ENTHALPY (H)
Enthalpy is defined as total heat content of a system and is equal to the sum of internal energy and pressure volume work.
Mathematically
H = U + W
H = U + PV
It is state function
At constant Pressure, we know of First law of Thermodynamics
ΔU= q
p +W
Or
q
p = ΔU – W = ΔU + P(V
2 – V
1) = U
2 - U
1 + PV
2 -PV
1
= (U
2 + PV
2) – (U
1 + PV
1 ) = ΔH
So, Change is Enthalpy is the heat absorbed or evolved by the system at constant pressure.
ΔH = +ve (Endothermic reaction)
ΔH = -ve (Exothermic process)
Relation between ΔH and ΔU
H
1 = U
1 + PV
1
H
2 = U
2 + PV
2
H
2 – H
1 = U
2 + PV
2 – (U
1 + PV
1)
ΔH = U
2 – U
1 + PV
2 – PV
1
=U
2 – U
1 + P(V
2 – V
1)
= ΔU + PΔV
Now Let us consider a reaction involving gases. If V
1 is the total volume of the gaseous reactants, V
2 is the total volume of the gaseous products, n
1 is the number of moles of gaseous reactants and n
2 is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we
write,
pV
1 = n
1RT
pV
2 = n
2RT
pΔV =Δn RT
So, Enthalpy of reaction can be written as
ΔH =ΔU + ΔngRT
Heat Capacity
It is the increase in temperature (1 degree) of any substance.
So, $C =\frac {Q}{\Delta T}$
Molar Heat Capacity
It is the heat capacity for 1 mole of a substance
$C =\frac {Q}{n\Delta T}$
Specific Heat Capacity
It is defined as the heat required to raise the temperature of unit mass of a substance by 1 degree or Kelvin.
$C =\frac {Q}{m\Delta T}$
Relation between Cp & Cv
At constant volume Molar heat capacity is C
v. At constant pressure, Molar heat capacity is C
p.
Now At constant volume q
v =ΔU= C
v ΔT
Now At constant pressure q
p = C
p. ΔT=ΔH
ΔH = ΔU + PΔV
C
pΔT = C
v ΔT+ PΔV
C
pΔT = C
v ΔT+ R ΔT
C
p – C
v = R
BOMB CALORIMETER
Working
The bomb colorimeter consists of a strong vessel called bomb which can with stand at very high pressure. It is surrounded by water both to ensure that no heat is lost to the surrounding.
Procedure
A known mass of combustible substance is burnt in presence of pure O
2 in the steel bomb. Heat evolved during the reaction is transferred to the water and its temperature is monitored during the process, volume remains constant & ΔV = 0
At constant volume, q
v =ΔU
Δ T = T
final – T
initial
CHANGE IN ENTHALPY IN REACTION(Δ H)
In a chemical reaction, reactants are converted into product .The enthalpy change in the chemical reaction is called reaction enthalpy
ΔH = H
product – H
reactants
H
product -> Sum of enthalpy's of the products
H
reactants -> Sum of enthalpy's of the reactants
STANDARD ENTHALPY OF A REACTION [ H*]
When reaction takes place at standard state of a substance, the enthalpy of reaction is known as Δ H
o
ENTHALPY OF FUSION
It is the heat energy or change in Enthalpy (Δ H ) when one mole of a solid at its melting point is converted into a liquid state.
Example
H
2O (s) -> H
2O (l)
Δ
fusH
o = + 6 K joule/mol
Melting of a solid is endothermic, so all enthalpies of fusion are positive
ENTHALPY OF VAPOURISATION
It is Δ H when one mole of liquid substance at its boiling point changes to gas.
Eg:- H2O (l) -> H2O (g)
Δ
vapH
o =40.8 Kj/mol
ENTHALPY OF SUBLIMATION
It is ΔH when one mole of solid substance changes to vapour.
Eg:- I2 (s) -> I2 (g)
Δ
subH
o = 62.4 Kj/mol
STANDRAD ENTHALPY OF FORMATION
It is ΔH for the formation of one mole of a substance from its constituting elements under standard condition of temperature & pressure.
The reference state of an element is its most stable state of aggregation at 25°C and 1 bar pressure. For example, the reference state of Dihydrogen
is H
2 gas and those of dioxygen, carbon and sulphur are O
2 gas, C
graphiteand S
rhombic
respectively
C + 2H
2 -> CH
4
Δ
f H
o = - 74.8 Kj/mol
Important notes
a) It is important to understand that a standard molar enthalpy of formation, Δ
f H
o is just a special case of Δ r
H
o, where one mole of a compound is formed from its constituent elements
b) Also standard enthalpy for formation, Δ
f H
o , of an element in reference state, i.e., its most stable state of aggregation is taken as zero
c) Standard enthalpy of formation can be used to calculate the enthalpy change for the reaction. The following general equation can be used for the enthalpy change calculation
Δ
r H
o =∑ aΔ
f H
o (product) - ∑ bΔ
f H
o (reactant)
where a and b represent the coefficients of the products and reactants in the balanced equation
ENTHALPY OF COMBUSTION
It is ΔH for the combustion of 1 mole of a substance in presence of air.
