Thermochemistry | Physical Chemistry for NEET PDF Download

Thermochemistry
Thermochemistry is the branch of physical chemistry which deals with the thermal or heat changes caused by chemical reactions.

Specific heat (s)
Amount of energy required to raise the temp by 1ºC of 1 gm of a substance.
Unit → J/kg-K

Heat Capacity (ms)
The amount of heat required to raise the temperature by 1ºC or 1K of a given amount of a substance.
Units → J/K
Total heat given to increase the temperature by Δt.
q = msΔt

Molar Heat Capacity (C_m)
The amount of heat required to raise the temp by 1º of 1 mole of a substance.

Classification of Molar heat Capacity
(a) Molar heat capacity at constant pressure (C_p,m)
(b) Molar heat capacity at constant volume (C_v,m)

Relation between C_p and C_v
C_p - C_v = R (Mayor's formula)

Rules for Thermochemical Equation
(1) It is necessary to mention physical state of all reactants and products.

(2) A → B, ΔH = H_B - H_A
If A → B x kJ/mole
ΔH < 0 → (Exothermic reaction)
ΔH = -x kJ/mole
If A x kJ/mole → B
ΔH > 0 → (Endothermic reaction)
ΔH = + x kJ/mole

(3) After reversing a thermochemical eqⁿ then sign of enthalpy also get changed.
Example:
A_(g) + B_(g) → C_(g) + D_(g) , ΔH = x kJ
C_(g) + D_(g) → A_(g)+ B_(g) , ΔH = - x kJ

(4) When two reactions are added their enthalpies are also get added with their sign.
Example:
A_(g) + B_(g) → C_(g)  D_(g) ΔH = x₁ kJ
C_(g) + E_(g) → F_(g) ΔH = - x₂ kJ
A_(g) + B_(g) + E_(g) → D_(g) + F_(g) , ΔH = x₁- x₂

(5) If a thermochemical equation is multiplied by a number then DH is multiplied by the same number.
Example:
A_(g) + B_(g) → C_(g) + D_(g) ΔH = x₁ kJ
2A_(g) + 2B_(g) → 2C_(g) + 2D_(g) ΔH = 2x₁ kJ

Intensive Property
The property which does not depend upon the mass of substance is called intensive property.
Example: density, refractive index, specific heat, etc.

Extensive Property
Mass dependent properties are called extensive properties
Example:
ΔH, ΔS, ΔG, V, U, Resistance, Number of moles etc.
* Two extensive property can be added
* Ratio of two extensive properties is Intensive.
* Intensive properties can not be added directly.
Example: We can not add the density of two liquids to get the density of the final mixture of the two.

Enthalpy (H)

ΔH = ΔU +Δ(PV) = ΔU (P₂V₂ - P₁V₁)
(i) At constant Pressure, ΔH = ΔU + PΔV
(ii) At Constant Volume, ΔH = ΔU + VΔP

• Enthalpy of ideal gas is function of temperature only.
  H = U +PV
  = U +nRT
• Enthalpy is always defined at constant temperature and it varies with variation in temperature.

For ideal gas, ΔH = ΔU + Δ(PV)
= ΔnC_VT +ΔnRT = ΔnT(C_V + R)
→ ΔH = Δn C_pT

Where Δng = no. of moles of gaseous product - no. of moles of gaseous reactant.

Kirchoff's Equation
This gives the relation between enthalpy and temperature.

• Physical state is changed at constant temperature.

According to Hess'Law
ΔH₂  + Cp(T₂-T₁) = ΔH₁ + Cp¹ (T₂- T₁)
ΔH₂ - ΔH₁ = (Cp¹ - Cp) (T₂- T₁)
= ΔCp (T₂- T₁)

Where
ΔCp = Molar heat capacity of product -- Molar heat capacity of reactant
Example:
N₂(g) + 3H₂(g) ⇒ 2NH₃(g)
ΔCp = 2Cp(NH₃) - Cp(N₂) - 3Cp(H₂)

• If the above formula (Kirchoff's eqⁿ) is to be written for molar heat capacity at constant volume then

