CHEMICAL REACTIONS 1

A CHEMICAL REACTION

A chemical reaction is the rearrangement of atoms to form a new substance.

It is the process where substances combine to form a new substance.

The substances that react with each other are called reactants.

The new substances formed from the reaction are called products.

For a chemical reaction to take place bond breaking and bond making (formation) occur. Old bonds
break and new bonds are formed.

A CHEMICAL EQUATION

A chemical equation is a statement expressing a chemical reaction.

BALANCING CHEMICAL EQUATIONS

Balancing a chemical equation means making sure that the number of atoms of each kind on one side
of an equation is equal to the number of atoms on the other side.

RULES

a) First of all, balance the number which appears only once on any side of the equation. Thereafter,
balance the number of the other atoms.

b) Multiply the reactants and products with some factors.

c) Do not change the chemical formula of any substance in the equation.

EXAMPLES

Balance the chemical equations below;

i) Na(s) + H2O(l) → NaOH(aq) + H2(g)

Look for atoms which are not balanced.

Hydrogen is not balanced.

2Na(s) + H2O(l) → 2NaOH(aq) + H2(g) (balanced eq.)

ii) CaCO3(s) + HCl(aq) → CaCl2(s) + CO2(g) + H2O(l)

Look for atoms not balanced.

Chlorine and hydrogen atoms are not balanced.

CaCO3(s) + 2HCl(aq) → CaCl2(s) + CO2(g) + H2O(l)

(balanced eq.)

iii) H2SO4 (aq) + NH3 (aq) → (NH4)2SO4 (aq)

Number of nitrogen and hydrogen atoms is not balanced.

Multiply NH3 by 2.

H2SO4 (aq) + 2NH3 (aq) → (NH4)2SO4 (aq) (balanced equation)

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RELATIVE ATOMIC MASS (RAM)

This is the average mass of all the isotopes of an element compared to the mass of 1 atom of carbon
12.

Isotopes : These are different forms of the same element with different mass numbers due to
difference in number of neutrons.

For instance, chlorine has two isotopes which are Cl-35 and Cl-37.

EXAMPLE

To find the Ram of Chlorine, we need to consider the percentage abundance of the isotopes. The
percentage abundance of Cl-35 is 75.771 and that of Cl-37 is 24.230. The atomic mass of chlorine atom:

= 35.48495

= 35.5

RAM of chlorine =

= 35.5

RELATIVE FORMULA MASS

This is the sum of all the relative atomic masses of the atoms found in a compound.

It is sometimes called relative molecular mass (RMM) when the compound is a molecule.

EXAMPLES

1. Find the relative formula mass of CuSO4.5H2O

To find the RFM of CuSO4.5H2O, one has to consider the atomic masses of each kind of atom present by
multiplying the mass by the number of atoms present to find the total mass of each kind present.

Type of atom No of atoms RAM Total

Cu 1 1×64= 64

S 1 1×32= 32

O 9 9×16= 144

H 10 10×1= 10

RAM for CuSO4.5H2O 250

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2. Find the relative formula mass of MgSO4.10H2O

Type of atom No of atoms RAM Total

Mg 1 1×24= 24

S 1 1×32= 32

O 14 14×16= 224

H 20 20×1= 20

RAM for MgSO4.10H2O 300

THE MOLE AND AVOGADRO NUMBER

23
A mole refers to amount of substance which contains 6.02 × 10 particles.

Is a ratio of mass of substance to its atomic mass unit.

Number of Moles =

23
1 mole represents 6.02 × 10 particles.
23
The value 6.02 × 10 is called the Avogadro number.

AVOGADRO NUMBER

The Avogadro number is the number of particles in an atom.

The RAM or RFM stands for 1 mole of that substance. In other words, if you weigh 24g of magnesium, it
means that you have 1 mole of magnesium.
23
In one mole of an element, there are 6.02 × 10 atoms of that element.
23
Thus in 24g of magnesium, there are 6.02 × 10 units or molecules of that compound.

