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THERMODYNAMICS Mittal Sir

BMC CLASSES Page No - 32 III-A/172, Nehru Nagar Ghaziabad. Ph : 0120-4102287, 9811212090

THERMODYNAMICS

• Limitations :-

(1) Not applicable on microscopic system like, change inside an atoms.

Or

system having few molecules only.

(2) deals only with initial & final state (does not deal with path )

(3) Deals only with spontaneity of a process does not deal with ‘how process occurred’ or by what rate

it occurred

(4) Laws of thermodynamics are applied only when a system is in equilibrium or moves form one

equilibrium to another equilibrium

• System & surrounding : –

The part of universe which is under observations is called system and remaining universe

constitutes the surrounding of that system.

The universe = The system + the surrounding

Type of system

OPEN CLOSED ISOLATED

� There is no perfectly isolated system in the universe,

• Boundary :- Anything which separate system & surrounding.

Boundary can be real / imaginary, rigid / non rigid, conductor (Diathermic) / non

conductor (adiavatic).

• Extensive property :- depend upon mass of system

Eg. Internal energy, enthalpy, entropy Gibbs free energy, heat capacity, volume etc.

If system is divided in to two part the properties which become half are extensive in nature.

• Intensive property :- Do not change on changing mass.

e.g. E°, density of Homogenous system, surface tension, reflective index, viscosity,

melting point, Boiling point, Molarity, Normality, etc.

Extensive properties which are measured for 1 mol also become intensive. eg Volume is

intensive but molar volume is extensive. Heat capacity is extensive but molar heat capacity is intensive.

Similarly enthalpy is extensive but molar enthalpy is intensive.

The ratio of two extensive variables gives intensive variable. Eg. Mass and volume are both

extensive but mass/volume that is density is intensive. Other examples are molarity, normality

• State functions : –

The state of a thermodynamic system is described by its measurable or macroscopic

properties. The state functions are those thermodynamic functions which depend only on the initial and

final states and not on how the state is reached. Examples of thermodynamics state functions are :

Both energy & matter

can be exchanged

between system and

surrounding eg open

beaker

Only energy can be

exchanged between

system & surrounding.

eg. closed vessel

No exchange of

energy or matter. eg.

Thermos flask or

Insulated flask



Internal energy (U), enthalpy (H), entropy (S), free energy (G), volume (V) etc. For any state function,

we can calculate the change in function (∆). For example, change in volume.

∆V = Vfinal – Vinitial

For a cyclic process, the change in state function is equal to zero because initial and final state

are same.

Path Dependent Functions. The thermodynamic functions which depend on the path are known as

path dependent functions. The important path dependent functions are heat(q) and work(w) . For

example, the amount of work done in climbing a mountain peak depends on the path chosen.

Reversible process. In a reversible process, the change is carried out so slowly in infinite number of

steps so that the system and the surroundings are always in equilibrium. The reversible processes are

ideal processes and are not realized in actual practice.

Irreversible Process. A process which is not exactly reversed i.e., the system does not pass through the

same intermediate states as in the direct process is known as irreversible process. An irreversible

process cannot be reversed without the help of an external agency. All natural processes are

irreversible. For example, falling of water from a hill top, germination of seeds, diffusion of gases etc,

are all irreversible processes.

1. IITJEE – 2010 (Multiple choice) Among the following, the intensive property is (properties are)

(a) molar conductivity (b) electromotive force

(c) resistance (d) heat capacity

Ans : (a), (b)

2. IITJEE – 1993 (Multiple choice) Identify the intensive property from the following

(a) Enthalpy (b) Temperature

(c) Volume (d) Refractive Index.

Ans : (d)

3. IITJEE – 2001 In thermodynamics a process is called reversible when

(a) surrounding and system change into each other

(b) there is no boundary between system and surroundings

(c) surroundings are in equilibrium with system

(d) system change into surroundings spontaneously

Ans : (c)

4. IITJEE – 2001 Which of the following statements is false?

(a) Work is a state function

(b) Temperature is a state function

(c) Change of state is completely defined when initial and final states are specified

(d) Work appears at the boundary of the solution. Ans : (a) work is a path function

Problems



Isothermal ∆T = 0, ∆U = 0, ∆q = -w

Isobaric :- ∆P = 0, ∆H = ∆q

Isochoric :- ∆V = 0, W = 0, ∆U = ∆q

Adiavatic :- ∆q = 0, ∆U = W

Adiavatic reversible –

T Vγ-1

= constant

P Vγ = constant

TYP

1-γ = constant

W = Cv / R (P1V1 –P2V2) or P1V1 – P2V2

γ - 1

W = Cv (T1-T2) or – Cv ∆T

Isothermal reversible ;- W = -2.303nRT log V2/V1

W = -2.303nRT log P1 /P2

Cyclic ⇒ ∆E = 0

∆H = 0

∆S = 0

∆G = 0

Exercise on type of Process

(1) :- An ideal diatomic gas is caused to pass through a cycle shown on the P – V diagram in

figure where V2 = 3V1. If P1, V1 and T1 specify state 1, then the temperature of state 3 is :-

Isothermal

Pressure Isometric

Adiavatic

Volume

(a) (T1/3)1.4

(b) T1/(3)1.4

(c) T1/(3)0.4

(d) can not be determined

Solution:- State 1 ⇒ P1 V1 T1

State 3 ⇒ V2 = 3V1

Adiavatic expansion

TVY – 1

= constant

T1/T2 = (V2/V1)Y – 1

(diatomic gas y = 1.40)

T1/T2 = (V2/V1)1.40 – 1

T1/T2 = (3V1/V1)0.4

T2 = T1/(3)0.4

hence, (c)

(2) :- A given mass of gas expands from the state A to the state B by three path 1, 2, 3 as shown in

figure, W1, W2 & W3 respectively are work done by the gas along three paths. What is relation

between W1, W2 & W3. A

3

P 1 2

B

V



Solution :- Work is a path function hence W1 ≠ W2 ≠ W3

Work is slop of PV graph hence it is clear W1 < W2 < W3

(3) :- A sample of 2 kg of helium (ideal) is taken through the process ABC and another sample of

2Kg of the same gas is taken through the process ADC, then the temperature of state A & B are

B C

10

P

(104 N/m

2) 5 A D

10 20

V (m3)

Solution :- At state A

PV = nRT

T = PV/nR = 5 x 104 x 10

2000 x 8.3

4

= 120.5 K

At stage B

P is double

Hence TB = 2 x TA

= 241 K

(4) :– Two mols of a perfect gas undergoes the following processes

(a) a isobaric expansion from 1.0 atm, 20 lit to 1.0 atm, 40lit

(b) a isochoric change of state from 1.0 atm, 40 lit to 0.5 atm, 40 lit

(c) a reversible isothermal compression from 0.5 atm, 40 lit to 1.0 atm 20 lit

(i) sketch with labels each of the process on the same P-V diagram

(ii) calculate the total work & total heat change ‘q’ invalved in the above Process.

