Thermodynamics 1

Level-I
Chapter 6
Thermodynamics
Solutions (Set-1)
Very Short Answer Type Questions :

1. What type of system does the following represents?
(a) A glass of water without lid
(b) A glass of water with lid
(c) A glass of water placed in a thermos flask
Sol. (a) Open system
(b) Closed system
(c) Isolated system
2. What is an adiabatic process?
Sol. A process during which no heat flows between the system and the surroundings is called an adiabatic process.
3. Why the heat changes are generally called enthalpy changes rather than internal energy?
Sol. Because most of the processes are carried out in open vessels as constant pressure (atmospheric pressure)
whereas internal energy is measured at constant volume.
4. What is the significance of free energy change?
Sol. Free energy change signifies the useful work done by a system.
5. When a system absorbs a certain amount of heat and does work, how the change in internal energy can be
thermodynamically related according to the first law of thermodynamics. Give the mathematical equation only.
Sol. ΔU = q + w, where w = negative as work is done by the system.
6. Under what condition is the enthalpy change equal to the heat of reaction?
Sol. The enthalpy change is equal to the heat of reaction at constant pressure.
ΔH = qp.
7. What do you understand by the reference state of a substance?
Sol. The reference state of the substance is the state in which the substance is stable at 1 bar pressure and
298 K. It is represented by the sign s in superscript.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
2 Thermodynamics Solutions of Assignment (Level-I) (Set-1)
8. State Hess’s law of constant heat summation.

Sol. Hess’s law states that the total enthalpy change of reaction is the same, regardless of whether the reaction
completes in one step or in several steps.
Example :
ΔH1
A B
ΔH4 ΔH2
D ΔH3
C
It means the change in enthalpy from A to D i.e., ΔH4 is equal to ΔH1 + ΔH2 + ΔH3 for A → B → C → D.
9. What is heat capacity?
Sol. It is the quantity of heat required to raise the temperature of the system by one degree.
10. What is enthalpy of solution at infinite dilution?
Sol. The change in enthalpy when a substance is dissolved in an infinite amount of solvent. At this point the
interactions between the ions are negligible.
Short Answer Type Questions :
11. What factors can bring about a change in the internal energy of the system?
Sol. The internal energy of a system changes when
(i) Heat is given or absorbed by the system
(ii) Work is done on or by the system
(iii) Matter enters or leave the system
12. Why a smooth curve is obtained when a graph is plotted for the work done by a system in the isothermal
reversible expansion of an ideal gas?
Sol. When reversible expansion of an ideal gas is taken into consideration, the change is brought about in number
of infinite steps. The change is infinitesimally smaller such that the difference between the driving force and
opposing forces are almost in equilibrium. The work done is –ΣpΔV. So, when the graph is plotted, they gives
a smooth curve.
13. How the first law of thermodynamics can be justified? Give an example in support of your answer.
Sol. First law of thermodynamics states that whenever a certain quantity of some form of energy is consumed, an
equivalent amount of energy is produced. For example when 4.184 joules of mechanical work is done 1 calorie
of heat is produced and vice-versa.
14. Calculate the change in internal energy of the system when it absorbs 20 kJ of heat from the surrounding and
does work equivalent to 50 kJ.
Sol. The change in internal energy is given by the expression
ΔU = q + w
Here, q = 20 kJ
w = –50 kJ (negative sign implies the work is done by the system on the surrounding)
Solutions of Assignment (Level-I) (Set-1) Thermodynamics 3
ΔU = 20 kJ + (–50) kJ
= 20 – 50 kJ
= –30 kJ
The internal energy of the system decreases by 30 kJ.
15. One mole of carbon dioxide undergoes an isothermal and reversible expansion at 27°C from 10 L to 50 L.
Calculate the work done. (R = 8.314 J)
V2
Sol. W = –2.303 nRT log
V1
50
= –2.303 × 1 × 8.314 × 300 × log
10
= –2.303 × 8.314 × 300 × 0.699
= –4015.16 J
16. State the conditions under which qp = qv or ΔH = ΔU.
Sol. (i) When the reaction is carried out in a closed vessel so that the volume remains constant, i.e., ΔV = 0.
(ii) When the reaction involves only solids or liquids or solutions but no gaseous reactants or products. This
is because the volume changes of the solids and liquids during a chemical reaction are negligible.
(iii) When the reaction involves gaseous reactants and products but their number of moles are equal (i.e.,
np = nr), e.g., in the reaction
H2(g) + Cl2(g) ⎯→ 2HCl(g)
17. A water storage tank placed on the top of a building contains 103 L of water in it. It is receiving energy from
the sun causing the heating effect of water. How much energy in joules would be required to raise the
temperature of water from 30°C to 35°C? (Given the specific heat of water is 4.184 JK–1g–1)
m
Sol. Mass of water = (103 × 103 mL) × (1 gmL–1) (∵ D = , considering density of water = 1 gmL–1)
V
= 106 g
q = m⋅C⋅ΔT (Where ΔT = 35 – 30 = 5)
= 106 g × 4.184 JK–1g–1 × 5
= 20.92 × 106 J
18. What is the basic difference between enthalpy of reaction and enthalpy of formation? Illustrate with a suitable
example.
Sol. Enthalpy of a reaction is the amount of heat evolved or absorbed at constant temperature and at constant
pressure, when number of moles of reactants represented by the chemical equation have completely reacted.
Example : CaO(s) + CO2(g) ⎯→ CaCO3(s); ΔrHs = –178.3 kJ/mol
Here we can see that CaCO3(s) has not been obtained from the constituent elements C, O2 and Ca.
Whereas, enthalpy of formation is the amount of heat absorbed or evolved when 1 mole of compound is formed
from its constituent elements under given conditions of temperature and pressure.
Example : C(s) + O (g) ⎯→ CO (g); Δ Hs = –393.5 kJ mol–1
2 2 f
19. What do you understand by extensive and intensive properties? Give example.
Sol. Extensive properties : Those properties which depends on the quantity of matter present in the system.
Example : mass, volume, internal energy, heat capacity etc.
Intensive properties : Those properties which is independent of the quantity of matter present in the system.
Example : density, melting point, boiling point, specific heat, refractive index etc.
20. Calculate ΔG° at 25°C for the reaction
2NO2(g) ⎯→ N2O4(g)
Given, ΔH° = –57.20 kJ mol –1 and ΔS° = –175.83 J mol–1 K –1. Is the reaction spontaneous or non-
spontaneous?
Sol. ΔG° = ΔH° – TΔS°
ΔG° = [–57.20 × 103] – [298(–175.83)]
= –57200 + 52397.34
= –4802.66 J
Since ΔG° is negative i.e., the process is spontaneous.
21. Calculate the standard enthalpy of formation of n-butane, given that the standard enthalpies of combustion of
n-butane, C(graphite) and H2(g) are –2878.5 kJ mol–1, –393.5 kJ mol–1 and –285.8 kJ mol–1 respectively.
Sol. 4C(graphite) + 5H2(g) ⎯→ C4H10(g)
ΔfHs = ΣΔrHsProducts – ΣΔrHsReactants
= [C4H10] – [4(C) + 5(H2)]
= [–2878.5] – [4(–393.5) + 5(–285.8)]
= [–2878.5] – [–1574 – 1429]
= –2878.5 + 3003
= –124.5 kJ mol–1
22. (a) Predict the direction of the given reaction
1
Ag2O(s) 2Ag(s) + O2(g)
2
ΔH, ΔS and T are 40.63 kJ mol–1, 108.8 JK–1 mol–1 and 373.4 K respectively.
(b) Define bond enthalpy.
Sol. (a) ΔG = ΔH – TΔS
= 40.63 – 373.4 × 108.8 × 10–3
= 40.63 × 40.63
=0
Since ΔG = 0, system remains at equilibrium.
(b) Bond enthalpy is the amount of energy necessary to break bonds in one mole of gaseous covalent
substance to form products in the gaseous state.
23. The enthalpy change (ΔH) for the reaction,
N2(g) + 3H2(g) ⎯→ 2NH3(g) is –92.38 kJ at 298 K. What is ΔU at 298 K?
Sol. N2(g) + 3H2(g) ⎯→ 2NH3(g)
ΔH = –92.38 kJ
Δng = 2 – 4 = –2
R = 8.314 × 10–3 kJ mol–1 K–1
We know that
ΔH = ΔU + ΔngRT
∴ ΔU = ΔH – ΔngRT
= [–92.38 kJ] – [(–2) × 8.314 × 10–3 × 298 kJ]
= –92.38 + 4.955
= –87.42 kJ
24. What do you understand by the following terms?
(a) Heat
(b) Work
Sol. (a) Heat is a form of energy that is transferred between the system and its surroundings as a result of
temperature difference. Its symbol is q.
(b) By work we mean the energy is spent during the act of moving an object against an opposing forces such
as gas inside the cylinder acting against the piston.
25. When does entropy increase in the reaction?
Sol. Entropy increases when
(i) Number of molecules of products are more than the number of molecules of reactants.
(ii) When solid state changes to liquid state.
(iii) When liquid state changes to gaseous state.
(iv) When the substance undergo sublimation.
26. What does the term state variable means?
Sol. The fundamental properties which determine the state of the system are termed state variables. The change
in state property depends only upon the initial and final states of the system i.e., independent of the path
followed. Example pressure, volume, temperature, entropy, force, etc.
27. Calculate the work done during isothermal reversible expansion of one mole of an ideal gas from 10 atm to
1 atm at 300 K.
Sol. Number of moles of ideal gas (n) = 1
Initial pressure (P1) = 10 atm
Final pressure (P2) = 1 atm
Temperature, T = 300 K
Gas constant R = 8.314 × 10–3 kJ mol–1 K–1
∴ Work done for isothermal reversible expansion of an ideal gas.
P1
W = –2.303 nRT log
P2
= –2.303 × 1 × 8.314 × 10–3 kJ mol–1 K–1 × 300 K
= –5744.14 × 10–3 kJ mol–1
28. What is meant by enthalpy of formation? Given its significance.
Sol. Enthalpy of formation is defined as the enthalpy accompanying the formation of one mole of a compound from
its constituent elements. It is denoted by ΔfHs.
The knowledge of standard enthalpies of various substances can help us to calculate standard enthalpy change
of any reaction. Where standard enthalpy change of a reaction is equal to the difference of the standard
enthalpies of all the products and standard enthalpies of reactants.
ΔrHs = [ΣΔfHsproducts] – [ΣΔfHsreactants]
29. The heat of atomization of PH3(g) is 228 kcal mol–1 and that of P2H4 is 355 kcal mol–1. Calculate the average
bond energy of P–P bond.
Sol. PH3(g) ⎯→ P(g) + 3H(g); ΔH1 = 228 kcal mol–1
PH2 – PH2 ⎯→ 2P(g) + 4H(g); ΔH2 = 355 kcal mol–1
∴ On the basis of the given data
ΔH2 = ΔHP – P + 4ΔHP – H
228  ΔH1 228 

