Chemistry

Chemical thermodynamics-II

Session Objectives

Session objectives

1. Enthalpy

2. Various types of enthalpy of reactions

3. Heat capacities of gases

4. Adiabatic process

5. Hess’s law

6. Bond energy

7. Lattice energy

8. Limitation of first law of thermodynamics

EnthalpyEnthalpy is the total heat contentsof the system at constant pressure. Enthalpy is shown by ‘H’.

H = E + PV

H2 – H1 = E2 – E1 + P(V2 – V1)

Where H1, E1 and V1 are the enthalpy, internal energy and volume respectively in initial state while H2, E2 and V2

are the enthalpy, internal energy and volume respectively in final state.

H = E + P V

Enthalpy change at constant pressure

Enthalpy

Enthalpy is a state function

At constant pressurePV1 = n1RT (for initial state)

PV2 = n2RT (for final state)

P(V2 – V1) = RT(n2 – n1)

gP V = n RT

Where ng = np–nr (gaseous moles only)

gH = E + n RT

Enthalpy of formation

It is the change in enthalpy when one mole of a compound is formed

from its elements in their naturally occuring physical states.

2 2 4 f2C(s) 2H (g) C H H 52kJ

Enthalpy of combustion

4 2 2 2CH g 2O g CO g 2H O(g) ; H 890.3 kJ

2 6 2 2 2

1

2C H g 7O g 4CO g 6H O l

H 745.6K calmole

1745.6– 372.8 Kcal mol

2

It is the change in enthalpy when one mole of the substance undergoes complete combustion.

Heat of combustion

Application of heat of combustion

Amount of heat produced per gram of a substance (food or fuel) is completely burnt.

Calorific value

4 2 2 2CH g 2O CO g 2H O l

Calorific value of CH4 (g) =– 890

– 55.6 kJ / g16

Hydrogen has the highest calorific value (150 kJ/g)

H = -890.3 kJ /mol

Enthalpy of neutralization

Difference in energy is used to ionize weak base.

4 4 2weak basestrong acid

HCl aq NH OH aq NH Cl aq H O l

H – 12.3Kcal

2HCl (aq) + NaOH (aq) NaCl (aq) + H O

H = -13.7 Kcal

Strong acid and strong base

Ionization energy of NH4OH =

What is the Ionization energy of NH4OH?

13.7 12.3 1.4KCal.

Enthalpy of solution

Amount of heat evolved or absorbed per mole of the substance in excess of water,

2

2

KCl s H O KCl aq H – 4.4 Kcal

KOH s H O KOH aq H – 13.3 Kcal

Enthalpy of fusion

One mole of solid substance changes

to its liquid state at its melting point.

Melting2 2

Freezing2 2

H O s H O l H 1.44Kcal

H O l H O s H – 1.44 Kcal

Enthalpy of vaporization

One mole of the substance changesfrom liquid state to gaseous state atits boiling point.

Boiling2 2

Cooling2 2

H O l H O g H 10.5 Kcal

H O g H O l H – 10.5 Kcal

Enthalpy change per mole of a solid converts directly to its vapours

Enthalpy of sublimation

sublimation4 4NH Cl (s) NH Cl g H 14.9 Kcal

Heat capacity

Quantity of heat required to raise thetemperature of the system by one degree.

Heat capacity = dq

dT

Heat capacity at constant pressure

pp

HC

T

(Since at constant pressure dq = dH)

Heat capacity

V

V

EC

T

(Since at constant volume dq = dE)

The difference between Cp and Cv is equal to the work done by 1 mole of gas in expansion when heated through 1° C.

Heat capacity at constant volume

p vC -C = R

Heat capacity

Specific heat capacity is the heatrequired to raise the temperature ofunit mass by one degree.

q = c × m × T

m = Mass of the substance

q = Heat required

= Temperature difference

c = Specific heat capacity. Specific heat capacity of water is 4.18 J/g K.

