Answer key

THERMODYNAMICS

Class: 12 Worksheet Number: 11/Chem/07/QB

Subject: Chemistry Chapter: Thermodynamics

First law of Thermodynamics

Section A
Each question carries 1 mark :
1. Give the mathematical expression of first law of thermodynamics.
∆U = q + w
Where ∆U = change in internal energy
q = heat absorbed or released
w = work done by the system or on the system
st
2. Derive the mathematical expression for 1 law of thermodynamics and state its limitations.
First law of Thermodynamics:
Let a system having internal energy ‘u’ absorb certain amount of energy ‘q’
Internal Energy = q + U1
If work done is ‘w’ on it, Internal Energy becomes ‘U2’
U2 = U1 + q +w
U2 - U1 = q +w
ΔU = q+w
st
Limitations of 1 law of thermodynamics
st
1 law merely indicates that there is an exact equivalence between various forms of energies involved but gives no information concerning the spontaneity or feasibility of the
process i.e. whether the process is possible or not.
3. Why is U = O for the isothermal expansion of an ideal gas?
As Change in internal energy U is a function of temperature and in isothermal expansion temperature remains constant as the system can exchange heat with the surroundings.
Internal energy hence remains unchanged or U = O.
4. 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 from 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?
(i) Δ U = w ad, wall is adiabatic
(ii) Δ U = – q, thermally conducting walls
(iii) Δ U = q – w, closed system.
5. What is an adiabatic process? Give one difference between isothermal and adiabatic processes?
A process during which no heat flows between the system and the surroundings is called an adiabatic process.

Isothermal process Adiabatic process

Temperature of the process remains constant. There is no exchange of heat between the system and surrounding.

Heat can be gained or lost in order to maintain a constant temperature Temperature of the system may vary.

6. What would be the work done on / by the system if the internal energy of the system falls by 200 J even when 100 J of heat is supplied to it?
ΔU = -200 J ; q = 100 J
ΔU = q + w
-200 = 100 + w
-100 -200 = w
W= -300j
7. Define molar heat capacity.
Molar heat capacity of a substance is defined as the amount of heat required to raise the temperature of one mole of the substance through 1°C.
8. Explain the thermodynamic equilibrium?
Thermodynamic equilibrium is achieved when any of its macroscopic properties such as T, P, and V do not undergo any change with time.
9. Define Enthalpy of a reaction or heat of reaction
Enthalpy change is the amount of heat evolved or absorbed in a chemical reaction when the number of moles of the reactants as represented by the chemical equation has been
completely reacted, is called enthalpy change of a reaction.
Reactants → Products

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10. State the Hess’s Law of constant heat summation?

If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reaction into which the overall reaction may be
divided at the same temperature.
11. Enthalpy of formation of graphite is zero. Why?
ΔH of graphite is zero as it is the standard state of carbon.
12. What is Gibbs’s Helmholtz equation?
∆G = ∆H - T∆S
Where ∆G ,∆H and ∆S are free energy change, enthalpy change and entropy change respectively.

Section B

Each question carries 2 marks :

13. Define a system. What are open, closed and isolated systems? Give examples.
System- the part of the universe chosen for thermodynamics consideration is called a system
Open system- system that can exchange both matter and energy with the surroundings.Eg: hot tea in a cup
Closed system-system that can exchange energy but not matter with the surroundings.Eg: hot tea in a closed pot
Isolated system-a system that can neither exchange energy nor matter with the surroundings. Eg: hot tea in thermos flask.
14. What are intensive and extensive properties? Give two examples of each.
Extensive properties- are those properties which depends upon the quantity of matter contained in the system e.g mass, enthalpy
Intensive properties are those which depends only upon the nature of the substance and are independent of the amount of substance percent in the system.eg dustily, boiling point
15. Change in internal energy is a state function while work is not. Why?
Internal energy depends on initial and final state but does not depend on path of the system. Work is not a state function as it does not depend merely on the initial and the final
states of the system but it depends upon the path followed
16. Derive the relation between Δ H and Δ U.
H =U +PV
H1 = U1 +P1V1
H2 = U2 +P2V2
ΔH =H2 –H1
ΔH =(U2 +P2V2) – (U1 +P1V1)
ΔH = (U2-U1) +( P2V2 –P1V1)
(P1 =P2 =P)
ΔH = ΔU +P(V2-V1)
ΔH = ΔU +PΔV
17. Under what conditions is (i) ΔH = ΔU (ii) ΔH < ΔU. Give example in each case.
i) ΔH = ΔU when the process involves only solids and liquids, i.e no gases. ΔV=0 or when Δng=0 e.g. H2 (g) +Cl2(g) 2 HCl(g)
ii) ΔH < ΔU. when no. of moles of gaseous products are less then no. of moles of reactants for a reaction
e.g. CH4(g) +2O2(g)→ CO2(g) + 2 H2O (l)Δn = [1-3] = -2
18. Is the bond energy of all the four C-H bonds in CH4 molecule equal? If not why?
All the four C-H bonds in CH4 molecule are not equal as the chemical environment will change.
19. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process
The first law of thermodynamics states that : ΔU = q + W (i)
Where,
ΔU = change in internal energy for a process
q = heat
W = work
Given in the question,
q = + 701 J (Positive since heat is absorbed)
W = –394 J (Negative since work is done by the system)
Substituting the values in equation (i), we get
ΔU = 701 J + (–394 J)
ΔU = 307 J
Therefore, the change in internal energy for the process is 307 J.

