Download Thermodynamics

Thermodynamics
SUBJECTIVE PROBLEMS:
Q 1.
The enthalpy for the following reaction (✄Ho) at 25oC are given below:
(i) 1/2 H2(g) + 1/2 O2(g) ✆ OH(g) 10.06 kcal
(ii) H2(g) ✆ 2H(g)
104.18 kcal
118.32 kcal
(iii) O2(g) ✆ ✁✒✂☎
Calculate the O-H bond energy in the hydroxyl radical.
Q 2.
The standard heats of formation at 298 K for CCI4(g), H2O(g), CO2(g) and HCI(g) are -25.5, ✣ for the reaction
57.8, -94.1 and - ✝✞ ✟✠✡☛☞✌✍☛ ✎✏✑✓✏✠✔✕✖✏☛✗✝ ✘✡☛✠✙☛✡✔✏ ✄ ✚✛✜✢
CCI4(g) + 2H2✁ ✆ ✘✁2(g) + 4HCI(g)
Q 3.
The molar heats of combustion of C2H2(g), C(graphite) and H2(g) are 310.62 kcal, 94.05
kcal and 68.32 kcal, respectively. Calculate the standard heat of formation of C2H2(g).
Q 4. ✤✥✏ ✥✏✡✔ ✏✦✏✎✂✗✧ ★✧ ✡✩✑✍✎✩✏✪ ✩✗ ✡ ✂✡✑ ✄✫✧ ✕✑ ✔✎✙✏ ✡✔ ✬✥✡✔ ✠✍✦✪✕✔✕✍✦✒✑☎✝
Q 5.
Give the following standard heats of reaction :
(i) heat of formation of water = -68.3 kcal;
(ii) heat of combustion of acetylene = - 310.6 kcal;
(iii) heat of combustion of ethylene = -337.2 kcal;
Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25oC).
Q 6.
The bond dissociation energies of gaseous H2, CI2 and HCI are 104, 58 and 103 kcal/mole
respectively. Calculate the enthyalpy of formation of HCI gas.
Q 7.
The standard molar heats of formation of ethane, carbon dioxide and liquid water are -21.1,
-94.1 and -68.3 kcal respectively. Calculate the standard molar heat of combustion of
ethane.
Q 8.
An intimate mixture of ferric oxide, Fe2O3, and aluminium, Al, is used in solid fuel rockets.
Calculate the fuel value per gram and fuel value per cc of the mixture. Heats of formation
and densities are as slow :
Hf (AI2O3) = 399 kcal/mole;
✭✭✭✮✯✰✱✲✳✱✴✯✮✴✵✲
Hf (Fe2O3) = 199 kcal/mole;
Density of Fe2O3 = 5.2 g/cc;
Density of Al = 2.7 g/cc.
Q 9.
An athlete is given 100 gm of glucose (C6H12O6) of energy equivalent to 1560 kJ. He utilizes
50 percent fo this gained energy in the event. In order to avoids storage of energy in the
body, calculate the weight of water he would need to perspire. The enthalpy of evaporation
of water is 44 kJ/mole.
Q 10.
The standard enthalpy of combustion at 25oC of hydrogen, cyclohexene (C6H10) and
cyclohexane (C6H12) are -241 -3800 and -3920 kJ/mole respectively. Calculate the heat of
hydrogenation of cyclohexene.
Q 11.
Using the data (all values are in kcal mol-1 at 25oC) given below, calculate the bond energy
of C-C and C-H bonds.
=
-372.0
✄Hocombustion (ethane)
o
=
-530.0
✄H combustion (propane)
o
=
172.0
✄H C(s) ✆ ✒✁✂
Bond energy of H-H
=
104.0
o
=
-68.0
✄H f of H2O(l)
o
=
-94.0
✄H f of CO2(g)
Q 12.
A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25oC
produces 6.11 litres of CO2. Find out the amount of heat evolved on burning one litre of the
gas mixture. The heats of combustion of ethylene and methane are -1423 and -891 kJ mol-1
at 25oC.
Q 13.
Determine the enthalpy change of the reaction.
C3H8(g) + H2✒✁✂ ✆ 2H6(g) + CH4(g), at 25o, using the given heat of combustion values
under standard conditions
Compound H2(g)
CH4(g)
C2H6(g)
C(graphite)
o
-890.0
-1560.0
-393.5
✄H (kJ/mol) -285.8
The standard heat of formation of C3H8(g) is -103.8 kJ/mol.
Q 14.
In order to get maximum calorific output, a burner should have an optimum fuel to oxygen
ratio which corresponds to 3 times as much oxygen as is required theoretically for
complete combustion of the fuel. A burner which has been adjusted for methane as fuel
(with x litre/hour of CH4 and 6x litre/hour of O2) is to be readjusted for butane, C4H10. In
order to get the same calorific output, what should be the rate of supply of butane and oxygen?
☎☎☎✝✞✟✠✡☛✠☞✞✝☞✌✡
Assume that losses due to incomplete combustion, etc, are the same for both the fuels and the
gases behave ideally.
Heats of combustion :
CH4 = 809 kJ/mol; C4H10 = 2878 kJ/mol
Q 15.
The polymerization of ethylene to linear polyethylene is represent by the reaction
nCH2 = CH2 ✆ ✁✂2 ✄ CH2]
where n has a large integral value. Given that the average enthalpies of bond dissociation for C
= C and C-C at 298 K are + 590 and + 331 kJ mol-1, respectively, calculate the enthalpy of
polymerisation per mole of ethylene at 298 K.
Q 16.
The standard molar enthalpies of formation of cyclohexane(l)and benzene(l) at 25°C are- 156
and+ 49 kJ mol-1 respectively. The standard enthalpy of hydrogenation of cyclohexene(l) at 25oC
is -1 19 kJ mol-1. Use these data to estimate the magnitude of the resonance energy of benzene
Q 17.
The enthalpy change involved in the oxidation of glucose is 72880 kJ mol-1, Twenty five percent
of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one
kilometer, what is the maximum distance that a person will be able to walk after eating 120 g of
glucose.
Q 18.
