Solubility and Enthalpy Changes

This assignment examines the connection between enthalpy changes and the solubility of substances, focusing on sodium chloride (NaCl) and cesium chloride (CsCl). It analyzes the enthalpy values associated with the formation and dissociation of these compounds' lattices. The discussion delves into how lattice enthalpy differences influence the solubility properties of NaCl in water and CsCl in ethanol. Additionally, the assignment touches upon various factors affecting solubility and provides a framework for understanding this crucial chemical concept.

Chemical Energetics

Table of Contents
Measuring Enthalpy Changes that take place in Endothermic and exothermic reactions through
experimentation................................................................................................................................1
Abstract:.................................................................................................................................1
Key terms:..............................................................................................................................1
Background Information:.......................................................................................................1
Hypothesis:.............................................................................................................................2
Apparatus Required:...............................................................................................................2
Assessment of Risk:...............................................................................................................2
Methods:.................................................................................................................................2
Results:...................................................................................................................................2
Calculations:...........................................................................................................................3
Conclusions:...........................................................................................................................3
Evaluation:..............................................................................................................................4
TASK 1............................................................................................................................................5
1.1 Exothermic and endothermic reactions ...........................................................................5
1.2, 1.3 Standard enthalpy changes .......................................................................................5
TASK 2............................................................................................................................................6
2.1 Conservation of Energy and Hess's Law (1)....................................................................6
2.2 Enthalpy cycles and use Hess's Law ...............................................................................6
2 a) Combustion reactions......................................................................................................6
2 b) Enthalpy of combustion..................................................................................................6
3) Enthalpy for given reaction................................................................................................7
4) Enthalpy of formation for NH3..........................................................................................7
5) Enthalpy of combustion.....................................................................................................8
6) Combustion of ethanol.......................................................................................................9
2.3 Developing enthalpy cycles using bond enthalpy data ....................................................9
7) Data ...................................................................................................................................9
8) Enthalpy for N-H bond.....................................................................................................10
9) Average bond dissociation enthalpy for C-Cl bond.........................................................11

10) Bond dissociation enthalpy............................................................................................11
11) Bond dissociation enthalpy ...........................................................................................12
12) Energy level diagram.....................................................................................................12
3.1 Energy level and reaction pathway diagrams for the combustion of ethane (13)..........14
4.1 Lattice enthalpy (a & b)..................................................................................................15
4.2 Enthalpy of ion hydration (c)........................................................................................15
4.3 Solutions (d)...................................................................................................................15
4.4 Relations between enthalpy changes and solubility of a substance...............................16
REFERENCES..............................................................................................................................18

Measuring Enthalpy Changes that take place in Endothermic and exothermic
reactions through experimentation
Abstract:
The experiment which takes place between copper sulphate solution and zinc is
conducted for gaining insight regarding the enthalpy changes which take place in a displacement
reaction. The changes will be tracked using a calorimeter so that necessary enthalpy moderations
can be obtained.
Key terms:
The major terms related with this experimentation are enthalpy changes, displacement
reaction, calorimeter and exothermic/endothermic.
Background Information:
There are certain changes which take place when reactions take place. These changes
occur in the bond structure especially in case of displacement reactions. As per the Universal law
on Conservation of Energy, the energy cannot be created or destroyed. It can only be transferred
from one form to another. Hence, in every chemical reaction, the heat transfer takes place either
in the form of absorption or release. Copper (II) sulphate is a blue colour solution in aqueous
form. When standard conditions prevail i.e. pressure is 1 atm and temperature is 298K, then
enthalpy changes are considered to be standard.
The exothermic reactions are one in which energy is released while endothermic
reactions are based on absorption of energy. In terms of products and reactants, when energy is
quite less when reaction completes as compared to the initial state, then it is said to have
provided the energy to surroundings. However, in case of endothermic reactions the energy is
high in completion of products as compared to the initial state. This depicts that reaction is
endothermic in nature.
Bond dissociation enthalpy is derived from the energy which is required for breaking a
bond between two components. In case of this experiment, the ionic bond shared between
Copper and sulphate ions. When Zinc is introduced in this solution, then displacement takes
place and new product formed is zinc sulphate in a colourless solution.
1