or 355 = ΔHP – P + 4 ×
3 ∵ ΔHP − H = 3 = 3 
∴ ΔHP – P = 51 kcal mol–1
30. When an ideal gas expands into vaccum, there is neither absorption nor evolution of heat. Why?
Sol. In an ideal gas, there are no intermolecular forces of attraction. Hence, no energy is required to overcome these
forces. Moreover, when a gas expands against vacuum, work done is zero (because pext = 0). Hence, internal
energy of the system does not change i.e., there is no absorption or evolution of heat.
Long Answer Type Questions :
31. What is meant by average bond energy? In what way is it different from bond energy of a diatomic molecule?
Give a suitable example.
Sol. Average bond energy is used in the case of polyatomic molecules because even though the molecule may
be containing identical bonds with the identical bond length and energy, the energy required to break the
individual bonds differ in each successive step. So in such cases average bond energy is used. For example
in CH4 molecule, all the four C–H bonds are equal in bond length and energy but differs in terms of energy
required to break the bonds.
CH4(g) ⎯→ C(g) + 4H(g); ΔaHs = 1665 kJ mol–1
CH4(g) ⎯→ CH3(g) + H(g); ΔbondHs = +427 kJ mol–1
CH3(g) ⎯→ CH2(g) + H(g); ΔbondHs = +439 kJ mol–1
CH2(g) ⎯→ CH(g) + H(g); ΔbondHs = +452 kJ mol–1
CH(g) ⎯→ C(g) + H(g); ΔbondHs = +347 kJ mol–1
427 + 439 + 452 + 347