Compute the heat of formation of liquid methyl alcohol in kilojoules per mole, using the
following data. Heat of vaporization of liquid methyl alcohol = 38 kJ/mol. Heat of formation of
gaseous atoms from the elements in their standard states; H, 218 kJ1mol; C, 715 kJ1mol;
0,249kJ 1mol. Average bond energies :
C-H = 415kJ/mol, C-O =365 kJ/mol, O-H=463 kJ/mol
(IIT JEE 1997- 5 Marks)
Q 19.
Anhydrous AlCI3 is covalent. From the data given below, predict whether it would remain
covalent or become ionic in aqueous solution. (Ionisation energy for Al = 5137 kJ mol-1;
3+
-1
-1
(IIT JEE 1997- 2 Marks)
☎Hhydration for Al = 4665 kJ mol ✝ ☎✞hydration for CI = -381 kJ mol .)
Q 20.
From the following data, 'calculate the enthalpy change for the combustion of cyclopropane at
298 K. The enthalpy of formation of CO2(g), H2O(1) and propene(g) are - 393.5, - 285.8 and
20.42 kJ mol-I respectively. The enthalpy of isomerisation ofcyclopropane to propene is - 33.0kJ
mol-1.
(IIT JEE 1998 - 5 Marks)
Q 21.
Estimate the average S-F bond energy in SF6. The values of standard enthalpy of formation of
SF6(g), S(g)and F(g) are : - 1100,275 and 80 kJ mol-1 respectively.
(IIT JEE 1999 - 3 Marks)
✟✟✟✠✡☛☞✌✍☞✎✡✠✎✏✌
Q 22.
A sample of argon gas at 1 am pressure and 27oC expands reversibly and adiabatically from 1.25
dm3 to 2.50 dm3. Calculate the enthalpy change in this process.CV m for argon is 12.48 JK-1 mol-1.
(IIT JEE 2000 - 4 Marks)
Q 23.
Show that the reaction CO(g) + 1/2 O2✄ ✁ ✂ ☎✆2 (g) at 300 K, is spontaneous and exothermic,
when the standard entropy change is4.094 kJ mol-l K-l. The standard Gibbs free energies of
formation for CO2 and CO are -394.4 and, -137.2kJ mol-I, respectively.
(IIT JEE 2000 - 3 Marks)
Q 24.
Diborane is a potential rocket fuel which undergoes combustion according to the reaction.
(IIT JEE 2000 - 2 Marks)
B2H6 (g) +3 O2 ✄ ✁ ✂ ✝2O3 (s) + 3.H2O (g)
From the following data, calculate the enthalpy change for the combustion of diborane.
2B(s) + 3/2 O2 (g) ✞ B2O3(s) ✟H = - 1273 kJ mol-1
H2(g) + 1/2 O2(g) ✞ H2O(✠) ✟H = 44kJ mol-1
2B(s) + 3H2 (g) ✞ ✡2H6(g)
✟H = 36 kJ mol-1
Q 25.
When 1-pentyne (A) is treated with 4 N alcoholic KOH at 17S°C, it is converted slowly into an
equilibrium mixture of 1.3% I-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of 1,2- pentadiene (C).
The equilibrium was maintained at 175°C. Calculate AGO for the following equilibria :
B ☛ ☞ ✌✍✎✏ =? B ☛ ☎ ✌✍✑✏ =?
✒✓✔✕ ✖✗✘ ✙✚✛✙✜✛✚✖✘✢ ✣✚✛✜✘ ✔✤ ✌✍✎✏ ✚✥✢ ✌✍✑✏ indicate the order of stability of (A), (B) and (C). Write
a reasonable reaction mechanism showing all intermediates leading to (A), (B) and (C).
(IIT JEE 2001 - 10 Marks)
Q 26.
Two moles of a perfect gas undergo the following processes
(IIT JEE 2002 -5 Marks)
(a) a reversible isobaric expansion from (1.0 am, 20.0L) to (l.0 atm40.0 L);
(b) a reversible isochoric change of state from (1.0 atm, 40.0 L) to (0.5 atm, 40.0 L);
(c) a reversible isothermal compression from (0.5 am, 40.0 L) to (1.0 atm, 20.0 L).
(i) Sketch with labels each. of the processes on the same P-V diagram.
(ii) Calculate the total work (w) and the total heat change (q) involved in the above processes.
(iii) What will be the values of AU, AH and AS for the overall process?
Q 27.
Cv value of He is always 3R/2 but Cv value-of H2 is 3R/2 at low temperature and 5R/2 at
moderate temperature and more than 5R/2 at higher temperature explain in two to three lines.
(IIT JEE 2003 - 2 Marks)
Q 28.
An insulated container contains 1 mol of a liquid, molar volume 100 ml, at 1 bar. When liquid is
✦✖✘✘✧✛★ ✧✓✘✦✦✘✢ ✖✔ ✩✪✪ ✫✚✓✬ ✣✔✛✜✕✘ ✢✘✙✓✘✚✦✘✦ ✖✔ ✭✭ ✕✛✮ ✒✯✥✢✮ ✌✰ ✚✥✢ ✌✱ ✤✔✓ the process.
(IIT JEE 2004 - 2 Marks)
✲✲✲✳✴✵✶✷✸✶✹✴✳✹✺✷
Q 29. In the following equilibrium N2O4 (g) ✆ 2NO2(g)
(IIT JEE 2004 - 2 Marks)
When 5 moles of each is taken and the temperature is kept at 298 K, the total pressure was
found to be 20 bar.
Given : ✄✞☛ (N2O4) = 100kJ; ✄✞☛ (NO2) = 50 kJ
✁✂☎ ✝✂✟✠ ✡☞ ✌✍ ✎✏✑ ✒✑✓✔✎✂✌✟ ✓✎ ✕✖✗ ✘✙
(ii) Find the direction of the reaction
Q 30.
For the reaction, 2CO + o2 ✚ 2CO2✛ ✡✜ ✢ -560kJ. Two moles of CO and one mole of 0, are taken
in a container of volume 1 L. They completely form two moles of CO2, the gases, deviate
appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm, find
✎✏✑ ✣✓✤✟✂✎✥✠✑ ✁✓✦✧✌★✥✎✑ ✩✓★✥✑☎ ✌✍ ✡✪ ✓✎ ✫✬✬✘✙ ✁✭ ✮ ✓✎✣ ✢ ✬✙★ ✯✰☎
(IIT JEE 2006 - 6Marks)
✱✱✱✲✳✴✵✶✷✵✸✳✲✸✹✶
Thermodynamics-solutions
SUBJECTIVE PROBLEMS
Sol 1.
