Thermodynamics - II Three Marks

1. ΔH and ΔS values of a reaction at 300K are –10 k.cal mole-1 & 20 cal.deg-1mole-l respectively. Calculate ΔG value. Or For a chemical reaction the values of ΔH and ΔS at 300 K are –10 k cal mol-1 and 20 cal.deg–1mol–1 respectively. What is the value of ΔG of the reaction? Predict the nature of the reaction.

     ΔG = ΔH – TΔS

At 300 K; ΔG = –10,000 – (20 x 300)

                 ΔG = –16,000 cals.mole–1

Since ΔG is a negative value, the reaction is spontaneous (feasible) at 300 K.

2. Calculate the change of entropy for the process, water (liquid) → water (vapour, 373K) involving ΔH(vap), 373K = 40850J mol-1

H2O(l)                 →               H2O(g)

Water liquid      273 K            Water vapour

ΔSvap = ΔHvap /  Tb (K)

     = 40850 J.mol–1 / 373 K

ΔSvap = 109.52 Jmole–1K–1

3. Calculate the maximum % efficiency possible from a thermal engine operating between 110°C and 25°C.

Initial temperature T1 = 110°C = (273 + 110) K = 383 K

Final temperature T2 = 25°C = (273 + 25) K = 298 K

% efficiency η = [T1 – T2 / T1] x 100

      = [383 – 298 / 383] x 100

      = [85 / 383] x 100

% efficiency η = 22.19 %

4. Calculate the molar heat of vaporisation of the ideal liquid CCl4 (Boiling point of CCl4 is 76.7°C and ΔS = 87.864 J).

ΔSvap = ΔHvap / Tb (K)

ΔHvap = ΔS vap x  Tb (K)

ΔHvap = 87.864 J x (273 + 76.7) K

          = 87.864 J x 349.7 K = 30726 J.K–1

          ΔHvap = 30.726 kJ.K–1.mol–1

5. Give the entropy statement of second law of thermodynamics. Mention the unit of entropy also.

A process accompanied by increase in entropy tends to be spontaneous.

Units of entropy

Unit of entropy is calories per degree per mole (or) eu.per mole.

cgs units of entropy is cal.K–1mole–1denoted as eu

SI unit of entropy is JK–1mole–1 and denoted as EU. 1 eu = 4.184 EU.

6. Give the Kelvin-Planck statement of second law of thermodynamics.

It is impossible to construct an engine which operated in a complete cycle will absorb heat from a single body and convert it completely to work without leaving some changes in the working system

7. How is ΔG related to ΔH and ΔS? What is the meaning of ΔG = 0?

ΔG = ΔH – TΔS

Where,

ΔG = Free energy change

ΔH = Change in Enthalpy of the system

T = Temperature in Kelvin

ΔS = Change in Entropy of the system

When,

ΔG = 0, the process is in equilibrium

8. State Clausius statement of second law of thermodynamics.

It is impossible to transfer heat from a cold body to a hot body by a machine without doing some work

9. State Trouton's rule.

The heat of vaporisation (ΔHvap) in calories per mole divided by the boiling point (Tb) of the liquid in Kelvin is a constant equal to 21 cal.deg–1.mole–1 and is known as the entropy of vapourisaiton.

ΔSvap = ΔHvap / Tb (K) = 21cal.deg–1_.mole–1

10. The normal boiling point of CHCl3 is 61.5°C. Calculate the molar heat of vapourisation of CHCl3 assuming ideal behaviour.

       ΔSvap = ΔH vap / Tb (K)

       ΔHvap = ΔS vap x  Tb (K)

       ΔHvap = 87.864 J x (273 + 61.5) K

                   = 87.864 J x 334.5 K = 29390 J.K–1

        ΔHvap = 29.39 kJ.K–1.mol–1

11. What are the substances that deviate from Trouton's rule? Or What types of liquids or substances deviate from Trouton's Rule?

Substances that deviate from Trouton's rule are:

1. Low boiling liquids such as Hydrogen and Helium which boil only a little above 0K.

2.  Polar substances like water, alcohol which form hydrogen bonded liquids and exhibit very high boiling points as well as high ΔHvap.

3.  Liquids such as acetic acid whose molecules are partially associated in the vapor phase.

12. What is entropy? What is its unit?

Entropy is a measure of randomness or disorder of the molecules of a system.

Or

Entropy function ‘S’ represents the ratio of the heat involved (q) to the temperature (T) of the process. That is, S = q / T

Units of entropy

Unit of entropy is calories per degree per mole (or) eu. per mole.

cgs units of entropy is cal.K–1mole–1 denoted as eu

SI unit of entropy is JK–1mole-1 and denoted EU. 1 eu = 4.184 EU.

13. What is Gibbs' free energy?

Gibbs free energy function, denoted by the symbol ‘G’ is mathematically defined as,

G = H - TS

Where,

H = Enthalpy or Heat content of the system.

T = Temperature in Kelvin.

S = Entropy of the system.

14. What is the entropy change of an engine that operates at 100°C when 453.6 kcal of heat is supplied to it?

Heat supplied Δq = 453.6 kcal

Temperature T = 100°C = (273 + 100) K = 373 K

Entropy change ΔS = ?

Entropy change ΔS = Δq / T (K)

  ΔS = 453.6 / 373 kcal K–1

  ΔS = 1.216 kcal K–1

15. What is the nature of the reaction when - i) ΔG > 0? ii) ΔG < 0? iii) ΔG = 0?

When,

i.   ΔG > 0, ΔG is + ve, the process is non spontaneous and non feasible.

ii.  ΔG < 0, ΔG is – ve, the process is spontaneous and feasible.

iii. ΔG = 0, the process is in equilibrium.
Or

ΔG = ΔH – TΔS
ΔG < 0	Spontaneous in forward direction. Not Spontaneous in reverse→
ΔG = 0	At Equilibrium. No net change will occur. ↔
ΔG > 0	Not Spontaneous forward. Spontaneous in reverse. ←

16. Calculate the entropy change involved in the conversion of 1 mole of ice at 0°C and 1 atm to liquid at 0°C and 1 atm. The enthalpy of fusion per mole of ice is 6008 Jmol-l.

                        0°C

H2O(s) → H2O(l)

Ice                Water liquid

ΔSfusion = ΔHfusion / Tm (K)

    = 6008 J.mol–1 / (0 + 273) K

    = 6008 / 273 J.mol–1 K–1

ΔSfusion = 22.007 J.mol–1 K–1.
17. For a chemical reaction the values of ΔH and ΔS at 300 K are –10 k cal mol–1 and 20 cal.deg–1 mol–1 respectively. Calculate the ΔG of reaction

Given, ΔH at 300 K = – 10 k cal mol–1 = – 10, 000 cal mol–1

             ΔS at 300 K = 20 cal.deg–1

    ΔG = ΔH – TΔS

at 300 K; ΔG = – 10, 000 – (20 x 300)

                      = – 10, 000 – 6000

    ΔG = – 16, 000 cals. mole–1

18. Calculate the entropy increase in the evaporation of 1 mole of a liquid when it boils at 100°C having heat of vaporisation at 100°C as 540 Cal / gm.

ΔHvap = 540 Cal / gm

            = 540 x 18 Cal / mol

      Tb = 100°C = (100 + 273) K = 373 K

ΔSvap = ?

ΔSvap = ΔHvap / Tb (K)

            = 540 x 18 Cal / mol / 373 K

ΔSvap = 26.06 J.mol–1 K–1

19. Mention the essential condition for spontaneity in a chemical reaction.
 ONE MARKS  FIVE MARKS