Solution : (i) PV = nRT

1.0 atm A B T = PV/ nR

= (1 x 20) / (2 x 0.082)

0.5 = 121.8 k

atm

20 lit 40 lit

(ii) work from A → B

W = - P ∆V

= - 1 (20) = -20 lit atm = -20 x 101.3

= - 2026 J

Work from B →C

= 0 (isobaric )

Work from C →A PV = nRT

= - 2.303 nRT log V2/ V1

= - 2.303 x 2 x 8.31 x 121.8 log 20/40



= +1404 J

Total work = W1 + W2 + W3

= - 2026 + 0 + 1404

= - 622 J

Cyclic hence ∆E = 0

∆E = ∆q + w

∆q = - w

= 622 J

(5) IITJEE – 2010 (integer type)

One mole of an ideal gas is taken from a to b along two paths denoted by the solid and he dashed lines as

shown in the graph below. If the work done along the solid line path is ws and that along the dotted line

path is wd, then the integer closest to the ratio wd/ws is

Solution :

Wd ⇒ irreversible ⇒ –Pexternal ∆V = W1 + W2 + W3 + - - - - -

= (– 4 × 1.5) + ( – 1 × 1) + (– 0.75 × 2.5) = – 8.67 lit. atm

WS ⇒ reversible ⇒ – 2.303 nRT log V2/V1 = Pinitial × Vinitial (as PV =nRT )

= – 2.303 × 4 × 0.5 log 5.5/0.5 = – 4.8 lit. atm

Wd/WS = – 8.67 / –4.8 = 2

Internal Energy : – ‘U’ or ‘E’

The sum of all kinds of energies of a system ie. Chemical, electrical, mechanical (kinetic & potential) etc,

is called as internal energy.

Internal energy of a system may change when –

(a) Heat passing into or out of the system

(b) Work is done on or by the system

(c) Matter leaves or enters the system.

First law of thermodynamics – ‘Energy can be neither created nor destroyed’. If both the system and the

surroundings are taken into account it also be stated as.

‘The internal energy ‘U’ of an isolated system is constant’.

∆U

∆UTotal = ∆USystem + ∆USurrounding = 0

Means according to first law

∆USystem = – ∆USurrounding

‘U’ for a gas can be increased by heating it in a flame or by doing compression work on it.

∆U = q + w



∆∆∆∆E or ∆∆∆∆U : - Change in internal energy : –

* HEAT OF REACTION MEASURED AT CONSTANT VOLUME.

* For

(i) Isothermal process (But the phase of substance should not change.)

(ii) Cyclic process

∆E or ∆U = 0

* According to first law of thermodynamics ∆E = ∆q + w

[Using this formula the most important considerations are –

(i) sign convension

(ii) unit of q & W ]

eg.

Q :- 500 J of heat was given to a system by which its volume increased from 10 lit ot 20 lit at 1

atmospheric pressure calculate change in interal energy .

Solution :- ∆q = +500 (Heat gives to system hence +ve)

∆E = ∆q + W ( W = - P ∆V)

= + 500 + ( -1 ×××× {20 - 10})

= + 500 – 10 = 490

This method can not be taken as W = - P ∆V

= - 1 × (20 - 10 )

= -10 lit atm not -10 J while ∆q is in Joule.

Hence when ∆E = ∆q + W is applied W should be calculated separately and converted into Joule.

W = - P ∆ V

= - 1 ×××× 10 = - 10 lit atm

= -10 × 101.3 J = - 1013 J

∆E = ∆q + W

= + 500 – 1013 = - 513 J

* In Bomb calorimeter the heat released is also called ∆E as bomb calorimeter works at constant

volume.

Q : - When 3.2 gm of C2H6 (g) was combusted in a bomb calorimeter the temprature of

calorimeter increased by 0.5°°°°C if thermal capacity of calorimeter is 200 kJ / k find heat of

combustion of C2H6 at constant volume.

Solution :- Heat released = Thermal capacity ×××× ∆T (Rise in temp)

= 200 × 0.5 = 100 kJ.

∆E & ∆H should be answerd in kJ / mol hence

3.2 gm = 3.2 / 32 = 0.1 mol

0.1 mol gives = 100 kJ

1 mol = 100 / 0.1 = 1000 kJ

∆E = - 1000 kJ / mol

Remember heat released is always taken –ve.

ISOTHERMAL PROCESS AND ∆∆∆∆E : - Isothermal means constant temperature. At constant

temperature ∆E is considered to be zero but only when th

state of substance does not chane eg.

Q : -Pressure of 2 gms of H2(g) is reduced from 10 atm to 1 atm at a constant temprature of 300

k, the gas behave idealy find (i) W (ii) q (iii) ∆∆∆∆E



Solution :- Pressure is decreasing continuously hence

Wrev = - 2.303 nRT log 2

1

P

P

= - 2.303 × 1 × 8.314 × 300 log 1

10

(n = 2/2 = 1 mol. Because work is measured in J or in cal hence . R = 8.314 (if in s)

or 1.98 (if in cal)).

= - 5744.1 J

W = - 5.74 kJ

As Isothermal process and state is not changing hence

∆∆∆∆E = 0

∆E = ∆q + w

∆ q = - W

= - ( - 5.74 kJ) = + 5.74 kJ

q = 5.74 kJ

If In Isothermal conditions state of substance get change than ∆∆∆∆E ≠≠≠≠ 0 eg.

Q : - Calculate (i) W (ii) ∆∆∆∆E for the conversion of 36 gm of water into steam at a tempature of

100°°°°C and at a pressure of 1 atmostpher. Latent heat of vaporisation of water is 40.7 kJ / mol.

Solution :- Asd H2O (l) → H2O (g)

Inspite of isothermal conditions ∆E ≠ 0

W = - P ∆V = - ∆ n RT

∆n = number of gaseous mols of product – number of gasesous mols of reactant

= 1 -0 = 1

As 2 mols of gas (36 gm ) hence ∆n = 1 ×××× 2 = 2

W = - ∆n RT

= - 2 ×××× 8.314 ×××× 373

= - 6202.2 J

= - 6.2kJ

q = 2 ×××× 40.7 kJ = 81.4 kJ

∆E = ∆Q + W

= 81.4 + (- 6.2)

= 75.2 kJ

∆∆∆∆H ⇒⇒⇒⇒ CHANGE IN ENTHALPY (Heat of reaction at constant pressure)

* Heat given to a substance at costant pressure & temperature to change the state (phase) is considered

∆H.