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20. Two moles of an ideal gas initially at 27 C and one atmospheric pressure are compressed isothermally and reversibly till the final pressure of the gas is 10 atm. Calculate q, w
and U for the process.
-1 -1
n=2, R= 8.314 jk mol ;T=(27 +273) = 300K
W max= 2.303 nRT log 10 p1/p2
= 2.303 x 2 x 8.314 x 300 x log 10/1
= 11488j
ΔU = q + w
The condition is isothermal and reversible ΔT= 0;ΔU=0
0 = q +W
-q =w
q= -11488j ; w = 11488j ;ΔU=0
21. Explain the applications of Born Haber Cycle with an example?

Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber Cycle.
22. A gas absorbs 120 J of heat and expands against external pressure 1.1 atm from volume of 0.5 L to 2.0 L. What is the change in internal energy?
W = -PΔV = -1.1 X (2 – 0.5)
=-1.1 X 1.5 = -1.65 L-atm
1L-atm = 101.3 J
-1.65 L-atm = -167.145 j
ΔU = q + w
= 120 + (-167.145)
= -47.145 J
-11
23. Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol . Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas?

24. Given N2 + H2 → NH3

What is the standard enthalpy of formation of NH3 gas?

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–1
25. Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol . Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas?

0
26. The heat liberated on complete combustion of 7.8 g benzene is 327 kJ. The heat has been measured at constant volume at 27 C. Calculate the heat of combustion of benzene at
-1 -1
constant pressure. ( R= 8.314 JK mol )

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27. Calculate the enthalpy of formation of HF on the basis of the following data :
Bond energy of H-H = 434 kJ/mol
Bond energy of F-F = 158 kJ/mol
Bond Energy of H-F= 565 kJ/mol
H-H + F-F → 2 H-F
ΔH = ∑ B.E (reactant )- ∑ B.E (product)
= (434 + 158) – (2 x 565)
=592 -1130
= -538 kj for 2 moles of HF
= -538/2 = -269 KJ/mol
28. Calculate the bond enthalpy of Cl – Cl bond from the following data :
-1
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) H = -100.3 kJmol
-1
Also the bond enthalpies of C – H, C – Cl and H – Cl bonds are 413, 326 and 431 kJ mol respectively.
Let B.E. of Cl-Cl bond be X kj/ mol
ΔH = ∑ B.E. (R) - ∑ B.E. (P)
100.3 = (413 x 4 x X) - (431 + 431 x 3 + 326)
-100.3 = 1652 + X – 1996
X = -100.3 – 1652 + 1996
X = 243.7 kj/ mol
-100.3 = 1652 + X – 1996
X = -100.3 – 1652 + 1996
X = 243.7 kj/ mol
0
39. Calculate the ΔrH for the reaction
H2(g) + Br2(g) → 2HBr(g)
-1 -1 -1
Bond enthalpy are given as : H – H = 436 kJ mol , Br – Br = 192 kJ mol , H – Br =368 kJ mol
H2 (g) + Br2 (g) → 2 HBr(g)
-1
B.E. OF H-H = 436 kj/ mol; Br- Br = 192 kj/ mol, H-Br = 368 kj/ mol
ΔH = ∑ B.E. (Reactants) - ∑ B.E. (Products)
= (436 + 192) – (368 x 2)
= (436 + 192) – (736)

