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CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – Energetics

CHEMISTRY FORM 5-PHYSICAL CHEMISTRY - Energetics

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CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – Energetics

 

 

THERMOCHEMISTRY
This is also termed as Energetics.
Definition
Is the branch of physical chemistry which deals with energy changes that occurs during the chemical reaction.
The energy change during the chemical reaction is either positive or negative.
It is denoted by H.

ENTHALPY OF REACTION/HEAT OF REACTION
Definition
Enthalpy is the energy change which takes place during chemical reaction.





THE COMMON ENTHALPIES OF REACTIONS
The following are the common Enthalpies of reactions:-

 

1. STANDARD ENTHALPY OF COMBUSTION
This is the heat energy given out when 1mole of a certain substance is completely burnt in a given mole of oxygen at standard state.
If H for combustion reaction is negative i.e. The reaction is exothermic.
Example of combustion reaction

 

HOW TO BALANCE THE COMBUSTION REACTIONS
In balancing the combustion reactions we normally balance the other materials apart from hydrogen and oxygen first.
Secondly balance hydrogen then finally oxygen.

 

2. STANDARD ENTHALPY OF FORMATION
This deal with the formation of compounds. It is donated by ΔH°f.

Definition
Standard enthalpy of formation is the heat change which occur when 1mole of substance is formed from its element at standard state. Standard enthalpy of formation can be positive or negative.
ΔH°f may be +Ve or -Ve.

ΔH°f = +Ve Endothermic.
or
ΔH°f = -Ve Exothermic.

Example
Standard enthalpy of formation of CO2.
C+ O2 → CO2 ΔH°f.
ΔH°f of CO2 = ΔH comb of C

Example
ΔH°f of CH3COOH

Example
ΔH°f of CH3 – O – CH3

 

3. STANDARD ENTHALPY OF NEUTRALIZATION (EN)
Definition
This is the heat energy given out when one mole of water is formed from the reaction between acid and base at standard state. Or

This is the Enthalpy change which take place when one mole of water is formed from reaction between acid and base at standard condition.
Example
HCl + NaOH NaCl + H2O H = – Ve

4. ATOMIZATION ENERGY
Definition
Atomization energy is the energy absorbed when a molecule or electron is converted to gaseous atoms. Atomization energy applied for non-compound and H is positive.
Atomization of energy is denoted by “Eat”.
Example

Br2(g) 2Br(g)
O2(g) 2O(g)

 

5. SUBLIMATION ENERGY (Es)
It is also known as dissociation energy.
Definition
Sublimation energy is the energy absorbed when one mole of solid atom is converted into gaseous atom.
Sublimation energy is applied for metallic atom and H is positive.
Sublimation energy is denoted by ‘Es’.
Example

Al(s) Al(g)
Mg (g) Mg(g)
Ca (s) Ca(g)




6. IONIZATION ENERGY (EI)
Is the energy used to remove an electron from the outer most shell of an atom, gaseous atom or ions to form an ion(s).
It is denoted by “EI ”.
Ionization energy can be categorized as first 1st, 2nd and 3rd depending on the nature of stable ions that can be formed from an atom.
H is positive.
Example1

Ca +(g) Ca2+ +
Example 2
Al (g) Al + +
Al + Al2+ +
Al + Al 3+ +

7. ELECTRON AFFINITY ( Eaff)
Definition
Is the energy required when one mole of non- metallic gaseous atom combine with one mole of electron.
It is denoted by Eaff.
Example
Cl (g) + Cl-
O (g) + O-

8. LATTICE ENERGY (EL)
Definition
Is the energy given out when one mole of ionic compound is formed from its ions.
It is denoted by EL.
Example
Na + + Cl- 1 NaCl
Al 3+ + 3Cl-1 AlCl3
Mg 2+ + 2Br- MgBr2
H = -ve

9. STANDARD ENTHALPY OF SOLUTION
Definition
Is the heat change when one mole of a compound is dissolved in a given moles of water at standard conditions.
Na Cl + Aq NaCl (aq)

10. STANDARD ENTHALPY OF FORMATION
Is the heat change which occur when one mole of a substance is formed from its element at standard conditions.
eg. C+ O2→CO2(g) ΔHf = -393KJmol-1.

