3
THERMODYNAMICS
Type I
:
Remember
:
(1)
(2)
(3)
(4)
First law of Thermodynamics
∆U = q + W
Heat absorbed by the system then q
Heat released by the system then q
Work done on the system w
Work done by the system w
1 bar – lit = 101.32 Joule
=
=
=
=
+ Ve
–Ve
+ Ve
– Ve
1.
A system by getting 468.61 calorie heat from a surrounding does the work
equivalent to 200.83 calories. Calculate the change in internal energy. (Ans :
267.78 joule)
2.
One system lost 523 joule heat and did 1317.96 joule work, calculate the change
in its internal energy. (Ans : – 1845.14 joule)
3.
One system absorbed 711.28 joule heat and did the work change in its internal
energy is 460.24 joule. Calculate how much will be the work done. (Ans : – 251.04
joule)
4.
When argon gas expands, 3347.2 joule heat is absorbed. Work done by the system
is 1255.2 joule. Calculate change in internal energy. (Ans : 2092 joule)
5.
A system receives 224 Joule heat and does work of 156 Joule. Calculate the
change in the internal energy. (Text. Book Illus. - 1 Page 69)
6.
A system did the work of 785 joule after loss of 525 joule heat. Find the change in
its internal energy. (Text. Book Exe – 3 (7) Page 90)
7.
A system absorbed 650 joule heat and did the work. Its internal energy change is
440 joules, then find out how much work has been done. (Text. Book Exe – 3 (8)
Page 90)
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 ?
8.
Type II
Remember
(1)
:
Work due to change in volume at constant T
:
W = P∆V
= P (V2 – V1)
Where V2
V1
1
= Final volume
= Initial volume
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
(2)
(3)
1 bar – lit
= 101.32 Joule
= 24.216 calorie
If gas expands then work done by the system…w = –Ve
If gas contracts then work done on the system…w = +Ve
9.
Calculate work done by the system when 100 lit of an ideal gas expands to 120 lit
at 10 bar pressure. (Ans : – 20264 joule)
10.
Calculate work done on system when 10 lit. of an ideal gas compressed to 5 lit.
by 2 bar external pressure. (Ans : + 1012.95 joule)
11.
A cylinder is filled with 5 moles of an ideal gas at 300 k and 5 bar pressure. Due
to leakage in the valve gas diffuses in the atmosphere. How much work is done
by the gas ? (Ans : – 9966.29 joule)
12.
A cylinder is filled with 1 mole an ideal gas at 300 k & 10 bar pressure. Due to
leakage gas diffuses in the atmosphere. How much work is done by expansion of
gas. (Ans : – 2246.39 joule)
Type III
:
Work & First law of Thermodynamics
Remember : (1)
(2)
∆U
= q + W = q + P∆V
1 bar – lit
= 101.32 Joule
= 24.216 calorie
13.
The volume of gas at 1 bar pressure was 0.5 lit. If this gas obtains 121.34 joule
heat, its volume becomes 2.0 lit at 1 bar pressure. Calculate its internal energy.
(Ans : – 30.68 joule)
14.
A gas contracts from 8.4 lit to 4.2 lit by an external pressure of 1.5 bar. During
this process 830.10 joule heat is evolved calculate change in internal energy.
Ans. w = + 638.06 joule – because gas compressed and ∆U = – 192.04 joule)
15.
At one bar pressure, the volume of a gas is 0.6 litre. If the gas receives 122 Joules
of heat at one atmosphere pressure, the volume becomes 2 litres, then calculate
its internal energy (1 litre bar = 101.32 Joule). (Text. Book Illus. - 2 Page 69)
Type IV : Enthalpy and 1st law of Thermodynamics
Remember : (1)
(2)
2
∆H = ∆U + P∆V
∆H = ∆U + ∆n(g) RT
Where ∆n(g) = (no. of moles of gaseous product) – (no. of moles of gaseous
reactant)
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
16.
The specific volume of ice and water are 1.089 and 1.0 ml gm–1 at 273 K
temperature. The molecular heat of fusion (melting) of ice is 6.025 kJ mole–1 at
273 K temperature and 1 bar pressure. If 90 gram ice melts at 273 K temperature
and 1 bar pressure, water is produced at that temperature, then, calculate the
values of ∆H and ∆U for the process. (1 litre bar = 101.32 joule). (Text. Book Illus. 3 Page 71)
17.
The enthalpy change of the following reaction at 1 bar pressure and 27 °C temp.
∆H = – 83.68 kilojoule / mole. Calculate its change in ∆U.(R = 8.314 joule).
Reaction is… 2 C(s) + 3 H2(g) → C2H6(g).
(Ans : ∆U = – 78.70 kilojoule)
18.
The internal energy change of the following reaction at 27 °C is
∆E = – 908 kjoule/mole. Calculate its change in enthalpy. (R = 8.314 joule) (Text.
Book Exe – 7 (1) Page 91)
4 NH3(g) + 5 O2(g) → 6H2O(l) + 4NO(g). (Ans : ∆H = – 920.40 kilojoule)
19.
The heat associated with combustion of liquid benzene, at constant volume is –
3268 kilojoule mole–1. Calculate the change in enthalpy, when this reaction
occurs at 300 K temperature (R = 8.314 joule). (Text. Book Illus. - 4 Page 71)
20.
Calculate the change in internal energy of the following reaction at 300 K
temperature ( R = 8.314 Joule) (Text. Book Illus. - 6 Page 72)
2 C(s) + O2(g) → 2CO(g) ∆H = 110.53 kilojoule
21.
