Thermodynamics Part-I: Prepared by V. Aditya vardhan
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NOTE: PART 2 CONTAINS QUESTIONS FROM ENTROPY AND FREE ENERGY
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INTRODUCTION & FIRST LAW OF THERMODYNAMICS Which of the following statement is true about thermodynamics a) Thermodynamics will help in predicting whether a physical or chemical change is possible under given conditions. b) Thermodynamics only deals with the initial and final states of the system and is not helpful in evolving the mechanism of the process. c) The rate of a reaction can be evolved from thermodynamics. d) The mechanism of a reaction can be evolved from thermodynamics. 1) a only 2) a & b 3) a,b & c 4) a & d Choose the incorrect statement 1) Open systems can exchange both energy and matter with its surroundings. 2) Closed systems can only exchange energy and do not exchange matter with its surroundings. 3) Isolated systems can exchange energy and matter with its surroundings. 4) None. Which of the following is not an intensive property? 1) Entropy 2) Density 3) Temperature 4) Pressure Which of the following is an extensive property? 1) Entropy 2) Enthalpy 3) Volume 4) All H Extensive properties, out of a) boiling point, b) viscosity, c) p , d) emf of a cell and e) molar heat capacity; are 1) a, b & c 2) b & e 3) c & e 4) None Which of the following is not a state function? 1) Internal energy 2) Enthalpy 3) Work 4) Free energy Which of the following is a state function and not an intensive property? 1) Temperature 2) Volume 3) Pressure 4) None In an adiabatic process, 1) dT = 0 2) dU = 0 3) q 0 4) All Note: In an adiabatic process, there is no exchange of heat between system and surroundings. Hence dq = 0 and dU = w. However there is change in temperature and internal energy during this process.
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In which of the following processes, dT =0? 1) Adiabatic process 2) Isobaric process 10) In an isochoric process, 1) dT = 0 2) dP = 0
3) Isothermal process 4) Isochoric process 3) dV = 0
4) dH = 0
Note: in isochoric (dV=0) and isobaric (dP=0) processes, the pressure-volume work is always zero.
11) In a cyclic process, 1) dU = 0 2) dH = 0 3) dT = 0 4) All 12) If the systems A and B are in thermal equilibrium with system C, then system A is also in thermal equilibrium with system B. In thermodynamics, this statement is known as
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Thermodynamics Part-I: Prepared by V. Aditya vardhan
14) 15)
16) 17) 18)
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1) 1st law 2) 2nd law 3) 3rd law 4) zeroeth law Thermos flask is an example of 1) Closed system 2) Open system 3) Isolated system 4) All The correct statement about isothermal process is 1) dT = 0 2) dU = 0 3) q = -w 4) All The statement which is not true according to 1st law of thermodynamics is 1) Energy can neither be created nor destroyed although it can be changed from one form to another. 2) The total energy of the universe is constant. 3) The change in the internal energy of a closed system is equal to the sum of heat lost or absorbed by the system and work either done by the system or done on the system 4) The change in internal energy of a system is zero in all the processes The work done during the free expansion of an ideal gas is equal to 1) Pext 2) dV 3) dU 4) zero Which of the following is true about isothermal free expansion of a gas? 1) dU = 0 2) dT = 0 3) PextdV = 0 4) All The work done during isothermal irreversible change will be given by
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1) -Pext(Vf-Vi)
2) 2.303nRT log
Vf Vi
3) nRT log
Vf Vi
4) zero
pressure.
19) The work done during an isothermal reversible change will be given by 1) -Pext(Vf-Vi)
2) 2.303nRT log
Vf Vi
3) nRT log
Vf Vi
4) zero
Note: In case of reversible process, the internal pressure is almost equal to the external pressure. This internal pressure is given by nRT / V.
20) Two litres of an ideal gas at a pressure of 10 atm expands isothermally into vacuum until its total volume becomes 100 litres.The amount of heat absorbed in the expansion is 1) 2.303 J 2) 100 J 3) 90 J 4) None Note: As the process is isothermal, q=-w=PdV; And as the expansion is done against zero pressure(Pext=0), the work done as well as the heat absorbed will be zero.
