Thermochemistry, Class 11, Chemistry Detailed Chapter Notes 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 1º C 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 Dt.

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)

 = g (Poison's Ratio)

Rules for Thermochemical Equation

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

(2) A → B, DH = H_B - H_A

If A → B x kJ/mole

DH < 0 → (Exothermic reaction)

DH = -x kJ/mole

If A x kJ/mole → B

DH > 0 → (Endothermic reaction)

DH = x kJ/mole

(3) After reversing a thermochemical eqⁿ then sign of enthalpy also get changed.

e.g. A_{(g) +}  B_(g) → C_(g) + D_(g) , DH = x kJ

C_(g) + D_(g) → A_(g)+ B_(g) , DH = - x kJ

(4) When two reactions are added their enthalpies are also get added with their sign.

e.g.

A_(g) + B_(g) → C_(g)  D_(g) DH = x₁ kJ

C_(g) + E_(g) → F_(g) DH = - x₂ kJ

A_(g) + B_(g) + E_(g) → D_(g) + F_(g) , DH = x₁- x₂

(5) If a thermochemical equation is multiplied by a number then DH is multiplied by the same number.

e.g.

A_(g) + B_(g) → C_(g) + D_(g) DH = x₁ kJ

2A_(g) + 2B_(g) → 2C_(g) + 2D_(g) DH = 2x₁ kJ

Intensive Property

The property which does not depend upon the mass of substance is called intensive property.

e.g. density, refractive index, specific heat, etc.

Extensive Property

Mass dependent properties are called extensive properties

e.g. DH, DS, DG, 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.

e.g. We can not add the density of two liquids to get the density of the final mixture of the two.

Enthalpy (H)

H = U + PV

Internal energy

DH = DU +D(PV) = DU (P₂V₂ - P₁V₁)

(i) At constant Pressure, DH = DU + PV

(ii) At Constant Volume, DH = DU + VDP

* 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, DH = DU + D(PV)

= DnC_VT +DnRT = DnT(C_V + R)

→ DH = Dn C_pT

DH = DE + DnRT

Where Dng = 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

DH₂  + Cp(T₂-T₁) = DH₁ + Cp¹ (T₂- T₁)

DH₂ - DH₁ = (Cp¹ - Cp) (T₂- T₁)

= DCp (T₂- T₁)

→ DCp =

Where

DCp = Molar heat capacity of product -- Molar heat capacity of reactant

e.g.

N₂(g) + 3H₂(g) → 2NH₃(g)

DCp = 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 DCp is function of temperature

DCp = 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)

DH = DH_f(CO₂)

CO_(g)  + 1/2O_2(g) → CO_2(g)

DH ¹ DH_fCO₂(g)

(because CO₂ has not been formed form its element in their most stable form)

Similarly

CH₂-CHO + H₂ → C₂H₅OH

DH ¹ DH_f(C₂H₅OH)

Heat of reaction

Element                                       Most stable form

H                                                  H₂(gas)

O                                                 O₂(gas)

N                                                 N₂(gas)

F                                                  F₂(gas)

Cl                                                Cl₂(gas)

Br                                                Br₂(gas)

I                                                  I₂(solid)

C                                                C(grapnite)

P                                                P(white)

S                                                S(rhombic)

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

Enthalpy At Standard State :-(DHº)

T = 25ºC = 298 K

P = 1 atm

Conc = 1M

DHº = 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

DHº = DH_fº(product) - DH_fº(Reactant)

If we use the above concept for the above given reaction then

DHº = DH_fº_(c) +  DH_fº_(D) - DH_fº_(A) + DH_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)

DHº = DH_fº(CO₂) - DH_fºC_(s) - DH_fº(O₂)(g)

→ DHº = DH_fº(CO₂) (As DH_fºC_(s) = 0 and DH_fºO₂(g) = 0)

Another example can be taken as

H₂(g) + 1/2O₂(g) → H₂O_(l)