EXAMPLES

1. Work out the number of moles:

a) 16g of NaOH

RFM of NaOH = 40

Number of moles =

= 0.4 Moles.

b) 175.5g of NaCl

RFM of NaCl = 58.5

Number of moles = = 3 Moles

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 c) 19.6g of H2SO4

Number of moles=

= 0.2 Moles.

2. How many particles are there in 0.5 Moles of Oxygen?

Number of particles = Moles × Avogadro constant

23
= 0.5 × 6.02 × 10
23
= 3.01 × 10 particles.

THE EMPIRICAL FORMULA

The empirical formula is the simplest formula of a compound that shows the lowest ration of atoms in
the compound.

For instance, the molecular formula of glucose is C6H12O6. The formula can be shown in its simplest
form by dividing each number of atoms by the lowest number i.e. C6/6H12/6 O6/6 = CH2O

Therefore the empirical formula of glucose is CH2O.

For most substances, the empirical formula is not necessarily equal to the molecular formula. The
molecular formula shows all the atoms that make up a compound.

EXAMPLES

a) Find the empirical formula of a compound that has the following percentage composition by mass of
elements; C= 40%, H= 6.67%, O= 53.33%.

Element Ratio of atoms Lowest ratio of atoms

Atomic proportion (%) =

C 40/12 3.33 3.33/3.33= 1

H 6.67/1 6.67 6.67/3.33= 2

O 53.33/16 3.33 3.33/3.33= 1

Empirical formula = C1H2O1= CH2O.

b) In an experiment 2.6g of chromium were heated in an excess of chlorine gas. It was found that 7.93g
of chromium chloride was formed. If the relative formula mass of chromium chloride is 158, what is
molecular formula of the compound?

Mass of chromium chloride = 7.93g

Mass of chromium = 2.6g

So, mass of chlorine = (7.93-2.6)

= 5.33g

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 Atomic proportion (%) =
Lowest ratio of
Element Mass (g) Ratio of atoms
atoms

Cr 2.6 2.6/52 0.05 0.05/0.05 = 1

Cl 5.33 5.33/35.5 0.15 0.15/0.05 = 3

The empirical formula is therefore CrCl3

But RFM of CrCl3 = 52 + 3(35.5) = 158.5

The true RFM of chromium chloride is 158.5

So the molecular formula must be the same as the empirical formula = CrCl3

c) Find the empirical formula of a substance that has the following percentage composition; C= 42.87%,
H= 2.36%, N= 16.67%, O= 38.10%. If the RFM of the compound is 168, then find the correct molecular
formula.

Element Mass (g) Number of moles Simplest mole ratio

C 42.87 42.87/12 = 3.57 3.57/1.19 = 3

H 2.36 2.36/1 = 2.36 2.36/1.19 = 2

N 16.67 16.67/14 = 1.19 1.19/1.19 = 1

O 38.10 38.10/16 = 2.37 2.37/1.19 = 2

The empirical formula is C3H2NO2

Mass of this formula is = (3 × 12) + (1 × 2) + 14 + (16 × 2)

Mass of the empirical formula = 84

But correct molecular mass = 168 (which is twice mass of the empirical)

So molecular formula = (empirical formula × 2)

= 2(C3H2NO2)

= C6H4N2O4

PERCENTAGE COMPOSITION CALCULATIONS

To work out the percentage of an element in a compound, the steps that should be taken to do this
calculation are

 From the molecular formula, work out the Relative Formula Mass (RFM) using RAMs.

 Work out the mass of the element in the compound from the RFM.

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  Express the percentage mass of the element X in the compound by;

o % mass of X =

EXAMPLES

I) Work out the % of C in propene.

The molecular formula of propene is C3H6.

RFM of C3H6 = (3 × 12) + (6 × 1)

= 42

The mass of C in propene is 36

% mass of C = × 100%

= × 100%

= 85.7%

The percentage of C in propene is 85.7%.

II) Calculate the % of oxygen in Cu(NO3)2.