Therefore, ΔaHs =
4
= 416.25 kJ mol–1
It is different from bond energy of a diatomic molecule because the energy required in diatomic molecules is
to completely dissociate the molecules into atoms by breaking the covalent bonds of a gaseous covalent
compound to form the products in the gaseous phase.
32. A 0.150 mol sample of an ideal gas is allowed to expand at 294 K from 10.00 atm to 1.00 atm. If external
pressure is kept constant at 1.00 atm. Calculate the work done.
nRT
Sol. Initial volume V1 =
p
0.150 × 0.0821 × 294
=
10
= 0.362 dm3
nRT
Final volume V2 =
p
0.150 × 0.0821 × 294
=
1
= 3.62 dm3
∴ w = –pΔV
= –1 × (V2 – V1)
= –1(3.62 – 0.362)
= –1(3.26)
= –3.26 atm dm3
or –3.26 L-atm
= –3.26 × 101.3 J (1 L-atm = 101.3 J)
= –330.2 J
33. Calculate the difference in the work done when one mole of Al4C3(s) reacts with water in a closed vessel at
27°C against atmospheric pressure and that in an open vessel under the same condition.
Sol. Al4C3(s) + 12H2O(l) ⎯→ 4Al(OH)3 + 3CH4(g)
1 mol 3 mol
When vessel is closed, volume = constant i.e., ΔV = 0. Hence W = –PΔV = 0.
When the vessel is open, initial volume, V1 = 0 (as no gas is present).
Final volume, V2 = Volume of 3 mole of CH4
pV2 = nRT
nRT
∴ V2 =
P
∴ w = –p(V2 – V1)
nRT
= –pV2 = −p
p
= –3 × 8.314 × 300
= –7482.6 J
The work is done by the system on the surroundings.
34. Define the relation
ΔH = ΔU + ΔngRT
Sol. The difference between ΔH and ΔU is not usually significant for the system consisting of only solids and/or
liquids.
If VR = total volume of the gaseous reactants
VP = total volume of the gaseous products
nR = number of moles of gaseous reactants
nP = number of moles of gaseous products
Then at constant temperature and pressure, assuming ideal gas behaviour, we can write
RT
VR = nR
p
RT
and VR = nP
p
and ΔV = VP – VR
RT
ΔV = (nP – nR)
p
RT
ΔV = Δng
p
We know that
ΔH = ΔU + pΔV
RT
ΔH = ΔU + Δng
p
and ΔH = ΔU + ΔngRT
35. What is the origin of energy change in a chemical reaction?
Sol. Energy is released in the formation of a chemical bond between two atoms and energy is required to break
the bond. Chemical reactions involve the making and breaking of chemical bonds. Energy changes in a chemical
reaction is the net difference between the energy needed to break the old chemical bonds in the reactant
molecules and the energy released when new bonds are formed in the products.
For example :
H2(g) + Cl2(g) ⎯→ 2HCl(g)
H – H(g) ⎯⎯⎯⎯⎯⎯⎯
Energy absorbed
→ 2H(g); ΔH = +436 kJ
in breaking bonds
Cl – Cl(g) ⎯⎯⎯⎯⎯⎯⎯
Energy absorbed
→ 2Cl(g); ΔH = 242 kJ
in breaking bonds
2H(g) + 2Cl(g) ⎯⎯⎯⎯⎯⎯⎯⎯

Energy released when
→ 2H – Cl(g); ΔH = –866 kJ (For two moles of HCl formed)
new bonds are formed
Thus, for the reaction

H2(g) + Cl2(g) ⎯→ 2HCl(g)
Enthalpy change,
ΔH = (436 + 242) – 866
= 678 – 866
= –188 kJ
36. For oxidation of iron
4Fe(s) + 3O2(g) ⎯→ 2Fe2O3(s)
entropy change is –549.4 J K–1 mol–1 at 298 K. Inspite of negative entropy change of this reaction, why is
the reaction spontaneous? (ΔrHs for this reaction is –1648 × 103 J mol–1)
Sol. The spontaneity of a reaction can be decided by considering
ΔStotal = ΔS(sys) + ΔSsurroundings
For calculating ΔSsurr, we have to consider the heat absorbed by the surroundings which is equal to ΔrHs. At
temperature T, entropy change of the surroundings is
Δr Hs
ΔSsurr = (at constant pressure)
T
−1648 × 103 J mol−1

=
298 K
= 5530 J K–1 mol–1

Thus, total entropy change for this reaction
ΔrStotal = 5530 J K–1 mol–1 + (–549.4 J K–1 mol–1)
= 4980.6 J K–1 mol–1
This shows that the reaction is spontaneous.
37. Derive an expression for pressure volume work in a irreversible process.
Sol. Suppose an ideal gas is enclosed in a cylinder fitted with weightless and frictionless piston of area of cross-
section A. If the pressure of the gas inside the cylinder is greater than the pressure outside the cylinder
(external pressure) i.e., pint > pext. Piston will move outside. This will bring a change in the volume of the gas.
The process is considered to be happening at constant temperature i.e., isothermal process. The change in
volume of the gas, ΔV = V2 − V1 .
(Final) (Initial)
V2
ΔV l
V2
Area A V1
V1
During this expansion procedure the piston moves by a certain length ‘Δl’ in a single step.
The volume change = Δl × A
= ΔV = (V2 – V1)
F
We know, p =
A
F = P.A.
If w = force × displacement
= pex × A × Δl
= pex(ΔV)
= pexΔV
And w = –pex(V2 – V1)
The negative sign implies that work is done by the system on the surroundings.
38. At 60°, dinitrogen tetroxide is 50% dissociated. Calculate the standard free energy change at this temperature
and at one atmosphere.
Sol. N2O4(g) 2NO2(g)
If N2O4 is 50% dissociated, the given equilibrium is
N2O4(g) 2NO2(g)
Initial moles 1.0 0
Equilibrium moles 0.5 2 × 0.5
Total mole = 1 + 0.5
Pressure = 1 atm
0.5
Mole fraction of N2O4, xN2O4 =
1.5
1
Mole fraction of NO2, xNO2 =
1.5
0.5 1
pN2O4 = × 1 atm, pNO2 = × 1 atm
1.5 1.5
The equilibrium constant Kp is given by

2
 1 
2
(pNO2 )  1.5 
Kp = =  = 1.33 atm
(pN2O4 )  0.5 
 1.5 
Since,
ΔrGs = –RT lnKp
= (–8.314 J K–1 mol–1) × (333 K) × (2.303) × (0.1239)
= –763.8 kJ mol–1

39. The reaction CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l) was carried out at 300 K by taking
2
1 mol each of reactants. The reaction reached equilibrium when rd of reactants were consumed. Calculate
3
Gibb’s energy change for the reaction.
Sol. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

Initial mole 1 mol 1 mol 0 0
1 1 2 2
At equilibrium
3 3 3 3
The equilibrium constant,
[CH3 COOC2H5 ][H2O]