Q. : 40.67 kJ of Heat is given to vaporise 1 mol of H2O at 100°°°°C and 1 atmospheric pressure,

calculate ∆∆∆∆H & ∆∆∆∆E of the process.

Solution :- ∆H = 40.67 kJ

∆H = ∆E + ∆ nRT

∆E = ∆H - ∆nRT

(going for this formula remember there are several chances of mistakes eg.

∆H = in kJ

R = 8.314 J mol-1

k-1

)



H2 O (l) → H2O (g) ∆n = 1

∆E = 40.67 – 1 × 8.314 × 373

1000

= 40.67 – 3.10

∆∆∆∆E = 37.57 kJ

∆∆∆∆H in Bomb calorimeter

Heat change in bomb calorimeter is ∆E (as calorimeter works on constant volume)

∆∆∆∆H = ∆∆∆∆E + ∆∆∆∆n RT Before using this formula remember

(i) sign of ∆E (-ve if released)

(ii) ∆n = number of gaseous mols of products – number of gaseous mols of reactants.

[in calculating ∆n, H2O should be considered liquid if temp is < 373 k.]

(iii) most important is unit of ∆E & ∆nRT should be same.

Q :- 11.6 gm of isobutane (C4H10)gas was combusted in a bomb calorimeter at 300 k.The

temperature of calorimeter increases by 0.4°°°°C , if thermal capacity of calorimeter is 80 k cal k-1

find. (i) Heat of combustion of isobutane at 300 k and constant volume.

(ii) Heat of combustion of isobutane at 300 k and constant pressure.

Solution :- n = 11.6 / 58 = 0.2 mols

Heat released = 0.4 ×××× 80 = 32 k cal

∆E = - {32 × 1}/ 0.2 = - 160 kCal (Heat relesased )

C4H10 + 13/ 2 O2 → 4 CO2 + 5 H2O

(g) (g) (g) (l)

(Beacuse at 300 K)

∆n = 4 – ( 1 + 6.5) = - 3.5

∆H = ∆E + ∆n RT

= - 160 + ( -3.5 x 1.98 x 300 )

1000

= - 160 – 2.1

= 160 – 2.1

∆∆∆∆H = - 162 .1 k cal / mol

ENTHALPY AND TEMPRATURE :_

Kirchoff equation

∆Cp = H2 – H1

T2 – T1

Where H1 & H2 are enthalpies at T1 & T2 temprature respetively .

Q : - Enthalpy of vaporization of SO2 at -10°°°°C is 5950 cal/ mol calculate its value at - 25°°°°C.

CP SO2(l) = 206 cal / mol : Cp SO2 (g) = 93 cal / mol

Solution :- SO2(l) → SO2(g)

∆CP = CP SO2(g) – CPSO2(l)

= 93 – 206 = - 113

∆Cp = H2 – H1

T2 – T1

T1 = - 25°C = - 25 + 273 = 248 K

H1 = ?

T2 = -10°C = -10 + 273 = 263 K



H2 = 5950

-113 = 5950 – H1

263 – 248

- 113 = 5950 – H1

15

- 1695 = 5950 – H1 ; H1 = 7645 cal / mol.

Standard enthalpy of formation : – ∆H°f standard enthalpy of formation (∆H°f) is considered to be

zero for an element in its most stable naturally existing state at room temperature. eg –

∆H°f O2 (g) = 0 (naturally existing form at 298 of oxygen)

∆H°f O3 (g) ≠ 0 (naturally existing stable state of oxygen is O2)

∆H°f Cgraphite = 0

∆H°f CDiamond ≠ 0 (naturally existing stable of carbon is diamond)

∆H°f Br2(l) = 0

∆H°f Br2(g) ≠ 0 (Br2 exist as liquid at room temperature)

∆H°f Hg(s) ≠ 0

∆H°f Hg(l) = 0 (Hg exist as liquid at room temperature)

Similarly ∆H°f = 0 for Al(s), As grey (s), Ba(s), B(s), Ca(s), Ce(s), Cl2(g), Cu(s), D2(g), F2(g), H2(g),

H+(aq), I2(s), Pb(s), Mg(s), N2(g), P(s) white, K(s), Si(s), Na(s), S(s)rhombic, Sn(s)white, Zn(s)

* For H+(aq) all ∆H°f, ∆G° & S° are considered zero.

* For all the element in which ∆H°f is zero ∆G°f is also zero but S° is not equal to zero.

* ∆H°f H2O ≠ 0 Because it is a compound and ∆H°f is zero only for element not for compound.

* To form equation of standard enthalpy of formation of a substance

1. First write one molecule of product.

2. Write reaction in the form of element only

eg –

∆∆∆∆H°°°°f CH4

Cgraphite + 2H2 → CH4(g) ∆H = ∆H°f CH4 (g)

∆∆∆∆H°°°°f N2O3(g)

N2 + 3/2 O2 → N2O3(g) ∆H = ∆H°f N2O3 (g) (g)

Not

N2 + O3(g) → N2O3(g) ∆H ≠ ∆H°f N2O3 as naturally existing form of oxygen is O2 not O3

O3 O2

Not

2N2 + 3O2 → 2 N2O3 ∆H ≠ ∆H°f N2O3

Because here two moles of N2O3 are forming

2 1

1. IITJEE – 2010 (Multiple choice) The species which by definition has zero standard molar enthalpy of formation is 298 K is

(a) Br2 (g) (b) Cl2(g) (c) H2O (g) (d) CH4 (g)

Ans : (b)

Problem



RESONANCE ENERGY

In case of certain compound, the observed heat of formation differs significantly from the calculated

bond energy data. The difference between the observed value of heat of formation and the value

obtained from bond energy data is called the resonance energy of the compound. Resonance energy is

measure of extra stability of the compound. Resonance energy is measure of extra stability of the

compound. For example, the observed heat of formation of CO2 is 94 kcal and the value calculated

from bond energy data is 64 kcal, therefore, resonance energy of CO2 = 94 – 64 = 30 kcal

Q :- Calculate the resonance energy of N2O from the following data

∆∆∆∆Ho

f of N2O = 82 kJmol-1

Bond energies of N ≡≡≡≡ N, N = N, O = O and N = O bonds are 946, 418, 498 and 607 kJ-1

respectively.