Example
CH
4 + 2O
2 -> CO
2 + 2H
2O
Δ
c H
o = -390.3 Kj/mol
ENTHALPY OF SOLUTION
It is ΔH for dissolving one mole of a substance in given amount of solvent
NaCl (s) + H
2O -> NaCl (q)
Δ
s H
o = +5.35 Kj/mol
ENTHALPY OF HYDRATION
It is ΔH when one mole of anhydrous salt changes to hydrated salt.
CuSO
4 + 5H
2O -> CuSO
4 . 5 H
2O
Δ
h H
o = 78.2 Kj/mol
ENTHALPY OF ATOMISATION
It is ΔH for breaking one mole of a substance completely to obtain isolated atoms is gas phase.
Eg: H
2 (g) -> 2H (g)
Solved Examples
Question 1
The Enthalpy change (Δ H) for the reaction;
N2 + 3H2 -> 2NH3
is -92.38 KJ at 298 Kelvin. What is ΔU
Solution
So number of moles of Product=4
Number of Moles of reactant =2
Therefore Δn=-2
Δ H =ΔU + ΔnRT
ΔU=Δ H - ΔnRT
=-.92.38X1000 + 2 X 8.314 X 298
=-87424.856 J
Question 2
Enthalpy of combustion of cyclohexane (C6 H12) is 3920 KJ/mol at 25o C Calculate its heat of combustion at constant volume.
Solution
Δc Ho = -3920 Kj/mol
T = 25 + 273 = 298K
Chemical Equation
C6 H12+ 9O2 -> 6 CO2 + 6H2O
So number of moles of Product=10
Number of Moles of reactant =12
Therefore Δn=2
Δ H =ΔU + ΔnRT
ΔU=Δ H - ΔnRT
=-3924955.144 J
Question 3
Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100°C and 1 bar pressure is converted to ice at 0°C.
Given Enthalpy of fusion of ice is 6.00 kJ/ mol and heat capacity of water is 4.2 J/g°C
Solution
The change take place as follows:
Step - 1 1 mol H2O (l, 100°C) to 1 mole H2O (l, 00 C)
change ΔH1
Step - 2 1 mol H2O (l, 0°C) to 1 mole of ice
H2O( S, 0°C) Enthalpy
change ΔH2
Total enthalpy change will be -
ΔH = ΔH1 + ΔH2
ΔH1 = - (18 x 4.2 x 100) J mol-1
= - 7560 J mol-1 = - 7.56 k J mol-1
ΔH2 = - 6.00 kJ mol-1
Therefore,
ΔH = - 7.56 kJ mol-1 + (-6.00 kJ mol-1)
= -13.56 kJ mol-1
There is negligible change in the volume uring the change form liquid to solid tate.
Therefore, pΔv = Δng RT = 0
ΔH = ΔU = - 13.56kJ mol-1
BOND ENTHALPY
[ ΔH = ΔH
reactant – ΔH
product]
Bond dissociation enthalpy is the change in Enthalpy when one mole of covalent bond of a gaseous covalent compound is broken to form product in gas phase.
Example Cl
2 -> 2Cl
H
Cl – Cl = 2.42 Kj/mol
Standard enthalpy of reaction is related to bond enthalpy as
Δ
r H
o =∑ bond enthalpies (product) - ∑ bond enthalpies (reactant)
Solved Example
Question 1
Calculate ΔH of the reaction:-
CH2 Cl2 (g) -> C (s) + 2H (g) + 2Cl (g)
Bond energies for C – H is 415 Kj
Bond energies for C – Cl is 326 Kj
Solution
ΔH = Hr – Hp
=2(HC-H +HC-CL) -0
=1482 KJ
Question 2
Give the ΔH for the below reaction
CH2 = CH2 + H2 -> CH2 – CH3
C- H à 99K cal; C – C à 83 K cal
C = C à 147 K cal; H – H à 104 K cal
A. H = Hr – Hp
HESS LAW {LAW OF CONSTANT HEAT SUBMISSION}
The total amount of heat evolved or absorbed in a reaction is same whether the reaction takes place in one step or more than one.