The required reaction in terms of dissociation energy is OH(g) ✆ O(g) + H(g); ✄H = ?
This equation can be achieved by (a) reversing the equation (i), (b) dividing equation (ii) and (iii) each by
2, and (c) adding the three resulting equations.
OH(g) ✆ 1/2 H2(g) + 1/2 O2(g); ✄H = + 10.06 kcal [Reversing eq (i)]
1/2 H2(g) ✆ ✒✁✂☎✝
✄H = - 52.09 kcal [ 1/2 Eq (ii)]
✄H = - 59.16 kcal [1/2 Eq (iii)]
1/2 O2(g) ✆ ✞✒✁✂☎
OH(g) ✆ ✞✒✁✂ ✟ ✒✁✂☎ ✄H = - 101.19 kcal (adding)
Thus one mole of OH(g) needs 101.19 kcal of energy to break into oxygen and hydrogen gaseous atom.
Hence the bond energy of O-H bond is 101.19 kcal.
Sol 2.
Since we know that
Heat content of the compound
✌ ✍
= Heat of formation ✄H, ✄✠☛✡☞
= Total heat contents of the products
- Total heat contents of the reactants
Writing the given chemical reaction,
✌ =?
CCI4(g) + 2H2O(g) ✆ CO2(g) + 4HCI(g); ✄ ✠☛✡☞
✌
✄✠☛✡☞ = [✄✠✎✏✑ + 4 *✄HHCI] ✓ [✄✠✎✎✔✕ + 2 *✄✠✖✑✏ ]
Given, ✄ ✠✎✎✔✕ = - 25.5 kcal/mole ✄✠✖✑✏
= - 57.8 kcal/mole
✄ ✠✎✏✑ = - 94.1 kcal/mole ✄HHCI = - 22.1 kcal/mole
Substituting the values in the above equation,
✌ = [-94.1 + 4 * - 22.1] ✓ [-25.5 + 2 * - 57.8]
✄✠☛✡☞
= [-94.1 ✓ 88.4] ✓ [-25.5 ✓ 115.6]
= - 182.5 + 141.1 = -41.4 kcal
Sol 3.
The required equation is :
✄H = ?
2C(s) + H2(g) ✆ C2H2;
Write the thermochemical equations for the given data
(i) C2H2 + 5/2 O2(g) ✆ 2CO2(g) + H2O(✗);
(ii) C(s) + O2(g) ✆ CO2(g) ; ✄H = - 94.05 kcal
(iii) H2(g) + 1/2 O2(g) ✆ H2O(✗);
✄H = - 68.32 kcal
For getting the above required reaction, we will have to
NOTE:
(a) Bring C2H2 in the product that can be done by reversing the equation (i) to give equation (iv).
✘✘✘✙✚✛✜✢✣✜✤✚✙✤✥✢
(b) Multiply equation (ii) by 2 to get 2C atoms in the reactant and thus equation (v) is obtained.
(c) Keep equation (iii) as such.
(d) Add equations (iv), (v) and (iii).
(iv) 2CO2 + H2O ✆ 2H2 + 5/2 O2✁ ✂✄ ☎ ✝✞✟✠✡☛ ☞✌✍✎
(v) 2C + 2O2 ✆ ☛ ✏2; ✂✄ ☎ - 188.10 kcal
(iii) H2 + 1/2 O2 ✆ ✄ 2O;
✂✄ ☎ - 68.32 kcal
On adding, 2C + H2 ✆ 2H2✁ ✂✄ = 54.20 kcal
Hence the standard heat of formation of C2H2(g) =
Sol 4.
If heat is absorbed at constant pressure, then
qp= ✑E -'(- ✒✑✓✔
or qp= E2 - El - [-P(V2 ✕ V1)]
or qp=(E2 + PV2)-(E1 + PV1) = H2 ✕ H1 ✖ ✑✗
Sol 5.
The given data can be written as follows
(i) H2(g) + 1/2 O2(g) ✆ H2O(l); ✂ = - 68.3 kcal
(ii) C2H2(g) + 5/2 O2(g) ✆ ✄ 2O(l) + 2CO2(g);
✂✄ ☎ - 310.6 kcal
(iii) C2H4(g) + 3O2 ✆ ☛✄ 2O(l) + 2CO2✘✙✚✁ ✂✄ ☎ - 337.2kcal
The required thermochemical equation is
C2H2(g) + H2✘✙✚ ✆ 2H4(g)
The required equation can be obtained by subtracting equation (iii) from the sum of equations (i)
✍✛✜ ✘✢✢✚✣ ✤✥✦✧ ✂✄ ★✩ ✤✥✪ ✫✪✬✦✢✫✪✜ ✪✬✦✍✤✢★✛ ✌✍✛ ✭✪ ✌✍✎✌✦✎✍✤✪✜ ✍✧ ✭✪✎★✮✠
✂H = [-68.3 + (-310.6)] ✯ (-337.2)
= [-68.3 ✯ 310.6] + 337.2
= - 378.9 + 337.2 = - 41.7 kcal
✂✰ ☎ ✂✄ - ✂✛✱✲
✄✪✫✪ ✂✛ ☎ ✳★✎✪✧ ★✩ ✤✥✪ ✙✍✧✪★✦✧ ✴✫★✜✦✌✤✧ ✯ Moles of the gaseous reactants
= 1 ✯ (1 + 1) = - 1
✵✦✭✧✤✢✤✦✤✢✛✙ ✤✥✪ ✶✍✎✦✪ ★✩ ✂✄✣ ✂✛✣ ✱ ✍✛✜ ✲ ✢✛
✂✰ ☎ ✂✄ - ✂✛✱✲
✂E = - 41.7 ✯ (- 1 * 2 * 10-3 * 298)
[✷ R = 2cal/ degree/ mole = 2 * 10-3 kcal/ deg/ mole]
= - 41.7 + 2 * 10-3 * 298
= - 41.7 + 0.596 = 41.104 kcal
Sol 6.