Hypothesis:
The estimated changes in enthalpy indicate prevalence of exothermic reaction.
Apparatus Required:
The apparatus or equipment which are required for this experiment include Zn powder
and Copper (II) Sulphate solution. A calorimeter with thermometer is also required for
conducting this experiment. Stop clock and weighing scale is required for making appropriate
measures.
Assessment of Risk:
Risks involved in this reaction include spilling of the solution when temperature shall
rise. Hence, use of lab coat is preferable for avoiding any sort of such accidents.
Methods:
After measuring 6g of zinc powder accurately over the weighing scale, the thermometer
is placed in the cup as a medium of stirring and measuring the temperature on a continuous basis.
Solution present in the cup is aqueous copper sulphate in which 2.5g of CuSO4 was mixed with
water in limited amount through measuring cylinder. Soon after introducing zinc powder to the
solution, the cup is sealed with a lid and stirred occasionally for 9 minutes. This time is tracked
on a stop clock which helps in recording the changes that take place in the chemical reaction.
Results:
Following are the results obtained through this experimentation:
Time elapsing (s) Temperature recordings
0 20
20 21
40 23
60 25
80 25
100 26
120 26
140 26
2

160 26
180 27
200 27
220 28
240 29
260 29
280 29
300 30
320 31
340 32
360 32
380 33
400 34
420 36
440 38
460 41
480 45
500 50
520 51
540 50
Calculations:
Change in temperature ΔT = Final – Initial
ΔT = 50-20
3

ΔT = 30°C
H = Heat generated = m.c. ΔT
Here, m = mass of the material = 25g
c = specific heat capacity = 4.18
H = 25* 4.18 * 30
H = 3135 kJ/mol
Conclusions:
From the above experimentation, it is inferred that when zinc reacts with copper(II)
sulphate solution, then displacement takes place followed by release of energy and completion of
an exothermic reaction. The temperature difference between initial and final time is ΔT = 50-20
ΔT = 30°C.
And the heat generated is H = 3135 kJ/mol
Evaluation:
There are possibilities of occurrence of flaws or defects in the calorimeter which
indicates that temperature changes might get affected. However, when using a bomb calorimeter
there is an automatic stirrer involved which results in greater changes in the temperature. The
heat loss which occurs on a minute scale also affects the readings and results on a minor scale.
On the other hand, the material insulation of the cup is also a major factor which regulates heat
flow. But polystyrene cups are more favourable as compared to other material equipment.
4

TASK 1
1.1 Exothermic and endothermic reactions
Every chemical reaction involves certain transformation of energy. The release or
absorption of energy is categorised as exothermic and endothermic reactions. These can be
understood in a better form with following examples:
Example 1: Reaction between zinc and copper sulphate
Zn(s) + Cu2+(aq) + SO42-(aq) → Cu(s) + Zn2+(aq) + SO42- (aq)
Zn(s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
The above equation is an ionic equation when zinc and copper(II) sulphate react. The
oxidation of Zn ions and reduction of Cu ions depicts that this is a redox reaction. The reactivity
of Zinc is more than Copper. In this reaction, the aqueous solution of copper sulphate has less
reactive metal Cu which is displaced by Zn with release of energy (Fernández and Bickelhaupt,
2014). The overall temperature of this solution increases which indicates it is a exothermic
reaction.
Example 2: Reaction of ethanoic acid and calcium carbonate
The following equation depicts the respective reaction :
2CH3COOH + CaCO3 → Ca(CH3COO)2 + CO2 + H2O
The carbonate ions that are present in calcium carbonate react with acid (ethanoic acid)
which is followed by production of carbon dioxide. The proton ions released from acid form
water and the calcium ions combine with remaining ions of acid. Since, acetic or ethanoic acid is
quite weak, there is absorption of energy. The overall reaction is endothermic.
1.2, 1.3 Standard enthalpy changes
The change in temperature during the reaction of zinc and copper sulphate takes place
from 20°C to 50°C. The aqueous solution turns out to be colourless and the colour of zinc
powder changes to orange. On the other hand, a fizzing sound is generated when ethanoic acid
and calcium carbonate react. This is because of generation of carbon dioxide.
5