KC =
[CH3 COOH][C2H5 OH]
2 2
3 3
=    = 4
 1  1
 3   3 
Since,
ΔrGs = –2.303 RT logK
= –2.303 × 8.314 × 300 × log4
= –2.303 × 8.314 × 300 × 0.6020
= –3457.97 J mol–1
= –3.458 kJ mol–1
40. List differences between reversible and irreversible process.
Sol. Reversible process Irreversible process
1. It is a process which follows reversible path. 1. It is a process which follows irreversible path.
2. It is an ideal process and takes infinite time. 2. It is a spontaneous process and takes finite time.
3. In this process, the opposing force and 3. There is large difference between driving and
driving force differ only by infinitesimally small opposing force.
magnitude.
4. Work obtained is maximum. 4. Work obtained is not maximum.
5. It is only an imaginary process and cannot 5. It is a natural process and occurs in a particular
be realised in actual practice. directions under given set of conditions.
6. It occurs through infinite number of steps 6. Equilibrium exist only in the beginning and at the
under equilibrium conditions at each step. completion stage.
41. What are the conditions for ΔG to be negative by considering Gibb’s Helmholtz equation?
Sol. Gibb’s Helmholtz equation is
ΔG = ΔH – TΔS
The conditions for ΔG to be negative are :
(i) Both the energy factor as well as entropy factor are favourable i.e., ΔH is negative and ΔS is positive.
Under these conditions ΔG would be certainly negative and the process will be spontaneous.
(ii) Energy factor favours but entropy factor opposes i.e., ΔH is negative and ΔS is also negative. Under these
conditions ΔG would be negative if ΔH is greater than TΔS in magnitude and that would be possible at
low temperature.
(iii) Energy factor opposes but entropy factor favours i.e., ΔH is positive and ΔS is also positive. Under these
conditions ΔG would be negative if TΔS is greater than ΔH in magnitude and that would be possible at
high temperature.
42. Standard enthalpies of combustion of C2H4(g), C2H6(g) and H2(g) are –1410.8, –1559.8 and –285.9 kJ mol–1
at constant pressure and 298 K respectively. What is the enthalpy of hydrogenation of C2H4(g) at constant
volume and at 298 K?
Sol. (i) C2H4(g) + 3O2(g) ⎯→ 2CO2(g) + 2H2O; ΔcHs = –1410.8 kJ mol–1
7
(ii) C2H6(g) + O (g) ⎯→ 2CO2(g) + 3H2O; ΔcHs = –1559.8 kJ mol–1
2 2
1
(iii) H2(g) + O (g) ⎯→ H2O(l); ΔcHs = –285.9 kJ mol–1
2 2
Our aim is :
(iv) C2H4(g) + 3H2(g) ⎯→ C2H6(g); ΔcHs = ?
Adding equations (i) & (iii),
C2H4(g) + 3O2(g) ⎯→ 2CO2(g) + 2H2O; ΔcHs = –1410.8 kJ mol–1
1
H2(g) + O (g) ⎯→ H2O(l); ΔcHs = –285.9 kJ mol–1
2 2
7
(v) C2H4(g) + H2(g) + O ⎯→ 2CO2(g) + 3H2O; ΔcHs = –1696.7 kJ
2 2
Subtract equation (ii) from equation (v)
7
C2H4(g) + H2(g) + O ⎯→ 2CO2(g) + 3H2O; ΔrHs = –1696.7 kJ
2 2
7
C2H6(g) + O (g) ⎯→ 2CO2(g) + 3H2O; ΔcHs = –1559.8 kJ
2 2
– – – – +
C2H4(g) + H2(g) – C2H6(g) ⎯→ ΔrHs = –136.9 kJ

or C2H4(g) + H2(g) ⎯→ C2H6(g); ΔrHs = –136.9 kJ
ΔH = ΔU + ΔngRT
–136.9 = ΔU – 1 × 8.314 × 10–3 × 298
–136.9 = ΔU – 2.48
ΔU = –136.9 + 2.48
= –134.42 kJ mol–1
43. Calculate standard heat of formation of CS2. Given that standard heat of combustion of C, S and CS2 are
–393.3, –293.72 and –1108.76 kJ mol–1.
Sol. Our aim :
(i) C(s) + 2S(s) ⎯→ CS2(l); ΔH = ?
The data provided is :
(ii) C(s) + O2(g) ⎯→ CO2(g); ΔH = –393.3 kJ mol–1
(iii) S(s) + O2(g) ⎯→ SO2(g); ΔH = –293.72 kJ mol–1
(iv) CS2(l) + 3O2(g) ⎯→ CO2(g) + 2SO2(g); ΔH = –1108.76 kJ mol–1
Multiply equation (iii) by 2, we get
(v) 2S(s) + 2O2(g) ⎯→ 2SO2(g); ΔH = –587.44 kJ
Add equation (ii) with equation (v), we get
(vi) C(s) + 2S(s) + 3O2(g) ⎯→ 2SO2(g) + CO2(g); ΔH = –980.74 kJ
Subtract equation (iv) from equation (vi), we get
C(s) + 2S(s) ⎯→ CS2(l); ΔH = +128.02 kJ mol–1
ΔfHsCS2 = +128.02 kJ
44. Give the appropriate reasons for the following

(a) Is it preferable to determine a change in enthalpy than change in internal energy?
(b) Is it necessary to define the ‘standard state’?
(c) Is it necessary to specify the phases of the reactants and products in the thermochemical equation?
Sol. (a) Normally the changes takes place at constant pressure (atmospheric pressure) and thus it is preferable
to measure enthalpy changes.
(b) Yes, it is necessary to define ‘standard state’ because it holds significance for being at constant pressure
of 1 bar and a temperature of 298 K. This state represents the most stable state of the substances.
(c) Yes, it is necessary to specify the phases of the reactants and products in the thermochemical equation
because heat changes during the course of reaction depends on the physical states of reactants and
products.
45. (a) Hess’s law is just an extension of the law of conservation of energy. Justify the statement.
(b) Bond formation is always exothermic, comment.
Sol. (a) The first law of thermodynamics states that energy can neither be created nor destroyed but it changes
its forms. This holds true for the law of conservation of mass. Hess’s law also predicts the similar fact
that total energy change during the same course of reaction is independent of the path by which reaction
is made.
(b) Formation of a bond involves decrease in the potential energy of atoms forming the bond and thus energy
is released when a bond is formed. This decrease in potential energy is evolved as heat in the reaction
and brings about stability of the atoms.
The energy released (or change in enthalpy) when a bond is formed from isolated gaseous atoms is known
as bond enthalpy.