Bond
H✸H
CI ✸ CI
H ✸ CI
104 kcal
58 kcal
103 kcal
✂H disso.
Formation of hydrogen chloride can be represented as
H ✯ H + CI ✯ ✹ ✆ ☛✄ ✯ CI
Thus the reaction involves
Cleavage of one H ✯ ✄ ✭★✛✜✣ ✂✄ ☎ ✞✟✺ ☞✌✍✎
Cleavage of one CI ✯ ✹ ✭★✛✜✣ ✂✄ ☎ ✻✼ ☞✌✍✎
✽✽✽✾✿❀❁❂❃❁❄✿✾❄❅❂
Formation of two H ✂ ✁ ✄☎✆✝✞✟ ✠✡ ☛ ☞ ✌ - 103 kcal
✍ ✠✡ ☎✎ ✏✑✒ ✓✒✔✕✏✖☎✆ ☛ ✗✘✙✚ ✛ ✜✢✣ ✂ 2(103) = 162 ✂ 206 = - 44 kcal
Now since the enthalpy of formation of a compound is the change in heat content accompanied in
the formation of one mole of the compound, the enthalpy of formation of
HCI gas = - 44/2 = -
Sol 7.
The required chemical reaction.
2C2H6 + 7O2 ✤ ✚ ✥2 + 6H2✥✦ ✠✡ ☛ ✧
★☎✏✒ ✏✑✔✏ ✞✖✆✕✒ ☞ ✩☎✪✒✞ ☎✎ ✒✏✑✔✆✒ ✔✓✒ ✓✒✔✕✏✖✆✫✟ ✏✑✒ ✠✡ ☎✎ ✏✑✒ ✓✒✔✕✏✖☎✆ ✬✖✪✪ ✄✒ ✭ ✧✮
The thermochemical equations for the given data are written as below.
(i) C(s) + O2✗✫✣ ✤ ☎2(g);
✠✡ ☛ - 94.1 kcal
(ii) H2(g) ½O2✗✫✣ ✤ ✡2O(g); ✠✡ ☛ - 68.3 kcal
(iii) 2C(s) + 3H2✗✫✣ ✤ 2H6(g);
✠✡ ☛ - 21.1 kcal
✯✒ ✰✆☎✬ ✏✑✔✏ ✠✡ ☛ ✡Products ✂ HReactants
✠H = ✚✠✱✲✳✴ + ✵✠✱✶✴ ✳ ✂ (☞✠✱✲✴✶✷ + ✸✠✱✳✴ )
✠H = 4 * - 94.1 + 6 * - 68.3 ✂ (2 * - 21.1 + 0)
= - 376.4 ✂ 409.8 + 42.2 = - 744.0 kcal/2 mole of ethane
=
Sol 8.
Fe2O3 +
2AI
2Fe
+
AI2O3
✤
2 * 56 + 48 = 160
2 * 27 = 54
Heat of reaction = 399 ✂ 199 = 200 kcal [Al & Fe are in their standard states]
Total weight of reactans = 160 + 54 = 214 g
✍ ✹✺✒✪ ✻✔✪✺✒✼✫✓✔✩ ☛ ☞✙✙✼☞✘✚ ☛ ✙✮✽✾✚✵ kcal/g
Volume of Al = 54/2.7 = 20 cc
Volume of Fe2O3 = 160/5.2 = 30.77 cc
Total volume = 20 + 30.77 = 50.77 cc
✍ ✹✺✒✪ ✻✔✪✺✒ ✿✒✓ ✕✕ ☛ ☞✙✙✼✜✙✮✸✸ ☛
Sol 9.
100 g of glucose = 1560 kJ
Energy utilized in body = 50/100 * 1560 = 780 kJ
Energy left unutilized in body = 1560 ✂ 780 = 780 kJ
Energy to be given out = 1560 ✂ 780 = 780 kJ
Enthalpy of evaporation of water = 44 kJ/mole = 44 kJ/18 g
Of water [1 mole H2O = 18g water]
Hence amount water to be perspired to avoid storage of energy = 18/44 * 780 =
Sol 10.
The required reaction is
C6H10(g)
+ H2✗✫✣ ✤
C2H12✟✗✫✣✟ ✠✡1 = ?
Cyclohexene
Cyclohexane
The given facts can be written as :
H2(g) + ½O2✗✫✣ ✤ ✡2O,
✠✡2 = - 241 kJ/mol
❀✮✗✘✣
❀✗☞✣
❁❁❁❂❃❄❅❆❇❅❈❃❂❈❉❆
C2H10(g) + 17/2 O2 ✆ ✄ ✁2, (g) + 5H2✁✂ ☎✝3 = - 3800 kJ/mol ✞✟✠✡
C6H10(g) + 902✟✒✡ ✆ ✄ ✁2(g) + 6H2O, ☎✝ 4 = - 3920 kJ/mol ✞☛✟☞✡
The required reaction (1) can be obtained by adding equations (2) and (3) and subtracing (4) from
the sum of (2) and (3).
C2H10(g) + H2✟✒✡ ✆ 6H12(g)
☎H1 ✌ ✟☎✝ 2 ✍ ☎✝3) -☎✝4
= [-241 + (-3800)] ✎ (-3920)
= (-241 - 3800) ✎ (- 3920)
= - 4041 + 3920 = -
Sol 11.
For C3H8 : 3C + 4H2 ✆ 3H8✂ ☎✝1 = ?
For C2H6 : 2C + 3H2 ✆ 2✝✂ ☎✝2 = ?