11. HEAT OF DILUTION
Is the heat change when one mole of a substance is dissolved in a given mole of water

METHODS OF FINDING HEAT OF FORMATION OF A GIVEN COMPOUNDS
The calculation in thermochemistry are categorized in the following;-
(i). Calculation based on combustion Data.
(ii). Calculation based on bond energies.
(iii). Calculation based on atomization Data.
(iv). Calculation based on calorimetry information.
(v). Determination of molecular formular by combustion Data.
(vi). Calculation based on Born–Haber cycle.




CALCULATIONS BASED ON COMBUSTION DATA
Calculations involving combustion data has got the following Steps:-
(i). Identify the required equation.
(ii). Data presentation.
(iii). Data manipulation.
(iv). Conclusion.

Example 1
a) With one example in each briefly define the following terms;-
(i). Standard enthalpy of combustion.
(ii). Sublimation energy.
(iii). Standard enthalpy of formation.
(iv). Atomization energy.
(v). Lattice energy.

Solution
(i). Standard enthalpy of combustion: is the heat energy given out when 1mole of a certain substance is completely burnt in a given moles of oxygen at standard state.
(ii). Sublimation energy: Is the energy absorbed when one mole of solid atom is converted to gaseous atom.
(iii). Standard enthalpy of formation: is the heat change which occur when 1 mole of substance is formed from its element.
eg. C + O2 → CO2(g) ΔH°f = – 393 KJmol-1
(iv). Atomization energy: is the energy absorbed when a given molecule or element is converted into gaseous atom.
(v). Lattice energy: Is the energy given out when 1 mole of ionic compound is formed from it ions.
eg. Na+(g) + Cl-(g) → NaCl

b) Calculate the enthalpy of formation of methane given that,
Enthalpy of combustion are;-
i. Carbon 394 KJmol-1.
ii. Hydrogen 286 KJmol-1.
iii. Methane 891 KJmol-1.
Solution
Required equation
C + 2H2 CH4

Data presentation KJ mol -1
i. C + O2 CO2 – 394
ii. H2 + O2 H2O – 286
iii.CH4 + 2O2 CO2 + 2H2O 891

Data manipulation KJmol-1
i. C + O2 CO2 – 394
ii. H2 + O2 H2O – 286) 2
2H2 + O2 2H2O – 572
iii.2H2 + O2 CH4 + 2O2 891
C + 2H2 CH4 – 75
… The enthalpy of formation of methane is -75 KJmol-1.
Example 2
Calculate the standard enthalpy of formation of ethane given that enthalpy of combustion.
i) C = -394 KJmol-1
ii) hydrogen = -216 KJmol-1
iii) Ethane = -1561 KJmol-1

Solution
Required equation
2C + 3H2 C2H6
Given KJmol-1
(C + O2 CO2 -394
H2 + O2 2CO2 – 788
C6H6 + 7/2 O2 2CO2 + 3H2O -1561

Data Manipulation

(C + O2 CO2 -394 )×2
2C + 2O2 2CO2 -788
(H2 + O2 H2O -216) 3
3H2 + O2 3H2O -648
2CO2 + 3H2O CH3 CH3 + O2 1561

2C +3H2 C2H6 + 125
… The standard enthalpy of formation of C2H6 is 125 KJmol-1.