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ΔH = -108 kj/ mol

30. Calculate the enthalpy change for the process
CCl4(g) → C(g) + 4Cl(g)
And calculate bond enthalpy of C–Cl in CCl4(g).
θ –1
ΔvapH (CCl4) = 30.5 kJ mol .
θ –1
ΔfH (CCl4) = –135.5 kJ mol .
θ –1 θ
ΔaH (C) = 715.0 kJ mol , where ΔaH is enthalpy of atomisation
θ –1
ΔaH (Cl2) = 242 kJ mol

31. Calculate the enthalpy of formation of benzoic acid (C6H5COO(l))) from the following data :
-1 -1
ΔfH of CO2(g) = --393.5 kJ mol , ΔfH of H2O(l) = --285.8 kJ mol ,
-1
ΔcHof C6H5COOH(l) = --3227 kJ mol .
C6H5COOH(l) + 15/2 O2 (g) → 7CO2 (g) + 3H2O (l)
ΔrH = ∑ ΔfH (Products) - ∑ ΔfH (Reactants)
-3227 = {7 x (-393.5) + 3 x (-285.8)} - ΔfH C6H5COOH(l) + 15/2 x O2}
-3227 = - 2754.5 – 857.4 – x + 15/2(0)
x = - 2754.5 – 857.4 + 3227
x = -384.9 kj/ mol
32. Calculate the enthalpy of formation of benzene (C6H6) from the following data :
-1 -1
ΔfH of CO2(g) = -393.5 kJ mol , ΔfH of H2O(l) = -285.8 kJ mol ,
-1
ΔcH of C6H6 (l) = -3266 kJ mol

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1
33 The heat of combustion of naphthalene (C10H8) at constant volume is -5133 kJ mol .Calculate the value of enthalpy change at 298 K.

34 Calculate the enthalpy of formation of methane from the following data :

-1 -1
ΔfH of CO2(g) = -393.5 kJ mol , ΔfH of H2O(l) = -285.8 kJ mol ,
-1
ΔcH of CH4 (g) = -890 kJ mol
-1
1) Cgraphite + O2 (g) → CO2(g) ∆H= -393.5 kJ mol
-1
2) H2(g) + ½ O2 (g) →H2O(l)∆H = -285.8 kJ mol
-1
3) CH4 (g) + 2O2 (g) →CO2(g) + 2 H2O(l)∆H =-890 kJ mol
Required equation
Cgraphite+ H2(g) →CH4 (g)
Eq(1) +eq(2) x2 +eq (3)
ΔfH = -393.5 + 2 x -285.8 +890
= -393.5 – 571.6 + 890
= -75.1 kj/mol
35 Calculate the enthalpy of formation of acetic acid (CH3COOH(l)) from the following data :
-1 -1
ΔfH of CO2(g) = -393.5 kJ mol , ΔfH of H2O(l) = -285.8 kJ mol ,
-1
ΔcHof CH3COOH(l) = -875 kJ mol .
CH3COOH(l) + 2 O2 → 2CO2(g)+ 2 H2O(l)

-875 = 2x (-393.5) + 2x(-285.8) –ΔfH of CH3COOH(l) -2 ΔfH O2x 0(as oxygen is in its standard rate)
-875 =-787 – 571.6 -∆f H CH3COOH
X = -787 -571.6 + 875
X = -483.6 kj/mol
36 Comment on the thermodynamic stability of NO(g), given
θ –1
½ N2(g) + ½ O2(g) → NO(g) ; ΔrH = 90 kJ mol
θ –1
NO(g) + ½ O2(g) → NO2(g) : ΔrH = –74 kJ mol

Hence unstable NO(g) changes to stable NO2(g).

Section C
Each question carries 3 marks :
–1
37. If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ mol . Calculate the internal energy change,
when
(i) 1 mol of water is vaporized at 1 bar pressure and 100°C.
(ii) 1 mol of water is converted into ice.

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38. Both ethane and methane are used as fuel. Predict which of these is a better fuel? Given that heat of formation of CH4 (g), C2H6 (g), CO2 (g) and H2O (l) are -74.85, -84.68, -
393.5 and -286 kJ mol-1 respectively.
CH4 (g) + 2O2 (g) → CO2(g) + 2 H2O(l)

= [-393.5 + (- 286 x2)] – [(-74.85 +0)]

= -393.5 – 572 +74.85
= -890.65 kj/mol
Per gram = -890.65/16 = -55.66 kj/g
C2H6 (g) +7 /2 O2 (g) → 2CO2(g) + 3 H2O(l)

= [2 x -393.5 +3 x -286] –[7/2 x (0) + ( -84.68)]