1 mole of an ideal gas is filled in a cylinder at 300 K temperature and 10 bar
pressure. This gas leaks in atmosphere due to leakage in the valve of the
cylinder, then What work should be done in the diffusion of gas under this
condition ? Calculate the values of ∆H and ∆U for this process. (1 litre bar =
101.32 joule) (Text. Book Illus. - 5 Page 71~72)
22.
The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb
calorimeter, and ∆U was found to be – 742.7 kJ mol–1 at 298 K. Calculate enthalpy
change for the reaction at 298 K. (Text. Book Exe – 7 (2) Page 91)
→ N2(g) + CO2(g) + H2O(l)
NH2CN(s) + O2(g)
Type V
:
Remember
:
(1)
(2)
(3)
3
Heat capacity
The heat required to raise the temp of any substance by 1 °C is called the heat
capacity (c) of that, substance.
q
Heat absorbed
Heat capacity (C)
=
=
joule / °C
Temp. difference
∆T
Heat absorbed
Specific heat capacity =
Temp. difference × weight of sub in gm
Unit of specific heat capacity is… joule / °C gm
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
(4)
Heat absorbed
Temp. difference × (mole of substance)
Unit of molar heat capacity is joule / °C mol.
Molar heat capacity
=
23.
The temperature of a sample of ethanol is increased by 2 °C when 98.03 joule
heat is supplied. Calculate the heat capacity of that sample. (Ans : 49.04 joule/°C)
24.
Calculate the heat gained by water when 2 kg water kept at temp 25 °C is heated till
it boils. The specific heat capacity of water is 4.184 joule / °C gm.
(Ans : 627.6 joule)
25.
Calculate increase in temp of water (kept at 25 °C), when 1.5 kg water absorbs
418.4 kjoule energy. The specific heat capacity of water is 4.184 joule / °C gm.
(Ans : Final temp of water is 91.67 °C and increase in temp of water is 66.67 °C)
26.
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.
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–1K–1
27.
Type VI
:
Remember
:
(1)
(2)
(3)
Heat of Reaction.
Consider a following hypothetical reaction…
aA + bB → cC + dD
∆H° = H°p — H°R
∴ ∆H°
= [ c H°C + d H°D ] — [ a H°A + b H°B ]
Enthalpy of elements in their standard state is zero.
Unit of Standard Enthalpy change ∆H° is kJ/mole.
28.
In the following reaction, the standard enthalpies of Al2O3(s) and Fe3O4(s) are
–1669.4 kilojoule mole–1 and –117 kilojoule mole–1 respectively. What is the
enthalpy change of the reaction ? 8Al + 3 Fe3O4 → 4 Al2O3 + 9 Fe.
(Text. Book Illus. - 7 Page 75)
29.
The enthalpies of gases NH3, H2O and NO are –192.56, –241.83 and 90.37
kilojoule/mole respectively. Calculate enthalpy change in following reaction.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
30.
Enthalpies of formation of CO(g), CO2(g), N2O(g), N2O4(g) are –110, –393, 81, and 9.7
kJ mol–1 respectively. Find the value of ∆rH for the reaction :
N2O4(g) + 3 CO(g) → N2O(g) + 3 CO2(g)
4
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
Type VII
:
Remember
:
+ OH–
Heat of Neutralization.
(1)
H+
31.
How much heat will evolve on mixing a dilute soln. of 0.5 mole of HNO3 with a
dilute soln. of 0.2 mole KOH ? (Ans : 11.18 kilojoule)
32.
How much heat will evolve on adding 600 ml 0.1 M NaOH soln. to 400 ml 0.5 M
H2SO4 soln. (Ans : 3.35 kilojoule)
→
H2O
∆H°
= 55.90 = 56 kilojoule/mole
Type VIII
:
Remember
:
(1)
= H°p — H°R
= [Std. enthalpies of product] – [Std. enthalpies of product]
33.
∆H°
Heat of Formation
Given…N2(g) + 3H2(g) → 2 NH3(g) ; ∆rH = – 92.4 kJ mol–1
What is the standard enthalpy of formation of NH3 gas ?
Type IX
Remember
:
Heat of Combustion.
:
(1)
∆H°
34.
C6H6(l) + 7.5 O2(g) → 6CO2(g) + 3H2O(l) ∆H = – 3267.7 Kjoule.
In this rean. std. enthalpy change of formation of CO2(g) and H2O(l) are –393.5 and
– 285.85 Kjoule/mole respectively. Calculate using above combustion reaction the
std. heat of formation of benzene. (Text. Book Illus. - 8 Page 75)
35.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l). In this reaction, the heat of formation
of ethanol(l), CO2(g) and H2O(l) are – 185.35, – 393.5 and – 285.85 kcal/mole
respectively. Calculate the enthalpy of combustion of ethanol. (Text. Book Illus. 9 Page 76)
36.
Calculate std. enthalpy of formation of propane (C3H8) if its enthalpy of
combustion is – 2220.2 Kjoule/mole. The enthalpies of formation of CO2(g) and
H2O(l) are – 393.5 Kjoule/mole and – 285.85 Kjoule/mole respectively. (Ans : 103.51
kilojoule)
37.
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. Calculate
enthalpy of formation of CH4(g). (Ans : –74.8 kilojoule mole–1)
5
= H°p — H°R
= [Std. Enthalpies of products] — [Std. Enthalpies of Reaction]
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
38.
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.
Type X
:
Remember
:
1.
2.
Hess’s Law
If a given reaction passes through several steps, the net enthalpy change is equal to
the sum of enthalpy changes of theses steps.
Significance of Hess’s law is that thermochemical equation can be added or
substracted.
39.
Using the information given below, calculate the heat of combustion of glucose.