Example 3
a) Define;-
i.Heat of combustion
Is the heat given out when one mole of a substance is completely burnt in given moles of O2 at standard state.
ii. Heat of neutralization
Is the heat energy given out when one mole of water is formed the reaction between acid and base at standard state.
iii.Heat of dilution
Is the heat change when one mole of a compound is dissolved in a given moles of water.

b) Calculate the heat of formation of ethanoic acid , if the enthalpy of combustion are C = -394 H2 = – 284 ethanoic acid -876 ( All in KJ mol-1).
Solution
Required equation
2C + 2H2 +O2 CH3COOH

Data presentation KJ mol-1
C + O2 CO2 -394
H2 + O2 H2O -284
CH3 + 2O2 2CO2 + 2H2O -876

Data manipulation
2C + 2O2 2CO2 – 788
2H2 + O2 2H2O – 568
2CO2 + 2H2O CH3COOH + 2O2 876
The equation obtained
2C + 2H2 + O2 CH3COOH -480
… The heat of formation of ethanoic acid is -480 KJmol-1.

Example 4
From the following Thermodata at 298k.
(i) +
(ii)
(iii)
Calculate for the reaction

Solution
Required equation

Data presentation
+

Data manipulation

+ – 288

For the reaction is




Example 5
Calculate the enthalpy change for the reaction.

Given
Enthalpies of combustions are;-

Heat of formation of water is

Solution
Required equation

Data presentation

Data manipulation

Example 6
Calculate the enthalpy change for the reaction.

Given
Heat of combustion of
Heat of combustion of
Heat of combustion of
Heat of combustion of

Solution
Required equation

Data presentation

Data manipulation

Example 7
(a) Define the following terms;-
(i) Enthalpy of sublimation.
Is the energy absorbed when one mole of solid atom is converted to gaseous atom.
(ii) Enthalpy of atomization.
Is the energy absorbed when a certain molecule or element is covered into gaseous atom.
(iii) Standard enthalpy of combustion.
Is the head given out when one mole of a substance is burnt in a given moles of Oxygen.

(b) The combustion of carbon disulphide is exothermic and the enthalpy of combustion of the compound is 1180 .
Given that carbon dioxide and sulphur dioxide are exothermic compounds with enthalpies of formation of 405 and 293 respectively.
(i) Calculate the heat of formation of carbon disulphide.
(ii) comment on the stability of this compound at various temperature considering the results obtained in the light of Le-chaterlier`s principle.
Solution
Required equation
2S + C
Data presentation
– 1108
C +
S +

Data manipulation

The heat of formation of C is 117
ii) In high temperature the compound will be move stable.

Example 8
Given the following reaction;-

Calculate the enthalpy change for the reaction
Solution
Required manipulation

 

2. CALCULATION OF ENTHALPIES
BASED ON BOND ENERGIES

BOND ENERGIES
Is the energy change which is obtained when one mole covalent bond is formed or broken of an atom.
Any reaction involves bond breaking and bond formation. Reactants bonds are normally broken while products bonds are formed.

Since the bonds energies are known then of the reaction can be calculated as the difference between broken bond energies and formed bond energies.

= B.B.E – F.B.E
Where by;-
Is the heat change of reaction.
B.B.E is the broken bond energies.
F.B.E is the formed bond energies.
Example 1
(a) Define
(i) Bond energy.
Is the energy which is obtained when one mole of covalent bond is formed or broken of an atom.
(ii) Enthalpy of neutralization.
Is the heat given out when one mole of water is formed from the reaction between acid and base at standard state.

(b) Calculate the heat of formation of ethane given that:

Solution
Required equation

Broken bond energies

Example 2
Calculate the enthalpy of hydrogenation of ethane to ethane.
Given

Solution

B.B.E F.B.E

= 348

Total FBE = 2496
348 +
= 2844

Total BBE = 436
612 + 1664
= 2712

= BBE – FBE
= 2712 – 2844
= -132




Example 3
Calculate the enthalpy of hydrogenation of prop -1 – yne to saturated if mean bond energies are: –

Solution
Required equation

BBE

1664

Total BBE =
= 3721

FBE

Total FBE

Then,

Example 4
Benzene under harsh condition undergoes hydrogenation recitation. Calculate the heat of hydrogenation of benzene if mean bond energies are;-
i)

Solution

Therefore,

FBE

Heat of formation

Example 5
Use the data in example 4 above to calculate the enthalpy of hydrogenation of the following compounds.
i) (iii)

ii) (iv)

Solution
i)

BBE

872
BBE

FBE

F B E

BBE

F.B.E

BBE

FBE

HESS`S LAW OF CONSTANT HEAT SUMMATION
It State that,
“The total heat change of a chemical reaction is independent of the route taken”.

