

0.03 kg + 0.003 kg = 0.033 kg

Specific heat capacity, c:

4.2 kJ kg-1 C-1
Temperature change, ∆T:

6 °C

Quantity of heat absorbed:
Q = mc∆T

= (0.033) x (4.2) x (6)

= 0.832 kJ

Number of mol of HCl:
N =

1000

= 2 30

1000

= 0.06 mol

Number of mol of Na2CO3:

N =


= 2.4 3)
(23)+(1)+(12)+(16

= 0.036 mol

Hence, it is inferred that 0.036 mol of Na2CO3 reacted with 0.06 mol of

HCl to create 0.036 mol of water molecule while absorbed 0.832 kJ of

heat energy.

From this, we are able to calculate its reaction heat, whereby

∆H =



= 0.832

0.036

= 23.11 kJ mol-1

∴ The reaction heat between NaHCO3 and HCl is 23.11 kJ mol-1.

Table 3
Calculation of Enthalpy Change of Reaction in Conversion of Sodium Hydrogen Carbonate to
Sodium Carbonate Reaction

Reaction equation:
2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (l)

NaHCO3 (s) + HCl (aq)  NaCl (aq) + H2O (l) + CO2(g) ΔH1 = 23.11 kJ mol-1 (1)
Na2CO3 (s) + 2HCl (aq)  2NaCl (aq) + H2O (l) + CO2(g) ΔH2 = 5.91 kJ mol-1 (2)
Reaction (1) times 2 to balance the equation 1 and equation 2.
Reaction (2) is inversed to obtain the correct changing.

2NaHCO3 (s) + 2HCl (aq)  2NaCl (aq) + 2H2O (l) + 2CO2(g) ΔH1 = 46.22 kJ mol-1 (3)
2NaCl (aq) + H2O (l) + CO2(g)  Na2CO3 (s) + 2HCl (aq) ΔH2 = -5.91 kJ mol-1 (4)

According to Hess Law,

[2x ΔH1 NaHCO3 (s)] + ΔH = ΔH2 [Na2CO3 (s) + CO2 (g) + H2O (l)]

ΔH = ΔH2 [Na2CO3 (s) + CO2 (g) + H2O (l)] - [2x ΔH1 NaHCO3 ]

ΔH = ΔH2 – ΔH1

2NaHCO3 (s) + 2HCl (aq) → 2NaCl (aq) + 2H2O (l) + 2CO2 (g) ΔH1 = 46.22 kJ mol-1 (3)

2NaCl (aq) + H2O (l)+CO2 (g)  Na2CO3 (s) + 2HCl (aq) ΔH2 = -5.91 kJ mol-1 (4)

2NaHCO3 (s)  Na2CO3 (s) + CO2 (g) + H2O (l) ΔH = -52.13 kJ mol-1

ΔH = ΔH2 (3) - ΔH1 (4)
= -5.91 – (46.22)
= -52.13 kJ mol-1

Discussion
The heat of reaction acquired from this experiment is -52.13 kJ mol-1. However, this is opposite
with the theoretical value, which is -1130.77 kJ mol-1 (Dawkins, 2004). This happens because
there are several factors which cannot be controlled well in a laboratory environment. In other
word, factors such as surrounding temperature and pressure is indefinite. It is important to fix
these two values in order to get a precise value which is as close as the theoretical number
as possible. Also, since the concentration for the hydrochloric acid used in this experiment is
2.0 mol dm-3 instead of 1.0 mol dm-3, it is clashed with the requirement for creating a standard

condition when calculating the heat of reaction. Moreover, random error has occurred on the
constant pressure condition and also the loss of heat energy towards the surrounding which
potentially leads to changes of temperature of the system. According to Goos & Petrucci
(2007), enthalpy is a state function that measures the heat transferred from a system to its
surroundings or vice versa at constant pressure. Therefore, it can be concluded that other
than the standard condition factor, factor such as the apparatus used plays an important role
as well.