21) Which of the following statements is true 1) The work done in reversible expansion is less than the work done in irreversible expansion. 2) The work done in reversible expansion is equal to the work done in irreversible expansion. 3) The work done in reversible expansion is greater than the work done in irreversible expansion. 4) All. Note: But in case of compression, |wrev| < |wirrev|
22) Ten litres of an ideal gas at a pressure of 10 atm expands isothermally against a constant pressure of 1 atm. until its total volume becomes 100 litres.The work done in the expansion is 1) -90 L-atm 2) -10 L-atm 3) -23.03 L-atm 4) -230.3 L-atm Hint: This is the case of irreversible expansion.
23) Ten litres of an ideal gas at a pressure of 10 atm expands reversibly under isothermal conditions until its total volume becomes 100 litres.The work done in the expansion is 1) -90 L-atm 2) -10 L-atm 3) -230.3 L-atm 4) -100 L-atm Note: The work done is maximum in case of reversible expansion. (negative sign only indicates the work is done by the system)
24) The amount of heat absorbed, when 2.5 L of an ideal gas is compressed isothermally under constant atmospheric pressure until the volume becomes 0.5 L is
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Note: In the irreversible process, the work is done against constant external pressure which differs largely from the internal
Thermodynamics Part-I: Prepared by V. Aditya vardhan
1) 2 J Note:
2) -202.65 J
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3) +202.6 J
3
4) -2 J
1) During isothermal compression, heat is liberated by the system and hence negative with respect to it. 2) 1 L-atm = 101.325 J
28) 25 L of an ideal gas is compressed isothermally under constant atmospheric pressure until the volume becomes 5 L. The change in internal energy is 1) 20 J 2) -2026.5 J 3) +2026 J 4) 0 J 29) There is a balloon of given volume, V1, containing a gas at temperature, T1. When the balloon is placed in a colder room at temperature, T2, the balloon’s temperature starts to drop. What are the signs of the system’s q, w, and E for this process? 1) +q, +w, +E 2) -q, -w, -E 3) -q, -w, +E 4) -q, +w, -E 30) 11.2 L of a hydrogen gas at 273 K temperature and 1 atm of pressure in a sealed rigid container is heated to double its temperature. The change in internal energy dU will be equal to 1) +w 2) +q 3) -w 4) -q Note: As the process is isochoric, the change in internal energy is only due to exchange of heat
qv 31) An ideal gas at 10 atm pressure and occupying 0.1 L is expanded to 1.1 L by supplying 101.325 J of heat against constant atmospheric pressure irreversibly. The change in the temperature of the gas during this process is 1) 10 K 2) 1.1 K 3) 1K 4) No change dU =
Hint: dU = q+w = 101.325 J + (-101.325 J) = 0
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25) 2 moles of an ideal gas at STP is expanded isothermally in infinitisimally small steps until the volume is doubled. The amount of heat absorbed during this process is 1) 1.37 kJ 2) 8.3 kJ 3) -1.37 kJ 4) None 26) 1 liter of an ideal gas, initially at a pressure of 10 atm, is allowed to expand at constant temperature to 10 liters by reducing the external pressure to 1 atm in a single step. The work done during this process is 1) 10 L-atm 2) -100 L-atm 3) -9 L-atm 4) 0 L-atm 27) The area of which of the following graph shows PV work done during reversible expansion of ideal gas at constant temperature?
