Enthalpy , Change in enthalpy in reaction And Hess's Law

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(V₂ – V₁) = U₂ - U₁ +   PV₂ -PV₁
= (U₂ + PV₂) – (U₁ + PV₁ ) = 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
H₁ = U₁ + PV₁
H₂ = U₂ + PV₂
H₂– H₁ = U₂+ PV₂ – (U₁ + PV₁)
ΔH = U₂ – U₁ + PV₂ – PV₁
=U₂– U₁ + P(V₂– V₁)
= ΔU + PΔV
Now Let us consider a reaction involving gases. If V₁ is the total volume of the gaseous reactants, V₂ is the total volume of the gaseous products, n₁ is the number of moles of gaseous reactants and n₂ is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we
write,
pV₁ = n₁RT
pV₂ = n₂RT
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 C_p & C_v
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 O₂ 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
Enthalpy is Exothermic and Endothermic reactions

Enthalpy is Exothermic and Endothermic reactions

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
H₂O (s)  ->      H₂O (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 H₂ gas and those of dioxygen, carbon and sulphur are O₂ gas, C_graphiteand S_rhombic
respectively
    C + 2H₂         ->        CH₄
    Δ _fH^o = - 74.8 Kj/mol
Important notes
a) It is important to understand that a standard molar enthalpy of formation, Δ _fH^o is just a special case of Δ rH^o, where one mole of a compound is formed from its constituent elements
b) Also standard enthalpy for formation, Δ _fH^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 _fH^o (product) - ∑ bΔ _fH^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
CH₄+ 2O₂ ->    CO₂+ 2H₂O
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) + H₂O -> 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.
    CuSO₄+ 5H₂O        ->      CuSO₄ . 5 H₂O
    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:        H₂ (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 25^o C Calculate its heat of combustion at constant volume.
Solution
     Δ_c H^o = -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 Cl₂ -> 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 = H_r – H_p
=2(H_C-H +H_C-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 = H_r – H_p

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=ΔH₁ + ΔH₂ +ΔH₃
In general, if enthalpy of an overall reaction along one route is ΔrH and ΔrH1, ΔrH2, rH3..... representing enthalpies of reactions
leading to same product, B along another route then we have
ΔrH = ΔrH1 + ΔrH2 + ΔrH3 .
Hess law

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                    Δ_fH ^o ?
C₆ H₆ +    15/2 O₂ ->         6CO₂ + 3H₂ O       Δ_rH^o = -3267 KJ
  C(graphite) + O₂ ->           CO₂                  Δ_rH^o =– 393.5 KJ
H₂ + ½ O₂    ->           1 H₂O                             Δ_rH^o =-285.83 KJ
Now
C(graphite) + O₂ ->           CO₂                  Δ_rH^o =– 393.5 KJ
can be written as
6C(graphite) + 6O₂ ->           6CO₂                  Δ_rH^o =– 2361 KJ
Similarly
H₂ + ½ O₂    ->           1 H₂O                             Δ_rH^o =-285.83 KJ
can be written as
3 H₂ + 3/2O2 ->           3H₂O                             Δ_rH^o =-857.49 KJ
Summing above CO₂ and H₂O formation equation we get
6C(graphite) + 3 H₂ + 15/2 O₂ -> 6CO₂ + 3H₂O        Δ_rH^o =-3218.49 KJ
Now for
C₆ H₆ +    15/2 O₂ ->         6CO₂ + 3H₂ O       Δ_rH^o =-3267 KJ
Reversing it
6CO₂ + 3H₂ O -> C₆ H₆ +    15/2 O₂    Δ_rH^o =3267 KJ
Adding both the reverse and combing CO₂ and H₂O equation ,we get
6C + 3H₂    ->          C₆ H₆                    Δ_fH ^o = 48.49 KJ/mole

Question 2
calculate Enthalpy of formation of Ethanol using following data.
i.      C₂ H₅ OH + 3O₂        ->         2CO₂ + 3H₂O                   Δ_rH^o = -1380.7 Kj/mol
(ii)     C + O₂    ->  CO₂                         Δ_rH^o = -394.5 Kj/mol
(iii)     H₂ + ½ O₂    ->           1 H₂O                             Δ_rH^o =-285.83 KJ
A.       2C + 3H₂ + ½O₂         ->        C₂ H₅ OH                Δ_rH^o = ?
Solution
C + O₂    ->  CO₂                         Δ_rH^o = -394.5 Kj/mol
So,
2C + 2O₂    ->  2CO₂                         Δ_rH^o = -789 KJ /mol
H₂ + ½ O₂    ->           1 H₂O                             Δ_rH^o =-285.83 KJ/mol
So,
3H₂ + 3/2O₂    ->           3H₂O                             Δ_rH^o =-857.49 KJ/mol
Summing above CO₂ and H₂O formation equation we get
2 C + 3 H₂ + 7/2 O₂ -> 2CO₂ + 3H₂O        Δ_rH^o =-1646.49 KJ
Now
C₂ H₅ OH + 3O₂        ->         2CO₂ + 3H₂O                   Δ_rH^o = -1380.7 Kj/mol
Reversing this
2CO₂ + 3H₂O                   -> C₂ H₅ OH + 3O₂        Δ_rH^o =1380.7 Kj/mol
Adding both these we get
2C + 3H₂ + ½O₂         ->        C₂ H₅ OH                Δ_rH^o = -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   Δ_rH^o =-890.3 KJ/mol
       C + O₂    ->              CO₂   Δ_rH^o =393.5 KJ/mol
      H₂   + ½ O2    ->       H2O   Δ_rH^o =-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
Born Haber cycle for calculating lattice enthalpy

One Route
1) Na (s) -> Na(g) [Sublimation]
2) Na(g) --> Na⁺ +e [ionization]
3) (1/2) Cl₂ (g) -> Cl (g) [dissociation]
4) Cl(g) +e -> Cl^- [Ionization]
5) Na⁺ (g) + Cl^-(g) -> NaCl(s)
Another Route
Na(s) + (1/2) CL₂ (g) -> NaCl(s)
As per Hess,law
Δ_latticeH^o
= 411.2 + 108.4 + 121 + 496 – 348.6 =788 KJ

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Notes

Enthalpy , Change in enthalpy in reaction And Hess's Law

ENTHALPY (H)

BOMB CALORIMETER

CHANGE IN ENTHALPY IN REACTION(Δ H)

STANDARD ENTHALPY OF A REACTION [ **H*]**

BOND ENTHALPY

HESS LAW {LAW OF CONSTANT HEAT SUBMISSION}

BORN HABER CYCLE

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Enthalpy , Change in enthalpy in reaction And Hess's Law

ENTHALPY (H)

BOMB CALORIMETER

CHANGE IN ENTHALPY IN REACTION(Δ H)

STANDARD ENTHALPY OF A REACTION [ H*]

BOND ENTHALPY

HESS LAW {LAW OF CONSTANT HEAT SUBMISSION}

BORN HABER CYCLE

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Other Links

Legal

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STANDARD ENTHALPY OF A REACTION [ **H*]**