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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 (Cm)

 

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 (Cp,m)

(B) Molar heat capacity at constant volume (Cv,m)

 

Relation between Cp and Cv

Cp - Cv = R (Mayor's formula)

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = 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 = HB - HA

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)

= DnCVT +DnRT = DnT(CV + R)

→ DH = Dn CpT

DH = DE + DnRT

 

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

 

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.

 

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

 

According to Hess'Law

DH2  + Cp(T2-T1) = DH1 + Cp1 (T2- T1)

DH2 - DH1 = (Cp1 - Cp) (T2- T1)

= DCp (T2- T1)

→ DCp =CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

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

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

* If DCp is function of temperature

DCp = T2 + T

Then CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

 

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)  + O2(g) → CO2(g)

DH = DHf(CO2)

CO(g)  + 1/2O2(g) → CO2(g)

DH ¹ DHfCO2(g)

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

Similarly

CH2-CHO + H2 → C2H5OH

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, ThermochemistryDH ¹ DHf(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 = DHfº(product) - DHfº(Reactant)

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

DHº = DHfº(c) +  DHfº(D) - DHfº(A) + DHfº(B)

Assumption :-

The heat of formation of most stable form of an element is taken as zero.

C(graphite)  + O2(g) → CO2(g)

DHº = DHfº(CO2) - DHfºC(s) - DHfº(O2)(g)

→ DHº = DHfº(CO2) (As DHfºC(s) = 0 and DHfºO2(g) = 0)

Another example can be taken as

H2(g) + 1/2O2(g) → H2O(l)

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

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

DHº = DHfº(H2O)

 

Ex.1 From the following data,

C(s, graphite)  O2(g) → CO2(g) DHº = -393.5 kJ/mole

H2(g) 1/2O2(g) → H2O(l) DHº = -286 kJ/mole

2C2H6(g) 7O2(g) → 4CO2(g)  6H2O(l) DHº = -3120 kJ/mole

Calculate the standard enthalpy of formation of

C2H6(g) (in kJ/mole)

 

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

From eqn(2) DHfº(H2O) = -286 kJ/mole

From eqn (3) DrHº = 4 DHfº(CO2) 6DHfº(H2O) -2 ×DHfº(C2H6 ) - 7×DHfºO2(g)

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

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

→ -3120 + 3290 = - 2 ×DHfº(C2H6 )

→ 170 = - 2 ×DHfº(C2H6 )

→ DHfº = - 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 = DHCºC(s) = DHfº(CO2) - DHfºC(s) - DHfº(O2)

DHº = = DHC ºC(s) = DHfº(CO2)

Other example

H2 + 1/2 O2 → H2O(g)

DHº = DHCº (H2) = DHfº (H2O) - DHfº(H2) - 1/2DHfº (O2)

DHº = DHfº (H2) = DHfº (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)

DHrº = Heat of combustion of reactant - Heat of combustion of reactant .

DHrº = DHCº (Reactant) - DHCº (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 C2H6(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

DHrº = DHcº (Reactant) - DHCº (Product)

DHrº = DHcº (C3H8) DHCº (H2) - {DHcº(C2H6) DHCº(CH4) }

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

→ DHcº(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)  + 3H2(g)+ O2(g) → C6H5COOH(s) DHº = -408 kJ

→ DHfº(C6H5COOH) = - 408 kJ

C(s)  + O2(g) → CO2(g) DHº = -393 kJ

→ DHfº(CO2) = -393 kJ

H2(g)  + 1/2O2(g) → H2O(l) DHº = -286 kJ

→ DHfº(H2O) = -286 kJ

C6H5COOH(s)  + 15/2O2(g) → 7CO2(g) + 3H2O(l)

→ DHCº(C6H5COOH) = 7 DHfºCO2 + 3DHfº(H2O) - 4DHfº(C6H5COOH)

= 7 ×(-393) 3× (-286) 408

DHCº(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 = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

DHº = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = 2DHfºH(g) - DHfº(H2)

→ CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = 2DHfºH(g) → DHfº H(g) = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

 

Similarly

DHfºO(g) = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry, DHfºN(g) = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

* Let us consider the similar bond breaking In CH4

CH4(g) → C(g)  +  4 H(g)