Level-I
Chapter 6
Thermodynamics
Solutions (Set-2)
First Law of Thermodynamics

1. Select the correct statement about heat change at constant pressure.
(1) Path function and extensive property (2) State function and intensive property
(3) State function and extensive property (4) Path function and intensive property
Sol. Answer (3)
Heat at constant pressure = ΔH
2. What is the work done in cyclic process as shown in given P-V diagram?
3P0
P P
0
V0 4V0
V
(1) –6 P0V0 (2) –12 P0V0 (3) 9 P0V0 (4) – P0V0
Sol. Answer (1)
Magnitude of work = | Area of rectangle |
= (3P0 – P0) (4V0 – V0)
= (2P0) (3V0)
= 6P0V0
3. A gas is expanded isothermally from 2 L to 5 L volume at constant pressure of 1.5 atm. What is the change in
internal energy of system? (in atm L)
(1) – 3 (2) +3 (3) –6 (4) Zero
Sol. Answer (4)
ΔU = nCVΔT
ΔT = 0 ⇒ ΔU = 0
4. 2 moles of ideal gas is expanded reversibly from 4 atm to 3 atm at constant temperature of 300 K. Calculate
the work done (approx.). [log4 = 0.6, log3 = 0.48]
(1) – 273 cal (2) – 332 cal (3) – 402 cal (4) – 315 cal
Sol. Answer (2)
V2 P1
W = – 2.303 nRT log = – 2.303 nRT log
V1 P2
5. What is the change in internal energy of 1 mole of monoatomic gas when temperature is increased by 200°C
adiabatically and reversibly?
(1) 2.50 kJ (2) 5.9 kJ (3) 1.93 kJ (4) 3.10 kJ
Sol. Answer (1)
nR
Wrev = [T – T1]
y −1 2
1× 8.314 × 200
= = 2.5 kJ
5
−1
3
Thermochemistry
6. The heat of reaction for P4 (white, s) + 6Cl2(g) ⎯→ 4PCl3(l) is – 1280 kJ. What is the heat of formation of
PCl3(l) (in kJ/mol)?
(1) – 640 (2) – 540 (3) – 320 (4) – 290
Sol. Answer (3)
1 3
P4 (white, s) + Cl2 (g) → PCl3 , ΔH°f
4 2
7. Calculate standard heat of combustion of ethanol (C 2 H 5 OH(l)). Given that Δ f H° (C 2 H 5 OH, l) =

– 278 kJ/mol, ΔfH° (CO2, g) = – 394 kJ and ΔfH° (H2O, l) = – 286 kJ/mol.
(1) – 4713 kJ/mol (2) – 1368 kJ/mol (3) – 2848 kJ/mol (4) – 1794 kJ/mol
Sol. Answer (2)
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
ΔCH° = 2ΔfH°(CO2) + 3ΔfH°(H2O) – ΔfH°(H2O)

= 2(– 394) + 3(– 286) – (– 278)
= – 1,368 kJ/mol
8. Among following standard enthalpy of formation is non-zero for
(1) C (graphite) (2) Br2(l) (3) H2(g) (4) I2(g)

Sol. Answer (4)
For I2(s), ΔfH° = 0
9. What is the heat of reaction of following reaction?
3C(g) + 4H2(g) → H3C — C — CH3(g)
H
Given that BE of H–H = x, H–C = y and C–C = z.
(1) 4x – 2z – 8y (2) x–y–z (3) 4x – 2y – 2z (4) 4y + 5x – z
Sol. Answer (1)
ΔHr = 4(BE of H2) – 2BE(C – C) – 8 BE(C – H)
10. The heat of formation of CO2 is –407 kJ/mol. The energy required for the process
3CO2(g) → 3C(g) + 2O3(g) is
(1) Less than 1221 kJ (2) More than 1221 kJ
(3) Is equal to 1221 kJ (4) Cannot be predicted
Sol. Answer (2)
ΔHf(CO2 ) for 3 moles will be = 3 × 407 = 1221 kJ
∴ With O3 it should be greater than 1221 kJ

11. The heat of combustion of sucrose C12H22O11 (s) at constant volume is –1348.9 kcal mol–1 at 25°C, then the heat
of reaction at constant pressure, is
(1) – 1348.9 kcal (2) – 1342.344 kcal (3) 1250 kcal (4) – 1250 kcal
Sol. Answer (1)
C12H22O11 + 12O2 ⎯⎯→ 12CO2 + 11H2O

(s) (g) (g) (l)
12. For the reaction,
C7H8 (l) + 9O2 (g) ⎯⎯→ 7CO2 (g) + 4H2O(l),
the calculated heat of reaction is 232 kJ mol–1 and observed heat of reaction is 50.4 kJ mol–1, then the resonance
energy is
(1) – 181.6 kJ mol–1 (2) + 181.6 kJ mol–1 (3) 172 kJ mol–1 (4) – 172 kJ mol–1
Sol. Answer (1)
Resonance Energy = –232 – (–50.4) = –182.2 kJ mol–1
13. Given that
A(s) ⎯⎯→ A(l) ΔH = x
A(l) ⎯⎯→ A(g) ΔH = y
The heat of sublimation of A will be

(1) x – y (2) x+y (3) x or y (4) –x+y
Sol. Answer (2)
According to Hess’s law ΔHsub = x + y
14. AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB and B2 are in the ratio 1 : 1 : 0.5 and
enthalpy of formation of AB from A2 and B2 is –100 kJ mol–1. What is the bond enthalpy of A2?
(1) 400 kJ mol–1 (2) 200 kJ mol–1 (3) 300 kJ mol–1 (4) 100 kJ mol–1
Sol. Answer (1)
1 1 1
A 2 + B2 ⎯⎯→ AB
2 2 2
1 1
ΔH = B.E. of H2 + B.E. of B2 − B.E. of A–B
2 2
[Second Law of Thermodynamics]

15. For the reaction 2HgO(s) → 2Hg(l ) + O2(g)
(1) ΔH > 0 and ΔS < 0 (2) ΔH > 0 and ΔS > 0
(3) ΔH < 0 and ΔS < 0 (4) ΔH < 0 and ΔS > 0
Sol. Answer (2)
2HgO(s) → 2Hg(l) + O2(g)
It is endothermic [∵ Δ ng > 0]
∴ ΔH > 0 and ΔS > 0
16. If the entropy of vaporisation of a liquid is 110 JK–1 mol–1 and its enthalpy of vaporisation is 50 kJmol–1. The
boiling point of the liquid is
(1) 354.5 K (2) 454.5 K (3) 554.5 K (4) 445.5 K
Sol. Answer (2)
ΔSvap = 110 J/K/mol
ΔHvap = 50 × 103 J/mol
ΔH
ΔS = T
B.P.
50 × 103
TB.P. = = 454.54 K
110
17. 1 mole of an ideal gas is expanded from an initial pressure of 1 bar to final pressure of 0.1 bar at constant
temperature of 273 K. Predict which of the following is not true.
(1) ΔE = 0 (2) ΔH = 0 (3) PV is constant (4) ΔS < 0
Sol. Answer (4)
For an isothermal process, ΔE = 0, PV = constant, ΔH = 0 and ΔS > 0.
18. Which of the following conditions should be satisfied for the given reaction to be spontaneous at 0°C and 1 atm?
H2O(s) H2O(l )
(1) ΔH = ΔG (2) ΔH < TΔS (3) ΔH > TΔS (4) ΔH = TΔS
Sol. Answer (2)
If ΔH < TΔS
The ΔG < 0 (–ve) and reaction will be spontaneous
19. The free energy change due to a reaction is zero when
(1) The reactants are initially mixed (2) A catalyst is added
(3) The system is at equilibrium (4) The reactants are completely consumed
Sol. Answer (3)
At equilibrium ΔG = 0
20. From the given graph
A
Free energy
B
Reactant Product
Which of the following statement is correct?