T

Adiabatic work

For adiabatic process, q = 0 it means no heat is exchanged with the surrounding.

v

v

E q w

i.e. E w q 0

For a finite change of an ideal gas,

E C . T

w E C T

Reversible Adiabatic expansion

Relations for reversible adiabatic expansion of an ideal gas

1

1

v

2 1v

PV cons tant ...(i)

TV cons tant ...(ii)

T V cons tant ...(iii)

Now, considering this relation

Work done, W C T

R R(T T ) [ 1]

( 1) C

(For 1 mole of gas)

Irreversible Adiabatic expansion

(i) For free expansion, w = 0. Since Pext = 0

(ii) For intermediate expansion,

ext 2 1

2 1ext

2 1

2 1v 2 1 ext

2 1

w P (V V )

T TP R

P P

T Ti.e. w C (T T ) R P

P P

Hess’s Law

Cq1 q2

A Bq

According to Hess’s law

q = q1 + q2

Applications of Hess’s law

Determination of heat changes of

transformation of rhombic sulphur

into monoclinic sulphur.Given

Subtracting (ii) from (i), we get

S (rhombic) – S (monoclinic)

H = +2.5 kJ

2 2

2 2

S (rhombic) + O (g) SO (g); H = - 297.5 kJ -----(i)

S (monoclinic) + O (g) SO (g); H = - 300.0 kJ -----(ii)

Resonance energy

Resonance energy = Enthalpy of formation calculated from bond energy – Experimental value of enthalpy

Question

Illustrative example

Calculate resonance energy of benzene from the following data:

(i) benzene = –358.5 kJ mol–1

(ii) Heat of atomization of carbon is 716.8 kJ mol-1

(iii) Bond energies of C–H, C–C, C = C and H–H bonds are 490, 340, 620 and 436.9 kJ mol–1 respectively.

0fH

Solution

The required equation is

2 6 66C(s) + 3H (g) C H , H = -358.5 kJ

Bond energies of reactants – Bond energies of products

fH =

= {6 [ HC(s) c(g)] + 3 ( HH–H) – 3{[ HC–C] + 3 ( HC =C) + 6 ( HC–H )

=6716.8 + (3 436.9) – (3 340) – (3620) – (6490) kJ mol–1

Resonance energy = Hf (obs.) – Hf (cal)

= – 358.5 – ( – 208.5)

= – 150.0 kJ mol–1

Determination of lattice energy

sub diss X M X MXMX M 2

1H H ( H ) (IP) ( EA) ( U)

2

Figure

Determination of bond energies or bond enthalpiesEnergy required to break the bondor energy released during the bond formation is called bond energy.

2H S H g SH g H 100 kJ / mole

SH g S g H g H 200 kJ / mole

The average of these two bond dissociation energies gives the value of bond energy of S — H.

Bond energy of S — H bond100 200

150 kJ / mole2

Limitations of first law

1. The first law of thermodynamics states that one form of energy disappears, an equivalent amount of another form of energy is produced. But it is salient about the extent to which such conversion can take place.

2. It does not tell about the direction of flow of heat.

3. It does not tell about spontaneity of reaction.

Class exercise

Class exercise 1

For the reaction

for the reaction will be

(a) – 781.80 Kcal (b) – 780.009 Kcal

(c) – 780.75 Kcal (d) 780.05 Kcal

= -780.9 Kcal, E

6 6 2 2 21

C H (l) 7 O (g) 6CO (g) 3H O(l)2

0, H(27 )

SolutionH = E + nRT

15 3n 6 1.5

2 2

E = H - nRT

= –780.9 – (–1.5 RT)

1.5 1.987 300780.9

1000

= – 780.9 + 0.891

= – 780.009 Kcal

Hence, the answer is (b).

Class exercise 2

When 1 gram of methane (CH4) burnsin O2 the heat evolved (measured under standard conditions) is 13.3 kcal. What is the heat of combustion?