ΔH=ΔH
1 + ΔH
2 +ΔH
3
In general, if enthalpy of an overall reaction along one route is Δ
rH and Δ
rH1, Δ
rH2,
rH3..... representing enthalpies of reactions
leading to same product, B along another route then we have
Δ
rH = Δ
rH1 + Δ
rH2 + Δ
rH3 .
Question 1
The combustion of 1 mole of benzene takes place at 298 K an atm. After combustion CO2 and water are formed and 3267 Kj heat to liberate. Calculate enthalpy of formation of benzene. Standard enthalpy of formation of CO2 & H2O is – 393.5 Kj/mol and -285.83 Kj/mol respectively
Solution
Given
6C + 3H2 -> C6 H6 ΔfH o ?
C6 H6 + 15/2 O2 -> 6CO2 + 3H2 O Δr Ho = -3267 KJ
C(graphite) + O2 -> CO2 Δr Ho =– 393.5 KJ
H2 + ½ O2 -> 1 H2O Δr Ho =-285.83 KJ
Now
C(graphite) + O2 -> CO2 Δr Ho =– 393.5 KJ
can be written as
6C(graphite) + 6O2 -> 6CO2 Δr Ho =– 2361 KJ
Similarly
H2 + ½ O2 -> 1 H2O Δr Ho =-285.83 KJ
can be written as
3 H2 + 3/2 O2 -> 3H2O Δr Ho =-857.49 KJ
Summing above CO2 and H2O formation equation we get
6C(graphite) + 3 H2 + 15/2 O2 -> 6CO2 + 3H2O Δr Ho =-3218.49 KJ
Now for
C6 H6 + 15/2 O2 -> 6CO2 + 3H2 O Δr Ho =-3267 KJ
Reversing it
6CO2 + 3H2 O -> C6 H6 + 15/2 O2 Δr Ho =3267 KJ
Adding both the reverse and combing CO2 and H2O equation ,we get
6C + 3H2 -> C6 H6 ΔfH o = 48.49 KJ/mole
Question 2
calculate Enthalpy of formation of Ethanol using following data.
i. C2 H5 OH + 3O2 -> 2CO2 + 3H2O Δr Ho = -1380.7 Kj/mol
(ii) C + O2 -> CO2 Δr Ho = -394.5 Kj/mol
(iii) H2 + ½ O2 -> 1 H2O Δr Ho =-285.83 KJ
A. 2C + 3H2 + ½O2 -> C2 H5 OH Δr Ho = ?
Solution
C + O2 -> CO2 Δr Ho = -394.5 Kj/mol
So,
2C + 2O2 -> 2CO2 Δr Ho = -789 KJ /mol
H2 + ½ O2 -> 1 H2O Δr Ho =-285.83 KJ/mol
So,
3H2 + 3/2O2 -> 3H2O Δr Ho =-857.49 KJ/mol
Summing above CO2 and H2O formation equation we get
2 C + 3 H2 + 7/2 O2 -> 2CO2 + 3H2O Δr Ho =-1646.49 KJ
Now
C2 H5 OH + 3O2 -> 2CO2 + 3H2O Δr Ho = -1380.7 Kj/mol
Reversing this
2CO2 + 3H2O -> C2 H5 OH + 3O2 Δr Ho =1380.7 Kj/mol
Adding both these we get
2C + 3H2 + ½O2 -> C2 H5 OH Δr Ho = -265.49 KJ/mol
Question 3
The Enthalpy of combustion of Ethane, Graphite, Dihydrogen are – 890.3, 393.5 and -285.8 Kj/mol. Give Enthalpy of formation of Methane CH4
Solution
CH4 + 2O2 -> CO2 + 2H2O Δr Ho =-890.3 KJ/mol
C + O2 -> CO2 Δr Ho =393.5 KJ/mol
H2 + ½ O2 -> H2O Δr Ho =-258.8 KJ/mol
C+2H2 -> CH4
ΔH = 2(-258.8) -393.5 + 890.3
=20.8KJ/mol
BORN HABER CYCLE
lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its
ions in gaseous state
It is not possible to determine Lattice energy of ionic compound by direct experiment.So we construct a enthalpy diagram called Born Haber cycle
One Route
1) Na (s) -> Na(g) [Sublimation]
2) Na(g) --> Na
+ +e [ionization]
3) (1/2) Cl
2 (g) -> Cl (g) [dissociation]
4) Cl(g) +e -> Cl
- [Ionization]
5) Na
+ (g) + Cl
-(g) -> NaCl(s)
Another Route
Na(s) + (1/2) CL
2 (g) -> NaCl(s)
As per Hess,law
Δ
lattice H
o
= 411.2 + 108.4 + 121 + 496 – 348.6 =788 KJ
link to this page by copying the following text