For calculation purposes Cu(NO3)2 becomes;

Cu + 2N + 6(O)

= 64 + (2 × 14) + (6 × 16) = 188

RFM of Cu(NO3)2 is 188.

% mass of Oxygen =

= 51.1%

The percentage by mass of oxygen in Cu(NO3)2 is 51.1%

III) What is the % compound of potassium nitrate.

For calculation purposes KNO3 becomes;

K + N + 3(O)

= 39 + 14 + (3 × 16) + 101

RFM of KNO3 is 101

Express the mass of each element as a percentage of the RFM.

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 %K =

= 38.62%

%N =

= 13.86%

%O =

= 47.52%

IV) Calculate the mass of each element in 5.0g of potassium chromate.

For calculation purposes, K2CrO4 becomes 2K + Cr + 4(O)

(2 × 39) + 52 + (4 × 16) = 78 + 52 + 64

= 194

The RFM of K2CrO4 is 194

By proportion,

Mass of K =

= 2. 040g

Mass of Cr =

= 1.341g

Mass of O =

= 1.649g

CONCENTRATION
Concentration is the amount of solute present in a given/specific quantity of solution.

The solute in most cases is in form of solid or liquid. The unit for concentration is molarity.

Molarity is the number of moles per litre of solution i.e.

Molarity =

Example of concentration;

1.46M C6H12O6 means we have 1.46 mole of glucose (C6H12O6) in a solution of 1l. etc.

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WAYS OF EXPRESSING CONCENTRATION

Concentration can be expressed as;

 In moles per litre

 In grams(g) per litre

 As a percentage.

EXAMPLES

Calculate the concentration in mol/litre when 19.6g of sulphuric acid is dissolved in 250ml water, RAM
of H=1, S=32 and O=16.

No of moles =

RFM of H2SO4 = H= 1 × 2 =2

S = 32 × 1 =32 O= 16 × 4 = 64

RFM of H2SO4 = 98g

Number of moles =

= 0.2 moles.

Concentration = , but 250ml = 0.25l

Concentration = = 0.8 mol/ litre.

NOTE:

Some Acids have their concentration in percentage e.g. 50% sulphuric acid. Molarity of this % can be
worked out by taking the % as mass per 100ml.

E.g: 50% sulphuric acid means 50g of H2SO4 dissolved in 100ml of water.

EXAMPLE

Calculate the molarity of 5% ethanoic acid.

5% CH3COOH is the same as 5g of CH3COOH in 100ml in H2O.

1000ml = 1 litre

100ml = 0.1 litre.

Number of moles = = = 0.083 moles

Molarity = =

= 0.83mol/l

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CONCENTRATION IN GRAMS PER LITRE

Example 1

Calculate the concentration of NaCl solution in g/l when 5g of it is dissolved in 50ml.

NaCL concentration =

50ml = × 1l

= 0.05 litres

Concentration =

= 100g/l

Example 2

Explain how you can make a concentration of 20g/litre from sugar solution whose concentration is
50g/litre.

50g of sugar in 1000ml

20g of sugar = × 1000ml

= 400ml

Therefore collect 400ml from 50g/l solution and dilute it to 1 litre.

Example 3

Calculate the mass of solutes when 100ml of salt solution is removed from a stack of 200g/l salt
solution.

In 1000ml =200g

100ml = × 200g

= 20g

TITRATION
Titration is the gradual addition of one solution to another in order to determine the unknown
concentration of the other solution, OR

It is a gradual addition of a base to an acid or an acid to base until end point is reached.

END POINT: It is a balanced point whereby an acid is completely neutralized by a base.

An end point is deduced by colour change using an indicator (phenolphthalein or methyl orange) which
is usually a dye that has different colour in acidic and basic solutions.

A STANDARD SOLUTION: Is the solution whose concentration is known.

A TITRE: Is a liquid which is gradually added to a standard solution.

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A simple acid-base titration involves the addition of an acid solution to an alkali (a soluble base)
solution until the reaction is complete (the reverse addition of alkali to acid is also possible).