•  If ΔCp is function of temperature
  ΔCp = T² + T
  Then
   

Heat of formation
Enthalpy change during the formation of 1 mole of a compound form its most stable common occurring form (also called reference states) of elements is called heat of formation.
C_(s)  + O_2(g) ⇒ CO_2(g)
ΔH = ΔH_f(CO₂)
CO_(g)  + 1/2O_2(g) ⇒ CO_2(g)
ΔH ≠ ΔH_f CO₂(g)
(because CO₂ has not been formed form its element in their most stable form)
Similarly

Heat of reaction

• All metal excpet Hg exist in solid form (reference states)

Enthalpy At Standard State :-(ΔHº)
T = 25ºC = 298 K
P = 1 atm
Conc = 1M
ΔHº = Heat of formation at standard state
If A(g) + B_(g) ⇒ C_(g +D_(g) is any reaction, Heat of reaction for any thermochemical equation can be written as

ΔHº = DH_fº(product) - DH_fº(Reactant)
If we use the above concept for the above given reaction then
ΔHº = DH_fº_(c) + ΔH_fº_(D) - ΔH_fº_(A) + ΔH_fº_(B)

Assumption :-
The heat of formation of most stable form of an element is taken as zero.
C_(graphite)  + O_2(g) ⇒ CO_2(g)
ΔHº = ΔH_fº(CO₂) - ΔH_fºC_(s) - ΔH_fº(O₂)(g)
⇒ ΔHº = ΔH_fº(CO₂) (As ΔH_fºC_(s) = 0 and Δ H_fºO₂(g) = 0)
Another example can be taken as
H₂(g) + 1/2O₂(g) ⇒ H₂O_(l)

Example 1. From the following data,
C_{(s, graphite)}  O_2(g) ⇒ CO_2(g)   DHº = -393.5 kJ/mole
H₂(g) + 1/2O₂(g) ⇒ H₂O(l)   DHº = -286 kJ/mole
2C₂H₆(g) + 7O₂(g) ⇒ 4CO_2(g)  6H₂O(l) ΔHº = -3120 kJ/mole
Calculate the standard enthalpy of formation of C₂H₆(g) (in kJ/mole)
Solution. From eqⁿ(1) ΔH_fº(CO₂) = -393.5 kJ/mole
From eqⁿ(2) ΔH_fº(H₂O) = -286 kJ/mole
From eqⁿ (3) Δ_rHº = 4 ΔH_fº(CO₂) + 6ΔH_fº(H₂O) -2 ×ΔH_fº(C₂H₆) - 7×ΔH_fºO₂(g) - 3120
= 4 ×(-393.5) + 6×(-286) - 2×ΔH_fº(C₂H₆) (As ΔH_fº O₂(g) = 0)
⇒ -3120 = -1574 -1716 - 2 ×ΔH_fº(C₂H₆)
⇒ -3120 + 3290 = - 2 ×ΔH_fº(C₂H₆)
⇒ 170 = - 2 ×ΔH_fº(C₂H₆)
⇒ ΔH_fº = - 85 kJ/mole

Heat of Combustion
It is the enthalpy change (always -ve) when One mole of the substance undergo complete combustion.
C_(s)  + O₂ ⇒ CO₂(g)
ΔHº = ΔH_CºC(s) = ΔH_fº(CO₂) - ΔH_fºC(s) - ΔH_fº(O₂)
ΔHº = = ΔH_C ºC(s) = ΔH_fº(CO₂)
Other example

Note:
Heat of combustion is always exothermic

•  N₂ + O₂ ⇒ 2NO

(It is an endothermic reaction)

• O₂  + F₂ ⇒ OF₂

(Since O has normally tendency to accept electron and opposite is happening above hence reaction is is considered endothermic)

•  Heat of reaction for any thermochemical equation can be written as (in form of heat of combustion)

ΔH_rº = Heat of combustion of reactant - Heat of combustion of reactant.
ΔH_rº = ΔH_Cº (Reactant) - ΔH_Cº (Product)