(i)
C(graphite) + O2(g)
= CO2(g)
∆H1 = −94.05 Kcal
(ii)
H2(g) + ½ O2(g) = H2O(l)
∆H2 = −68.32 Kcal
(iii) 6C(graphite) + 6 H2(g) + 3 O2(g) = C6H12O6(s)
∆H3 = –279.45 Kcal
Ans : Combustion of glucose will occur as follows.
+ 6 O2 = 6 CO2 + 6 H2O
C6H12O6
To obtain above equation, given three thermo chemical equation can be sum up in
the following way.
((i) × 6) + ((ii) × 6) – (iii)
Therefore,
+ 6 O2(g)
= 6 CO2(g)
6 C(graphite)
6 H2(g)
+ 3 O2(g)
= 6 H2O(l)
−6 C(graphite) − 6 H2(g) − 3 O2(g) = −C6H12O6(s)
C6H12O6
+ 6 O2
= 6 CO2 + 6 H2O
Therefore, ∆H of the above reaction is,
∆H
= (∆H1 × 6) + (∆H2 × 6) – ∆H3
= (−94.05 × 6) + (−68.32 × 6) – (–279.45)
= −564.3 + –409.92 + 279.45
= –694.77 Kcal
Heat of combustion = –694.77 Kcal/mole
40.
Using the information given below, calculate the heat of combustion of benzene.
(i)
6C(s) + 3 H2(g)
= C6H6(l)
∆H1 = 11.72 Kcal
(ii)
H2(g) + ½ O2(g) = H2O(l)
∆H2 = −68.32 Kcal
∆H3 = –94.05 Kcal
(iii) C(s) + O2(g) = CO2(g)
Ans : Combustion of benzene will occur as follows.
15
C6H6(l)
+
O2(g) = 6 CO2(g) + 3 H2O(l)
2
To obtain above equation, given three thermo chemical equation can be sum up in
the following way.
((ii) × 3) + ((iii) × 6) – (i)
6
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
Therefore,
3
O2(g)
2
+ 6 O2(g)
− 3 H2(g)
3 H2(g) +
6 C(s)
−6C(s)
=
3 H2O(l)
= 6 CO2(g)
= −C6H6(l)
15
O2(g) = 6 CO2(g) + 3 H2O(l)
2
Therefore, ∆H of the above reaction is,
∆H
= (∆H2 × 3) + (∆H3 × 6) – ∆H1
= (−68.32 × 3) + (−94.05 × 6) – (11.72)
= (–204.96) + (–564.3) – 11.72
= –780.98 Kcal
C6H6(l)
+
Heat of combustion = –780.98 Kcal/mole
41.
Calculate the enthalpy change in a reaction of conversion of graphite into
diamond using following reaction.
(i)
C(graphite) + O2(g)
= CO2(g)
∆H1 = –94.05 Kcal
(ii)
C(diamond) + O2(g)
= CO2(g)
∆H2 = –94.50 Kcal
Ans : To obtain reaction of conversion of graphite into diamond, given two thermo
chemical equation can be sum up in the following way.
(i) − (ii)
Therefore,
C(graphite) + O2(g)
= CO2(g)
− C(diamond) − O2(g)
= −CO2(g)
C(graphite)
=
C(Diamond)
Therefore, ∆H of the above reaction is,
∆H
= (∆H1) − (∆H2)
= (−94.05) − (−94.50)
= 0.45 Kcal
Enthalpy change of reaction = –780.98 Kcal/mole
42.
7
Calculate the C−H bond energy by calculating
given information.
CH4(g)
=
C(g) + 4 H(g)
(i)
CH4 + 2 O2 = CO2 + 2 H2O
(ii)
C
+ O2
= CO2
(iii) H2
+ ½ O2 = H2O
(iv)
C(s)
= C(g)
(v)
H2(g)
= 2 H(g)
∆H of the following reaction by
∆H
∆H1
∆H2
∆H3
∆H4
∆H5
=
=
=
=
=
=
?
–212.85 Kcal
–94.05 Kcal
–68.32 Kcal
+170 Kcal
+ 103 Kcal
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
Ans : To calculate ∆H of the methane, given five thermochemical equation can be sum up
in the following way.
(i) − (ii) – (2 × (iii)) + (iv) + (2 × (v))
Therefore,
CH4 + 2 O2 = CO2 + 2 H2O
−C
− O2
= − CO2
− 2 H2 − O2
= − 2 H2O
C(s)
= C(g)
2 H2(g)
= 4 H(g)
CH4(g)
=
C(g)
+ 4 H(g)
Therefore, ∆H of the above reaction is,
∆H
= ∆H1 − ∆H2 − (2 × ∆H3) + ∆H4 + (2 × ∆H5)
= (−212.85) − (−94.05) − ( 2 × –68.32) + 170 + (2 × 103)
= –212.85 + 94.05 + 136.64 + 170 +206
∆H
= + 393.84 Kcal
∴ One C−H bond energy in CH4
= ¼ (393.84)
= 98.46 Kcal
C−H bond energy
= 98.46 Kcal/mole
43.
Calculate C−C bond energy on the basis of the following information.
(i)
2 C(graphite) + 3 H2(g) = C2H6(g)
∆H1 = –20.23 Kcal
(ii)
C(graphite) = C(g)
∆H2 = +170 Kcal
(iii) H2(g)
= 2 H(g)
∆H3 = +103 Kcal
The bond energy of C−H bond is 98 Kcal.
Ans : To calculate ∆H of the following decomposition reaction of ethane…,
+ 6 H2(g)
C2H6(g) = 2 C(g)
Given three thermo chemical equation can be sum up in the following way.