TASK 2
2.1 Conservation of Energy and Hess's Law (1)
According to the Hess's Law, irrespective of the number of steps involved in a chemical
reaction or process, the heat evolved or absorbed is absolutely the same. This is supportive for
the rule of conservation of energy that is universally applicable (Göstl, Senf and Hecht, 2014). It
determines that energy is neither created or destroyed during a particular process. It is only
conserved.
The Hess's Law is important in developing an understanding for the chemical process
because it helps in developing calculations for enthalpies regarding each reaction. The control
mechanisms can be activated for those reactions which have imbalanced flow of energy.
2.2 Enthalpy cycles and use Hess's Law
2 a) Combustion reactions
(I) Combustion of ethane
ethane + oxygen → water + carbon dioxide
C2H6 + 7/2O2 → 3H2O + 2CO2
(II) Combustion of ethene
C2H4 (g) + 3O2 (g)→ 2CO2 (g) + 2H2O (g)
(III) Combustion of hydrogen
2H2 (g)+ O2 (g)→ 2H2O (l)
2 b) Enthalpy of combustion
The enthalpy cycles and enthalpy of combustion for each case produced above is given as
follows:
Combustion of ethane
ΔcH° = 2ΔfH° (CO2) + 3ΔfH°(H2O) – ΔfH°(C2H6)
The value of ΔfH° (O2) is zero
ΔcH° = 2 * (-393) + 3 * (-285.5) - (-83.6)
ΔcH° = -786 – 856.5 + 83.6
ΔcH° = 83.6 – 1642.5
6

ΔcH° = -1558.9 kJ/mol
Combustion of ethene
C2H4 (g) + 3O2 (g)→ 2CO2 (g) + 2H2O (g)
The enthalpy of combustion for this reaction is calculated as follows:
ΔcH° = 2ΔfH° (CO2) + 2ΔfH°(H2O) – ΔfH°(C2H4)
ΔcH° = 2 * (-393) + 2* (-285.5) – 52.0
ΔcH° = -786 - 571 -52
ΔcH° = -1409 kJ/mol
Hydrogen's combustion
2H2 (g)+ O2 (g)→ 2H2O (l)
ΔcH° = 2ΔfH°(H2O) – ΔfH°(H2)
ΔcH° = -572 kJ/mol
3) Enthalpy for given reaction
ΔrH (CH4) = -74.8 kJ/mol
ΔrH (CH3Cl) = -134.5 kJ/mol
ΔrH (HCl) = -92.3 kJ/mol
CH4 (g) + Cl2 (g) → CH3Cl (g)
4) Enthalpy of formation for NH3
7

4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l), H = -1530kJmol-1 .. (i)
H2(g) + 1/2O2(g) →H2O(l), H = -288 kJmol-1 … (ii)
Flip and divide the equation (I) by 4,
New Reaction will be
NH3(g) + 3/4O2(g)→ ½ N2(g) + 3/2 H2O(l) H = +382.5 kJmol-1 … (iii)
Multiply equation (ii) by 2
2H2(g) + O2(g) →2H2O(l), H = -564 kJmol-1 … (iv)
Now we will comment on the oxygen part of the given equation
For equation (i), put 3/4O2 on the right side
For equation (ii), put O2 on the right side
the new equations will become
NH3(g) + 3/4O2(g) → ½ N2(g) + 3/2 H2O(l) + 3/4O2(g H = +382.5 kJmol-1
2H2(g) + O2(g)→ 2H2O(l),+ O2(g) H = -564 kJmol-1
By adding both the equations and cancel; out the oxygen present, the new equation will be;
NH3(g) + 2H2(g)→ ½ N2(g) + 3/2 H2O(l) + 2H2O(l),
NH3(g) + 2H2(g) → ½ N2(g) + 4H2O
The enthalpy will become
H = - 184.5 kJmol-1
5) Enthalpy of combustion
The combustion of gaseous diborane has following formula:
B2H6 (g) + 3O2 → B2O3 (s) + 3H2O
ΔcH° is the enthalpy of combustion for this combustion process.
Putting the values from the given data
ΔcH° = -1270 -(3*242) -31.4
8