✏ ☎✝1 = -[2 (C - C) + 8(C - H)] + [3C✑✓✔ + 4(H - H)] ✞☛✟✕✡
✏ ☎✝2 = -[1 (C - C) + 6(C - H)] + [2C✑✓g + 3(H - H)] ✞☛✟✖✡
Let bond energy of C ✎ C be x kcal and bond energy of C ✎ H by y kcal
✏ ✗✘ ✙✚☛ ✟✕✡ ☎✝1 = - (2x + 8y) + [3 * 172 + 4 * 104] ✞☛✟✠✡
☎H2 = - (x + 6y) + [2 * 172 + 3 * 104]
✞☛✟☞✡
Also given C + O2 ✆ ✁2; ☎✝ ✌ - 94.0 k cal
✞☛✟✛✡
H2 + ½O2 ✆ ✝ 2✁✂ ☎✝ ✌ - 68.0 k cal
✞☛☛✟✄✡
C2H6 + (7/2)O2 ✆ ✖ ✁2 + 3H2✁✂ ☎✝ ✌ - 372 k cal
✞☛✟✜✡
C3H8 + 5O2 ✆ ✠ ✁2 + 4H2✁✂ ☎✝ ✌ - 530 k cal ✞☛☛✟✢✡
By inspection method : 2 * (5) + 3 * (6) ✎ (7) gives
2C + 3H2 ✆ 2H6✂ ☎✝ 2 = - 20 k cal
✞☛✟✣✡
And 3 * (5) + 4 * (6) ✎ (8) gives
3C + 4H2 ✆ 3H8✂ ☎✝ 1 = - 20 k cal
✞☛✟✕✤✡
✏ ✗✘ ✙✚☛ ✟✠✡✥ ✟☞✡✥ ✟✣✡ ✦✧★ ✟✕✤✡
X + 6y = 676
2x + 8y = 956
✏ ✩ ✌ ✢✖ ✪ ✫✦✬ ✦✧★ ✘ ✌ ✣✣ ✪ ✫✦✬
Bond energy of C ✎ C bond =
and Bond energy of C ✎ h bond =
Sol 12.
Combustion of C2H4 and Ch4 takes place as follows :
C2H4 + 3O2 ✆ ✖ ✁2 + 2H2O
1 vol
2 vol.
CH4 + 2O2 ✆ ✁2 + 2H2O
1 vol.
1 vol
Let the vol. of CH4 in mixture = x l
✏ ✭✮✬☛ ✮✯ 2H4 in the mixture = (3.67 - x) l
Vol. of CO2 produced by x l of CH4 = x l and
Vol. of CO2 produced by (3.67 - x) l of C2H4 = 2(3.67 - x) l
✏ ✰✮✱✦✬ ✲✮✬☛ ✮✯ ✁2 produced = x + 2 (3.67 - x)
Or 6.11 = x + 2(3.67 - x) or x = 1.23 l
✳✳✳✴✵✶✷✸✹✷✺✵✴✺✻✸
✆ ✁✂✄ ✁☎ ✝✞4 in the mixture = 1.23 l
and Vol. of C2H4 in the mixture = 3.67 ✟ 1.23 = 2.44 l
Vol. of CH4 per litre of the mixture = 1.23/3.67 = 0.335 l
Vol. of C2H4 per litre of the mixture = 2.44/3.67 = 0.665 l
Now we know that volume of 1 mol. Of any gas at 25°C (298 K) = 22.4 * 298/273 = 24.45 l
✒✠ ✁✂✡☛☞ ✌✍ ✎✏✑ ✓ ✔✔✄✕✖✗
Heat evolved due to combustion of 0.335 l of CH4 = - 0.335 * 891/24.45 = - 12.20 kJ [given, heat
evolved by combustion of 1 l = 891 kJ]
Similarly, heat evolved due to combustion due to combustion of 0.665 l of C2H4
= - 0.665 * 1423/24.45 = - 38.70 kJ
✆ ✏✁✍✌✂ ✘☞✌✍ ☞✙✁✂✙☞✚ ✓ ✛✔✄✔✜ ✢ ✣✣✤✄✥✜ ✓
Sol 13.
From the given data, we can write :
(i) H2 + 1/2 O2 ✦ ✞2O; ✧✞ 1 = - 285.8 kJ/mol
(ii) CH4 + 2O2 ✦ ✝★2 + 2H2O; ✧✞ 2 = - 890 kJ mol
(iii) C2H6 + 7/2 O2 ✦ ✔✝★2 + 3H2★✩ ✧✞ 3 = - 1560 kJ/mol
(iv) C(s) + O2 ✦ ✝★2; ✧✞ 4 = - 393.5 kJ/mol
(v) 3C(s) + 4H2 ✦ C3H8(g);
✧✞5 = - 103.8 kJ/mol
The required reaction is C3H8(g) + H2(g) ✦ ✝2H6(g) + CH4(g), ✧H = ?
It can be obtained by the following calculation.
3 * (iv) ✟ (v) + 5(i) ✟ (iii) ✟ (ii)
✪✫ ✁✍✘☞✬ ✭✁✬✚✮✯ ✧✞ ✓ ✣✧✞4 - ✧✞5 ✢ ✰✧✞1 - ✧✞2 - ✧✞3
✆ ✧✞ ✓ ✣✱- 393.5) ✟ (-103.8) + 5(-285.8) + 890 + 1560
= - 2609.5 + 2553.8 =
Sol 14.
Combustion of CH4 and C4H10 takes place as follows
CH4 + 2O2 ✦ ✝★2 + 2H2★✯ ✧✞ ✓ - 809 kJ mol-1
C4H10 + 13/2O2 ✦ ✕✝★2 + 5H2=O, ✧H = - 2878 kJ mol-1
In order to get the same calorific output due to C4H10,
The rate of supply of butane = x *809/2878 =
Rate of supply of oxygen = 0.2803 x * 13/2 * 3 =
The reaction of combustion of CH4 and C4H10 can be written as follows :
CH4 + 2O2 ✦ ✝★2 + 2H2★✩ ✧✞ ✓ - 809
X
6X
Initial volume (in litre)
Let the temperature by T and assume volume of 1 mole of a gas is V litre at this condition.