Solution :- N2 + ½ O2 → N2O

N ≡ N + ½ (O = O) → N = N = O

∆H =[{(N ≡ N) + ½ ( O = O) – {(N = N) + ( N = 0)}]

= (946 + 249) – (418 + 60) = 170

R.E = ∆H°f Cal ~ ∆H°f ob = 88 kJ

Q : - The standard molar enthalpies of formation of cyclohexane (l) and benzene (l) at 25°°°°C are -

156 and +49 kJ mol-1

respectively. The standard enthalpy of hydrogenation of cyclohexene (l) at

25°°°°C is – 119 kJ mol-1

. Use these data to estimate the magnitude of the resonance energy of

benzene.

Solution :-Standard enthalpy of hydrogenation of cyclohexene = - 119 kJ mol-1

. This is the enthalpy of

hydrogenation of one double bond (present in cyclohexene). If benzene is considered as cyclohextriene,

the enthalpy of reaction

+ 3 H2(g) →

∆H°of this hydrogenation = 3 × (-119) = - 357 kJ mol-1

Thus, the observed heat of hydrogenation of benzene

(∆H°ob) = - 357 kJ mol-1

On the basis of above thermochemical equation

∆H°cal = [∆Hf° of cyclohexane (l) – (∆H°f of benzene (l) + 3 × ∆H°f of H2)]

= [-156 – (+49 + 3 ×××× 0)] = -205 kJ mol-1

Resonance energy = ∆H°ob - ∆H°cal = [-357 – (-205)] kJ mol-1

= - 152 kJ mol-1

Q : - Calculate the electronegativity of fluorine from the following data :

EH – H = 104 .2 kcal mol-1

, EF – F = 36.6 kcal mol-1

EH – F = 134.6 kcal mol-1

, XH = 2.1

Solution :- Let the electronegativity of fluorine be XF.

Applying Pauling’s equations

XF – XH = 0.208 [EH – F – (EF- F x EH - H)1/2

]1/2

In this equation, dissociation energies are taken in kcal mol-1

.

XF – 2.1 = 0.208 [ 134.6 – (104.2 x 36.6)1/2

]1/2

XF = 3.87

Q : - Calculate the electronegativity of carbon from the following data :

EH – H = 104 .2 kcal mol-1

, EC – C = 83.1 kcal mol-1

EC – H = 98.8 kcal mol-1

, XH = 2.1



Solution :- Let the electro negativity of carbon be XC.

Applying Pauling’s equations

XC – XH = 0.208 [EC – H – (EC - C x EH - H)1/2

]1/2

XC – 2.1 = 0.208 [ 98.8 – (83.1 x 104.23)1/2

]1/2

XC = 2.59

Q : - Ionisation potential and electron affinity of fluorine are 17.42 and 3.45 eV respectively.

Calculate the electronegativity of fluorine.

Solution : - According to Mulliken equation

X = IP + EA when both IP and EA are taken in eV.

5.6

XF = 17.42 + 3.45 = 3.726

5.6

Q : - The electron affinity of chlorine is 3. eV. How much energy in kcal is released when 2 g of

chlorine is completely converted to Cl- ion in a gaseous state?

Solution : - Cl + e → Cl- + 3.7 eV

35.5 3.7 x 23.06 kcal

∴ Energy released for conversion of 2 g gaseous chlorine into Cl- ions

= 3.7 x 23.06 x 2 = 4.8 kcal

35.5

Q : - The first ionistation potential of Li is 5.4 eV and the electron affinity of Cl is 3.6 eV.

Calculate ∆∆∆∆H in kcal mol-1

for the reaction

Li(g) + Cl (g) →→→→ Li+ + Cl

-

Carried out at such low pressures that resulting ions do not combine with each other.

Solution : - The overall reaction is written into two partial equations.

Li(g) → Li+ + e ∆E1 = 5.4 eV

Cl(g) + e → Cl- ∆E2 = - 3.6 eV

∆H = ∆E1 - ∆E2 = 5.4 – 3.6 = 1.8 eV

= 1.8 x 23.06 kcal mol-1

= 41.508 kcal mol-1



THERMODYNAMICS - II

SPANTANIOUS PRECESS

An irreversible process which can be reversed only by some external

agency. OR A process which either start by its own or by any external factor but remain continue till

the end is called a spontaneous process. Eg.

1. Melting of ice above 0°C

2. Combustion of L.P.G.

3. Combustion of coal

4. Dissolution of salt in water

5. Neutralization reaction when aced and base are mixed.

6. Mixing of two gases (diffusion)

7. Flow of heat from a region of higher temperature to a region of lower temp.

8. Evaporation of water etc.

NON SPONTANIUS PROCESS

A process which occur only when external work is carried out. A non

spontaneous process only remain continue till external factor is working. eg

1. In a refrigerator or air conditioner heat flow from a region of lower temp oto higher temp and it take

place only till electric current is passed.

2. electrolysis of water take place only in presence of electric work

WHAT WAS THE NEED OF SECOND LAW OF THERMODYNAMICS OR WHAT WAS

THE NEED OF TERM ENTROPY :

‘∆ H’ change in enthalpy can not be the criterion for

spontaneity of a process because

(I) THERE ARE SEVERAL SPONTANEOUS PROCESS IN WHICH ∆∆∆∆ H IS NEGATIVE. eg 1. Combustion of L.P.G., C.N.G. etc

CH4 + 2 O2 → CO2 + 2 H2O ∆H = –ve

2. Neutralization of Acid/Base

HCl + NaOH → NaCl +H2O ∆H = –ve

3. Formation of NH3

N2 + 3 H2 → 2NH3 ∆H = –ve

4. Combustion of Hydrogen

H2 + 2

1 O2 → H2O ∆H = –ve

BUT

(II) THERE ARE SEVERAL SPONTANEOUSLY PROCESS IN WHICH ∆H IS ‘+VE’ . Eg

1. Melting of ice above 0°C

H2O(s) → H2O(l) ∆H = +ve

2. Evaporation of H2O

H2O(l) → H2O(g) ∆H = +ve

3. Dissolution of salt in water

H2O

NaCl(s) NaCl(aq) ∆H = +ve

MEANS It becomes obvious that while decrease in enthalpy may be a contributory factor for

spontaneity but it is not true for all cases.

To explain spontaneity an another term was require – ENTROPY, certainly second law of

thermodynamics.



ENTROPY :– ‘S’ It is the measurement of the degree of randomness or disorder of a system.

