THERMODYNAMICS

Question 1

What are extensive and intensive properties? Give examples

Answer 1

Extensive properties are properties which depend on the amount of matter present in the system.
E.g.: Volume, length, height, internal energy (U), enthalpy (H), entropy (S), Gibb’s energy (G), heat capacity etc.

Intensive properties are properties which are independent of the amount of matter present in the system.
E.g. : Temperature, Pressure, molar volume, density, refractive index, molar heat capacity, viscosity, surface tension etc.

Question 2

Define state function and path function.

Answer 2

A function or a property that depends only on the initial and final state of a system and not on the path followed is called a state function.
E.g.: T, P, V, U, H, S, G etc.

Path functions are properties which depend on the path followed also.
E.g. heat and work

Question 3

State the first law of thermodynamics. Give its mathematical form

Answer 3

It states that energy can neither be created nor be destroyed. Or, the total energy in the universe is always a constant. Mathematically ΔU = q + w

Question 4

Give the relationship between ∆H and ∆U.

Answer 4

ΔH= ΔU + PΔV

Question 5

State and illustrate Hess’s law of constant summation.

Answer 5

The law states that the total enthalpy change for a process is the same whether the reaction taking place in a single step or in several steps. Or, the total enthalpy change for a process is independent of the path followed.

Illustration:
Consider a process in which the reactant A is converted to product D in a single step by involving heat change, ΔH. Let the same reactant A is first converted to B, then to C and finally to D involving heat changes ΔH1, ΔH2 and ΔH3 respectively.
Then according to Hess’s law:
ΔH = ΔH1 + ΔH2 + ΔH3

Question 6

Define lattice enthalpy.

Answer 6

The lattice enthalpy of an ionic compound is the enthalpy change when one mole of an ionic compound dissociates into gaseous ions.

Question 7

Construct an enthalpy diagram for the determination of lattice enthalpy of sodium chloride

Answer 7

Lattice enthalpy of NaCl is determined by Born–Haber cycle.
Na(s) + ½ Cl2(g) → NaCl(s)
This involves the following steps

By applying Hess’s law we can write:
ΔfH0= ΔsubH0 + ΔiH0 + ½ ΔbondH0 + ΔegH0 + ΔlatticeH0
From this lattice enthalpy can be determined as:
ΔlatticeH0= ΔfH0 – [ΔsubH0 + ΔiH0 + ½ ΔbondH0 + ΔegH0]

Question 8

What are spontaneous processes? Give examples

Answer 8

It is a process that takes place without the help of any external agency.
E.g.
flow of water from high level to low level, burning of fuels, melting of ice, evaporation of water etc.

Question 9

What are the criteria for spontaneous process?

Answer 9

Decrease in energy and increase in disorderness.

Question 10

Define entropy. What is its unit?

Answer 10

It is a measure of degree of disorderness or randomness of a system.
The unit of entropy and entropy change is J/K/mol.

Question 11

What happens to the entropy during the following changes?
a) A gas condenses into liquid
b) CaCO3(s) → CaO(s) + CO2 (g)

Answer 11

a) Entropy decreases.
b) Entropy increases

Question 12

State the second law of thermodynamics.

Answer 12

It states that the entropy of the universe always increases during every spontaneous process.

Question 13

Define Gibb’s energy. Give the criteria for spontaneity of a process in terms of free energy change (∆G).

Answer 13

It is defined as the maximum amount of available energy that can be converted to useful work. For a spontaneous process, ΔG should be negative.

Question 14

Calculate the work done for the reversible isothermal expansion of 1 mole of an ideal gas at 27°C, from a volume of 10 dm³ to a volume of 20 dm³.

Answer 14

: Here n = 1 mol, R = 8.314 J/K/mol, V1 = 10 dm3, V2 = 20 dm3 and T = 25 + 273 = 298 K
For isothermal reversible expansion, work done, W(exp)
= –2.303nRT log(V2/V1)
= –2.303 x 1 x 8.314 x 298 x log (20/10)
= –1717.46 J

Question 15

Enthalpy and entropy changes of a reaction are 40.63 kJ/mol and 108.8 J/K/mol. Predict the feasibility of the reaction at 27°C.

Answer 15

Given
ΔH = 40.63 kJ/mol = 40630 J/mol,
ΔS = 108.8 J/K/mol
T = 27 + 273 = 300K.

From Gibb’s equation,
ΔG = ΔH – TΔS
= 40630 – 300 x 108.8 = 7990 J/mol
Since ΔG0 is positive, the process is non –spontaneous at this temperature.

Question 16

Calculate the enthalpy of formation of CH4 from the following data:
(i) C(s) + O2(g) → CO2(g); ∆H = –393.7 kJ/mol
(ii) H2(g) + ½ O2(g) → H2O(l); ∆H = –285.8 kJ/mol
(iii) CH4(g) + 2 O2(g)→ CO2(g) + 2H2O(l); ∆H = –890.4 kJ/mol

Answer 16

The required equation is:
C(s) + 2H2(g) → CH4(g)
Given:
C(s) + O2(g) → CO2(g); ΔH = –393.7 kJ/mol

Multiply equation (ii) x 2;
2H2(g) + O2(g) → 2 H2O(l);
ΔH = –285.8 x 2= –571.6 kJ/mol
Reverse equation (iii)
CO2(g) + 2H2O(l) → CH4(g) + 2 O2(g);
ΔH = 890.4 kJ/mol
Add the above three equations we get
C(s) + 2H2(g) → CH4(g);
ΔH = –393.7 + –571.6 + 890.4
= –74.9 kJ/mol

Question 17

Calculate the standard free energy (∆G0) for the conversion of oxygen to ozone:
3/2 O2(g) → O3(g) at 298K, if the equilibrium constant for the conversion is 2.47 x 10-29.(Given R = 8.314 JK-1mol-1).

Answer 17

Here K = 2.47 x 10-29, R = 8.314 J/K/mol and T = 298K
ΔG0 = –2.303RTlogK
= –2.303 x 8.314 x 298 x log (2.47 x 10 –29)
= 163000 J/mol = 163 kJ/mol