Matriculation Chemistry ( Introduction to Organic Compound ) part 4
Matriculation Chemistry ( Thermochemistry )
description
Transcript of Matriculation Chemistry ( Thermochemistry )
![Page 1: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/1.jpg)
9.0 THERMOCHEMISTRY
![Page 2: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/2.jpg)
Concept of Enthalpy
![Page 3: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/3.jpg)
Important Terms Heat is energy transferred between two bodies of
different temperatures
System is any specific part of the universe
Surroundings is everything that lies outside the system
![Page 4: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/4.jpg)
Open system is a system that can exchange mass and energy with its surroundings
Closed system is a system that allows the exchange of energy with its surroundings
Isolated system is a system that does not allow the exchange of either mass or energy with its surroundings
![Page 5: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/5.jpg)
Energy is the ability to do work SI unit of energy is kg m2 s-2 or Joule (J) Non SI unit of energy is calorie (Cal) 1 Cal = 4.184 J
![Page 6: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/6.jpg)
Thermochemistry A study of heat change in chemical reactions. Two types of chemical reactions:
Exothermic Endothermic
![Page 7: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/7.jpg)
Exothermic reactions Enthalpy of products < Enthalpy of reactants, ΔH is
negative.
Energy is released from the system to the surroundings.
![Page 8: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/8.jpg)
Consider the following reaction: A (g) + B (g) → C (g) ΔH = ve (reactants) (product)
reactantsenthalpy
reaction pathway
= -veH
products
Energy profile diagram for exothermic reaction
![Page 9: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/9.jpg)
Enthalpy of products > enthalpy of reactants, ΔH is positive
Energy is absorbed by the system from the surrounding
Endothermic Reactions
![Page 10: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/10.jpg)
Consider the following reactionA (g) + B (g) → C(g) ΔH = + ve (reactants) (product)
Energy profile of diagram endothermic reactions
![Page 11: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/11.jpg)
Enthalpy, H The heat content of a system or total energy in the
system
Enthalpy, H of a system cannot be measured when there is a change in the system.
Example: system undergoes combustion or ionisation.
![Page 12: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/12.jpg)
Enthalpy of Reaction, ∆H and Standard Condition
Enthalpy of reaction: The enthalpy change associated with a chemical
reaction.
Standard enthalpy, ∆Hº The enthalpy change for a particular reaction that
occurs at 298K and 1 atm (standard state)
![Page 13: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/13.jpg)
Thermochemical Equation The thermochemical equation shows the enthalpy changes.
Example : H2O(s) → H2O(l) ΔH = +6.01 kJ
1 mole of H2O(l) is formed from 1 mole of H2O(s) at 0°C, ΔH = +6.01 kJ
However, when 1 mole of H2O(s) is formed from 1 mole of H2O(l), the magnitude of ΔH remains the same with the opposite sign of it.
H2O(l) → H2O(s) ΔH = 6.01 kJ
![Page 14: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/14.jpg)
Types of Enthalpies There are many kind of enthalpies such as:
Enthalpy of formation Enthalpy of combustion Enthalpy of atomisation Enthalpy neutralisation Enthalpy hydration Enthalpy solution
![Page 15: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/15.jpg)
Enthalpy of Formation, ∆Hf
The change of heat when 1 mole of a compound is formed from its elements at their standard states.
H2 (g) + ½ O2(g) → H2O (l) ∆Hf = 286 kJ mol1
The standard enthalpy of formation of any element in its most stable state form is ZERO.∆H (O2 ) = 0 ∆H (Cl2) = 0
![Page 16: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/16.jpg)
Enthalpy of Combustion, ∆Hc
The heat released when 1 mole of substance is burned completely in excess oxygen.
C(s) + O2(g) → CO2(g) ∆Hc = 393 kJ mol1
![Page 17: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/17.jpg)
Enthalpy of Atomisation, Ha The heat change when 1 mole of gaseous atoms is formed
from its element
Ha is always positive because it involves only breaking of bonds
e.g:
Na(s) Na(g) Ha = +109 kJ mol-1
½Cl2(g) Cl(g) Ha = +123 kJ mol-1
![Page 18: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/18.jpg)
Enthalpy of Neutralization, ∆Hn The heat change when 1 mole of water, H2O is formed from the
neutralization of acid and base .