Greater the entropy more the system is disordered (Chaotic)

* Entropy is a state function ∆S = Sf – Si

* When a system is in equilibrium its entropy is maximum and at equilibrium ∆S = O

∆Suniv = ∆STotal = ∆Ssystem + ∆SSurrounding

for Irreversible process ∆Suni > O

for Reversible process at equilibrium ∆Suni = O

� 2nd

Law of Thermodynamics :–

∆Suni = ∆Ssyst + ∆Ssurrounding

for a spontaneous process Entropy of universe always increases.

� 3rd

Law of Thermodynamics :– Entropy of a ‘perfectly crystalline solid’ is considered to be zero at

zero Kelvin temp (absolute zero)

From the 3rd

law it is obvious that S° can not be zero for any substance because S°

means slandered entropy i.e measured at 298 K and entropy can be considered zero at zero

Kelvin only.

� ∆∆∆∆S +ve :– The process in which randomness increases

1. Solid → liquid

2. Liquid → gas

3. Solid → gas

4. A reaction in which ∆n > O

5. Mixing of two or more gases/liquid/solid in which reaction is not taking place.

eg. Diffusion of gases

6. Increase in temperature/decreases in pressure

7. If temp and pressure both are changed find value of P

T if it increases than ∆S = +ve

8. Breaking of a bond.

9. Dissolution of a solid e NaCl(s) NaCl(aq)

10. Decomposition of a solid into two or more solid/ ‘l’’/where ∆n can not calculated

Pb3O4 (s) 2 PbO + pbO2

(s) (s)

eg (2) 2HCl H2 + Cl2

(g) (g) (g)

� ∆∆∆∆S –ve :– Apposite of process in which ∆S is +ve particularly

gas liquid or solid

liquid solid, mixing of two or more reacting gases/ substance.

eg. Mixing of H2 + He ⇒ ∆S = +ve

Mixing of H2 & Cl2 ⇒ ∆S = –ve

reacting

Crystallization

NaCl (aq) NaCl(s) etc.

* for an isothermal reversible process

∆S = T

qrev



GIBBS FREE ENERGY : –

If a reaction is carried out in a isolated system or thermally insulated

container than spontaneity means positive value of ∆Suni. BUT

Most of reaction are carried out in open or closed system then both enthalpy and entropy charge take place

hence both should be considered to explain spontaneity.

GIBBS FREE ENERGY (G) IS THE ENERGY OF A SYSTEM WHICH CAN BE

CONVERTED INTO USEFUL WORK AT A GIVEN TEMPERATURE

G = H – TS

∆G = Gf – Gin (state function, extensive property)

∆G = ∆H – ∆(TS)

at constant temp ∆G = ∆H – T∆S

* ∆G = –T ∆STOTAL

Let at constant temp (T) and constant pressure (P) ‘q’ foule of heat is given to system

Constant pressure hence

S ∆H = q

YST q ∆Hsystem = + q

EM ∆Hsystem = – q

∆ Ssystem = T

q =

T

H∆

∆Ssurrounding = T

H∆− – 1

∆STotal = ∆SSystem + ∆SSurrounding – 2

By equation 1 & 2

∆STotal = ∆SSystem + T

H∆−

T ∆STotal = ∆SSystem – ∆H (∆G = ∆H – T∆S)

T ∆STotal = –∆G

∆G = – T∆STotal

* ∆G = Wnon expansion or ∆G = Wmax

According to first law of thermodynamics ∆U = ∆q +W

W = Wexpansion + Wnon expansion

∆U = ∆q + Wexpansion + Wnon expansion

∆q = ∆U – Wexpansion – Wnon expansion

= ∆U – (– P ∆ V) – Wnon expansion

= ∆U + P ∆ V – Wnon expansion

∆H = ∆U + P∆V

∆q = ∆H – Wnon expansion – 1

∆S = T

q∆ , ∆q = T ∆ S

T ∆ S = ∆H – Wnon expansion Wnon expansion = ∆H – T ∆ S

Wnon expansion = ∆G



OVERALL CRITERION FOR SPONTANEITY

For spontaneous process ∆G = –ve

If ∆G = O system is in equilibrium

If ∆G < O Process is irreversible (Spontaneous)

If ∆G > O Process is non spontaneous

EXOTHERMIC REACTION (∆H = –ve)

(i) if ∆S = +ve reaction is spontaneous at all temp.

(ii) if ∆ S = –ve, ∆G = ∆H – T ∆ S

∆G = – ∆H – T (– ∆S)

∆G = – ∆H + T∆S

Can be negative when T∆S < ∆H

means if ∆H = –ve, ∆S = –ve reaction is spontaneous only when T∆S < ∆H certainly low temp

favour spontaneity for such reaction.

ENDOTHERMIC REACTION (∆H = +ve)

(i) if ∆S = +ve

∆G = +∆H – T (+∆S)

∆G = ∆H – T∆S

Can be negative only when T∆S > ∆H mean if for a reaction ∆H = +ve & ∆S = +ve

reaction can be spontaneous if T∆S > ∆H mean high temp will favour spontaneity

(ii) if ∆S = –ve

∆H = +ve, ∆S = –ve ⇒ ∆ G = +OH – T (–∆S)

∆ G = +∆H + T∆S

Reaction can not be

spontaneous at any Can not be negative

condition

Conclusion

(i) ∆H = –ve , ∆S = +ve ⇒ spontaneous always

(ii) ∆H = –ve , ∆S = –ve ⇒ spontaneous only when ∆H > T∆S (low temp)

(iii) ∆H = +ve , ∆S = +ve ⇒ spontaneous only when ∆H < T∆S (High temp)

(iv) ∆S = +ve , ∆S = –ve ⇒ can never be spontaneous

∆∆∆∆G & EQUILIBRIUM

At equilibrium ∆G = O but ∆G° ≠ O (usually)

∆G° = – 2.303 RT log KC

∆G° = – 2.303 RT log KP

∆∆∆∆G°°°° & emf

∆G° = – n F E°Cell

COUPLING OF REACTION : –

A → B ∆G1 = +ve (small value)

C → D ∆G2 = –ve (large value)

Coupling

A B

∆G = ∆G1 + ∆G2 = –ve

C D

Both reaction take place spontaneously

eg. Metallurgical operations ⇒ Reduction of metal oxide into metal (∆G = +ve) + Oxidation

of carbon/carbon mono oxide.

∆G = ∆G° + RTl n K



I.I.T. – 2006 (1) :- The process A →→→→ B is a difficult process hence carried out in several step as follows C D

A B

Given ∆S(A → C) = 50e.u. ; ∆S (C → D) = 30e.u.