(1) The point B represents the state of equilibrium
(2) The equilibrium composition strongly favours the reactant
(3) From the point B formation of product is equally spontaneous as of reactant
(4) From the point B formation of reactant is more spontaneous than that of product
Sol. Answer (1)
At equilibrium ΔG = 0
21. For which of the following reaction, entropy change is positive?
3
(1) H2(g) + O2(g) → H2O2(l) (2) CH3OH(l) + O (g) → CO2(g) + 2H2O(l)
2 2
(3) 2NO2(g) → N2O4(g) (4) 2NH3(g) → N2(g) + 3H2(g)
Sol. Answer (4)
Δng > 0 ⇒ Entropy change is positive.
22. What is the value of ΔSuniverse for a spontaneous reaction?
(1) Greater than zero (2) Less than zero (3) Zero (4) Always negative
Sol. Answer (1)
If ΔSuniverse = Positive
23. The reaction 4Al(s) + 3O2(g) → 2Al2O3(s) is exothermic in nature. The reaction would be
(1) Spontaneous at all temperature (2) Spontaneous at low temperature

(3) Spontaneous at high temperature (4) Non-spontaneous always
Sol. Answer (2)

ΔH = – ve, ΔS = – ve
ΔG = ΔH – TΔS
Reaction would be spontaneous only at low temperature.
24. The value of ΔG° is –1.2 kcal/mol at 600 K for a reaction at equilibrium. The value of its equilibrium constant
is
(1) e–1 (2) e2 (3) e (4) e–2

Sol. Answer (3)
ΔG° = – RT ln K
2 × 600
– 1.2 = – ln K
1000
⇒ ln K = 1
Keq = e
25. Select the correct statement.

(1) Exothermic reaction is always spontaneous
(2) Endothermic reaction is always Non-spontaneous

(3) Exothermic reaction is always spontaneous if entropy change of reaction is positive
(4) Endothermic reaction is always non-spontaneous if entropy change of reaction is positive

Sol. Answer (3)
For spontaneous process
ΔG = ΔH – T ΔS < 0
ΔH = –ve for exothermic process
26. 1 mol of diatomic ideal gas expands from 1 L to 10 L in reversible isothermal process. The entropy change ΔS
of surrounding for this process
(1) + 1.9 J/mol K (2) – 8.3 J/mol K (3) – 19.15 J/mol K (4) + 14.2 J/mol K
Sol. Answer (3)
ΔSSystem = n Rln(v2/v1) (Isothermal process)
ΔSSurr = –n Rln (v2/v1)
= – 1 × 8.31 × ln(10)
27. According to second law of thermodynamics,
(1) Heat can't flow spontaneously from a reservoir at lower temperature to a reservoir at higher temperature
(2) All spontaneous process leads to increase in entropy of universe
(3) Melting a solid increases entropy, therefore a spontaneous process
(4) All of these
Sol. Answer (4)
All statements are representing second law of thermodynamics.
28. Consider the following sequence of reaction at 300 K
Reaction :
I : M → N, ΔH° = x kJ
II : N → O, ΔH° = y kJ
III : O → P, ΔH° = z kJ
x, y and z are positive integer and x > y > z
S° : M > N > O > P (at 300 K)
Select the Nonspontaneous reaction at 300 K.
(1) P → N (2) O→M (3) M→P (4) P→O
Sol. Answer (3)
For spontaneous process ΔG = –ve
ΔH = –ve
ΔS = +ve
S:M>N>O>P
ΔH ΔS
P → N = –ve +ve
O → M = –ve +ve
M → P = +ve –ve
P → O = –ve +ve
29. In which of the following process, the entropy decreases?
(1) 4NO2 ( g) + O2 ( g) ⎯⎯⎯

→ 2N2O5 ( g) (2) 2HCl ( g) ⎯⎯⎯
→ H2 ( g) + Cl2 ( g)
(3) 2NH3 ( g) ⎯⎯⎯

→ N2 ( g) + 3H2 ( g) (4) CH4 ( g) ⎯⎯⎯
→ C ( g) + 4H ( g)
Sol. Answer (1)

In reaction 1, number of gaseous molecule decreases.
30. Conversion of diamond to graphite is
(1) Spontaneous only at high temperature (2) Spontaneous only at low temperature
(3) Always non-spontaneous (4) Always Spontaneous
Sol. Answer (4)

Diamond → Graphite
ΔS = +ve
ΔH = –ve
⇒ G = –ve (always)
31. One mole of monoatomic ideal gas at temperature T1 K is compressed adiabatically and final temperature
becomes T2 K . Select the correct relation.
(1) T1 > T2 (2) T1 = T2
(3) T1 < T2 (4) Cannot be compared
Sol. Answer (3)
For adiabatic process
ΔU = w
For compression w = +ve
ΔU = +ve
So T2 > T1
32. Calculate the average molar heat capacity at constant volume of a mixture containing 2 moles of monoatomic
and 3 moles of diatomic ideal gas.
(1) R (2) 2.1 R (3) 3.2 R (4) 4R
Sol. Answer (2)
n1CV1 + n2CV2
= CV (average)
n1 + n2
33. What is the change in entropy when 2.5 mole of water is heated from 27°C to 87°C?
Given CP,M(H2O) = 4.2 J/g [ln 1.2 = 0.18]
(1) 16.4 J K–1 (2) 34.02 J K–1 (3) 2.89 J K–1 (4) 18.2 J K–1
Sol. Answer (2)
T2
ΔS = nCP  dln T
T1
34. Which amount the following is most soluble in water?