= [-787-858 ] + 84.68
= - 1560.32 KJ/mol
Per gram = -1560.32/30 = -52.01 kj/g
On comparing the values of Heat of the reaction for combustion reaction of methane and ethane it is seen that methane is better fuel than ethane as ∆rH of CH 4 is more negative
it is a better fuel
39.. Write the Born – Haber Cycle to calculate the lattice enthalpy of MgBr2(solid).
Mg (s) → Mg (g) ΔH Sublimation
+
Mg (g) → Mg (g) + e- ΔH ionization enthalpy1
+ +2
Mg (g) → Mg (g) + e- ΔH ionization enthalpy2
Br 2 (l) → Br2(g) ΔH vapourization
Br 2 (g) → 2Br (g) ΔH dissociation energy
-
Br + e- → Br ΔH electron gain enthalpy x 2
+2 -
Mg + 2Br →MgBr2 ΔH lattice enthalpy

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40. At 60°C, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Second Law of Thermodynamics

Section A
Each question carries 1 mark :
1. Write the conditions in terms of ∆H and ∆S when a reaction would be always spontaneous.
The reaction would be always spontaneous when both energy factor and randomness factor favors it i.e. ∆H = -ve and ∆S = +ve
2. Predict the sign of S giving reasons for:
(i) Br2(g) → Br2(l) (ii) N2(g)(2 atm) → N2(g)(0.5atm)
(iii) Raw egg → Boiled egg (iv) Stretched rubber band → Loose rubber band
(v)Steam → Water (vi) C (graphite) → C (diamond)

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(vii) CH4(g) + H2O(g) → 3H2(g) + CO(g) (viii) AgNO3 (s) → AgNO3 (aq)
(ix) H2 (g) → 2H (g)
(i) –ve (ii) +ve (iii) -ve(iv) +ve (v) -ve (vi) -ve(vii) +ve (viii) +ve (xi) +ve
3. What is the effect of temperature on entropy?
Increase in temperature will lead to increase in entropy.
4. Why does entropy of a solid increase on fusion?
This is almost always positive since the degree of disorder increases in the transition from an organized crystalline solid to the disorganized structure of a liquid.
5. ΔH alone cannot be the sole criterion for the spontaneity of a reaction. Explain.
∆G = ∆H - T∆S
If ∆G = -ve (spontaneous)
∆G = 0 (equilibrium)
∆G = +ve (non-spontaneous)
6. 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.
(i)After freezing, the molecules attain an ordered state and therefore, entropy decreases.
(ii)At 0 K, the constituent particles are static and entropy is minimum. If temperature is raised to 115 K, these begin to move and oscillate about their equilibrium positions in the
lattice and system becomes more disordered, therefore entropy increases
7. What do you understand by spontaneity?
A reaction is said to be spontaneous if it occurs without being driven by some outside force.
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8. If ΔG for a reversible reaction is found to be zero, what is the value of its equilibrium constant?
0
ΔG = -2.303 RT log Kc
0=
ΔG 0, Kc =1
-1
9. Calculate the entropy change involved in conversion of one mole (18 g) of solid ice at 273 K to liquid water at the same temperature (latent heat of fusion = 6025 J mol )

2+
10. Does aqueous solution of Mg ions have larger entropy before or after hydration of ions?
Hydration is the term used for attachment of water molecules to ions when an ionic compound is dissolved in water. It is a special case of dissolution of energy, with the solvent
2+ 2+
being water. Before hydration an aqueous solution of Mg ions will have larger entropy as after hydration Mg has achieved an orderly lattice.
11. For a reaction there is an increase in both enthalpy and entropy. Under what conditions does the reaction occur spontaneously?
∆G = ∆H - T∆S
∆H =+ve and ∆S =+ve .At high temperature T∆S>∆H
∆G = -ve and spontaneous
12. Dissolution of NH4Cl in water is endothermic but it is spontaneous. Explain.
+ -
NH4Cl(s) + H2O → NH4 (aq) + Cl (aq)
∆G = ∆H –T∆S ; as the number of particles are increasing after the dissolution of NH4Cl as it ionizes ∆s becomes +ve and –T∆ S overweighs +ve ∆H factor, so ∆G=-ve, reaction
becomes spontaneous.
Since the entropy of system increases therefore the process becomes spontaneous.
13. Lifting of water to the top of a hill is quite possible. Why can’t this be considered as a spontaneous process?
Because it does not happen on its own and it will immediately stop once the initiation from the motor stops .Therefore it is a non-spontaneous process.
14. For the reaction 2Cl(g) → Cl2, what are the signs of ∆H and ∆S?