The diagram below is showing the energy level profile for the reaction that converts
sodium hydrogen carbonate to sodium carbonate according Hess Law.

Conclusion
In conclusion, the reaction heat between anhydrous sodium carbonate, Na2CO3 and
hydrochloric acid, HCl is 5.91 kJ mol-1, while the reaction heat between sodium hydrogen
carbonate, NaHCO3 and hydrochloric acid, HCl is 23.11 kJ mol-1. The enthalpy changes of
reaction in conversion of sodium hydrogen carbonate to sodium carbonate reaction using Hess
Law is -52.13 kJ mol-1.

References
Britannica, T. Editors of Encyclopaedia. (2022). Enthalpy. Encyclopedia Britannica. Retrieved

from https://www.britannica.com/science/enthalpy
Clark J. (2010). Hess's Law And Enthalpy Change Calculations. Retrieved from

http://www.chemguide.co.uk/physical/energetics
Dawkins R. (2004). Oxford Illustrated Science Encyclopaedia. New York: Oxford University

Press

Siddhesh S. (2022). Heat of Reaction Formula. Retrieved from
https://www.geeksforgeeks.org/heat-of-reaction-formula/

3. PRACTICAL REPORT 3

Investigating the effect of concentration effect on E.M.F cell

3.1. Example of task (Muhammad Zulhilmi bin Zulkifli)

1.0 INTRODUCTION

A subfield of chemistry called thermochemistry examines the variations in energy that take
place during chemical reactions. Enthalpy is a state function that calculates the heat
transferred from a system to its surroundings or vice versa at constant pressure (Bodner,
2021). When heat is transferred from a system to its surroundings, lowering the system's
enthalpy, the change in enthalpy (H) can be measured (exothermic reaction). In contrast, the
(H) value is positive when heat is transferred from the environment to a system, increasing
the system's enthalpy (endothermic reaction). Using the calorimetry technique, which
measures the change in temperature during the chemical reaction, the enthalpy change can
be determined.

In this experiment, the change in enthalpy that takes place when one equivalent of an
acid and a base undergoes a neutralisation process to generate water and a salt is measured.
The (H) value is negative because neutralisation reactions are often exothermic. Acids and
bases that totally ionised when dissolved in water are referred to as strong acids and bases.
Strong acids and alkalis react under standard conditions, and the enthalpy changes of
neutralisation values are always quite similar, with values ranging between -57 and -58 kJ mol-
1 (Wirkkla, 2014). The following formula can be used to calculate the heat emitted during this
reaction:

= ∆

Where:
m = mass of the solution
c = specific heat capacity of the solution
∆ = temperature change

2.0 AIM

To determine the enthalpy change of neutralization between strong acid and strong base.

3.0 MATERIALS AND APPARATUS

9) 1.0 mol dm-3 sodium hydroxide solution, NaOH
10) 1.0 mol dm-3 hydrochloric acid, HCl
11) 1.0 mol dm-3 sulfuric acid, H2SO4
12) 1.0 mol dm-3 nitric acid, HNO3
13) Plastic cups
14) Measuring cylinder
15) Thermometer
16) Beaker

4.0 Procedure

1. 50 cm3 of 1.0 mol dm-3 sodium hydroxide solution, NaOH was measured by using a
measuring cylinder then transferred into a polystyrene cup.

2. 50 cm3 of 1.0 mol dm-3 hydrochloric acid HCl was measured by using another
measuring cylinder then transferred into another polystyrene cup.

3. The initial temperature of both solutions in a polystyrene cup was recorded using a
thermometer after 5 minutes.

4. The hydrochloric acid was poured into a polystyrene cup containing sodium hydroxide.
5. The mixture was stirred for 30 seconds.
6. The maximum temperature was recorded in the Table 1.
7. Steps 1 to 5 were repeated using 25 cm3 of 1.0 mol dm-3 sulfuric acid, H2SO4 and 50

cm3 of 1.0 mol dm-3 nitric acid, HNO3 to replace 50 cm3 of 1.0 mol dm-3 hydrochloric
acid HCl.