∆S (B → D) = 20e.u.

where e.u. is entropy unit then ∆S(A → B) is



NCERT PROBLEMS

1. Express the change in internal energy of a system when

(i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system.

What type of wall does the system have?

(ii) No work is done on the system, but q amount of heat is taken out form the system and given to the

surroundings. What type of wall does the system have ?

(iii) w amount of work is done by the system and q amount of heat is supplied to the system. What

type of system would it be?

Solution : ∆ E system = ?

(i) q = 0, W = +Ve

∆E = q + W ∆ E = W, Adiavatic walls. (∆q = 0)

(ii) W= 0 , q = –ve ∆ E = – q (Negative)

∆ E = q + W Walls through which heat can flow ⇒ conducting walls.

(iii) W = –ve , q = +ve

∆ E = q + W Conducting walls, closed system.

= q – W

2. Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total

volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Solution : In vacuum Pext = 0

W = – Pext ∆V = 0, W = 0

Isothermal expansion ∆E = 0

∆E = q + W

0 = q + 0 , q = 0

3. Consider the same expansion, but this time against a constant external pressure of 1 atm.

Solution :

W = – Pex ∆V = – 1 × (10 – 2) = – 8 lit atm

4. Consider the same expansion, to a final volume of 10 litres conducted reversibly

Solution :

Wrev = – 2.303 nRT log 1

2

V

V as temperature and number of mols are not given

= – 2.303 × 20 log 2

10 but PV = nRT

= – 2.303 × 20 (0.6990) nRT = PV

= –32.196 lit atm = 10 × 2 = 20

∆E = 0

∆E = q + W, q = – W

= – ( – 32.196) = +32.2 lit atm

= 32.2 × 101.3 = 3261.86 J

5. If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol

water at 1 bar and 100°C is 41 kJ mol–1

. Calculate the internal energy change, when

(i) 1 mol of water is vaporised at 1 bar pressure and 100° C.

(ii) 1 mol of water is converted into ice.

Solution :

∆H = 41 kJ , ∆E = ?

H2O (l) H2O(g) ∆n = 1 – 0 = 1



∆H = ∆E + ∆nRT

∆E = ∆H – ∆nRT As ∆H is in kJ and

= 41 – 1×8.314×373 R = 8.314 J/mol/K

1000 hence R = 8.314 kJ mol–1

K–1

1000

= 37.904 kJ/mol

(ii) H2O (l) H2O(s) (It is an exothermic process)

∆H = – ve (Value of enthalpy of fusion is not given)

As volume H2O(l) is approximately same to H2O(s) means ∆nRT = –P∆V would be considered

zero

∆E = ∆H But ∆H should be –ve

6. 1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure

according to the equation

C (graphite) + O2 (g) → CO2 (g)

During the reaction, temperature rises from 298 to 299 K. if the heat capacity of the bomb calorimeter

is 20.7 kJ/K. What is the enthalpy change for the above reaction at 298 K and 1 atm?

Solution : A bomb calorimeter works at constant volume hence heat released per mol = ∆E

Heat released = Thermal capacity × ∆T

= 20.7 × 1 = 20.7 kJ

1 gm graphite = 12

1 mol

∆E = – 20.7 × 12 = –248 kJ (As heat is released ∆E = –ve)

Cgr + O2 CO2 (g) ∆n = 1 – 1 = 0 (g)

∆H = ∆E + ∆nRT

∆H = ∆E = – 248 kJ/mol

7. A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much

heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at

100°C.

∆vapH for water

at 373K = 40.66 kJ mol–1

Solution : ∆H vap of H2O = 40.66 kJ mol–1

Means

H2O (l) H2O (g) ∆H = 40.66 kJ

1 mol = 18 gm water will require 40.66 kJ heat

∆H = ∆E + ∆nRT

H2O (l) H2O(g)

∆n = 1 – 0 = 1

∆E = ∆H – ∆nRT

= 40.66 – 1 × 8.314 × 373 As ∆H is in

1000 kJ, R = 8.314

1000

= 37.56 kJ

8. The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and

H2O (1) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation,

∆f H of benzene. Standard enthalpies of formation of CO2(g) and H2O(1) are – 393.5 kJ mol–1

and

– 285.83 kJ mol–1

respectively.

Solution :

3267 kJ of heat is released

∆H = –3267 kJ



C6H6 + 2

15 O2 6 CO2 + 3H2O

∆H = (6 H°f CO2 + 3 H°f H2O) – (H°f C6H6)

–3267 = 6 × (–393.5) + 3 (–285.83) – (x)

x = 48.51 kJ

Alternatively

Enthalpy of combustion of benzene = –3267 kJ

1 C6H6 + 2

15 O2 6 CO2 + 3H2O ∆H = –3267

Enthalpy of formation of CO2 = –393.5 kJ

2 Cgr + O2(g) CO2(g) ∆H = –393.5 kJ

Enthalpy of formation of H2O(l) = –285.83 kJ

3 H2(g) + ½ O2(g) H2O(l) ∆H = –285.83 kJ

∆H°f C6H6 = ?

4 6 Cgr + 3H2(g) C6H6 ∆H = ?

To form equation No 4 from (1), (2) & (3) multiply equation No. 2 by six and equation number 3

by three and add them. Substract equation number one from resulting equation.

∆H°f benzene = 6 ∆H2 + 3 ∆H3 – ∆H1

= 48.51 kJ

9. Predict in which of the following, entropy increases/decreases :

(i) A liquid crystallizes into a solid.

(ii) Temperature of a crystalline solid is raised from 0 K to 115 K.

(iii) 2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (g)

(iv) H2 (g) → 2H (g)

Solution :

(i) liquid → solid (Entropy decreases ∆S = –ve)

(ii) Temperature is increased means randomness will increase ∆S = +ve

(iii) 2 NaHCO2(s) → Na2CO3(s) + CO2(g) + H2O(g)

∆n = 2 – 0 = 2

As ∆n is +ve entropy will increases (a substance is decomposing means ∆S = +ve)

(iv) H2(g) → 2H(g) ∆n = 2 – 1 = 1

∆S = +ve

10. For oxidation of iron.

4Fe (s) + 302 (g) → 2Fe2O3 (s)

Entropy change is – 549.4 JK–1

mol–1

at 298 K. Inspite of negative entropy change of this reaction,

why is the reaction spontaneous?