Compound ΔHhydration (kJ mol–1) ΔHlattice(K)
(1) A –400 +500
(2) B –300 +650
(3) C –200 +150
(4) D –100 +250
Sol. Answer (3)
Miscellaneous
35. A hungry man weighing 80 kg take quickly 20 g lunch, and then climbs up a mountain making it to a height
of 200 m. If 60% of food energy was wasted as heat and the rest was used as climbing work. The fuel intake
could have been any one of the following with given enthalpy of combustion?
(1) Glucose 16 kJ/g (2) Wheat bread 20 kJ/g
(3) Fructose syrup 13 kJ/g (4) Olive oil 35 kJ/g
Sol. Answer (2)
Work done = M × a × displacement
= 80 × 9.8 × 200 = 156.8 kJ
40% of total energy = 156.8 kJ
156.8 × 100
100% of total energy = = 392 kJ
40
36. If an element X has three allotropes A, B and C. If heat of combustion at standard state for A, B and C are –
100, –150 and –110 kJ/mol respectively. The correct order of stability is
(1) A > B > C (2) B>C>A (3) A>C>B (4) A=B=C
Sol. Answer (3)
|Heat of Combustion| decreases as Stability increases.
1
37. Considering the reaction C(s) + → CO(g) + 200 kJ . The signs of ΔH, ΔS and ΔG respectively are
O2 (g) ⎯⎯
2
(1) +, –, – (2) –, +, + (3) –, –, – (4) –, +, –
Sol. Answer (4)
38. For the reaction,
→ NH3 ( g) + H2S ( g)
NH4HS ( s) ⎯⎯⎯
The value of ΔH at 27°C is 12 kcal, what is the value of ΔU for the reaction? [R = 2.0 cal K–1mol–1]
(1) 10.8 kcal (2) 11.8 kcal (3) 9.8 kcal (4) 12.8 kcal
Sol. Answer (1)
ΔH = ΔU + Δng RT
2 × 2 × 300
ΔU = 12 – = 10.8 kcal
1000
39. The heat of combustion of ethene (C2H4) is –1409.3 kJ/mol. Calculate the weight of ethene required to
produce 470 kJ of heat on combustion.
(1) 12.5 g (2) 9.34 g (3) 23 g (4) 18.4 g
Sol. Answer (2)
1409.3 kJ heat is produced by = 28 g ethene
28
470 kJ heat will be produced by = × 470 = 9.34 g
1409.3
40. Two moles of an ideal gas is compressed isothermally and reversibly from a volume 2 L to 0.5 L at initial
pressure of 1 atm. The work done by gas is (Given 1 L atm = 101 J)
(1) 280 J (2) 150 J (3) 184 J (4) 250 J
Sol. Answer (1)
V2
w = – 2.303 nRT log V
1
V2
= – 2.303 P1V1log
V1
0.5
= –2.303 × 1× 2 log
2
= 2.76 L atm
280 J
41. During an adiabatic process, a gas obey TV0.4 = constant. The gas could be
(1) He (2) SO2 (3) N2 (4) NH3
Sol. Answer (3)
TVγ–1 = constant
γ – 1 = 0.4
γ = 1.4
⇒ Diatomic gas
42. In which of the following reaction, Kp ≥ Kc at 25°C?

(1) H2(g) + I2(g)
2HI(g) (2) PCl5(g) ⎯→ PCl3(g) + Cl2(g)
(3) 2H2O(g) ⎯→ 2H2(g) + O2(g) (4) All of these
Sol. Answer (4)
43. A system performs 5 J of work on its surroundings and absorbs 95 J heat from surroundings. The change in
internal energy of the system is
(1) +100 J (2) +90 J (3) –100 J (4) –90 J
Sol. Answer (2)
q = +95 J, w = – 5 J
ΔU = q + w = 90 J
44. Average bond energy of S–F and C–H is x and y. if y is 2 times of x
SF ⎯⎯
→ S(g) + 6F(g)
6 ΔHro = Q kJ
(2 mol)
CH4 (g) ⎯⎯
→ C(g) + 4H(g) ΔHro = P kJ
(1 mol)
Q
Then is
P
(1) 1 (2) 1.5 (3) 0.75 (4) 6
Sol. Answer (2)
Q = 12x
P = 4y
Q 12x 1
= = 3 × = 1.5
P 4y 2
45. The heat of neutralisation of a strong acid and strong base is 57 kJ/mol. The heat released when 0.4 moles
of HCl solution reacts with 0.3 moles of NaOH is
(1) 57.0 kJ (2) 11.4 kJ (3) 17.1 kJ (4) 22.8 kJ
Sol. Answer (3)
Heat released = 0.3 × 57 = 17.1 kJ
46. One mole of a perfect gas expands isothermally to ten times of its original volume. The change in entropy is
(1) 0.1 R (2) 10 R (3) 2R (4) 2.303 R
Sol. Answer (4)
V2
ΔS = nRln
V1
47. Bond energies of H – H bond is 80 kJ/mol, I – I bond is 100 kJ/mol and for H – I bond is
200 kJ/mol, the enthalpy of the reaction :
H2(g) + I2(g) → 2HI(g) is
(1) –120 kJ (2) –220 kJ (3) +100 kJ (4) +120 kJ
Sol. Answer (2)
H2 (g) + I2 (g) → 2HI (g)
ΔH = ∑B.Er – ∑B.Ep = 80 + 100 – 2 × 200 = – 220 kJ
48. The enthalpy of formation of H2O(l) is –280.70 kJ/mol and enthalpy of neutralisation of a strong acid and strong
base is –56.70 kJ/mol. What is the enthalpy of formation of OH– ions?
(1) –22.9 kJ/mol (2) –224 kJ/mol (3) –58.7 kJ/mol (4) –214 kJ/mol
Sol. Answer (2)
ΔHf(OH¯) = – 280 – (56.70) = – 224 kJ/mole
49. The heat of neutralisation of a strong dibasic acid in dilute solution by NaOH is nearly
(1) –27.4 kcal/eq (2) 13.7 kcal/mol (3) –13.7 kcal/eq (4) –13.7 kcal/mol
Sol. Answer (3)
H+ + OH– → H2O ΔH = – 13.7 kcal/eq
50. The heat released in neutralisation of HCl and NaOH is 13.7 kcal/mol, the heat released on neutralisation of NaOH
with CH3COOH is 3.7 kcal/mol. The ΔH° of ionisation of CH3COOH is
(1) 10.2 kcal (2) 10 kcal (3) 3.7 kcal (4) 9.5 kcal
Sol. Answer (2)

ΔH = 13.7 – 3.7 = 10 kcal
51. Which of the following reactions represents the enthalpy of formation of water?
(1) H+(aq) + OH–(aq) → H2O(l ) (2) H2(g) + ½O2(g) → H2O(l )
(3) 2H2(g) + O2(g) → 2H2O(l ) (4) 2H+(aq) + 2OH–(aq) → 2H2O(l )
Sol. Answer (2)
Enthalpy associated with formation of one mole of water from its constituent elements in their natural states.
52. The energy required to break 76 g gaseous fluorine into free gaseous atom is 180 kcal at 25°C. The bond
energy of F – F bond will be
(1) 180 kcal (2) 90 kcal (3) 45 kcal (4) 104 kcal
Sol. Answer (2)
F2 → 2F
38g
∵ 76 g F2 needs 180 kcal energy
180
∴ 38 g F2 needs energy = × 38 = 90 kcal
76
53. The heat of combustion of yellow phosphorus and red phosphorus are –9.91 kJ/mol and – 8.78 kJ/mol
respectively. Then the heat of transition of yellow phosphorus to red phosphorus is
(1) –18.69 kJ (2) +1.13 kJ (3) +18.69 kJ (4) –1.13 kJ
Sol. Answer (4)
P → P
(yellow ) (red)
ΔHTransition = – 9.91 – (–8.78) = – 1.13 kJ