Fig: Titration apparatus

TITRATION METHOD (PROCEDURE)

3
 An exact volume of 25cm of the standard alkali solution (solution of known molarity) M 1, is
measured out using a pipette.

 The alkali solution is then transferred into a conical flask where the neutralization reaction will
take place later.

 The acid of the unknown molarity M2 is poured into the burette up to the zero mark.

 A few drops of phenolphthalein indicator are added to the alkali in the conical flask.
Phenolphthalein would give a pink colour in alkali solution.

 The acid is then slowly added to the conical flask by opening the tap at the bottom of the
burette.

 The conical flask is holding the alkali and a reaction begins to take place. To help mixing, the
flask and contents are swirled around.

 Finally when the last drop of acid from the burette neutralizes the base, the colour changes
from pink to colourless at the end of the titration.
3
 The titre would be noted in the results as the volume of acid needed to just neutralise 25cm
of standard base solution.

EXAMPLES
3 3
1. What is the concentration of HCl if 20.0cm of it were use to neutralize 25.0cm of a 0.1M NaOH
solution.

From the balanced equation;

NaOH (aq) + HCl (aq) → NaCl (aq) + H2O(l)

1 mole of NaOH alkali reacts with 1 mole of HCl acid.

Changing moles into (molarity × volume)

(Base) M1 × V1 = M2 × V2 (acid)

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 Note: This formula only applies when you are working with a formula for 1 more ration 1:1.

The volume should be in dm3.

0.1 × = M2 ×

M2 =

M2 = 0.125M

2. The molarity of a solution of HCl is found by titrating it against 0.25M sodium carbonate solution,
3
using methyl orange as an indicator. The result was that 25cm of 0.25M Na2CO3 solution required
3 2
20.08cm of the acid of unknown molarity M . Work out the concentration of the acid in;
3
a) moles/dm .
3
b) in grams per dm .

The balanced equation of the reaction is;

Na2CO3 + 2HCl → NaCl + H2O + CO2.

1 mole of Na2CO3 will neutralize 2 moles of HCl.

Therefore changing moles into (molarity × volume) we have

(Base) (Acid)

This becomes;

M2 =

= 0.6M.

So the molarity of the HCl acid is 0.6M, but 0.6M means 0.6 moles per dm3.

a) The concentration of the acid is 0.6 moles per dm 3.

b) for HCl, the RFM = 1 +35.5 = 36.5

but concentration (g/dm3) = 0.6 × 36.5

= 21.9g/l.

3. What is the concentration of NaOH is 16.1ml of it were used to neutralize 20.0ml of a 0.245M H 2SO4
solutio?

2NaOH + H2SO4 → Na2SO4 + 2H2O

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M1=? V1=16.1ml or 0.016L N1= 2

M2= 0.245M V2= 20.0ml N2= 1

M1=

M1=

M1= 0.61M

PREPARING A SOLUTION OF A KNOWN CONCENTRATION AND VOLUME

A) By Dissolution Method.

To prepare a solution by dissolution method, we have to know the amount (mass) substance to dissolve
as follows;

Number of moles (n) =

m = n × RFM ... (i)

But molarity =

Number of moles (n) = molarity × litre of solution... (ii)

Substituting (molarity × litre of solution) for n in equation (i) gives us;

Mass= molarity × litre of solution × RFM

EXAMPLES

How many grams of NaOH must be dissolved in order to prepare 250ml of 0.5M NaOH solution?

Molarity = 0.5M, Volume (in litres) of solution= 250ml or 0.25L, RFM of NaOH = 40.

Mass = molarity × litre of solution × RFM

= 0.5M × 0.25L × 40

= 0.5mole/L × 0.25L × 40g/mole.

Mass = 5 grams.

After finding the solute to be dissolved, do the following;

 Measure 5g of NaOH in a beaker on a beam balance.

 Dissolve the solute in a small amount of water.

 Transfer the solution made into a 250ml volumetric flask.

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  Add more water to the solute in the beaker and transfer the solution into the volumetric flask
again.