(∆r H

for this reaction is – 1648 × 103 J mol

–1)

Solution :

∆ H = –1648 × 103 J/mol, ∆S = –549 J K

–1 mol

–1

∆ G = ∆H – T ∆ S

= –1648 × 103 – 298 (–549)

= –1648000 + 163602 = –1.48 × 106 J

As ∆G is –ve reaction is spontaneous.



Alternatively

∆STotal = ∆Ssystem + ∆Ssurrounding

∆Ssystem = –549 J K–1

mol–1

∆Ssurrounding = ∆H surr = – ∆H sys

T T

= ( )

298

101648 3×−− = 5.53 × 10

3 J

∆STotal = –549 + 5530 = 4980.6 J (+ve) means spontaneous

11. Calculate ∆r G for conversion of oxygen to ozone. 3/2 O2(g) → O3(g) at 298 K. if Kp for this

conversion is 2.47 × 10–29

.

Solution :

∆G° = –2.303 RT log Kp

= – 2.303 × 8.314 × 298 log 2.47 × 10–29

= 163 × 103 J

= 163 kJ

12. Find out the value of equilibrium constant for the following reaction at 298 K.

2NH3 (g) + CO2 (g) NH2 CONH2 (aq) + H2O (l) Standard Gibbs energy change, ∆rG at the

given temperature is – 13.6 kJ mol–1

.

Solution :

∆G° = –13.6 kJ Very important is

∆G° = –2.303 RT log Kc Unit system

–13.6 × 103 = –2.303 × 8.314 × 298 × log Kc • ∆G is kJ

log Kc = 2.38 R is in J

Kc = 2.4 × 102 • Just learn

2.303 × 8.314

= 19.15 J

13. At 60°C dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at

this temperature and at one atmosphere.

Solution : N2O4(g) 2 NO2(g)

1 0

1 – α 2 α

Total mols at equilibrium = 1 – α + 2 α = 1 + α

α = 50% = 0.5

P2NO =

α

α

+1

2× P = P×

×

5.1

5.02 = P×

5.1

1 =

5.1

1 =

3

2

P42ON =

α

α

+

−

1

1× P = P×

5.1

5.0 = P×

5.1

5.0 =

5.1

05 =

3

1

(P = 1atm)

Kp =

42

2)(

ON

NO

P

P =

3

13

2

3

2×

= 3

4 = 1.33

∆G° = –2.303 RT log Kp

= –2.303 × 8.314 × 333 × log 1.33

= –7.638 × 105 J = –763.8 kJ



NCERT EXERCISE

1. Choose the correct answer. A thermodynamic state function is a quantity

(i) used to determine heat changes

(ii) whose value is independent of path

(iii) used to determine pressure volume work

(iv) whose value depends on temperature only.

Solution : (ii) State functions are independent of path.

2. For the process to occur under adiabatic, conditions the correct condition is:

(i) ∆T = 0 (ii) ∆p = 0 (iii) q = 0 (iv) w = 0

Solution :

(iii) Adiavatic ⇒ ∆q = 0

3. The enthalpies of all elements in their standard states are :

(i) unity (ii) zero (iii) < 0 (iv) different for each element

Solution : (ii) Enthalpy of formation of all elements in their naturally existing most stable form = 0

4. ∆U of combustion of methane is – X kJ mol–1

. The value of ∆H is

(i) = ∆U (ii) > ∆U (iii) <∆U (iv) = 0

Solution : CH4(g) + 2 O2(g) → CO2(g) + 2 H2O

If Temp < 373 K H2O is taken as liquid then

∆n = 1 – 3 = –2

∆H = ∆E – 2 RT = –x – 2RT ∆∆∆∆H > ∆∆∆∆E

If temp ≥ 373 K H2O is taken as gas

∆n = 3 – 3 = 0 ∆H = ∆E

5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol–1

-393.5

kJ mol–1

, and – 285.8 kJ mol–1

respectively. Enthalpy of formation of CH4(g) will be

(i) – 74.8 kJ mol–1

(ii) – 52.27 kJ mol–1

(iii) + 74.8 kJ mol–1

(iv) + 52.26 kJ mol–1

Solution :

CH4 + 2O2 → CO2 + 2H2O ∆H = –890.3 kJ Enthalpy of combustion

–890.3 = (H°f CO2 + 2H°f H2O) – (H°f CH4) of ‘C’ graphite

–890.3 = [–393.5 + 2 (–285.8)] – (x) = H°f CH4

x = (–393.5 – 571.6) + 890.3 similarly enthalpy of combustion of

= – 74.8 kJ/mol (i) H2 = H°f H2O

6. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(iv) possible at any temperature Solution :

A + B → C + D + q Heat

means it is an exothermic reaction

∆H = –ve, ∆S = +ve ] reaction possible at any temperature (iv)

7. In a process, 701 J of heat is absorbed by a system an 394 J of work is done by the system. What is the

change in internal energy for the process? Solution :

q = 701 J , W = –394

∆E = q + W = 701 – 394 = 317 J



8. The reaction of cynamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U

was found to be – 742.4 kJ mol–1

at 298 K. Calculate enthalpy change for the reaction at 298/ K.

NH2CN(g) + 2

3 O2 (g) → N2(g) + CO2(g) + H2O(l)

Solution :

∆U = –742.7 kJ, ∆H = ?

NH2 CN(g) + 2

3O2(g) → N2(g) + CO2(g) + H2O(l)

∆n = 2 – 2.5 = –0.5

∆H = ∆U + ∆nRT = –742.7 + (–0.5)×8.314×298

1000

= –742.7 – 1.239

= –743.939 kJ

9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C

to 55°C. Molar heat capacity of Al is 24 J mol–1

K–1

.

Solution :

Molar heat capacity means heat required to raise the temperature of 1mol by 1°C

∆H = nCp ∆T = 60 × 24 × (55–35)

27

= 1066.7 J = 1.067 kJ

10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C.

∆fusH = 6.03 kJ mol–1

at 0°C.

Cp [H2O (l)] = 75.3 J mol–1

K–1

Cp [H2O (s)] = 36.8 J mol–1

K–1

Solution :

1 mol

Water (10°C) →− 1H

water (O°C) →− 2H

Ice (O°C)

⇓ Enthalpy – H3

of fusion

Ice (–10°C)

H1 = Cp H2O(l) ∆T

= 75.3 × –10 = –753 J

H2 = – 6.03 kJ = –6030 J

H3 = Cp H2O(g) ∆T = 36.8 × (–10) = –368

∆H = H1 + H2 + H3 = –753 + (–6030) + (–368)

= –7151 = –7.151 kJ

11. Enthalpy of combustion of carbon to CO2 is – 393.5 kJ mol–1

. Calculate the heat released upon

formation of 35.2 g of CO2 from carbon and dioxygen gas.