Therefore,
((ii) × 2) + ((iii) × 3) – (i)
=
2 C(g)
2 C(graphite)
=
6 H(g)
3 H2(g)
− 2 C(graphite) − 3 H2(g) = − C2H6(g)
C2H6(g)
= 2 C(g)
+ 6 H(g)
Therefore, ∆H of the above reaction is,
∆H
= (∆H2 × 2) + (∆H3 × 3) − (∆H1)
= (170 × 2) + (103 × 3) – (–20.23)
= 340 + 309 + 20.23
= + 669.23 Kcal
∴ One C−C bond energy in C2H6
= (C−C) + 6 (C−H)
∴ 669.23
= (C−C) + (6 × 98)
∴(C−C)
= 669.23 – 588
= 81.23 Kcal/mole
C−C bond energy
8
= 81.23 Kcal/mole
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
44.
Calculate the standard enthalpy of formation of CH3OH(l) from the following
data : (Text. Book Exe – 7 (3) Page 91)
3
CH3OH(l) + O2(g) → CO2(g) + 2H2O(l) ;
∆rH° = – 726 kJ mol–1
2
C(g) + O2(g)
→
CO2(g)
;
∆cH = – 393kJ mol–1
1
H2(g) +
O2(g)
→
H2O(l)
;
∆fH = – 286 kJ mol–1
2
45.
Calculate the enthalpy change for the process… CCl4(g) → C(g) + 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
Type XI
:
Remember
:
Bond Energy
(1)
∆H°
= [Bond Energy of Reactant] − [Bond Energy of product]
46.
The values of bond energies of C(s) to C(g), H – H, C – C and C – H are 170, 103, 80
and 98 Kcal/mole respectively. Calculate the value of ∆H for the following
reaction using these values of bond energies.
2 C(s)
+ 3H2(*g) → C2H6(g)
H
2 C(s)
+ 3H2(g) →
H—C—C—H
H
∴ ∆H =
=
∴ ∆H =
=
=
H
H
[2(C(s) → C(g)) + 3 (H — H )] —[C — C + 6 (C – H)]
[2 (170) + 3 (103)] — [80 + 6(98)]
[340 + 309] — [80 + 588]
[649] — [668]
– 19 kcal/mole.
47.
The values of bond energies of C(s) → C(g), H – H, C – H and C – C are 711.28,
430.95 and 410.03 and 334.72 Kjoule/mole respectively. Calculate the enthalpy
change of the following reaction using these values. (Ans : –92.05 kjoule mole–1)
3 C(s) + 4 H2(g)
→
C3H8(g)
48.
The bond energies of H–H, H–Cl and Cl–Cl are 103, 102 and 57 kcal/mol
respectively in the reaction…H2 + Cl2 → 2 HCl(g). Calculate the heat of reaction of
this reaction. (Ans : – 44 kcal mole–1 )
9
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
49.
Calculate the value of bond energy of C – C from the following information.
2 C(graphite) + 3H2(g) → C2H6(g)
∆H = – 20.23 Kcal
(i)
C(graphite) → C(g)
∆H1 = 170 Kcal
(ii)
H2(g) → 2H (g)
∆H2 = 103 Kcal
(iii) C — H bond energy = 98 Kcal.
50.
With the help of enthalpy change in the following reaction. Calculate the bond
energy of N – H bond in NH3. (N ≡ N = 945.58 Kjoule and H – H = 430.95 Kjoule)
(Ans : N — H = 389.11.0 Kjoule/mole)
51.
Thebond energies of O — H, H – H and O = O in the reaction H2(g) + ½O2(g) →
H2O(g) are 460.24, 430.95 and 493.71 kjoule/mole. Calculate the heat of formation
of H2O(g) using these values of bond energies. (Ans : ∆fH° = –242.67 Kjoule/mole)
Type XII
:
Remember
:
(1)
∆S
=
Change in Entropy (∆
∆S)
q rev
T
Where,
∆S
qrev
= Change in entropy
= Heat absorbed OR lost at temperature T K
∆H vap
(2)
∆Svap =
Where, ∆Hvap = Molar heat of vaporization
T
∆H fusion
Where, ∆Hfusion= Molar heat of fusion
T
∆H sublimation
Where, ∆Hsubli.= Molar heat of sublimation
T
+ ∆Ssurrounding = ∆Stotal
+Ve − Reaction is spontaneous.
−Ve − Reaction is not spontaneous.
0
− Reaction is in equilibrium.
(3)
∆Sfusion=
(4)
∆Ssubli. =
(5)
(6)
∆Ssystem
∆Stotal =
∆Stotal =
∆Stotal =
52.
What will be the change in entropy when two moles of water are converted into
its vapour at 100°C. The molar heat of vaporization of water is 40668.48 cal mol−1.
(Ans : 218.06 joule K–1)
53.
3 moles of water is boiled at 373 K and is changed to vapour state having the
same temperature, what will be the change in entropy of the system ? (The
molecular enthalpy of vaporization of water is 40.668 kjoule mole–1) (Text. Book
Illus. - 10 Page 79)
54.
The enthalpy of vaporization of benzene is 30.799 KJoule mol−1. If boiling point
is 353 K, find the change in entropy for the conversion of 1 mole liquid benzene
to its vapour at that temperature. (Text. Book Illus. - 11 Page 79)
55.
The molar heat of vaporization of benzene is 30.799 KJoule mol−1. Calculate the
change in entropy for converting 1 mole of gaseous benzene to liq. state at 353 K.
10
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
56.
When 600 gm of water at 30° C is placed in an atmosphere having temp 25° C,
251.04 joule heat is evolved. Prove that process is spontaneous.
57.
Amount of heat required to melt 1 g ice at 0° C is 334.72 joule.
(a) Calculate change in entropy when 9 g ice melts at 0°C.