ΔcH° = -1301.4 – 726
ΔcH° = -2027.4
The combustion of gaseous benzene is provided as follows:
C6H6 (g) + 7.5O2 (g) → 6CO2 (g) + 3H2O (g)
ΔcH° is the combustion enthalpy for the above formula is calculated as follows:
ΔcH° = 6 * (-393) + 3 (-242) - (+83.9)
ΔcH° = - 2358 -726 – 83.9
ΔcH° = -3167.9 kJ/mol
6) Combustion of ethanol
The combustion of ethanol in the presence of oxygen produces water and carbon-dioxide.
Furthermore, the energy produced = 1368 kJ/mol.
ΔfH° = Enthalpy of formation of ethanol
ΔfH° = -393.7 kJ/mol for CO2
ΔfH° = -285.9 kJ/mol for H2O
ΔcH° = 1368 kJ/mol
The chemical reaction is produced as follows:
C2H5OH (g) + 3O2 (g) → 2CO2 (g) + 3H2O (g)
ΔcH° = 1368 = 2 ΔfH°(CO2 ) + 3 ΔfH° (H2O) - ΔfH° (C2H5OH)
1368 = 2 (-393.7) + 3(-285.9) - ΔfH° ( C2H5OH)
ΔfH° ( C2H5OH) = 3013.1 kJ/mol
2.3 Developing enthalpy cycles using bond enthalpy data
7) Data
a) The H for the following reaction: CH4(g) + Br2(g) → CH3Br(g) + HBr(g)
H is the enthalpy which has been utilised for breaking the bonds between methane and
Br-Br
H = 193+ 413
H = 606 kJ/mol
b) The H for the following reaction: CH4(g) + F2(g)→ CH3F(g) + HF(g)
H = is the energy that has been required for breaking the bonds between C-H and F-F .
H = 413 + 158
9

H = 571 kJ/mol
c) The enthalpy of formation of HF
ΔfH° = 565 kJ/mol as mentioned in the table
d) The enthalpy of formation of propene
ΔfH° (propene) = 20.41 kJ/mol
e) The enthalpy of formation of 1,2-dibromopropane
ΔfH° = -113.6 kJ/mol
f) The enthalpy of formation of methane
ΔcH° = -882.0 kJ/mol is the enthalpy of combustion for ethane
ΔfH° = is the enthalpy of formation = -74.87 kJ/mol
g) The enthalpy of formation of ethane
ΔfH° = -83.8 kJ/mol
h) The enthalpy of combustion of ethane = ΔcH° = -1558.9 kJ/mol
8) Enthalpy for N-H bond
2*N-> N2= 2* 945
6*N-> 3N2=6* 436
average enthalpy for N-H bond=(945+436)/2
=2245 kJ/mol
10

9) Average bond dissociation enthalpy for C-Cl bond
C (s) →C(g) ΔH = +715 kJ/mol
Cl2 (g) → 2Cl(g) ΔH = +242 kJ/mol
C(s) + 2Cl2 (g) → CCl4 (g) ΔH = -135.5 kJ/mol
ΔH = -850.5 kJ/mol
Concerned enthalpy of the C-Cl bond is given below:
-850.5 = 2(+242) + 4(x)
x = -334 kJ/mol
This value of x is the answer the average bond dissociation enthalpy of the C-Cl bond.
10) Bond dissociation enthalpy
The enthalpy of formation for 1-iodobutane = ΔfH° = -52 kJ/mol
Bond dissociation enthalpy for I-I bond = +214 kJ/mol
The structure of 1-iodobutane is given below:
The energy is ΔH = +713 kJ/mol
11) Bond dissociation enthalpy
Formation enthalpy for ethanol denoted by -277 kJ/mol. The enthalpy of bond
dissociation of the C-O bond in this formation process is given below:
CH3CH2–H →CH3CH2+ + H+
ΔH = 1077 kJ/mol
11

12) Energy level diagram
Following diagrams depict the energy level changes which happen when zinc powder is
introduced in copper sulphate solution.
12

An exothermic reaction takes place which determines that Zn ions displaces copper ions
and they form ZnSO4 whereas the solution turns colourless. The temperature of the entire
situation changes from 20° to 50° . This indicates the release of energy and completion of an
exothermic reaction.
3.1 Energy level and reaction pathway diagrams for the combustion of ethane (13)
The above produced diagram depicts the reaction pathway of ethane's combustion in the
presence of oxygen. There is formation of carbon dioxide and water as the products of this
reaction. The energy of combustion so released is 1561 kJ/mol.
13