DHº = DH_fº(H₂O) - DH_fº(H₂) - 1/2 DH_fº(O₂)

DHº = DH_fº(H₂O)

Ex.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) DHº = -3120 kJ/mole

Calculate the standard enthalpy of formation of

C₂H₆(g) (in kJ/mole)

Sol. From eqⁿ(1) DH_fº(CO₂) = -393.5 kJ/mole

From eqⁿ(2) DH_fº(H₂O) = -286 kJ/mole

From eqⁿ (3) D_rHº = 4 DH_fº(CO₂) 6DH_fº(H₂O) -2 ×DH_fº(C₂H₆ ) - 7×DH_fºO₂(g)

- 3120 = 4 ×(-393.5) 6×(-286) - 2×DH_fº(C₂H₆ ) (As DH_fº O₂(g) = 0 )

→ -3120 = -1574 -1716 - 2 ×DH_fº(C₂H₆ )

→ -3120 + 3290 = - 2 ×DH_fº(C₂H₆ )

→ 170 = - 2 ×DH_fº(C₂H₆ )

→ DH_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)

DHº = DH_CºC(s) = DH_fº(CO₂) - DH_fºC(s) - DH_fº(O₂)

DHº = = DH_C ºC(s) = DH_fº(CO₂)

Other example

H₂ + 1/2 O₂ → H₂O(g)

DHº = DH_Cº (H₂) = DH_fº (H₂O) - DH_fº(H₂) - 1/2DH_fº (O₂)

DHº = DH_fº (H₂) = DH_fº (H₂O)

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)

DH_rº = Heat of combustion of reactant - Heat of combustion of reactant .

DH_rº = DH_Cº (Reactant) - DH_Cº (Product)

Ex.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 CH₄are - 285.8 and - 890.0 kJ/mole respectively. Enthalpy of combustion of propane is -2220 KJmol^-1.

Sol. As we know any thermochemical eqⁿ can be written in terms of heat of combustion as follows

DH_rº = DH_cº (Reactant) - DH_Cº (Product)

DH_rº = DH_cº (C₃H₈) DH_Cº (H₂) - {DH_cº(C₂H₆) DH_Cº(CH₄) }

- 55.7 = (-2220 - 285.8) - {-890 DH_cº(C₂H₆) }

→ DH_cº(C₂H₆) (g) = - 1560.1 kJmol^-1

Problems Based on Both HOC and HOF :

Ex.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.

Sol. 7C(s)  + 3H_2(g)+ O_2(g) → C₆H₅COOH_(s) DHº = -408 kJ

→ DH_fº(C₆H₅COOH) = - 408 kJ

C_(s)  + O_2(g) → CO₂(g) DHº = -393 kJ

→ DH_fº(CO₂) = -393 kJ

H_2(g)  + 1/2O_2(g) → H₂O(l) DHº = -286 kJ

→ DH_fº(H₂O) = -286 kJ

C₆H₅COOH_(s)  + 15/2O_2(g) → 7CO_2(g) + 3H₂O(l)

→ DH_Cº(C₆H₅COOH) = 7 DH_fºCO₂ + 3DH_fº(H₂O) - 4DH_fº(C₆H₅COOH)

= 7 ×(-393) 3× (-286) 408

DH_Cº(C₆H₅COOH) = - 3609 408

= -3201 kJ/mol

→ enthalpy of combustion at constant pressure = - 3201 kJ mol^-1

Also

DH = DU  DngRT

-3201 = DU  (-0.5) × 8.31 × 10^-3 × 300

(As Dn = 0.5, R = 8.314 × 10^-3 kJ)

→ DU = - 3201 1.2471

DU = -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"

e.g. H₂(g) → 2H(g) , DHº = 

DHº =  = 2DH_fºH(g) - DH_fº(H₂)