(b) Calculate change in entropy of atmosphere at 27°C.
(c) Prove that ice melts spontaneously.
58.
“Will ice having temperature 273 K placed in a surrounding having temperature
298 K will give water having temperature 273 K.” Prove this statement. The
molecular enthalpy of fusion of ice is 6.025 kjoule mole–1.” (Text. Book Illus. - 12
Page 79)
59.
Calculate change in entropy when 90 g of water are converted into its vapour at
100 °C. (Molar heat of vaporization of water is 40.67 Kjoule mole−1). (Ans : 0.545
joule K−1)
60.
Entropy change of vapourization of benzene is 87.25 joule K−1mol−1. Molar heat
of vapourization of benzene is 30.8 KJoule mol−1. Calculate boiling point of
benzene.
61.
Entropy change of vapourization of acetic acid is 1 joule mol−1 K−1. Boiling point
of acetic acid is 118.2 °C. How much heat is absorbed when 120 g acetic acid is
vapourized at its boiling point. (Ans : 782.32 joule)
Type XIII
:
Remember
:
Relation between Free Energy (∆
∆G), Enthalpy (∆
∆H) and
Entropy (∆
∆S)
(1)
∆G
= ∆H
−
Where, ∆G
∆G
∆G
T∆S
= − Ve − Reaction is spontaneous in forward direction
= + Ve − Reaction is non-spontaneous in forward direction
= 0
− Reaction is in equilibrium state
62.
If ∆H = 400 kjoule mole–1 and ∆S = 0.2 kjoule mole–1 for a reaction 2x + y → z, at
what minimum temperature the reaction will be spontaneous ? (Text. Book Exe –
7 (4) Page 91)
63.
For the following reaction 2P(g) + Q(g) → 2R(g) ∆U° = – 10.5 kjoule and ∆S° = – 44.2
joule K–1. Find ∆G° for the reaction. Will the reaction occur on its own ? Why ? (Text.
Book Exe – 7 (5) Page 91)
64.
From the given data mention which of the following reaction will occur on their
own at 298 K. (Text. Book Exe – 7 (6) Page 91)
Reaction : X : ∆H = – 52 kjoule ; ∆S = 956 joule K–1
Reaction : Y : ∆H = – 60 kjoule ; ∆S = – 65 joule K–1
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CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
65.
The changes in enthalpy and entropy for a reaction P + Q R + S occurring at
320 K temperature are 170 kilojoule and 26 joule K–1. Will this reaction be
spontaneous ? Why ? Find out. (Text. Book Exe – 7 (7) Page 91)
66.
Which reaction will occur spontaneously at 25 °C ?
Rean. A : ∆H
= 10 Kcal
∆S
= 30 cal K−1
Rean. B : ∆H
= − 15 Kcal
∆S
= −15 cal K−1
(Ans : Reaction A will not occur spontaneously but reaction B will occur
spontaneously)
67.
The vapourization of 1 gm of water at 373 K absorbs 2259.36 joule. What would
be the change in enthalpy, change in entropy and change in free energy if 9 gm
of water is converted into vapour at 373 K ?
68.
A + B → C + D; ∆H = − 40 Kcal. Above 400 K forward reaction is
spontaneous below 400 K reverse reaction is spontaneous. Calculate ∆S at 400 K.
69.
P + Q → R + S ∆H = 12 Kcal and ∆S = 24 cal K−1. At what temperature
reaction will be in equilibrium state. (Ans : At 500 K reaction is in equilibrium
state.)
70.
Calculate the value of ∆G at 25° C for the following reaction.
CaCO3(s) →
CaO(s)
+ CO2(g)
The value of ∆H and ∆S are 42.5 Kcal and 38.3 cal K−1. Will this reaction be
spontaneous ? Why ?
71.
At which temperature following reaction will be in equilibrium state. Predict the
nature of forward reaction above and below this temperature.
Ag2O(s) →
2 Ag(s)
+ ½ O2(g)
−1
−1
∆H = 7.3 kcal mol and ∆S = 15.76 cal K . (Ans : 463.19 K ; Above 463.19 K
forward reaction will be spontaneous)
Type XIV
:
Remember
:
Standard Free Energy of Formation of Compound
=
∑ ∆G
∑ ∆G
(1)
∆G °
(2)
∆G °f element = Zero
72.
The standard free energy of formation of C6H6(l), CO2(g) and H2O(l) are 124.52,
−394.38 and −237.13 Kjoule mole−1 respectively. Calculate the standard free
energy change ∆G° of the reaction. (Text. Book Illus. - 13 Page 82)
C6H6(l)
+ 7.5 O2(g)
→
6 CO2(g)
+ 3 H2O(l)
73.
∆G° of the following reaction is −966.50 Kjoule. Calculate ∆G°f CH4, if ∆G°f
CH3OH is −695.80 Kjoule mole−1. (Ans : 212.55 kjoule mol–1)
2 CH4
+ O2
→
2 CH3-OH
12
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
f
pro.
−
f
rea.
74.
Calculate ∆G° of the following reaction.
CH4 + 2 O2 →
CO2 + 2 H2O
∆G°f CH4 = −12.14 Kcal mole−1 ; ∆G°f CO2 = −94.05 Kcal mole−1 ; ∆G°f H2O = −56.76
Kcal mole−1. (Ans : ∆G ° = −195.43 Kcal)
Type XV
:
Remember
:
Relation between ∆G°° and Equilibrium Constant
(1)
(2)
∆G °
= − 2.303 R T log K
For gaseous system, Equilibrium constant K = Kp
For liquid system, Equilibrium constant K = Kc
75.
The equilibrium constant of the following reaction is 0.5 at 25° C.