HCl(aq)+ NaOH(aq) → NaCl(aq) +H2O(aq) ΔHn = 58 kJ mol1
![Page 19: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/19.jpg)
Enthalpy of Hydration, Hhyd
The heat change when 1 mole of gaseous ions is hydrated in water.
e.g:Na+
(g) Na+(aq) Hhyd = 406 kJ mol-1
Cl-(g) Cl-
(aq) Hhyd = 363 kJ mol-1
![Page 20: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/20.jpg)
Enthalpy of Solution, Hsoln The heat change when 1 mole of a substance is
dissolves in water. e.g: KCl(s) K+
(aq) + Cl(aq) Hsoln = +690 kJ mol-1
![Page 21: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/21.jpg)
Enthalpy of Sublimation, Hsubl
The heat change when one mole of a substance sublimes (solid into gas).
I2 (s) → I2(g)
sublH
Hsubl = +106 kJ mol1
![Page 22: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/22.jpg)
Calorimetry A method used in the laboratory to measure the
heat change of a reaction. Apparatus used is known as the calorimeter Examples of calorimeter
Simple calorimeter Bomb calorimeter
![Page 23: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/23.jpg)
The outer Styrofoam cup insulate the reaction mixture from the surroundings (it is assumed that no heat is lost to the surroundings)
Heat release by the reaction is absorbed by solution and the calorimeter
Simple calorimeter
![Page 24: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/24.jpg)
A bomb calorimeter
![Page 25: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/25.jpg)
Important Terms in Calorimeter Specific heat capacity, c
Specific heat capacity, c of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius (Jg 1C1).
Heat capacity, C Heat capacity,C is the amount of heat required to raise
the temperature of a given quantity of the substance by one degree Celsius (JC1)
![Page 26: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/26.jpg)
q = mc∆T
Heat released by substance =
Heat absorbed by calorimeter
q = heat released by substance
m= mass of substance
C= specific heat capacity
∆T = temperature change
![Page 27: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/27.jpg)
Basic Principle in Calorimeter
Heat released by a reaction = Heat absorbed
by surroundings
• Surroundings may refer to the:i. Calorimeter itself or;ii. The water and calorimeter • qreaction= mcΔT or CΔT
![Page 28: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/28.jpg)
Example 1
In an experiment, 0.100 g of H2 and excess of O2 were compressed into a 1.00 L bomb and placed into a calorimeter with heat capacity of 9.08 x 104 J0C1. The initial temperature of the calorimeter was 25.0000C and finally it increased to 25.155 0C. Calculate the amount of heat released in the reaction to form H2O, expressed in kJ per mole.
![Page 29: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/29.jpg)
SolutionHeat released = Heat absorbed by the calorimeter q = C∆T = (9.08 X 104 J0C-1) X (0.1550C) = 1.41 X 104 J = 14.1 kJ
H2(g) + ½O2(g) → H2O(c)
mole of H2 = 0.100 2.016 = 0.0496 mol
![Page 30: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/30.jpg)
moles of H2O = mole of H2
0.0496 mol of H2O released 14.1 kJ energy
1 mol H2O released = kJ
= 284 kJ∆Heat of reaction, ∆H = - 284 kJ mol1
0496.01.14
![Page 31: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/31.jpg)
Example 2
1. Calculate the amount of heat released in a reaction in an aluminum calorimeter with a mass of 3087.0 g and contains 1700.0 mL of water. The initial temperature of the calorimeter is 25.0°C and it increased to 27.8°C.