BORN HABER CYCLE
Is the cycle which is used to determine the heat of formation of a given compounds.

Example 1
a) Draw the born haber cycle for the formation of the following compounds;-
i) ii)

Solution
(i)

∴ ΔH°f = Es + EI + Eat + Eaff + EL

Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

(ii)

∴ ΔH°f = Es + EI + Eat + Eaff + EL

Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

Example 2
Draw the born Haber cycle for the formation of the following;-
i) Aluminium Chloride ( AlCl3).
Solution

∴ ΔH°f = Es + EI + Eat + Eaff + EL

Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

∴ ΔH°f = Es + EI + Eat + Eaff + EL
Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

(iii) Aluminium oxide (Al2O3).
Solution

∴ ΔH°f = Es + EI + Eat + Eaff + EL

But, EI = 1st EI + 2nd EI + 3rd EI

Eaff = 1st Eaff + 2nd Eaff




Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

iv) Calcium iodide (CaI2).

∴ ΔH°f = Es + EI + Eat + Eaff + EL
Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

v) Potassium Bromide ( KBr).

∴ ΔH°f = Es + EI + Eat + Eaff + EL
Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

Example 3
a) What is “Born Haber cycle” as applied in energetic?

Solution
Born Haber cycle is the cycle which is used to determine the heat of formation of ionic compound which involves intermediate changes.
b) Construct born haber cycle for the formation of KCl.

solution

∴ ΔH°f = Es + EI + Eat + Eaff + EL
Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.
( c ) Using data below calculate the heat of formation of KCl.

∴ ΔH°f = Es + EI + Eat + Eaff + EL

Given that,

∴ ΔH°f = Es + EI + Eat + Eaff + EL

Example 4
(a) Define the following;-
(i) Atomization energy.
Atomization energy is the energy absorbed when a certain molecule or element is converted to gaseous atom.
(ii) Ionization energy.
Ionization energy is the energy required by the gaseous atom to release electron from its outer most shell.
(iii) Electron affinity.
Electron affinity is the energy requires when 1 mole of gaseous atom combine with one mole of electron.
(b) Use the following data and calculate the electron affinity of chlorine.

Solution

∴ ΔH°f = Es + 1st EI + 2 nd EI + Eat + Eaff + EL
Where

From,
∴ ΔH°f = Es + 1st EI + 2 nd EI + Eat + Eaff + EL

Eaff = ΔH°f – (Es + 1st EI + 2 nd EI + Eat + EL )

(d) Account for the differences in energies between;-

Answer
The difference in energies is due the fact that when the first electron is removed, the force of attraction becomes greater in second electron. Finally greater energy is needed to remove it from the atom. The energy is greater so as to overcome the force of attraction between the electron and the nucleus.
Example 5
1. (a) Define the following terms;-
i) Bond dissociation energy.
Bond dissociation energy is the e required to remove electron from the outer must shell of an atom.
ii) Enthalpy of dilution.
Enthalpy of dilution is the heat change when one mole of a compound dissolved in a given moles of water.
iii) Lattice energy.
Lattice energy is the heat given out when one mole of ionic compound is formed from its respective ions.
2. (a) Enthalpy of atomization.
Enthalpy of atomization is the energy absorbed when a certain gaseous molecule or elements converted to gaseous ions.
(b) State Hess`s law of constant heat summation.
It state that,
“If the reaction can take place in more than on rout, then the overall heat change is the same for which ever rout may be taken”.
(c) (i) Draw a well labeled born Haber cycle for the reaction of formation of (Al2O3 ).