A
+
B
→
C
If the standard free energy of formation of A and B are −20.2 and −25.0 Kcal
mole−1 respectively. What will be the standard free energy of formation of C ?
Ans :
∆G °
∴ ∆G °
=
=
=
=
=
=
− 2.303 R T log K
− 2.303 × 1.987 × 10−3 × 298 × log 0.5
− 2.303 × 1.987 × 10−3 × 298 × ( 1 .6990)
− 2.303 × 1.987 × 10−3 × 298 × (−1.0 + 0.6990)
− 2.303 × 1.987 × 10−3 × 298 × (−0.3010)
+0.4105 Kcal
Now, ∆G °
=
∑ ∆G
∴ 0.4105
∴ ∆G°fC
∴ ∆G°fC
f
pro.
−
∑ ∆G
f
rea.
= ∆G°fC
− ( ∆G°f A + ∆G°f B )
= ∆G°fC
− ( −20.2
− 25.0 )
= 0.4105 − 45.2
= − 44.7896 Kcal mole−1
76.
The value of Equilibrium constant for the following reaction is 4.0. Calculate the
value of ∆G° for this reaction at 25°C. (Text. Book Illus. - 15 Page 82~83)
C2H5-OH(l) + CH3-COOH(l)
→
CH3COOC2H5(l)
+ H2O(l)
77.
2 NO(g)
+ O2(g)
→
2 NO2(g)
Following data is obtained for above reaction at 298 K temperature.
NO
NO2
O2
∆S°f JouleK–1mol–1
210.45
240.6
205
∆H°f Kjoulemole–1
90.37
33.85
-
Calculate ∆H°, ∆S°, ∆G° and equilibrium constant (K) for above reaction.
(Text. Book Illus. - 14 Page 82)
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CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
78.
Calculate the equilibrium constant for the following reaction at 25°C.
2 SO2(g)
+ O2(g)
→
2 SO3(g)
−1
∆G°f SO2 = −300.37 Kjoule mole
∆G°f SO3 = −370.37 Kjoule mole−1
Ans : ∆G° = − 140 Kjoule and
Kp
= 3.388 × 1024
79.
Find out the equilibrium constant of the following reaction at 298 K temperature
1
NO(g) + O2(g)→ NO2(g) The value of ∆fG° for NO and NO2 are 104.2 and 51.3
2
kjoule mole–1 respectively of 298 K temperature. (Text. Book Exe – 7 (8) Page 91)
The equilibrium constant Kp of the following reaction is 1.8 × 10−7. Calculate
value of ∆G° for this reaction at 25°C. ( PCl5(g)
→
PCl3(g)
+ Cl2(g) )
–1
Ans : ∆G° = 38.484 Kjoulemol
80.
81.
The equilibrium constant of the following given reaction is Kp = 2.4 × 10–5 at 298 K
temperature. Reaction : PCl5(g) → PCl3(g) + Cl2(g) . Calculate the value of ∆fG° for the
given reaction. (Text. Book Exe – 7 (11) Page 92)
82.
NO(g)
+ O3(g)
→
NO2(g)
+
O2(g)
Following data is obtained for above reaction at 298 K
∆G°f Kcal/mole
12.39
20.72
39.06
NO2
NO
O3
∆H°f Kcal/mole
8.09
21.60
34.0
Calculate ∆G°, K, ∆H° and ∆S° for above reaction.
(Ans : ∆G° = − 47.39 Kcal, ∆H ° = − 47.51 Kcal, Kp = 5.6481 × 1034,
∆S° = − 4.0268 × 10−4 Kcal K−1)
Type XVI
[A]
:
Common Problems
Electrochemistry
of
Thermodynamics
and
Relation between Free Energy of Reaction and Cell Potential :
Remember
:
(1)
∆G ° = − nFE°cell
(For standard electrochemical cell)
Where, ∆G ° = Standard free energy change
E°cell = Standard cell potential
n
= Change in oxidation number in balanced redox
reaction
83.
In one electrochemical cell following cell reaction occur…
Cu(s) + 2 Ag+(aq)
→
Cu2+(aq)
+ 2 Ag(s)
E°cell of the cell in which above reaction occurs is 0.46 V. Calculate standard
change in free energy of above reaction .
14
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
Ans : Cu(s) + 2 Ag+(aq)
→
Cu2+(aq)
∴ n = 2
∆ G° = −nFE°cell
+ 2 Ag(s)
C
× 0.46 V
F
= − 88780 CV OR Joule
88780
= −
4.186
∴ ∆ G° = −21208.8 Calorie
= − 2 F × 96500
84.
Find the value of ∆G° for the following reaction occurring in standard cell. The
value of standard cell potential (E°cell) is 1.1 volt at 298 K temperature.
Zn(s) + Cu2+(aq)
→
Zn2+(aq)
+ Cu(s)
(F = 96500 coulomb and 4.184 J = 1 calorie) (Text. Book Illus. - 16 Page 83)
85.
The value of change in free energy for the following cell reaction at 298 K
temperature is – 76322 calories. Calculate the potential of the electrochemical cell.
F = 96500 coulomb , 1 Calorie = 4.184 joule. (Text. Book Exe – 7 (10) Page 92)
Reaction : A (s) + B(2+aq) → A (2+aq) + B(s)
[B]
Relation between Maximum Electrical Work done by the Cell and Cell
Potential:
Remember
(1)
:
Wele = − nFE°cell
Where, Wele = Maximum electrical work done by the cell
E°cell = Standard cell potential
nF
= Quantity of electrical charge in Faraday which is
drawn from the cell
86.
Calculate maximum amount of electrical work that can be done by drawing 0.5 F
electricity from the cell. The standard cell potential is 1.1 V.