Example 2. The enthalpy change for the reaction
C₃H₈(g) + H₂(g) ⇒ C₂H₆(g) + CH₄(g) at 25ºC is - 55.7 kJ/mole calculate the enthalpy of combustion of C₂H_6(g). The enthalpy of combustion of H₂, and CH₄ are - 285.8 and - 890.0 kJ/mole respectively. Enthalpy of combustion of propane is -2220 KJmol^-1.
Solution. As we know any thermochemical eqⁿ can be written in terms of heat of combustion as follows
ΔH_rº = ΔH_cº (Reactant) - ΔH_Cº (Product)
ΔH_rº = ΔH_cº (C₃H₈) +ΔH_Cº (H₂) - {ΔH_cº(C₂H₆)+ ΔH_Cº(CH₄)}
- 55.7 = (-2220 - 285.8) - {-890 +ΔH_cº(C₂H₆)}
⇒ ΔH_cº(C₂H₆) (g) = - 1560.1 kJmol^-1

Problems Based on Both HOC and HOF:
Example 3. At 300K, the standard enthalpies of formation of C₆H₅COOH_(s), CO₂(g) and H₂O(l) are -408, -393 and -286 kJmol^-1 respectively. Calculate the enthalpy of combustion of benzoic acid at
(i) constant pressure
(ii) constant volume.
Solution. 7C(s)  + 3H_2(g)+ O_2(g) ⇒ C₆H₅COOH_(s)  ΔHº = -408 kJ
⇒ ΔH_fº(C₆H₅COOH) = - 408 kJ
C_(s)  + O_2(g) ⇒ CO₂(g) ΔHº = -393 kJ
⇒ ΔH_fº(CO₂) = -393 kJ
H_2(g)  + 1/2O_2(g) ⇒ H₂O(l) ΔHº = -286 kJ
⇒ ΔH_fº(H₂O) = -286 kJ
C₆H₅COOH_(s)  + 15/2O_2(g) ⇒ 7CO_2(g) + 3H₂O(l)
⇒ ΔH_Cº(C₆H₅COOH) = 7 ΔH_fºCO₂ + 3ΔH_fº(H₂O) - 4ΔH_fº(C₆H₅COOH)
= 7 ×(-393) +3× (-286)- 4(408)
ΔH_Cº(C₆H₅COOH) = - 3609 408
= -3201 kJ/mol
⇒ enthalpy of combustion at constant pressure = - 3201 kJ mol^-1
Also
ΔH = ΔU  + ΔngRT
-3201 = ΔU + (-0.5) × 8.31 × 10^-3 × 300
(As Δn = 0.5, R = 8.314 × 10^-3 kJ)
⇒ ΔU = - 3201 + 1.2471
ΔU = -3199.7529
⇒ enthalpy of combustion at constant volume = - 3199.7529

Bond Energy
It is defined for gaseous molecules. "The enthalpy change during the breaking of one mole of bond into isolated gaseous atoms is called bond energy of the compound"
Example:
H₂(g) ⇒ 2H(g) , ΔHº = ∑H-H
ΔHº = ∑H-H = 2ΔH_fºH(g) - ΔH_fº(H₂)

Similarly

• Let us consider the similar bond breaking In CH₄
  CH₄(g) ⇒ C(g)  +  4 H_(g)
  ΔH = 4 ∑C-H =ΔH_fºC(g) 4ΔH_fºH(g) - ΔH_fº(CH₄)
• Enthalpy of reaction in terms of bond energy for a thermochemical eqⁿ can be written as
  ΔH_fº = B.E._(Reactant) - B.E. _(products)

Example 4. Using the bond enthalpy data given below, calculate the enthalpy change for the reaction
C₂H₄ (g) + H₂(g) ⇒ C₂H₆(g)
Data :

Solution.
Δ_rHº = Σreactant - Σproduct
ΔH_rº = ΣC = C+4 ΣC-H+ΣH-H -ΣC-C -6ΣC -H
ΔH_rº = 606.68 4×410.87 431.79 -336.81 -6×410.87 = 2681.95 -2802.03
ΔH_rº = -120.08 kJ/mol

Heat of hydrogenation
"Enthalpy change during the addition of 1 mole of H₂ to an unsaturated compound. is called heat of hydrogentation."
Hydrogenation is an exothermic process. of therefore heat of hytdrogentaiton is always -ve.

Example:
CH₂ = CH₂ + H₂ ⇒ CH₃ -CH₃
ΔH_rº = ΔHº (Hydrogenation of CH₂ = CH₂)

 
ΔH_rº = ΔHº _{(Hydrogenation)} of cyclohexene

Example 5. Compare the heat of hydrogenation of the following alkene

Solution. 