 Continue washing away the solution from the beaker into the volumetric flask until the final
volume of the solution is 250ml.

 Shake the solution in the flask in order to evenly distribute its concentration.

B) By Dilution Method.

Dilution is the process of adding more solvent to a solution in order to reduce its concentration.

In order to prepare a solution of specific concentration and volume by dilution method, you must know
the amount (volume) of the original solution to be diluted. To do this, we use the equation below;

C1V1=C2V2

Where C1= initial concentration of solution.

V1= initial volution of solution.

C2= final concentration of solution

V2= final volume of solution.

C1V1= number of moles before dilution= number of moles after dilution.

EXAMPLE

1. How many ml of 0.7M CH3COOH solution are required to prepare a 250ml solution of 0.3M CH3COOH?

C1= 0.7M V1= ?

C2= 0.3M V2= 250ml

C1V1=C2V2

V1= =

= 107.1ml

This means we must;

a) Measure 107.1ml of the original solution using a measuring cylinder.

b) Put 50ml of distilled water into a 250 volumetric flask. Add 107.1ml of the acid little by
little to the conical flask.

c) Add water to the solution in the flask until the volume reaches 250ml.

d) Shake the solution thoroughly in order to distribute the concentration evenly.

MOLAR VOLUME AND REACTING MASSES

MOLAR VOLUME

The molar volume of any gas is the volume occupied by 1 mole of that gas at standard temperature and
pressure (s.t.p).

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 3
The molar volume of all the gases is the same and it is 22.4dm . This means that 1 mole of any gas
3
occupies a volume of 22.4dm .
3
The molar volume of any gas at room temperature and pressure (r.t.p) is 24dm .

EXAMPLES

1. What is the mass of 1 mole of each of the following gases?

a) Oxygen (O2)

RFM of O2= 2O = 2 × 16

Mass of 1 mole O2= 32g.

b) Hydrogen (H2)

RFM of H2 = 2H= 2 × 1

Mass of 1 mole H2= 2g

c) Sulphur dioxide (SO2)

1S = 1 × 32 = 32

2O = 2 × 16 = 32

RFM of SO2= 64

Mass of 1 mole SO2=64g.

2. At s.t.p, what is,

3
a) The mass of 100dm of hydrogen gas (H2)?
3
Mass of 22.4dm of H2 =2g
3
Mass of 100dm of H2=? (More)

3
Mass of 100dm of H2 =

= 8.928g.

b) The volume occupied by 12g of 0xygen gas (O2)?

3
Mass of 32g of O2= 22.4dm .

Mass of 12g of O2=? (less)

Mass of 12g of O2=

3
= 8.4dm

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 c) The volume occupied by 146g of HCl gas?
3
Mass of 36.5g of HCl = 22.4dm .

Mass of 146g of HCl = ?

Mass of 146g of HCl =

3
= 89.6dm .

REACTING MASSES

Reacting masses is about how substances react with each other based on their masses.

Are the proportions in which substances react with each other by mass.

EXAMPLES

1. Reaction between carbon and oxygen gas.

C(s) + O2(g) → CO2(g)

1 atom + 1 molecule → 1 molecule.

Of C of O2 of CO2

1 mole + 1 mole → 1 mole

Of C of O2 of CO2

12g of C + 32g of O2 → 44g of CO2

12g of C react with 32g of O2 to form 44g of CO2.

2. Reaction between Magnesium and Oxygen gas.

2Mg + O2 → 2MgO

2 atoms + 1 molecule → 2 units of MgO

Of Mg of O2

2 mole of Mg + 1 mole of O2 → 2 moles of MgO

48g of Mg react with 32g of O2 to form 80g of MgO.

HEATS OF REACTION
Heat is the form of energy measured in joules.

Often when a chemical reaction takes place heat energy is given out, causing the container to get hot,
or heat can be taken in, causing the container to get cold. Heat is taken from the container.

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TYPES OF REACTIONS

1. EXOTHERMIC REACTION

This is a type of reaction in which heat is given out to the surrounding by the reactants.