Solution :

35.2 gm CO2 = 44

2.35 mol

∆H = –393.5 × 35.2 = –314.8 kJ

44



12. Enthalpy of formation of CO(g),CO2(g) N2O(g) and N2O4 (g) are – 110, - 393, 81 and 9.7 kJ mol–1

respectively. Find the value of ∆rH for the reaction :

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Solution : N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

∆ H = (H°f N2O + 3H°f CO2) – (H°f N2O4 + 3H°f CO)

= (81 + 3 (–393)) – (9.7 + 3 (–110))

= (81 – 1179) – (9.7 – 330)

= – 777.7 kJ

13. Given

N2(g) + 3H2(g) → 2NH3(g); ∆rH = - 92.4 kJ mol–1

What is the standard enthalpy of formation of NH3 gas?

Solution :

For two mols of NH3 ⇒ – 92.4 kJ

∆H°f NH3 = –92.4 = –46.2 kJ

2

14. Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

CH3OH (l) + 2

3 O2(g) → CO2(g) + 2H2O(l) ; ∆rH = - 726 kJ mol

–1

C(graphite) + O2(g) → CO2(g) ; ∆cH = - 393 kJ mol

–1

H2(g) +

2

1O2(g) → H2O(l) ; ∆f H = - 286 kJ mol

–1.

Solution :

∆H°f CH3OH(l)

Cgr + 2H2 + ½ O2 → CH3 OH ∆H = ?

(1) CH3OH(l) 2

3O2(g) → CO2(g) + 2H2O (l)

∆H1 = –726 kJ

(2) Cgr + O2(g) → CO2(g)

∆H2 = –393 kJ

(3) H2(g) + ½ O2(g) → H2O(l)

∆H3 = –286 kJ

To obtain above equation from equation (1), (2) & (3) Multiply equation No. (3) by two and add it

with equation No. (2), substract equation No. (1) from this reaction

∆ H = (2(∆H3) + ∆H2) – ∆H2) – ∆H1

= –572 + (–393) – (–726)

= –239 kJ

15. Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl4(g) .

∆vapH (CCl4) = 30.5 kJ mol

–1.

∆fH (CCl4) = - 135.5 kJ mol

–1.

∆aH

(C) = 715.0 kJ mol–1

, where ∆aH

is enthalpy of atomization

∆aH (Cl2) = 242 kJ mol–1

Solution :

First find ∆H of reaction

CCl4(g) → C(g) + 4Cl(g)



(1) ∆Hvap of CCl4

CCl4(l) → CCl4(g)

∆H = 30.5 kJ

(2) ∆H°f CCl4(l)

Cgr + 2 Cl2(g) → CCl4 (l)

∆H = –135.5 kJ

(3) ∆H°fatom C(gr)

Cgr → C(g) ∆H = 715 kJ

(4) ∆Hatom Cl2

Cl2(g) → 2 Cl(g)

∆H = 342 kJ

To obtain reaction

CCl4(g) → C(g) + 4Cl(g)

Multiply equation No. 4 by two and add it with equation No. 3.

(5) Cgr + 2Cl2 → C(g) + 4Cl(g)

∆H = 2 (242) + 715 kJ

Add equation No. (1) & (2)

(6) C(gr) + 2Cl2(g) → CCl4(g)

∆H = 30.5 – 135.5 kJ

Subtract equation No. 6 from equation No. 5

∆H = 2(242) + 715 – (30.5 – 135.5)

= 484 + 715 – 30.5 + 135.5

= 1304 kJ/mol.

B. Energy C – Cl bond

= 4

1304 = 326 kJ/mol

(As the equation obtained is enthalpy of atomisation of CCl4)

16. For an isolated system, ∆U = 0, what will be ∆S ?

Solution :

For a spontaneous process carrying out in a isolated system if ∆U = 0 then certainly ∆S would be

driving force ∆S = +ve means ∆S > 0

17. For the reaction at 298 K,

2A + B → C

∆H = 400 kJ mol–1

and ∆S = 0.2 kJ K–1

mol–1

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over

the temperature range.

Solution :

For spontaneity

∆G = –ve

∆G = ∆H – T∆S

O (minimum condition for spontaneity)

T∆S = ∆H

T = S

H

∆

∆ =

2.0

400

= 2000 K



18. For the reaction

2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?

Solution : 2 Cl(g) → Cl2(g)

Bond formation is exothermic ∆H = –ve

∆n = 1 –2 = – 1 hence

∆S = –ve

19. For the reaction

2 A(g) + B(g) → 2D(g)

∆U

= - 10.5 kJ and ∆S = - 44.1 JK–1

.

Calculate ∆G

for the reaction, and predict whether the reaction may occur spontaneously.

Solution : 2 A(g) + B(g) → 2 D(g) = –10.5 – 2.48

∆n = 2 – 3 = –1 = –12.98 kJ

∆H°f = ∆U° + ∆nRT ∆G° = ∆H° – T∆S°

= –10.5 + (–1)×8.314×298 = –12.98 – 298 – 44.1

1000 1000

= –12.98 + 13.14

= +0.16 kJ

Reaction is non spontaneous

20. The equilibrium constant for a reaction is is 10. What will be the value of ∆G ? R = 8.314 JK–1

mol–1

,

T = 300 K. Solution : Kc = 10

∆G° = –2.303 RT log Kc

= –2.303 × 8.314 × 300 log 10

= –19.15 × 300

= –5745 J

= –5.745 kJ

21. Comment on the thermodynamic stability of NO(g), given

2

1 N2(g) +

2

1 O2(g) → NO(G) ; ∆rH = 90 kJ mol

–1

NO(g) +

2

1 O2(g) → NO2(g) ; ∆rH = - 74 kJ mol

–1

Solution :

As conversion of NO into NO2 is exothermic ‘NO’ would not stabilize in air.

22. Calculate the entropy charge in surroundings when 1.00 mol of H2O(l) is formed under standard

condition. ∆fH

= - 286 kJ mol–1

.

Solution : ∆H°f H2O = –286 kJ

Means ∆Hsurr = +286 kJ

T = 298

∆S = ∆H

T

= 298

286× 1000

= 959.7 J mol–1

K–1

∆Ssurr = 959.7 J mol–1

K–1

THERMODYNAMICS Mittal Sir - BMC CHEMISTRYbmcchemistry.com/temp/1411270731thermodynamics2014.pdf ·...

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