✆ ✂✲✍✬☞ ✁☎ ✛ mole CH4 gives energy on combustion = 809 kJ
✆ ✳ ✂✲✍✬☞ ✁☎ ✝✞4 gives energy on combustion = 808(X)/V kJ
✆ ✔✤✥✤ ✴✵ ☞✫☞✬✶✷ ✲✮ ✁✸✍✌✲✫☞✚ ✸✷ ✛ ☛✁✂☞ ✁✬ ✂✲✍✬☞ ✝4H10
✆ ✤✜✹✱✳✺ ✻ ✼ ✻ ✔✤✥✤ ✂✲✍✬☞ ✝4H10 = 0.281 (X) litre C4H10
Thus, butane supplied for same calorific output = 0.281 (X) litre
✠ ✝4H10 + 13/10 O2 ✦ ✕✝★2 + 5H2★✩ ✧✞ ✓ - 2878 kJ/mol
✽✽✽✾✿❀❁❂❃❁❄✿✾❄❅❂
Volume of O2 required = 3 * volume of O2 for combustion of C4H10
= 3 * 13/2 * volume of C4H10
= 3 * 13/2 * 0.281 (X) litre O2
Sol 15.
nCH2 = CH2 ✆ ✒ ✁2 ✂ CH2)n
During the polymerization of ethylene, one mole of ethylene breaks i.e. one C = C double
bond breaks and the two CH2 ✂ groups are linked with C ✂ C single bonds thus forming three single
bonds (two single bonds are formed when each CH2 ✂ group of ethylene links with one CH2 ✂ group
of another ethylene molecule). But in the whole unit of polymer, number of single C ✂ C bonds
formed/mole of ethylene is 2.
Energy released = Energy due to formation of 2 C ✂ C single bonds
= 2 * 331 = 662 kJ/mol of ethylene
Energy absorbed = energy due to dissociation of 1 C = C double bond
= 590 kJ/mol of ethylene
✄ ☎✝✞✟✠✡☛☞ ✌✍ ☛✌✡☞✎✏✑✓✔✠✞✓✌✝✕✎✌✡ ✌✍ ✏✞✟☞✡✏✝✏ ✌✑ ✖✁polymerisation = 590 ✂ 662 kJ/mol =
Sol 16.
Standard enthalpy of hydrogenation of cyclohexen (- 119kJ mol-1) means the enthalpy of
hydrogenation of one C = C double bond. Now benzene has three C = C double bonds, the enthalpy
of the reaction would be = 3 * - 119 = - 357 kJ mol-1
Actual enthalpy of the reaction can be evaluated as follows.
✖H(Reaction) = ✖✗✘✙ (Product) - ✖✗✘✙ (Reactants)
= - 156 ✂ (49 + 0)
= - 205 kJ mol-1
✄ ✚✏✛✌✝✠✝✜✏ ✏✝✏✑✢☞ ✣ ✖✁Exp - ✖cal
= - 357 ✂ (-205) = C6H10 + H2 ✆ 6H12; ✖H = - 119kJ
(involves breaking up of three double bond and addition of three H2 molecule)
✄ 6H6 + 3H2 ✆ 6H12 ✤ ✖✁ ✣ ✥ ✦ ✒-119) = - 357 kJ
(involves breaking up of three double bond and addition of three H2 molecule)
Also given 6C + 6H2 ✆ 6H12(l)✤ ✖✁ ✣ - 156
we have C6H6 + 3H2 ✆ 6H12(l)✤ ✖✁ ✣ - 357
6C + 3H2 ✆ 6H6✤ ✖✁ ✣ ✧ ★✩✪ ✫✬
Therefore, resonance energy = 49 ✂ 201 = - 152 kJ
Sol 17.
Energy available for muscular work by 1 mole of glucose = 2880 *25/100 = 720 kJ mol-1
Thus 180 g of glucose (mol. Wt. of glucose) supplies 720 kJ
120 g of glucose will supply = 720/180 * 120 = 480 kJ
100 kJ is needed to walk 1 km, Hence, 480 kJ is needed to walk 1/100 * 480 =
✭✭✭✮✯✰✱✲✳✱✴✯✮✴✵✲
Sol 18.
the required thermochemical equation is
✄Hf ✌ ✄✁✂☎✆✝✞✂☎✟✝ ✠ ✡✄✁H-H ✠ ☛☞✡ ✄✁O=O]
- ✒✄✁C-H ✠ ✄✁C-O ✠ ✄✁O-H + ✄✍✎✏✑✓✔✕✖ ✗✕ ]
=[715 + 2 * 436 + 249] ✘ [3 * 415 + 356 + 463 + 38]
The diven data is as follows :
(i) CH3OH(l) ✙ ✚✛3OH(g), ✄H = 38 kJ mol-1
(ii) 1/2 H2✜✢✣ ✙ ✁✜✢✣✤ ✄✁ ✌ ✡☛✥ ✦✧ ★✩✪-1
(iii) C(✢✫✬✭✛✮✯✰✣ ✙ ✚✜✢✣✤ ✄✁ ✌ ✱☛✲ ✦✧ ★✩✪-1
(iv) 1/2 O2✜✢✣ ✙ ✳✜✢✣✤ ✄✁ ✌ ✡✴✵ ✦✧ ★✩✪-1
(v) C ✘ ✁✜✢✣ ✙ ✚✜✢✣ ✠ ✁✜✢✣✤ ✄✁ ✌ ✴☛✲ ✦✧ ★✩✪ -1
(vi) C ✘ ✳✜✢✣ ✙ ✚✜✢✣ ✠ ✳✜✢✣✤ ✄✁ ✌ ✒✲✶ ✦✧ ★✩✪-1
(vii) O ✘ ✁✜✢✣ ✙ ✳✜✢✣ ✠ ✁✜✢✣✤ ✄✁ ✌ ✴✶✒ ✦✧ ★✩✪ -1
✷✩✫ ✯✛✰ ✸✬✪✸✹✪✬✯✮✩✺✻ ✩✼ ✄✁f of CH3OH with the help of above data, following thermochemical
equation is written :
C(s) + 2H2(g) + 1/2 O2✜✢✣ ✙ ✚✁3✳✁✜✢✣✤ ✄✁ ✌ ✽
✳✺ ✯✛✰ ✾✬✻✮✻ ✩✼ ✾✩✺✿ ✰✺✯✛✬✪✭❀ ✸✩✺✸✰✭✯ ✄✁ ✩✼ ✬✾✩❁✰ ✫✰✬✸✯✮✩✺ ✮✻ ✸✬✪✸✹✪✬✯✰✿ ✬✻ ✼✩✪✪✩❂✻ ❃
✄H
= - [(sum of bond enthalpies of all bonds of products ✘ Sum of bonds enthalpies of all bonds of
reactants.)]