5.0 Data Collection

Table 1: Temperature Change Record

Solutions NaOH + HCl NaOH + HNO3 NaOH + H2SO4

Initial θ of alkaline, (oC) 29.0 29.0 29.0

Initial θ of acid, (oC) 29.0 29.0 29.0

Highest θ of mixture, (oC) 32.0 34.0 38.0

Increase in θ, ∆ (oC) 3.0 5.0 9.0

6.0 Data analysis

6.1 Enthalpy change of neutralization between NaOH and HCl

Equation NaOH + HCl –> NaCl + H2O
Net ionic equation H+(aq) + OH- (aq) –> H2O(I)

+ = ×

= 1.0 mol dm−3 × 50 −3
1000

= 0.05 mol

− = ×

= 1.0 mol dm−3 × 50 −3
1000

= 0.05 mol

From the calculation and equation, 1 mol H+ ions react with 1 mol of OH- ions to form 1 mol
of water. 0.05 mol H+ ions react with 0.05 mol of OH- ions to form 0.05 mol of water.

= ×

= (50 + 50) 3 × 1

= 100

= ∆

= 100 × 4.2 −1 −1 × 3℃

= 1260

0.05 = 1260

1 1260
= 0.05
= 25.2 −1

∆ + = −25.2 −1

6.2 Enthalpy change of neutralization between NaOH and HNO3

Equation NaOH + HNO3 –> NaNO3 + H2O

Net ionic equation H+(aq) + OH- (aq) –> H2O(I)

+ = ×

= 1.0 mol dm−3 × 50 −3
1000

= 0.05 mol

− = ×

= 1.0 mol dm−3 × 50 −3
1000

= 0.05 mol

From the calculation and equation, 1 mol H+ ions react with 1 mol of OH- ions to form 1 mol
of water. 0.05 mol H+ ions react with 0.05 mol of OH- ions to form 0.05 mol of water.

= ×

= (50 + 50) 3 × 1

= 100

= ∆

= 100 × 4.2 −1 −1 × 5℃

= 2100

0.05 = 2100

1 2100
= 0.05
= 42.0 −1

∆ + 3 = −42.0 −1

6.3 Enthalpy change of neutralization between NaOH and H2SO4

Equation NaOH + H2SO4 –> Na2SO4 + H2O

Net ionic equation H+(aq) + OH- (aq) –> H2O(I)

+ = ×

= 2 × 1.0 mol dm−3 × 50 −3
1000

= 0.05 mol

− = ×

= 1.0 mol dm−3 × 50 −3
1000

= 0.05 mol

From the calculation and equation, 1 mol H+ ions react with 1 mol of OH- ions to form 1 mol
of water. 0.05 mol H+ ions react with 0.05 mol of OH- ions to form 0.05 mol of water.

= ×

= (25 + 50) 3 × 1

= 75

= ∆

= 75 × 4.2 −1 −1 × 9℃

= 2835

0.05 = 2835

1 2835
= 0.05
= 56.7 −1

∆ + 2 4 = −56.7. −1

7.0 Experiment Questions

These questions will be discussed in the discussion part of the report:

1. Why is the temperature of initial solution of acid and alkali measured after 5 minutes?
2. How to improve the practical procedure to obtain a more accurate value of the heat of

neutralization.
3. Compare the value of heat of neutralization between a strong acid and a strong base

obtained with the actual value from theory. Give an explanation.
4. Give inference of the value of heat of neutralization between ethanoic acid and sodium

hydroxide. Give justification.