![Page 32: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/32.jpg)
Given: Specific heat capacity of aluminum = 0.553Jg-1 °C-1
Specific heat capacity of water = 4.18 Jg-1 °C-1
Water density = 1.0 g mL-1ΔT = (27.8 -25.0 )°C = 2.8°C
Solution
q = mwcwΔT + mcccΔT
= (1700.0 g)(4.18 Jg-1 °C-1)(2.8 °C) + (3087.0 g)(0.553 Jg-1 °C-1)(2.8°C)= 24676.71 J= 24.7 kJ
Heat absorbed by water
Heat absorbed by aluminium calorimeter
Heat released = +
![Page 33: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/33.jpg)
HESS’S LAW
![Page 34: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/34.jpg)
Hess Law Hess’s Law states that when reactants are converted to
products, the change in enthalpy is the same whether the reaction takes place in one step or in the series of steps.
The enthalpy change depends only on the nature of the reactants and products and is independent of the route taken.
1HA B
C
3H
2H
321 H H H
![Page 35: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/35.jpg)
i. List all the thermochemical equations involved
Algebraic Method
1--1560kJmol H )(2
3)(2
2)(22
7)(62
.
1--286kJmol H )(2)(22
1)(2
.
1-393kJmol- H )(2)(2)(
.
gOH
gCO
gO
gHCiii
gOH
gO
gHii
gCO
gO
SCi
Step 1
![Page 36: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/36.jpg)
i. List all the thermochemical equations involved
Algebraic Method
1--1560kJmol H )(2
3)(2
2)(22
7)(62
.
1--286kJmol H )(2)(22
1)(2
.
1-393kJmol- H )(2)(2)(
.
gOH
gCO
gO
gHCiii
gOH
gO
gHii
gCO
gO
SCi
Step 1
![Page 37: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/37.jpg)
ii. Write the enthalpy of formation reaction for C2H6
)(62
?
)(23
)( gHCf
H
gH
sC
![Page 38: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/38.jpg)
iii. Add the given reactions so that the result is the desired reaction.
-84kJ 3H2H1HfH
84kJ- )(62?
)(23)(2
_______________________________________________________________
1560kJ3H )(227
)(62)(232(g)2CO (iii)
-286kJ32H )(23)(223
)(23 3)(
kJ -3932 1H )(22)(22)(2 2)(
gHCfHgHsC
gOgHCgOHreverse
gOHgOgHii
gCOgOSCi
![Page 39: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/39.jpg)
Energy Cycle MethodDraw the energy cycle and apply Hess’s Law to calculate the unknown value.
2 C ( s ) + 3 H2 ( g ) C2 H6 ( g )
2 C O2 ( g ) + 3 H2 O ( g )
2 O2 ( g ) 3 / 2 O2 ( g ) H
H
H
H
7 / 2 O2 ( g ) O
O
O
O f
1 2
3 = 2 ( - 3 9 3 )
= 3 ( - 2 8 6 )
= - ( - 1 5 6 0 )
![Page 40: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/40.jpg)
1-kJmol -84
1560858 - -786
°3ΔH + )°
2H3( + )°1H 2( = °
fΔH
![Page 41: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/41.jpg)
Example 1The thermochemical equation of combustion of carbon monoxide is shown as below.
C(s) + ½ O2(g) CO(g) = ?
given : C(s) + O2(g) CO2(g) ∆H= -394 kJ mol-1
CO(s) + ½ O2(g) CO2(g) ∆H= -283 kJ mol-1
Calculate the enthalpy change of the combustion of carbon to carbon monoxide.
H
H H
![Page 42: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/42.jpg)
Example 2Calculate the standard enthalpy of formation of methane if the enthalpy of combustion of carbon, hydrogen and methane are as follows:
∆H [C(s)] = -393 kJ mol-1
∆H [H2(s)] = -293 kJ mol-1
∆H [CH4(s)] = -753 kJ mol-1
cH cH cH
![Page 43: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/43.jpg)
Example 3 Standard enthalpy of formation of ammonia, hydrogen
chloride and ammonium chloride is -46.1 kJ mol-1, -92.3 kJ mol-1, 314.4 kJ mol-1 respectively. Write the thermochemical equation for the formation of each substance and calculate the enthalpy change for the following reaction.