DH = 4CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry =DHfºC(g) 4DHfºH(g) - DHfº(CH4)

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

DHfº = 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

 

SolDrHº = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

DHrº = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

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

DHrº = -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

DHrº = DHº (Hydrogenation of CH2 = CH2)

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry  DHrº = 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 - CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = C (5) C = C - C = C

 

Sol. Stability of alkene a CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

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) CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, ThermochemistryC = 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 CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry CH3-CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry -CH = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

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

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

of therefore decreasing order of heat of hydrogenation of alkene

5 > 1 > 2 > 3 > 4 >

 

Ex.6 Find DHf 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)

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = DHfºH(g) HfºCl(g) - DHfºHCl(g)

103 = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry  + CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry - DHfºHCl(g)

103 = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry×104 + CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry×58 - DHfº HCl(g)

→ DHfºHCl(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 = DHsubºC(s) = DHfº C(g) - DHfº C(s)

→ DHsubºC(s) = DHfº C(g)

→ DHsubºC(s) = DHfº C(g) = DHºatmosationC(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 DsubH of C = 718.4

 

Sol. CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry 3C(g) 6H(g) O(g)

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry   CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry  2CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = 3DHfºC(g) 6DHfºH(g) DHfºO(g) - DHfº(CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry)

 

6 × 413.4 728 2 × 347 = 3 × 718.4  CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry × 435.8  CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry×495.0 - DHfº(CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry)

→ DHfº(CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry) = - 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 CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, ThermochemistryO2 → 2CO2    +  H2O

DHCº(CH º CH) = 2DHfº(CO2) DHfº(H2O) - DHfº(C2H2)

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

DHfº(C2H2) = 54.86

CH º CH → 2C(g)  + 2H(g)

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry  2CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = 2DHfºC(g) 2DHfºH(g) - DHfº(CH º CH)

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry  2 × 93.64 = 2 × 150 2 × CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry × 103 -54.86 → CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = 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

 

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

Heat evolved = msDt CDt

¯

Heat capacity of container

Heat of combustion = - CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

= - CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = -CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = -CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

→ DH = -CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry→ 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

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

DHC = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = -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

H2SO4(aq)  2NaOH(aq) → Na2SO4  2H2O

2H   SO42-  2Na   2OH→ 2Na   SO42-  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 = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry = 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

DHlattice CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

 

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

 

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

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 DHf0 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= SDHfº(products) - SD Hfº(Reactants)

CBSE, Class 11, IIT JEE, Syllabus, Preparation, AIPMT, NCERT, Important, Thermochemistry

DHº = [-241.8 - 110.5] - [-393.5 0]

= - 352.3 393.5 = 41.2 KJ

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FAQs on Thermochemistry, Class 11, Chemistry Detailed Chapter Notes

1. What is thermochemistry?
Ans. Thermochemistry is the branch of chemistry that deals with the study of the relationship between heat and chemical reactions. It involves the measurement and calculation of the amount of heat absorbed or released during chemical reactions or physical changes.
2. What is enthalpy?
Ans. Enthalpy is the thermodynamic quantity that represents the total heat content of a system at constant pressure. It is denoted by the symbol 'H' and is expressed in joules (J) or kilojoules (kJ).
3. How is enthalpy change calculated?
Ans. Enthalpy change is calculated using the formula: ΔH = H(products) - H(reactants), where ΔH is the enthalpy change, H(products) is the enthalpy of the products, and H(reactants) is the enthalpy of the reactants. The enthalpy change can be either positive (endothermic) or negative (exothermic).
4. What is Hess's Law?
Ans. Hess's Law states that the enthalpy change of a chemical reaction is independent of the pathway between the initial and final states. In other words, the enthalpy change of a reaction is the same whether the reaction takes place in one step or multiple steps.
5. What is the difference between exothermic and endothermic reactions?
Ans. Exothermic reactions release heat energy to the surroundings, causing an increase in temperature. Endothermic reactions absorb heat energy from the surroundings, causing a decrease in temperature. Exothermic reactions have a negative enthalpy change, while endothermic reactions have a positive enthalpy change.
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