Ans :
Wele = ∆G° = − nFE°cell
C
= − 0.5 F × 96500
× 1.1 V
F
= − 53075 CV
= − 53075 Joule
53075
= −
Calorie
4.186
= − 12679.2 Calori = − 12.68 Kcal
15
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
[C]
Relation between Equilibrium constant (Kc) and Std. Cell Potential :
Remember
(1)
(2)
:
∆G° = − nFE°cell
∆G° = − 2.303 RT log Kc
From equation (1) and (2)…
− nFE°cell
=
− 2.303 RT log Kc
nFE cell
∴
log Kc
=
2.303 R T
RT
But, at 25° C ( 298 K )… 2.303
= 0.0592
F
nFE cell
∴
log Kc
=
2.303 R T
n × E cell
∴
log Kc
=
0.0592
87.
Following reaction occurs in Electrochemical cell…
Zn(s) + Cu2+(aq)
→
Zn2+(aq)
+ Cu(s)
Standard cell potential is 1.1 V at 25° C. Calculate Kc of the reaction at 25° C.
Ans : Zn(s) + Cu2+(aq)
→
Zn2+(aq)
+ Cu(s)
For above reaction… n = 2
and E°cell = 1.1 V
(1)
∆G° = − nFE°cell
(2)
∆G° = − 2.303 RT log Kc
From equation (1) and (2)…
− nFE°cell
− 2.303 RT log Kc
nFE cell
∴
log Kc
=
2.303 R T
RT
= 0.0592
But, at 25° C ( 298 K )… 2.303
F
nFE cell
∴
log Kc
=
2.303 R T
n × E cell
∴
log Kc
=
0.0592
Putting values of n and E°cell in above equation…
2 × 1.1
log Kc
=
0.0592
=
37.1622
∴
Kc
=
anti( 37.1622)
∴
Kc
=
1.4528 × 1037
88.
=
Find the change in standard free energy of formation and equilibrium constant of
the reaction Fe(s) + Cu (2+aq) → Fe(2+aq) + Cu (s) The standard electrochemical cell
potential is 0.78 volt and F = 96500 coulombs. (Text. Book Exe – 7 (9) Page 92)
(Ans : Kc
= 2.239 × 1026)
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CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
89.
Calculate the std. cell potential of galvanic cell in which the following reaction
takes place. Calculate ∆G° and equilibrium constant of the reaction.
2 Cr(s)
+ 3 Cd2+(aq)
→
2 Cr3+(aq)
+ 3 Cd(s)
34
Ans : Kc
= 2.88 × 10
90.
Calculate the std. cell potential of galvanic cell in which the following reaction
takes place. Calculate ∆G° and equilibrium constant of the reaction.
Fe2+(aq)
+ Ag+(aq)
→
Fe3+(aq)
+ Ag(s)
+
2+
3+
E° Fe / Fe = − 0.77 V
E° Ag/Ag = −0.80 V
Ans : Kc
= 3.211
91.
In the button cell widely used in the watches and other devices the following
reaction takes place. Determine ∆G° and E° for the reaction.
Zn(s)
+ Ag2O(s) + H2O(l)
→
Zn2+(aq) + 2 Ag(s)
+ 2 OH−(aq)
E° Zn / Zn 2+ = + 0.76 V
E° Ag+/Ag = +0.80 V
92.
Calculate the change in standard free energy for the reaction taking place in
standard electrochemical cell at 25°. The standard potential of the cell is 0.05 V.
F = 96500 Q and 4.184 joule = 1 calorie.
Co(s)
+
Ni2+(aq)
→
Co2+(aq)
+
Ni(s)
Ans : ∆G° = −9650 Joule = − 2306.4 calorie
93.
The value of change in free energy is −74266 cal at 25°C temperature. Calculate
the standard potential of electrochemical cell. ( F = 96500 Q and 4.184 joule = 1
calorie ) Reaction is : Mg(s) + Zn2+(aq) → Mg2+(aq)
+
Zn(s)
Ans : E°cell = 1.61 V
Problems Asked in Board Examination :
March − 1996
1.
The standard free energy of formation of NO2(g), NO(g) and O3(g) are 12.39, 20.72 and
39.06 Kcal mol−1 respectively at 25°C. Calculate the equilibrium constant of the
reaction at 25°.
NO(g) + O3(g)
→
NO2(g) + O2(g)
October − 1996
2.
The standard free energy of formation of NO2(g), NO(g) and O3(g) are 12.39, 20.72 and
39.06 Kcal mol−1 respectively at 25°C. Calculate the equilibrium constant of the
reaction at 25°C.
NO(g) + O3(g)
→
NO2(g) + O2(g)
March − 1997
3.
The standard heat of combustion of liquid benzene is −781.0 Kcal mole−1 at 25°C.
How much heat will evolve by combustion of 1 Kg of liquid benzene ? Calculate
the change in internal energy, when 1 mole of liquid benzene is burnt.
17
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
October − 1997
4.
The standard enthalpies of formation of NO(g) and NO2(g) are 21.60 and 8.09 Kcal
mole−1 respectively at 25°C. The entropies of NO(g), O2(g) and NO2(g) at this
temperature are 50.3, 49.0 and 57.5 cal K−1 mole−1 respectively. Calculate ∆H°, ∆S°,
∆G° and the equilibrium constant Kp for the following reaction at 25°C.
2 NO(g)
+ O2(g) →
2 NO2(g)
March − 1998
5.
A cylinder is filled with 1 mole of an ideal gas at 300°K and 10 atmosphere
pressure. Due to leakage in valve, the gas diffuses in atmosphere. How much work
is done in expansion of the gas ? Calculate ∆H and ∆E for the process. ( 1
lit−atmosphere = 24.21 cal )
October − 1998
6.