NH3(g) + HCl (g) NH4Cl(s)
![Page 44: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/44.jpg)
Exercise1.Calculate the enthalpy of formation of benzene if :
∆H (CO2(g) ) = -393.3 kJ mol-1
∆H (H2O(l) ) = -285.5 kJ mol-1
∆H (C6H6(l) ) = -3265.3 kJ mol-1 fH fH
fH
![Page 45: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/45.jpg)
Born-Haber Cycle
![Page 46: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/46.jpg)
Lattice Energy, Hlattice is the energy required to completely separate one mole of a
solid (ionic compound) into gaseous ions
e.g:NaCl(s) Na+(g) + Cl-(g) Hlattice = +771 kJ mol-1
(lattice dissociation)
Na+(g) + Cl-(g) NaCl(s) Hlattice = -771 kJ mol-1 (lattice formation) The magnitude of lattice energy increases as
the ionic charges increase the ionic radii decrease
![Page 47: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/47.jpg)
There is a strong attraction between small ions and highly charged ions so the H is more negative.
H for MgO is more negative than H for Na2O because Mg2+ is smaller in size and has bigger charge than Na+, therefore Hºlattice (MgO) > Hºlattice (Na2O)
![Page 48: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/48.jpg)
Na+ and Cl- ions in the solid crystal are separated from each other and converted to the gaseous state (Hlattice)
The electrostatic forces between gaseous ions and polar water molecules cause the ions to be surrounded by water molecules (Hhydr)
Hsoln = Hlattice + Hhdyr
Hydration Process of Ionic Solid
![Page 49: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/49.jpg)
Lattice Energy Heat of Hydration
Heat of Solution
Na+ and Cl- ion in the gaseous state
Na+ and Cl- ion in the solid state
Hydrated Na+ and Cl- ion
![Page 50: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/50.jpg)
Born-Haber Cycle
The process of ionic bond formation occurs in a few stages. At each stage the enthalpy changes are considered.
The Born Haber cycle is often used to calculate the lattice energy of an ionic compound.
In the Born-Haber cycle energy diagram, by convention, positive values are denoted as going upwards, negative values as going downwards.
Consider the enthalpy changes in the formation of sodium chloride.
![Page 51: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/51.jpg)
Example :
Given;i. Enthalpy of formation NaCl = -411 kJmol-1
ii. Enthalpy of sublimation of Na = +108 kJmol-1
iii. First ionization energy of Na = +500 kJmol-1
iv. Enthalpy of atomization of Cl = +122 kJmol-1
v. Electron affinity of Cl = -364 kJmol-1
vi. Lattice energy of NaCl = ?
(s)2(g)21
(s) NaClCl Na
![Page 52: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/52.jpg)
Example: A Born-Haber cycle for NaCl
Na(s) + ½ Cl2(g)
Na(g) + ½ Cl2(g)
NaCl(s)
energy
E=0
Na(g) + Cl(g)
Na+(g) + e + Cl(g)
Na+(g) + Cl- (g)
HaNa
HaCl
Ionisation Energy of Na
Electron Affinity of Cl
Lattice energy
Hf NaCl
From Hess’s Law:Hf NaCl = HaNa + HaCl +IENa + EACl + Lattice Energy -ve
+ve
![Page 53: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/53.jpg)
Calculation:
kJ777HkJ364kJ122kJ500kJ108kJ411H
EAHIEHHH
HEAHIEHH
lattice
lattice
)Cl(aS0flattice
lattice)Cl(aS0f
![Page 54: Matriculation Chemistry ( Thermochemistry )](https://reader033.fdocuments.in/reader033/viewer/2022061119/546b0423b4af9f662c8b4a38/html5/thumbnails/54.jpg)
Exercise: Construct a Born-Haber cycle to explain why ionic compound
NaCl2 cannot form under standard conditions. Use the data below:
i. Enthalpy of sublimation of sodium = +108 kJmol-1
ii. First ionization energy of sodium = +500 kJmol-1
iii. Second ionization energy of sodium = +4562 kJmol-1
iv. Enthalpy of atomization of chlorine = +121kJmol-1
v. Electron affinity of chlorine = -364 kJmol-1
vi. Lattice energy of NaCl2 = -2489 kJmol-1