✄H
=-
(3* EC ❆ H ❇ EC ❆O ❇ EO ❆ H )
❈
❊
●
❊
■ ❏ E C (g) toC(g) E ❇ 4* E 1
❇
❊
❄ H ❆ H ❅❄ g ❅
2
◆ ▲
❉
❋
❍
❋
E1
❑
O2 ❄ g ❅ ❋
2
▼❖
= - [(3 * 415 + 356 + 463) ✘ (715 + 4 218 + 249)] kJ mol-1
(viii) C(s) + 2H2(g) + 1/2 O2✜✢✣ ✙ ✚✁3✳✁✜✢✣✤ ✄✁ ✌ - 228 kJ mol-1
(ix)j CH3OH(g) ✙ ✚✁3OH(l✣✤ ✄✁ ✌ - 38 kJ mol-1
On adding, above two thermochemical equation
C(s) + 2H2(g) + 1/2 O2✜✢✣ ✙ ✚✁3✳✁✜P✪✣✤ ✄✁f
Hence standard heat of formation of CH3OH (l✣ ✜✄✁f)
Sol 19.
✷✩✫ ✮✩✺✮◗✬✯✮✩✺ ✄✁ionisation ❘ ✄✁Hydration
Total hydration energy of AI3+ & 3CI- ions of AICI3 ✜✄✁hydration)
= (Hydration energy of AI3+ + 3 * Hydration energy of CI-)
= [- 4665 + 3 (-381)] kJ mole-1 = - 5808 kJ mole-1
This amount of energy is more than that required for the ionization of AI into AI3+
(Ionisation energy of AI to AI3+ = 5137 kJ mol-1). Due to this reason, AICI3 becomes ionic in aqueous
solution. In aqueous solution it exists in ionic form as below
AICI3 + 6H2✳ ✙ ❙P✜✁2O)6]3+ + 3CI-
❚❚❚❯❱❲❳❨❩❳❬❱❯❬❭❨
AICI3 ✌ ✁✂ ✆ ✄☎✝☎3(aq.); ✞✟ ✠ ✡
✞H = Energy released during hydration
- Energy used during ionization
= - 4665 ☛ 3 * 381 + 5137 = - 671 kJ/mol
Thus formation of ions will take place.
Sol 20.
✞✟ ✠ - 33.0 kJ
☞✂✒✍✎
C + O2 ✆ ✝✏2✒✑✎✓ ✞✟ ✠ - 393.5 kJ
☞✒✍✍✎
H2 + 1/2O2 ✆ H2O(l); ✞H = - 285.8 kJ
☞✂✂✒✍✍✍✎
3C + 3H2 ✆ ✝✟3 ✆ ✝✟3 ☛CH = CH2(g)✓ ✞✟ ✠ ✔✕✂✖✔ ✗✘
The required reaction is
☞✂✒✍✙✎
To calculate the value of ✞H follow the following steps.
(iv) ☛ (i) yields
[3 * (ii) + 3 * (iii)] ✚ (v) yields]
✞H =
Sol 21.
Given S(s) + 3F2(g) ✆ ✛✜6(g); ✞ H = - 1100 kJ
S(s) ✆ ✛(g)✓ ✞ ✟ ✠ ✔✢✣ ✗✘
☞✂✒✍✎
☞✒✍✍✎
1/2 F2(g) ✆ ✜(g)✓ ✞ ✟ ✠ ✤✕ ✗✘
☞✒✍✍✍✎
To get SF6(g) ✆ ✛(g) + 6F(g) we can proceed as
(ii) + 6 * (iii) - (i)
✥ SF6(g) ✆ S(g) + 6F(g); ✞ H = 1855 kJ
Thus average bond energy for S-F bond = 1855/6 =
Sol 22.
For adiabatic expansion, we have
✦✧ ★1/T2 = R/Cv ✦✧ ✩2/V1
✧✪ ✞✟ ✠ ✧✝P ✞★✂
✫✫✫✬✭✮✯✰✱✯✲✭✬✲✳✰
✄ ✁✂✂☎✆2 ✌ ✝✞✁✟☎✟✠✞✡✝ ✄ ✠✞☛✂☎✟✞✠☛
solving, we get, T2 = 188.5 K
No. of moles of argon gas, N = PV/RT = 1 *1.25/0.082 *300 = 0.05
now we know that
☞H = nCP☞T = 0.05 * 20.8(188.5 - 300) = ✒✍ ✎P = Cv + R = 12.48 + 3.314 = 20.8]
Sol 23.
For following reaction
CO(g) + 1/2 O2(g) ✏ ✎✑2(g)
☞G° can be calculated as follows :
☞G° = ☞Gp° - ☞✓°R ✌ ✒☞✓°CO2 ✔ ✕☞✓°✎✑ ✖ ✟☎✠ ☞✓°O2)]
= - 394.4 ✔ (- 137.2 + 1/2 * 0) = - 257.2 kJ mol-1
Since,
☞G° ✌ ☞✗° - ✆☞✘° or -✠☛✙✞✠ ✌ ☞✗ ✔ 300(0.094)
✚ ☞✗° = - 285.4 kJ/mol
Since,
☞✗° ✛✜ ✢✣✤✥✛✦✢✧ ✜★ ✥✩✢ ✪✢✤✫✥✛★ ✛✜ ✢✬★✥✩✢✪✭✛✫ ✤ ✮ ✜✛ ✫✢ ☞✓ ° is negative so the reaction is
spontaneous.
Sol 24.
The chemical reaction for combustion of diborane is
B2H6(g) 3O2✕✣✯ ✏ ✰2O3 (s) + 3H2✑ ✕✣✯✧ ☞✗ ✌ ✱
For this the enthalpy change can be calculated in the following way.
☞H = [☞✲✳✴ ✵✶ ✷✸✹ + ✁☞✲✺✴ ✵✷✻✹ ] - ☞✲✳✴ ✵✼✷✻✹ ;
✕✍☞✲✽✾ of O2 = 0)
☞✲✺✴ ✵✷✻✹ can be obtained by adding ✁☞✲✺✴ ✵✷✿✹ and
☞✲✺✴ ✵✷✻✹ , i.e. ✔ 286 + 44 = - 242 kJ mol-1
☞H = [- 1273 + 3 * - 242] ✔ 36 kJ mol-1 = - 1273 ✔ 726 ✔ 36 =
Sol 25.