(a) –13.3 k cals (b) +213 k cals

(c) – 213 k cals (d) – 416 k cals

Solution

13.3 Kcal/gm evolved

ocombH 13.3 16 Kcal / mol

= – 213 Kcal/mol

Hence, the answer is (c).

Class exercise 3

When 4.184 J of heat is transferred to 1 g of water at 20° C, its temperature rises to 21° C. The molar heat capacity at this temperature is

(a) 18 JK–1 (b)

(c) 75.4 JK–1 (d) 4.184 JK–1

18

4 1841

.JK

-

C = (4.184) × 18 = 75.4 J/K

n C T = 4.184

Hence, the answer is (c).

Solution:

Class exercise 4

When 0.532 g of benzene (B.P 80° C) is burnt in a constant volume systemwith an excess of oxygen, 22.3 kJ of heat is given out. for the combustion process is given by

(a) – 21 kJ (b) – 1234.98 kJ

(c) – 221 kJ (d) – 3273.26 kJ

H

Solution

6 6Mol. wt C HH 22.3

0.532

22.3 780.532

= – 3269.5 kJ

Hence, the answer is (d).

Class exercise 5

Consider the reaction

2 2 31

SO (g) O (g) SO (g) H2

= – 98.3 kJ. If the enthalpy of formation of SO3(g) is – 395.4 kJ, then the enthalpy of formation of SO2(g) is

(a) 297.1 kJ (b) 493.7 kJ

(c) – 493.7 kJ(d) – 297.1 kJ

3 2

o of SO f SO

H H 98.3 kJ

2

of SO

H 98.3 395.4 kJ 297.1 kJ

Solution:

Hence, the answer is (d).

Class exercise 6

Calculate the heat change for the following reaction:

4 2 2 2CH (g) + 2O (g) CO (g) + 2H O(l)

for CH4 , H2O and CO2 are –17.89, –68.3 and–94.05 kcal/mole.

0fH

Solution

4 2 2 2CH g 2O g CO g 2H O l

2 2 4 2

o o o o oreaction f H O f CO f CH f OH 2 H H H 2 H

= – 2 × 68.3 – 94.05 + 17.89 – 0

= – 212.76 kcal/mol

Class exercise 7

Calculate the heat of combustion of benzene from the following data:

2 6 66C (s) + 3H (g) C H (l)

2 2 21

H (g) + O (g) H O(l)2

2 2C(s) + O (g) CO (g)

H = 11720 cal

H = -68320 cal

H =-93050 cal

Solution

The required reaction is

6 6 2 2 215

C H O 3H O 6CO (1)2

2 6 66C + 3H C H (2)

2 2C + O CO (3)

2 2 21

H O H O2

3 × (4) + 6 × (3) – (2) o o o o1 4 3 2H 3 H 6 H H

= 3(– 68320) + 6(– 93050) – (1720 Cal)

= – 774.980 kcal/mol

Class exercise 8

Calculate for the following reaction at 27o C.

Given = – 337 kcal R= 1.987 cal deg–1 mole–1

2 4 2 2 2C H (g) + 3O (g) 2CO (g) + 2H O(l)

H

H

Solution

H = E + PV

3=– 337 – (–2)(1.987)(300) × 10

E = H - nRT

= – 335.8078 kcal/mol

= E + nRT { n = -2}

Class exercise 9

Calculate the heat of combustion of acetic acid at 25o C if the heat of formation of CH3COOH(l),CO2(g) and H2O(l) are –116.4, –94.0 and –68.3 kcal mole–1 respectively.

3 2 2 2CH COOH l 3O g 2CO g 2H O l

reactions f 2 f 2

f 3 f 2

H = 2 . H (CO ) + 2 . H (H O)

- H (CH COOH) - 3 H (O )

= 2(– 94.0) + 2(– 68.3) – (–116.4)

= – 208.2 Kcal/mol

Solution:

Thank you