**1.**

**ΔH and ΔS values of a reaction at 300K are –10 k.cal mole**

**-1**

**& 20 cal.deg**

**-1**

**mole**

**-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**

**–1**

**mol**

**–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 CCl**

**4**

**(Boiling point of CCl**

**4**

**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.**

**ΔS**

**vap = 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

**ΔH**

**vap = 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**

**ONE MARKS FIVE MARKS**

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