→  = 2DH_fºH(g) → DH_fº H(g) = 

Similarly

DH_fºO(g) = , DH_fºN(g) = 

* Let us consider the similar bond breaking In CH₄

CH₄(g) → C(g)  +  4 H_(g)

DH = 4 =DH_fºC(g) 4DH_fºH(g) - DH_fº(CH₄)

** Enthalpy of reaction in terms of bond energy for a thermochemical eqⁿ can be written as

DH_fº = B.E._(Reactant) - B.E. _(products)
 

Ex.4 Using the bond enthalpy data given below, calculate the enthalpy change for the reaction

C₂H₄ (g) H₂(g) → C₂H₆(g)

Data :

Bond C - C C = C C - H H - H

Bond Enthalpy(Kg/mol) 336.81 606.68 410.87 431.79

Sol. D_rHº = 

DH_rº = 

DH_rº = 606.68 4×410.87 431.79 -336.81 -6×410.87 = 2681.95 -2802.03

DH_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.

e.g. CH₂ = CH₂ + H₂ → CH₃ -CH₃

DH_rº = DHº (Hydrogenation of CH₂ = CH₂)

  DH_rº = DHº _{(Hydrogenation)} of cyclohexene

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

(1) C - C - C = C (2) C - C = C - C (cis) (3) C - C = C - C (trans)

(4) C -  = C (5) C = C - C = C

Sol. Stability of alkene a 

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.

(1) C - C - C = C (2 × H)

(2) C - C = C - C ( 6 × H)

(cis)

(3) C - C = C - C (6 × H)

(trans)

(4) C = C (6 × H)

(5) C = C - C = C (2 double bonds)

As we know trans > cis (stability)

→ Heat of hydrogen (trans) < Heat of hydrogenation (cis)

CH₃ - CH = CH - CH₃  CH₃- -CH = 

(less stable) ( I of CH₃ reduce the stability)

of therefore decreasing order of heat of hydrogenation of alkene

5 > 1 > 2 > 3 > 4 >

Ex.6 Find DH_f of HCl(g) if bond energies of H₂, Cl₂ and HCl are 104, 58, 103 kcal/mole respectively.

Sol. H - Cl → H(g)  + Cl(g)

 = DH_fºH(g) H_fºCl(g) - DH_fºHCl(g)

103 =   +  - DH_fºHCl(g)

103 = ×104 + ×58 - DH_fº HCl(g)

→ DH_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.

e.g. C_(s) → C_(g)

DHº = DH_subºC(s) = DH_fº C(g) - DH_fº C(s)

→ DH_subºC(s) = DH_fº C(g)

→ DH_subºC(s) = DH_fº C(g) = DHº_atmosationC(g)

Ex.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

Sol.  3C(g) 6H(g) O(g)

6      2 = 3DH_fºC(g) 6DH_fºH(g) DH_fºO(g) - DH_fº()

6 × 413.4 728 2 × 347 = 3 × 718.4   × 435.8  ×495.0 - DH_fº()

→ DH_fº() = - 192.3 kJmol^-1

Ex.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.

Sol. HC º CH O₂ → 2CO₂    +  H₂O

DH_Cº(CH º CH) = 2DH_fº(CO₂) DH_fº(H₂O) - DH_fº(C₂H₂)

- 312 = 2 × (-94.38) (-68.38) - DH_fº(C₂H₂)

DH_fº(C₂H₂) = 54.86

CH º CH → 2C(g)  + 2H(g)

  2 = 2DH_fºC(g) 2DH_fºH(g) - DH_fº(CH º CH)

  2 × 93.64 = 2 × 150 2 ×  × 103 -54.86 →  = 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.

Ex.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.