54. Which of the following represents heat of formation (ΔHf)?
(1) C(diamond) + O2(g) → CO2(g) (2) 2CO(g) + O2(g) → 2CO2(g)
(3) ½H2(g) + ½F2(g) → HF(g) (4) N2(g) + 3H2(g) → 2NH3(g)
Sol. Answer (3)
There is formation of one mole of gas from its constituent elements in their standard states.
1 1
H2 (g) + F2 (g) → HF(g)
2 2
55. 10 g of each Pb, Al , Ag and Cu is given 100 kJ of heat and temperature rise is x, y, z and w kelvins
respectively.
Then select the correct relation. [given heat capacity (in J/g k) is 0.9, 0.386, 0.233 and 0.128 for Al, Cu, Ag
and Pb respectively]
(1) x < y (2) y<w (3) z<w (4) x < z
Sol. Answer (2)
1
ΔT ∝ so x > z > w > y
heat capacity
56. In a process, 120 J heat is given to O2 gas and 180 J work is done on the gas. The molar heat capacity for this
process is (Assume gas is ideal and gas constant is R)
2R 5R R
(1) (2) (3) R (4)
5 2 5
Sol. Answer (3)

q = 120 J
w = 180 J
5 120
ΔU = n × R × ΔT = 300 J  nΔT =
2 R
q 120
Cm = = R
nΔT 120 / R
57. 1 mol of Ca(OH)2 is completely neutralised by HCl then heat released in the reaction is (Only magnitude)
(1) More than 57.4 kJ (2) Less than 57.4 kJ
(3) Equal to 57.4 kJ (4) Cannot be determined or compared
Sol. Answer (1)
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
ΔH°r = 2 × Heat of neutrialisation (kJ/mol)
58. Which of the following is(are) state function as well as intensive property?
(A) Entropy
(B) Internal energy
(C) Pressure
(1) A, B and C (2) Only A and B (3) Only B (4) Only C
Sol. Answer (4)
H, U, Pressure are state function.
H,U are extensive properties.
59. C2H5OH(l) on complete combustion produce CO2(g) and H2O(l) at 300 k. The difference between ΔH and ΔU
at this temperature is
(1) – 2.49 kJ (2) + 2.49 kJ (3) + 4.98 kJ (4) – 4.98 kJ
Sol. Answer (1)
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
ΔH – ΔU = ΔngRT
Δng = – 1


Thermodynamics 1

Uploaded by

Copyright:

Available Formats

Thermodynamics 1

Uploaded by

Document Information

Original Description:

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Thermodynamics 1

Uploaded by

Copyright:

Available Formats

Level-I

Very Short Answer Type Questions :

8. State Hess’s law of constant heat summation.

The internal energy of the system decreases by 30 kJ.

= –2.303 × 8.314 × 300 × 0.699

16. State the conditions under which qp = qv or ΔH = ΔU.

H2(g) + Cl2(g) ⎯→ 2HCl(g)

= 106 g × 4.184 JK–1g–1 × 5

Example : CaO(s) + CO2(g) ⎯→ CaCO3(s); ΔrHs = –178.3 kJ/mol

Gas constant R = 8.314 × 10–3 kJ mol–1 K–1

∴ Work done for isothermal reversible expansion of an ideal gas.

= –5744.14 × 10–3 kJ mol–1

28. What is meant by enthalpy of formation? Given its significance.

ΔrHs = [ΣΔfHsproducts] – [ΣΔfHsreactants]

Sol. PH3(g) ⎯→ P(g) + 3H(g); ΔH1 = 228 kcal mol–1

PH2 – PH2 ⎯→ 2P(g) + 4H(g); ΔH2 = 355 kcal mol–1

∴ On the basis of the given data

ΔH2 = ΔHP – P + 4ΔHP – H

228  ΔH1 228 

∴ ΔHP – P = 51 kcal mol–1

Long Answer Type Questions :

CH4(g) ⎯→ C(g) + 4H(g); ΔaHs = 1665 kJ mol–1

CH4(g) ⎯→ CH3(g) + H(g); ΔbondHs = +427 kJ mol–1

427 + 439 + 452 + 347

2H(g) + 2Cl(g) ⎯⎯⎯⎯⎯⎯⎯⎯

Thus, for the reaction

−1648 × 103 J mol−1

= 5530 J K–1 mol–1

And w = –pex(V2 – V1)

Sol. N2O4(g) 2NO2(g)

If N2O4 is 50% dissociated, the given equilibrium is

Initial moles 1.0 0

Equilibrium moles 0.5 2 × 0.5

Total mole = 1 + 0.5

The equilibrium constant Kp is given by

Sol. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

The equilibrium constant,

[CH3 COOC2H5 ][H2O]

40. List differences between reversible and irreversible process.

Sol. Reversible process Irreversible process

4. Work obtained is maximum. 4. Work obtained is not maximum.

C2H4(g) + H2(g) – C2H6(g) ⎯→ ΔrHs = –136.9 kJ

44. Give the appropriate reasons for the following

First Law of Thermodynamics

Sol. Answer (3)

7. Calculate standard heat of combustion of ethanol (C 2 H 5 OH(l)). Given that Δ f H° (C 2 H 5 OH, l) =

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

ΔCH° = 2ΔfH°(CO2) + 3ΔfH°(H2O) – ΔfH°(H2O)

(1) C (graphite) (2) Br2(l) (3) H2(g) (4) I2(g)

For I2(s), ΔfH° = 0

3C(g) + 4H2(g) → H3C — C — CH3(g)

ΔHf(CO2 ) for 3 moles will be = 3 × 407 = 1221 kJ

∴ With O3 it should be greater than 1221 kJ

C12H22O11 + 12O2 ⎯⎯→ 12CO2 + 11H2O

12. For the reaction,

C7H8 (l) + 9O2 (g) ⎯⎯→ 7CO2 (g) + 4H2O(l),

A(s) ⎯⎯→ A(l) ΔH = x

A(l) ⎯⎯→ A(g) ΔH = y

The heat of sublimation of A will be