The above concept is true as long as no. of double bonds are equal as heat of hydrogenation is defined for per mole of double bond. It will be certainly larger for higher number of double bonds irrespective of their stability.

As we know trans > cis (stability)
⇒ Heat of hydrogen (trans) < Heat of hydrogenation (cis)

of therefore decreasing order of heat of hydrogenation of alkene 5 > 1 > 2 > 3 > 4 >

Example 6. Find DH_f of HCl(g) if bond energies of H₂, Cl₂ and HCl are 104, 58, 103 kcal/mole respectively. 
Solution. H - Cl ⇒ H(g)  + Cl(g)

∑H-Cl= ΔH_fºH(g) + H_fºCl(g) - ΔH_fºHCl(g)

⇒ ΔH_fºHCl(g) = - 22 kcal/mol.

Heat of Atomisation
" When one mole of any substance is converted into gaseous atoms enthalpy change during the process is called heat of atomisation." It is always +ve.

Heat of sublimation
Enthalpy change during the conversion of one mole of solid to 1 mole of gaseous phase directly without undergoing into liquid phase is called enthalpy of sublimation or heat of sublimation,
It is always +ve due to endothermic nature of the process.

Example:
C_(s) ⇒ C_(g)
ΔHº = ΔH_subºC(s) = ΔH_fº C(g) - ΔH_fº C(s)
⇒ ΔH_subºC(s) = ΔH_fº C(g)
⇒ ΔH_subºC(s) = ΔH_fº C(g) = ΔHº_atmosationC(g)

Example 7. Using the given data, calculate enthalpy of formation of acetone(g) [All values in kJmol^-1] bond enthalpy of C-H = 413.4, C - C = 347.0 (C = O) = 728.0
O = O = 495.0 , H - H = 435.8 D_subH of C = 718.4
Solution.

Example 8. The enthalpy of combustion of acetylene is 312 kcal. If enthalpy of formation of CO₂ and H₂O are -94.38 and -68.38 kcal respectively
Calculate C º C bond enthalpy.
Given that enthalpy of atomisation of 150 kcal and H - H bond enthalpy and C - H bond enthalpy are 103 kcal and 93.64 kcal respectively.
Solution. 
HC ≡ CH  + 5/2O₂ ⇒ 2CO₂    +  H₂O
ΔH_Cº(CH ≡ CH) = 2ΔH_fº(CO₂) + ΔH_fº(H₂O) - ΔH_fº(C₂H₂)
- 312 = 2 × (-94.38) + (-68.38) - ΔH_fº(C₂H₂)
ΔH_fº(C₂H₂) = 54.86
CH ≡ CH ⇒ 2C(g)  + 2H(g)
ΣC=C + 2ΣC-H = 2ΔH_fºC(g) + 2ΔH_fºH(g) - ΔH_fº(CH º CH)
ΣC=C + 2 × 93.64 = 2 × 150 2 × (1/2) × 103 -54.86
⇒ ΣC=C = 160.86 kJmol^-1

Resonance Energy
"The energy difference between resonance hybrid and most stable canonical structure is called resonance energy".Resonance energy is generally -ve as nature of the process is exothermic.

Example 9. The enthalpy of formation of ethane, ethylene and benzene from the gaseous atoms are -2839.2, -2275.2 and -5506 kJmol^-1 respectively. Calculate the resonance energy of benzene. The bond enthalpy of C - H bond is given as equal to 410.87 kJ/mol.
Solution. 2C(g) + 6H(g) ⇒ C₂H₆   ΔHº = -2839.2
C₂H₆ ⇒ 2C(g)  + 6H(g)     ΔHº = 2839.2
∑C - C + 6∑C - H = 2839.2    ΔHº = 2839.2
∑C - C + 6 × 410.87 = 2839.2
∑C - C = 373.98 ... (1)
2C(g) + 4H(g) ⇒ C₂H₄(g) ΔHº = -2275.2
C₂H₄ ⇒ 2C(g) + 4H(g) ΔHº = 2275.2
∑C = C + 4∑C - H = 2275.2
∑(C = C) = 631.72 ...(2)
6C(g) + 6H(g) ⇒ C₆H₆    ΔHº = -5506
⇒ C₆H₆ ⇒ 6C(g) + 6H(g)     ΔHº = 5506
⇒ 3∑C = C + 3∑C - C + 6∑C - H+ x  = 5506
Putting all the values from eqⁿ , 1, & (2) we get x = 23.68
⇒ Resonance energy of benzene = - 23.68 kJ/mole