∴ ΔH°f = Es + EI + Eat + Eaff + EL
Where by,
Es is the sublimation energy.
EI is the Ionization energy.
Eat is the Atomization energy.
Eaff is the Electronic affinity energy.
EL is the lattice energy.

(ii) Calculate the heat of formation of (Al2O3) from the following;-

First electron affinity of oxygen is –
Second electron affinity of oxygen is
Lattice energy of is
Atomization energy of oxygen is

∴ ΔH°f = Es + EI + Eat + Eaff + EL

Where,
Es = 560

(d) Draw a born cycle for the formation of Al2Cl6 and calculate ΔH for the process.
2Al +3Cl3 → Al2Cl6 ΔH = ?
Given that,
i)

∴ ΔH°f = Es + EI + Eat + Eaff + EL
From the data,

Data manipulation




CALCULATIONS OF ENTHALPIES BY USING ATOMIZATION DATA
During the reaction reactants are atomization (changed to atoms) while the product energy. Calculate from bond energy.

Example1

Example 2
Calculate the following of formation of CH4 given that,
Enthalpy of atomization of carbon =
Enthalpy of atomization of hydrogen

Solution

 

Example 3
Use the information provided in the above example. Then calculate the heat of formation of ethane.
Solution

F.B.E

Example 4
The energy required to atomize 6g of carbon is 357.5 KJ. Calculate the energy required to atomize hydrogen atom. If the heat of formation of methane is given that,

Solution

CALCULATION OF ENTHALPIES BY CALORIMETRY
Calorimetry Is the method determined the heat change reaction by the using calorimeter.
Example of enthalpies that can be determined by calorimetry method;-
i) Enthalpy of hydrogen.
ii) Enthalpy of neutralization.

DETERMINATION OF ENTHALPIES OF NEUTRALIZATION BY CALORIMETRY
What is neutralization reaction?
Neutralization reaction is the reaction between acid and base to give salt and water.
Example

In this method the density of salt solution formed is assumed to be equal to the density of water.

Quantity of that in this method can be calculate by using heat capacity ( C )and by using specific heat capacity ( c ) .
Let the quantity of heat be Q.
By using heat capacity

C = heat capacity.
By using specific heat capacity.

Where by is change in temperature.
c = specific heat capacity.
If quantity of heat calculated is the same and is also the same, then the relationship between c and C can be,

Enthalpy of neutralization is calculate with respect to the number of moles of water producer.
Reaction.

But number of moles of water depends on the moles of limiting reagent




QUESTION
What is limiting reagent?
Limiting reagent is the reactant compound in the neutralization reaction which have small number of moles.

Example 1
250c of NaOH of 0.4M were added to 250cc of the HCl of 0.4M in a calorimeter. The temperature of the two solutions and the calorimeter was . The mass of calorimeter was 50g and its specific heat capacity was after the reaction the temperature rose to 19.5
Assuming the specific heat capacity of all the solution is . Calculate the standard enthalpy of neutralization.
Solution
Data

Reaction

Limiting reagent
Base

Acid
The same 0.1 mol
Both are limiting reagent.
Moles of water produced

.

Example 2
Define
i) Heat of neutralization.
Is the heat given out when one mole of water is formed from the reaction between base and acid at standard state.
ii) Heat of Ionization.
Is the heat energy required to remove an electron from the outer most shell of a gaseous atom.
iii) Heat of sublimation.
Is the heating absorbed when one mole solid atom is converted to gaseous atom.
iv) Limiting reagent.
Is the reacted compound in the neutralization reaction which have small number of moles.
b) State Hess`s Law of heat summation.
It state’s that,
“If a reaction take place by more than are route the overall heat change is the same for which ever route may be taken”.

Question
For neutralization. The reading was absorbed to rose temperature of both the calorimeter and the solution by 3.4k. Calculate the standard enthalpy of neutralization of . Given that capacity of calorimeter is and specific heat capacity of the solution was 4.2 .

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