☞G° = -2.303RT log [Product]/[Reactant]
✎✤❀✫❁❀✤✥✛★ ★❂ ☞✓ ✦✤❀❁✢✜ ❃
✆✩❁✜ ❂★✪ ✥✩✢ ✢❄❁✛❀✛❅✪✛❁✭ ✰ ❆ ❇
☞❈❉❊ (-2.303 * 8.314 * 448) log 1.3/95.2
Or ☞❈❉❊ =
✘✛✭✛❀✤✪❀❋ ❂★✪ ✥✩✢ ✢❄❁✛❀✛❅✪✛❁✭ ✰ ❆ ●
☞❈❍❊ = (-2.303 * 8.314 * 448) log 3.5/95.2
Or ☞❈❍❊ =
✘✛✭✛❀✤✪❀❋ ❂★✪ ✢❄❁✛❀✛❅✪✛❁✭✧ ❇ ❆ ✎
☞❈■❊ = - 8.314 * 448 * 2.303 * log10 3.5/1.3 = - 3.688 kJ mole-1
Hence, we have that
✰ ❆ ❇✧
☞❈❉❊ = + 15.992 kJ mole-1
✰ ❆✎✧
☞❈❍❊ = + 12.312 kJ mole-1
❏❏❏❑▲▼◆❖P◆◗▲❑◗❘❖
✁ ✂✄
☎✆☛✝ = - 3.688 kJ mole-1
Thus, the correct order of stability,
Mechanism of isomerisation
Sol 26.
(ii) Total work (W) = W1 + W2 +
W3= = - ✞☎✟ ✠ ✡ ✠ ☞✌✍✡✍✎✏✑ ✒✓✔ ✟2/V1
= - 1 * 20 + 2.303 * 2 * 0.082 * 121.95 log 2
= - 20 + 13.87
PV ✕ nRT
✖
✗
✘
✙ = 121.95k
✘T ✕ PV ✕ 5* 40 ✙
nR 2*0.082 ✛
✚
Since the ✜✢✜✣✤✥ ✦✧✜ ★✤✣✩★✎✤✪ ✣✓ ✫✣✜ ✫✎✫✣✫✧✒ ✜✣✧✣✤ ✫✌✤✌ ✣✦✤ ✬★✓✭✤✜✜ ✫✜ ✭✢✭✒✫✭✄ ✜✓ ☎✮ ✯ ✡
☎U = q + W = 0, so q = - W = - (-6.13) L.atm =
In a cyclic process heat absorbed is completely converted into work.
(iii) Entropy is a state function and since the system has returned to its initial state, so
.
and
for the same reason i.e. U and H are also state functions having definite
Similarly
value in a given state of a system.
Sol 27.
Helium molecule is monoatomic so it has just three degrees of freedom corresponding to the three
translational motion at all temperature and hence Cv value is always 3/2 R. Hydrogen molecule is
diatomic which are not rigidly held so they vibrate about a well defined average separation. For
hydrogen molecule we have rotational and vibrational motion both besides translation motion.
✰✰✰✱✲✳✴✵✶✴✷✲✱✷✸✵
These two additional contributions increase its total heat capacity. Contribution from vibrational
motion is not appreciable at low temperature but increases from 0 to R on raising temperature.
Sol 28.
For adiabatic process, W = P (V2 ✂ V1)
Here P1 = 1 bar, P2= 100 bar, V1 = 100 mL, V2 ✌
✖U = q + W
= q ✂ P(V2 ✂ V1) since W = - P(V2= - V1)
= 0 ✂ 100 (99 - 100) =
✖✙ ✌ ✖✗ ✚ ✖✛✜✢✣ ✌ ✖✗ ✚ ✛✜ 2V2 ✂ P1V1)
100 + [(100 * 99) ✂ (1 * 100)]
= 100 + (9900 - 100) =
✁✄☎ ✆✝✞ ✟✠✡✟☛✟☞✡✍ ✎✞✝✍✏✑✑✒ ✓ ✌ ✔ ✕ ✖✗ ✌ ✘
Sol 29.
(i) N2O4✛✤✣ ✥ ✦✧★2(g)
Initially ✩✪✫ ✬✭ ✌ ✩✪✬✫ = 10
✮✯✰✱✫ ✲✫
✯✰✫✱✭ = 100/10 = 10
✵
✷
✖G° = ✦✖✳✴✮✪✬
✫ ✲ ✶✖✳✴✮✪✫ ✬✭ ✲ = 100 ✂ 100 = 0
✖✸ ✌ ✖✸✹ - 2.303 RT log Kp
= 0 ✂ 2.303 * 298 log 10
Reaction quotient =
✛✡✡✣ ✺✻✏ ✼✏✤✟☞✡✽✏ ✽✟✾✿✏ ✝❀ ✖✸ ✡✼✠✡✍✟☞✏✑ ☞✻✟☞ ☞✻✏ ✞✏✟✍☞✡✝✼ ✡✑ ✑✎✝✼☞✟✼✏✝✿✑ ❁ ✘✡✾✾ ✾✡✏ ✡✼ ☞✻✏ ✞✡✤✻☞
direction, (forward).
Sol 30.
✖H = ✖✗ ✚ ✖✛✜✢✣ ✌ ✖✗ ✚ ✢ ✖✜
✛❂ ✖✢ ✌ ✔✣
✝✞ ✖✗ ✌ ✖✙ ✂ ✢ ✖✜ ✌ - 560 ✂ [1(40 - 70) * 0.1]
= - 560 + 3 = - 557 kJ mol-1
So the magnitude is 557 kJ mol-1.
❂ ✖✸✹ ✌ - 2.303RT log Kp ✟☞ ✏✓✿✡✾✡☛✞✡✿✁ ✖✸✹ ✌ ✔
❃ - 2.303RT log Kp = 0
Log Kp = 0 or Kp = 1
❄❄❄❅❆❇❈❉❊❈❋❆❅❋●❉