Bomb Calorimeter

Heat evolved = msΔt CΔt
                   ↓  
Heat capacity of container
Heat of combustion 

= -

Example 10. When 1.0 gm of fructose C₆H₁₂O₆ (s) is burnt in oxygen in a bomb calorimeter, the temperature of the calorimeter water increases by 1.56ºC. If the heat capacity of the calorimeter and its contents is 10.0 kJ/ºC. Calculate the enthalpy of combustion of fructose at 298 K.
Solution.
Heat capacity of the system
                 ↑

Heat of solution
Enthalpy change during the dissolution of 1 mole of salt in excess of solvent.
KCl(s) aq ⇒ KCl (aq)
ΔHº = Heat of solⁿ of KCl (s)

Note :-
(1) Heat of solution is always exothermic for the anhydrous form of salts which can form their hydrates.
Example:
CuSO₄, Na₂SO₄, FeSO₄, ZnSO₄, CaCl₂, LiCl etc.
CuSO₄(s) aq ⇒ CuSO₄(aq) ΔH < 0

(2) Heat of solution is endothermic for the hdyrated form of the salt.
CuSO₄ . 5H₂O aq ⇒ CuSO₄(aq)ΔH > 0

(3) Heat of solⁿ is endothermic for the salts which do not form their hydrates.
Example:  NaCl, NaNO₃, KCl etc.

Integral Heat of solution
Enthalpy change when 1 mole of salt is dissolved in given amount of solvent.
Example:
KCl(s) + 20 H₂O ⇒ KCl (20H₂O) -------------(1) DH₂ = x
KCl(s) + 100 H₂O ⇒ KCl (100H₂O) -------------(2) DH₂ = y

Heat of dilution
Reversing the eqⁿ (1) and adding in (2)
KCl(20H₂O)  + 80H₂O ⇒ KCl(100H₂O)
ΔH = y - x
enthalpy change when the conc. of salt changes from one to another on the basis of dilution
⇒ ΔH = y - x = Heat of dilution

Heat of hydration
Ethalpy change during the formation of hdyrated form of salt from its anhydrous form. It is always exothermic.
CuSO₄(s)  + 5H₂O ⇒ CuSO₄.5H₂O
ΔH = Heat of hydration of CuSO₄(s)

Example 11. Heat of solⁿ of CuSO₄(s) and CuSO₄.5H₂O is 15.9 and 19.3 kJ/mol respectively. Find the heat of hydration of CuSO₄(s)
Solution. CuSO₄(s) + aq ⇒ CuSO₄(aq) ......(1)
ΔH = -15.9
CuSO₄.5H₂O + aq ⇒ CuSO₄(aq) ......(2)
ΔH = 19.3
Reversing eqⁿ (1) and adding (2)
CuSO₄(s)  + aq ⇒ CuSO₄(aq)
ΔH = 15.9
CuSO₄.5H₂O  + aq ⇒ CuSO₄(aq)
ΔH = 19.3
CuSO₄.5H₂O ⇒ CuSO₄(s)
ΔH = 35.2
⇒ CuSO₄(s) ⇒ CuSO₄.5H₂O
ΔH = -35.2
⇒ Heat of hydration of CuSO₄(s) = -35.2 kJ/mol

Heat of neutralisation
Enthalpy change during neutralisation of 1 gm equivelant of Acid with 1 gm equivelant of base in dilute solⁿ is called heat of neutralisation.
HCl(aq)  + NaoH(aq) ⇒ NaCl + H₂O
H  + Cl^-  + Na  + OH^- ⇒ Na  + Cl^- + H₂O
H _(aq) + OH^-_(aq) ⇒ H₂O_(aq)
ΔHº = -13.7 Kcal
H₂SO_4(aq)  + 2NaOH_(aq) ⇒ Na₂SO₄  + 2H₂O
2H _⁺ SO₄^2-  + 2Na _⁺  2OH^- ⇒ 2Na  + SO₄^2- + 2H₂O
2H  +  2OH^- ⇒ 2H₂O
⇒ ΔH = -13.7 × 2

Note:-
In case of weak Acid or weak bases the observed value is little lower because of a part of it is used in dissociating weak Acid or weak base which is not at all completely ionised at dilute solution unditions.
These are however, completely ionised at infinite dilution.