8.0 Discussion

A common practice among scientists, whenever recording initial temperature, it is very
recommended to leave the subject matter for a short time before recording the temperature.
This is because, if we take the temperature too early from where the subject matter was being
produced, its temperature might still be affected by the reaction it was undergoing while
producing it (Davis, 2020). Other than that, giving some time for the subject matter to stabilize
its temperature and the temperature of its surroundings. This is due to the law of
thermodynamics where heat transfers from hot to cold matter and happens all the time until it
reaches 0 temperature difference among surrounding matter. In this experiment, the initial
temperature of acid and alkali was measured after 5 minutes for the exact reasons given
above. It is also because the whole reaction afterwards is also made in the same surrounding
which shares the same constant surrounding temperature. It is also to make sure the readings
of the experiment has less errors to deal with (Livonia, 2022).

For this experiment, a plastic cup is being used as the calorimeter in order to determine
the quantity of heat generated during the neutralisation process. The presence of random
errors means that the lab's ambient pressure may vary and not be constant, which will impair
the accuracy of the experiment's findings. In addition, heat loss to the environment may
possibly be influencing the test's accuracy. This is due to the calorimeter's lack of accuracy in
measuring the heat change of neutralisation since heat can be lost to the environment through
radiation (Maharashtra, et al, 2022). A calorimeter with constant pressure that prevents heat
loss while maintaining pressure. Therefore, to improve practical procedure to obtain a more
accurate value of the heat of neutralization, a plastic cup needs to be replaced with an actual
working calorimeter apparatus.

Table 2: Enthalpy change of neutralization between strong acid and strong base

Neutralization of strong acid Enthalpy change of neutralization, ∆ ( − )
and strong base

HCl + NaOH – NaCl + H2O −25.2

HNO3 + NaOH – NaNO3 + H2O −42.0

H2SO4 + NaOH – Na2SO4 + H2O −56.7

Based on the information in Table 2, three neutralisation reactions between three
potent acids; hydrochloric acid (HCl), nitric acid (HNO3), and sulfuric acid (H2SO4) and a potent
base; sodium hydroxide (NaOH) were studied. According to the Arrhenius (2020), acid-base
chemistry, where 1 mol of H+ and 1 mol of OH- ions combine to generate 1 mol of H2O, the
theoretical enthalpy changes of neutralisation between strong acid and strong base are -57.1
kJmol-1. As a result, in a strong acid–strong base reaction, neutralisation is essentially an
interaction between hydrogen ions and hydroxide ions to form water, with any additional ions
present not taking part in the reaction itself. The enthalpy change H ought to be the same.

Based on the information in Table 2, reveals that the enthalpy change of neutralisation,
or H, is different from the theoretical value of -57.1 kJmol-1 for each strong acid and strong
base, coming in at -25.2 kJmol-1, -42.0 kJmol-1 and -56.7 kJmol-1, respectively. These results
could be impacted by elements like a random error in the reaction's constant pressure
condition and the temperature change, T. Enthalpy is a state function that counts the amount
of heat that is transmitted at constant pressure from a system to its environment or the other
way around (Saylordo, 2017).

As for another case of another acid solution, the value of heat of neutralization between
ethanoic acid (CH3COOH) and sodium hydroxide (NaOH). From the theoretical value of heat
for normal neutralization is -57.1 kJmol-1 for each strong acid and strong base. However, for a
weak acid such as ethanoic acid, it is suggested to be around -50.6 kJmol-1 (Toppr, 2020).
The value of heat is lower because the rate of reaction between the ions OH- and H+ are not
as strong as strong acids such as HCl or HNO3. Therefore, the heat emitted is lower than -
57.1 kJmol-1.

9.0 Conclusion

In conclusion, the neutralisation of strong acid and strong base is a reaction between 1 mol of
H+ and 1 mol of OH- ions to form 1 mol of H2O, the enthalpy changes of the neutralisation

reaction, H between strong acid and strong base are almost the same for all strong acids and
bases, which is -57.1 kJmol-1. Due to additional circumstances, such as errors in the reaction's
constant pressure condition and the change in temperature, ∆T, the H values for each of the
strong acid and base reactions in this experiment are different.