Section B
Each question carries 2 marks :
15. Entropy of diamond is less than that of graphite. What conclusion do you draw from this?

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Less entropy means less disorder i.e., all the Carbon atoms are linked to form a network structure. Greater entropy of graphite implies some disorder which is due to presence of
free electrons and slipping of layers over each other.
3
In diamond each carbon atom undergoes sp hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The structure extends in space
and produces a rigid three-dimensional network of carbon atoms. It is very difficult to break extended covalent bonding and, therefore, diamond is a hardest substance on the earth
it implies less entropy in diamond
2
Graphite has a layered structure held by van der Waal’s forces, each C atom undergoes sp hybridisation and will make 3 sigma bond with 3 neighbouring C atom the fourth
electron forms a pi bond and the electrons are mobile which implies more entropy in graphite
16. Define entropy and free energy. How are they related to each other?
Entropy is the measure of the randomness or disorderliness of the system.
Whereas, free energy of a system is its capacity to do useful work.
They are related by Gibbs’s Helmholtz equation
∆G = ∆H –T∆S
17. Explain why the entropy of a pure crystalline substance is zero at 0 K? State the law on which it is based. Give the application of this law?
At 0 K, there is a perfectly ordered arrangement of the constituent particles of a pure crystalline substance and there is no disorder at all. Hence its entropy is taken a zero. This
statement is based on the third law of thermodynamics.The law is applied to findthe absolute entropy of a substance in any state at any temperature.
18.

-1 -1 -1
19. Enthalpy and entropy changes of a reaction are 40.63˚kJ mol and 108.8 JK mol respectively. Predict the feasibility of the reaction at 27˚ C
∆H =40.63˚kJ=40630 Jmol-1,T =27 + 273 =300K
∆G = ∆H - T∆S
= 40630 – 300x108.8
-1
= 7990 Jmol
As ∆G value is positive the reaction will not be spontaneous and feasible.
20. For the reaction at 298 K, 2A + B → C
-1 -1 ?
ΔH = 400 kJ mol and ΔS = 200 JK-1 mol . At what temperature will the reaction become spontaneous

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0 5
21. Calculate ΔG for the reaction, N2(g) + H2(g) → NH3(g) at 298 K. The value of Kc for the reaction is 6.6 × 10 .
-1 -1
[ R = 8.314 JK mol ]
ΔG = -2.303 RT Log Kc
5
= -2.303 X 8.314 X 298 X Log 6.6 X 10
= 33205.13J/mol = 33.205 kj/mol
0 -1
22. Calculate ΔS involved in the fusion of 3.6 g of ice at 0 C. Latent heat of ice is 334.72J g .
∆f S = ∆f H = 334.72 x 3.6/273 = 4.41 J/K
O O -1 —7
23. Calculate ΔS for a reaction X ↔ Y, if the value of ΔH = 28.40 kJ mol and Kc is 1.8 × 10 at 298 K.
ΔG = -2.303 RTLogKc
-3 -7
= -2.303 x 8.314 x 10 x 298 x log 1.8 x 10
= 38.4842 kj/mol
ΔG = ΔH – TΔS
38.4842 kj/mol = 28.40 -298 x ΔS
ΔS = - 0.03383kj/K
0 –1
24. Calculate the entropy change in surroundings when 1.00 mol of H2O (l) is formed under standard conditions. Δf H = –286 kJ mol .

0 –1 –1
25. The equilibrium constant for a reaction is 10. What will be the value of ΔG ? R = 8.314 JK mol , T = 300 K.

26. 200 J heat flow from the surroundings at 25 degree C into a large copper block at -10 degree C. What is the total entropy change?

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-1
27. The enthalpy change for the transition of liquid water to steam is 40.8 KJ mol AT 373 k.Calculate ΔS° for the process.

28.. For the reaction

2A(g) + B(g) → 2D(g)
θ θ –1
ΔU = –10.5 kJ and ΔS = –44.1 JK .
θ
Calculate ΔG for the reaction, and predict whether the reaction may occur spontaneously.

-1 -1
29. For the oxidation of iron, 4Fe (s) + 3O2(g) à 2Fe2O3(s), entropy change is –549.4 JK mol at 298 K. Inspite of negative entropy change of this reaction, why is the reaction
spontaneous?
3 -1 -1 -1
ΔrH = --1648 × 10 J mol , R = 8.314 JK mol

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0 –29
30. Calculate ΔrG for conversion of oxygen to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp for this conversion is 2.47 × 10 .

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