Example:
CH₃COOH  + NaOH ⇒ CH₃COONa + H₂O
ΔH = -13.7 + x
As we know, H  + OH^- ⇒ H₂O , ΔH = -13.7 kcal ----(1)
CH₃COOH ⇒ CH₃COO^-  + H ,  ΔHº = x ----(2)
Add eqⁿ (1) (2)
CH₃COOH + OH^- ⇒ CH₃COO^-  + H₂O ΔHº = y

Example 12. 100 ml 0.5M H₂SO₄(strong Acid) is neutralised by 200 ml 0.2 M NH₄OH. In a constant pressure calorimeter which results in temperature rise of 1.4ºC. If heat capacity of calorimeter constant is  1.5 kJ/ºC.
Which statement is/are correct.
Given: HCl + NaOH ⇒ NaCl  + H₂O 57 kJ
CH₃COOH + NH₄OH ⇒ CH₃COONH₄  + H₂O 48.1 kJ
(A) Ethalpy of neutralisation of HCl v/s NH₄OH is -52.5 kJ/mol
(B) Ethalpy of dissociation (ionisation) of NH₄OH is 4.5 kJ/mol
(C) Ethalpy of dissociation of CH₃COOH is 4.6 kJ/mol
(D) ΔH for 2H₂O(l) ⇒ 2H⁺ (aq) + 2OH^-(aq) is 114 kJ
Solution. 
(A) Total heat evolved due to the neutralization = C × Δt = 1.5 × 1.4 = 2.1
M. eq of H₂SO₄ = 100 ×0.5 = 50
M.eq of NH₄OH = 20 × 0.2 = 40
Since NH₄OH is limiting hence energy will evolved according to it.
⇒ 0.04 gm eq produces 2.1 kJ

⇒ Heat of neutralisation = -52.5 kJ

(B) -57 + x = - 52.5
⇒ x = - 52.5  + 57 = 4.5
⇒ Enthalpy of dissociation of NH₄OH = 4.5 kJ/mol

(C) 57-(x + y) = 48.1
⇒ x + y = 8.9
⇒ 4.5 y = 8.9
⇒ y = 4.4
⇒ enthalpy of dissociation of CH₃COOH = 4.4 kJ/mol

(D) As we know
H⁺ + OH^- ⇒ H₂O  , ΔH = -57
⇒ 2H  + 2OH^- ⇒ 2H₂O , ΔH = -57× 2
⇒ 2H₂O ⇒ 2H⁺+ 2OH^-  , ΔH = 114 kJ
⇒ Option A, B, and D are correct.

Born Haber Cycle
Ionisation Energy:-The minimum amount of energy required to remove one electron form the outermost shell of an isolated gaseous atom is called ionisation energy of the element.
Electron affinity:-Amount of energy released when an extra electron is added to an isolated gaseous atom.
Lattice Energy:- Amount of energy released when 1 mole of gaseous cation and 1 mole gaseous anion combine to each other and form 1 mole of ionic compound is called lattice energy.

Na (g) + Cl^-(g) →  NaCl(s)  +   heat
                             (Lattice energy)

Example 13. Calculate the standard enthalpy change for a reaction CO₂(g) + H₂(g) → CO(g) +H₂O (g) given that ΔH_f⁰ for CO₂(g), CO(g) and H₂O(g) as -393.5, -110.5 and -241.8 KJ/mol respectively.
(A) -31.2 KJ 
(B) - 21.2 KJ 
(C) -11.2 KJ 
(D) 41.2KJ
Ans. (D)
Solution. 
ΔHº= SΔH_fº_(products) - SΔH_fº_(Reactants)
=
ΔHº = [-241.8 - 110.5] - [-393.5 0]
= - 352.3 393.5 = 41.2 KJ