A gas expands from 100 liters to 120 liters against 10 atmosphere pressure at a
constant temperature. Calculate the work done by the system. ( 1 lit−atmosphere =
24.21 cal )
March − 1999
7.
The equilibrium constant of the reaction A + B = C is 0.50 liter−1 mole−1 at
25°C. The standard free energy of formation of A and B is −20.0 and –25.0 Kcal
mole−1 respectively. Calculate the standard free energy of formation of C at 25°C.
July − 1999
8.
A cylinder is filled with 1 mole of an ideal gas at 300°K and 10 atmosphere
pressure. Due to leakage in valve, the gas diffuses in atmosphere. How much work
is done in expansion of the gas ? Calculate ∆H and ∆E for the process. ( 1
lit−atmosphere = 24.21 cal )
March − 2000
9.
The specific volumes of ice and water at 0°C are 1.089 and 1 ml g−1 respectively.
The molar heat of fusion of ice at 0°C and 1 atmosphere is 1.440 Kcal mole−1. If 90
grams of ice melts at 0°C under 1 atmosphere to produce water at the same
temperature, what would the values of ∆H, ∆E, qp and qv ? (1 liter − atmosphere =
24.21 calories )
July − 2000
10.
A piece of ice kept in atmosphere at 25°C, melts spontaneously but liquid water
does not get converted into ice spontaneously at this temperature. Account for this
observation on the basis of the second law of thermodynamics.
March − 2001
11.
Using following reaction and other information calculate the heat of combustion of
C3H8(g).
(i)
3 C(s) + O2(g)
= C3H8(g)
∆Hi = − 24.82 Kcal
(ii)
C(s)
+ O2(g)
= CO2(g)
∆Hii = − 94.05 Kcal
(iii)
H2(g) + ½ O2(g)
= H2O(l)
∆Hiii = − 68.32 Kcal
May − 2001
12.
The boiling point of water is 100°C. The vaporization of 1.0 gram of water at this
temperature absorbs 540 cal. What would be the changes in enthalpy, entropy and
free energy, if 9 gram of water is converted into vapour ?
18
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
July − 2001
13.
Find out the change in free energy of the reaction :
Cu(s) + 2 Ag+(aq) = Cu2+(aq)
+ 2 Ag(s)
which occurs in a standard cell at 25°C. The standard potential of the cell is 0.54
volt.
March − 2002
14.
The cell potential of a standard electro−chemical cell is 0.32 volt at 25°C. Calculate
equilibrium constant value of the following reactions which occurs in it.
→ Zn2+(aq)
+ Fe(s)
Zn(s) + Fe2+(aq)
April − 2002
15.
At 27°C temperature and 1 atmosphere pressure, the complete combustion of 1.56
gm of benzene produces carbon dioxide and water vapour. During this reaction
15.62 Kcal heat is evolved. Calculate the heat evolved, if reactions occurs at
constant volume at 27°C temperature.
March − 2003
16.
For the reaction : NO(g) + O3(g) →
NO2(g)
+ O2(g)
Equilibrium constant is 5.648 × 1034 at 25° C temperature. The standard free energy
of formation of NO(g) and O3(g) are 20.72 and 39.06 Kcal mole−1 respectively at
25°C temperature. Calculate standard free energy of formation of NO2(g).
March − 2004
17.
For the reaction : 2 SO2(g)
+ O2(g) →
2 SO3(g)
Calculate equilibrium constant for the above reaction at 25° C temperature.
∆G°f SO3 = −88.52 Kcal mole−1
∆G°f SO2 = −71.79 Kcal mole−1
March − 2005
18.
At 25°C, ∆H° and ∆S° of the reaction… 2A
→
B + 2 C are −18.24 Kcal
−1
and 40 Cal K respectively. Calculate equilibrium constant of the reaction.
March − 2006
19.
Equilibrium constant of the following reaction at 25°C is Kp = 2.45 × 1040; if ∆G°f
SO2 = −71.8 Kcal mole−1 then calculate ∆G°f SO3. R = 1.987 cal mole−1 K−1.
2 SO2(g)
+ O2(g) →
2 SO3(g)
March − 2007
20.
In the fuel cell, the electro energy produced during the formation of 1 mole H2O(l) is
equal to 50 Kcal and standard free energy of water is 68.32 Kcal mole−1. Find out
the efficiency of the fuel cell. ( 1 Mark )
March − 2008
21.
The potential of a standard electrochemical cell is 1.1 volt at 25° C temperature.
Calculate the equilibrium constant and free energy change ∆G° (in coulombs) of the
following given reaction : Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
March − 2009
22.
At 25° C temperature : CuS(s) + H2(g) → Cu(s) + H2S(g) … For this reaction the
value of ∆H° and ∆S° is 7.77 Kcal and 10.03 cal/mol-K respectively.
(1)
Calculate the equilibrium constant of the reaction
(2)
On which temperature and at 1 atm pressure, the value of ∆G° will be zero.
19
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433
March − 2010
23.
24.
To convert 1 mole of water into its vapour at 100° C needs 9720 calorie of heat.
Calculate the entropy change of the system and surroundings for the process to
convert 100 gms of water to its vapour. Will this process occur by itself ?
(Temperature for the surroundings is 25° C)
If the pressure is raised from 8 atm to 20 atm in a closed vessel containing 4.28
moles of ideal gas at 25° C. Calculate the free energy change ∆G and also the
equilibrium constant for this process.
Accuracy is pathway of success.
20
CHEMTEST / CH-3 - Numericals / Sem-II / Ph : 30004433