Sol. 2C(g) + 6H(g) → C₂H₆ DHº = -2839.2

C₂H₆ → 2C(g)  + 6H(g) DHº = 2839.2

∑C - C 6∑C - H = 2839.2 DHº = 2839.2

∑C - C 6 × 410.87 = 2839.2

∑C - C = 373.98 ... (1)

2C(g) + 4H(g) → C₂H₄(g) DHº = -2275.2

C₂H₄ → 2C(g) + 4H(g) DHº = 2275.2

∑C = C 4∑C - H = 2275.2

∑(C = C) = 631.72 ...(2)

6C(g) + 6H(g) → C₆H₆ DHº = -5506

→ C₆H₆ → 6C(g) + 6H(g) DHº = 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 = msDt CDt

¯

Heat capacity of container

Heat of combustion = - 

= -  = - = -

→ DH = -→ Heat capacity of system

Ex.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.

Sol. Heat capacity of the system

DH_C = 

=  = -2808 kJ/mole

Heat of solution

Enthalpy change during the dissolution of 1 mole of salt in excess of solvent.

KCl(s) aq → KCl (aq)

DHº = 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.

e.g. CuSO₄, Na₂SO₄, FeSO₄, ZnSO₄, CaCl₂, LiCl etc.

CuSO₄(s) aq → CuSO₄(aq) DH < 0

(2) Heat of solution is endothermic for the hdyrated form of the salt.

CuSO₄ . 5H₂O aq → CuSO₄(aq) DH > 0

(3) Heat of solⁿ is endothermic for the salts which do not form their hydrates.

e.g. NaCl, NaNO₃, KCl etc.

Integral Heat of solution

Enthalpy change when 1 mole of salt is dissolved in given amount of solvent.

e.g. 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)

DH = y - x

enthalpy change when the conc. of salt changes from one to another on the basis of dilution

→ DH = 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

DH = Heat of hydration of CuSO₄(s)

Ex.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)

Sol. CuSO₄(s) aq → CuSO₄(aq) -----------(1)

DH = -15.9

CuSO₄.5H₂O aq → CuSO₄(aq) -----------(2)

DH = 19.3

Reversing eqⁿ (1) and adding (2)

CuSO₄(s) aq → CuSO₄(aq)

DH = 15.9

CuSO₄.5H₂O aq → CuSO₄(aq)

DH = 19.3

----------------------------------------------

CuSO₄.5H₂O → CuSO₄(s)

DH = 35.2

→ CuSO₄(s) → CuSO₄.5H₂O

DH = -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)

DHº = -13.7 Kcal

H₂SO_4(aq)  2NaOH_(aq) → Na₂SO₄  2H₂O

2H   SO₄^2-  2Na   2OH^- → 2Na   SO₄^2-  2H₂O

2H   2OH^- → 2H₂O

→ DH = -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. e.g.

CH₃COOH  + NaOH → CH₃COONa H₂O

D H = -13.7 x

As we know, H  + OH^- → H₂O DH = -13.7 kcal ----(1)

CH₃COOH → CH₃COO^-  + H  DHº = x ----(2)

Add eqⁿ (1) (2)

----------------------------------

CH₃COOH OH^- → CH₃COO^-  + H₂O DHº = y

Ex.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) DH for 2H₂O(l) → 2H (aq) 2OH^-(aq) is 114 kJ

Sol. (A) Total heat evolved due to the neutralization = C × Dt = 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

1 gm eq produces =  =  = 52.5

→ 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 DH = -57

→ 2H  + 2OH^- → 2H₂O DH = -57× 2

→ 2H₂O → 2H  + 2OH^- DH = 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

DH_lattice a 

Ex.13 The born-Haber cycle for formation of rubidium chloride ((RbCl) is given bellow (the enthalpies are in kcal mol^-1)

find the value of X?

Ex.14 Calculate the standard enthalpy change for a reaction CO₂(g) + H₂(g) → CO(g) +H₂O (g) given that DH_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

Sol. D

DHº= SDH_fº_(products) - SD H_fº_(Reactants)

= 

DHº = [-241.8 - 110.5] - [-393.5 0]

= - 352.3 393.5 = 41.2 KJ