4.1 Lattice enthalpy (a & b)
a) Lattice dissociation enthalpy
The energy required to break bonds or split a lattice is known as lattice dissociation
enthalpy. For NaCl , this energy has the value of +787 kJ/mol. The lattice is divided into separate
gaseous ions which is due to dissociation.
b) Lattice formation enthalpy
The energy which is released during the formation of lattice with help of gaseous ions is
known as lattice formation enthalpy (Ligorio and et.al., 2015). The value of this energy is -787
kJ/mol for NaCl. It can also be considered as the energy released during the conversion of
gaseous Na+ and Cl- ions into solid NaCl is lattice formation enthalpy.
4.2 Enthalpy of ion hydration (c)
The enthalpy of ion hydration is referred to the heat or energy which is released during
the formation of exceptionally new bonds amongst the molecules or water and ions which are
present. This is referred to as ionisation enthalpy or hydration enthalpy of ion. This is applicable
when certain gaseous ions are dissolved in water for formation of a solution (Truhlar, 2013). The
value of this enthalpy is always negative because energy is released.
4.3 Solutions (d)
The mixture which is present in liquid state due to amalgamation of different components
which are known as solvent and solute is known as solution. For instance, when NaCl is mixed
14

in water then NaCl is the solute and water is the solvent. The acquired solution is known as NaCl
solution.
4.4 Relations between enthalpy changes and solubility of a substance
C s (g ) + C l(g ) + e
C s ( g ) + C l (g )
C s ( g ) + C l (g ) + e
C s( g ) + C l (g )
C s( s ) + C l ( g )
C s C l(s )
H = + 7 9 k J m o l
H = – 4 3 3 k J m o l
H = + 3 7 6 k J m o l
H = + 1 2 1 k J m o l
H
H = – 3 6 4 k J m o l
6
1
3
4
5
– 1
– 1
– 1
– 1
– 1
2
+
+
+
2
2
2
2
2
2
1
1
1
–
–
–
(a)
ΔH1 = Enthalpy of formation of CsCl
ΔH2 = Lattice dissociation enthalpy for CsCl
ΔH3 = Lattice dissociation enthalpy for Cs
(b)
ΔH6 = Lattice enthalpy of formation
It is calculated as follows:
ΔH6 = +121+376+79-433-364
ΔH6 = -221 kJ/mol
(c)
ΔH3 has a lower value in terms of magnitude for Caesium as compared to that of sodium
because in case of NaCl the attraction bond is between one +1 ion of sodium and -1 ion of
chlorine. On the other hand, CsCl has quite different packing arrangement. There is least or small
15

impact of the lattice enthalpy and hence, it has less enthalpy value as compared to that of NaCl
(Ligorio and et.al., 2015).
(d)
The solubility of a particle or solute completely depends on the lattice structure. NaCl is
highly soluble in water. However, CsCl is soluble in ethanol. The crystal structure of both the
compounds are different and so is the lattice enthalpy. Four chlorine ions are surrounding one ion
of caesium whereas one on one ions are packed in structure of sodium chloride (Oteyza and
et.al., 2013). The lattice dissociation enthalpy is quite high in NaCl as compared to CsCl. This
automatically depicts the solubility properties.
16

REFERENCES
Books and Journals
Fernández, I. and Bickelhaupt, F. M., 2014. The activation strain model and molecular orbital
theory: understanding and designing chemical reactions. Chemical Society Reviews.
43(14). pp.4953-4967.
Göstl, R., Senf, A. and Hecht, S., 2014. Remote-controlling chemical reactions by light: Towards
chemistry with high spatio-temporal resolution. Chemical Society Reviews. 43(6). pp.1982-
1996.
Ligorio, G. and et.al., 2015. Organic Semiconductor/Gold Interface Interactions: From
Physisorption on Planar Surfaces to Chemical Reactions with Metal
Nanoparticles. ChemPhysChem. 16(12). pp.2602-2608.
Oteyza, D. G. and et.al., 2013. Direct imaging of covalent bond structure in single-molecule
chemical reactions. Science. 340(6139). pp.1434-1437.
Truhlar, D. ed., 2013. Potential Energy Surfaces and Dynamics Calculations: For Chemical
Reactions and Molecular Energy Transfer. Springer Science & Business Media.
17