This reaction causes the container habouring the reaction to get hot.

Examples of exothermic reaction are;

i) Reaction between any strong acid and base.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

2+ -
ii) CaCl2 + H2O → Ca + 2Cl

iii) The mixing of concentrated sulphuric acid (H 2SO4) with water.

2. ENDOTHERMIC REACTION

This is the type of reaction that takes in heat from the surrounding i.e. from the container.

This makes the container to get cold.

Examples of endothermic reactions are;

i) The dissolving of ammonium nitrate in water.

+ + -
NH4NO3 H2O → NH4 + NO3

ii) The evaporation of water to water vapour.

H2O(l) + heat → H2O(g)

BOND BREAKING AND BOND FORMATION

For a reaction to happen, two processes take place and these are bond breaking and bond making.

Bond breaking requires the absorption of heat from the surrounding because there is need to act
against attractive forces that bond atoms together. On the other hand bond making involves the
liberation of heat to the surrounding.

In the making of bonds energy is not needed or absorbed because the attractive forces between atoms
or between electrons and nuclei of atoms pull the atoms/ ions together. For instance, for HCl to react
with NaOH to form new substances (NaCl + H2O), bonds in HCl and NaOH must be broken first. Later,
bonds between the produced atoms/ions are formed i.e. NaCl and H2O.

A reaction is either endothermic or exothermic depending on the net (resultant) energy involved. To
find the energy involved, subtract the energy released in making of bonds from the energy absorbed in
the breaking of bonds i.e.

Net energy = energy (breaking the bonds) – energy (making of bonds)

In endothermic reaction, energy involved in bond breaking is greater than energy involved in bond
making.

In exothermic reaction energy involved in bond breaking is less than energy involved in bond making.

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HEAT ENERGY DIAGRAMS

The change in heat energy is ΔH

ΔH = Final heat energy (products) – Initial heat energy (reactants)

ΔH = H1 – H2

For example, reactant energy, H2= 28KJ, while the product energy H1= 10KJ, therefore;

ΔH = 10KJ – 28KJ

= -18KJ

Therefore ΔH is negative and the reaction is an exothermic reaction.

A) EXOTHERMIC REACTION

In exothermic reactions, the products have less energy than the reactants. This is so because energy is
lost to the surrounding by the reaction.

So when heat energy of reactants is subtracted from heat of products, the answer is negative.

EXAMPLES

The following diagrams show the energy level diagrams for exothermic reactions;

i) HCl(aq) + NaOH(aq) → NaCl(s) + H2O(l)

H2 HCl(aq) + NaOH(aq) Reactants

Heat energy

H1
NaCl(aq) +H2O (l) Products

Progress of reaction

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 ii) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH= -728KJ

H2 CH4(g) + 2O2(g) Initial

Heat energy

ΔH=-728KJ

H1
CO2(g) +2H2O (l) Final

Time

iii) Na2CO3 + 2HCl → 2NaCl + H2O + CO2

H2 Na2CO3 + 2HCl Initial

Heat energy

H1
2NaCl +2H2O + CO2 Final

Time

Note: The arrows in exothermic heat energy diagrams point downwards.

ENDOTHERMIC REACTION

In endothermic reaction, the products possess more energy than the reactants. This is so because the
reactants gain energy from the surrounding thereby giving products a high energy level.

So when heat energy from reactants is subtracted from heat energy of the product, the answer is
positive.

The arrow in the energy diagram point upwards.

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EXAMPLES
+ -
i) NH4NO3(s) + H2O(l) → NH4 (aq) + NO3 (aq)

H1 NH4+(aq) + NO3-(aq) Products

Heat energy

ΔH=+ve

H2
NH4NO3(s) +H2O (l) Reactants

Progress of reaction

ii) C(s) + H2O (l) → CO2(g) + H2(g) ΔH=+132KJ

H1 CO2(aq) + H(g) Final

Heat energy

ΔH=+132KJ

H2
C(s) +H2O (l) Initial

Time

He that was born to be hanged shall never be drowned…

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