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32) A gas is allowed to expand at constant temperature from a volume of 1.0 L to 10.1 L against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w, and E respectively? 1) 250 J, -461 J & -211 J 2) -250 J, -461 J & -711 J 3) 250 J, -461 J & -711 J 4) 250 J, -4.55 J & 245 J 33) In a process, 800 J of heat is absorbed by a system and 350 J of work is done by the system. The change in internal energy for the process is 1) 350 J 2) 450 J 3) -450 J 4) -350 J 34) When is dHsys= dEsys? 1) When qv= qp. 2) In reactions involving only liquids and solids. 3) In reactions running under a vacuum (P=0). 4) All 35) Which of the following reactions could do work of expansion on the surroundings? 1) 2CO(g) + O2(g) 2CO2(g)
2) Fe2O3(s) + 2Al(s) Al2O3(s) + 2Fe(s)
4) 2N2O(g) 2N2(g) + O2(g) 36) Suppose a gas in a piston is expanded at constant pressure and the temperature goes down. Which of the following correctly describes the signs for the work, the heat for the system and the energy change of the system? 1) Work is positive, q is negative, and dU is negative. 2) Work is negative, q is may be positive or negative, and dU is negative. 3) Work is positive, q is may be positive or negative, and dU is negative. 4) Work is positive, q is positive, and dU is negative. 37) In order to have E 0 for a process, which of the following conditions must be obeyed ? a) q > 0 b) w > 0 c) q + w > 0 d) q > 0 such that |q| > |w| (where |q| and |w| are absolute values) The correct options are 1) a,b,c & d 2) b &d 3) a & d 4) c & d 38) When 2.0 mol O2(g) is heated at a constant pressure of 4.25 atm, its temperature increases from 260 K to 285 K. Given that the isobaric molar heat capacity of O2 is 29.4 J.K-1 mol-1, the change in the internal energy during the process will be 1) 1054 J.K-1 mol-1 2) 416 J.K-1 mol-1 3) 1470 J.K-1 mol-1 4) 1887 J.K-1 mol-1
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ENTHALPY & HEAT CAPACITY The amount of heat exchanged between system and surroundings under constant pressure is called 1) Entropy 2) Enthalpy 3) Internal energy 4) Free energy Note: The amount of heat exchanged at constant pressure i.e., q p H U PV
2) 3)
Enthalpy(H) of a system can be represented by 1) U + PV 2) U-PV 3) q+w Consider the following statements about enthalpy. a) It is an intensive property b) It is a state function. c) It is an extensive property.
4) dU + PdV
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3) CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
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d) It is a path function. The correct statements are 1) a & d 2) b & c 3) a & c 4) b & d Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find U for the system when one mole of HCl dissolves in water under these conditions. 1) +2.48 kJ 2) -75.3 kJ 3) -72.82 kJ 4) +75.3 kJ Hint: The volume of the liquid is negligible. The process is isothermal and isobaric. Note: -72.82 kJ > -75.3 kJ; So the change in internal energy(dU) will be greater than the change in enthalpy(dH)i.e., dU > dH. But if absolute values are taken, 75.3 kJ > 72.3 kJ i.e., the heat liberated will be greater than decrement in internal energy. It is because, the compression of the gas which increases the internal energy. Also remember, the magnitude of work done on the system is less than heat liberated during the process.
One mole of an ideal gas is dissolved in a solvent by liberating 0.27 kJ of heat. What is the change in internal energy when one mole of gas is dissolved completely in this solvent at 273 K and 1 atm pressure. 1) -2 kJ 2) -0.27 kJ 3) +2 kJ 4) +0.53 kJ
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Note: Now +2 kJ > 0.27 kJ. Therefore dU > dH. This is true even when the absolute values are taken. Why? In this case the
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If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ mol-1. The internal energy change, when 1 mol of water is vaporised at 1 bar pressure and 100°C will be 1) 37.9 kJ mol–1 2) 41 kJ mol–1 3) 47.9 kJ mol–1 4) 18.3 kJ mol–1 Reaction: H2(liquid)
H2(gas)
Formula: U H ng RT Where n g = no. of moles of gaseous products - no. of moles of gaseous reactants
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Which of the following is not an intensive property? 1) Molar heat capacity 2) Specific heat capacity 3) Heat capacity 4) All o 8) The amount of heat required to raise the temperature of 1.00 g of aluminium by 1 C is called its 1) Enthalpy 2) Heat capacity 3) Specific heat 4) Molar heat -1 -1 9) The molar heat capacity for NaCl is 50.50 J mol K . What is the specific heat? 1) 0.8640 J g-1 K-1 2) 50.5 J g-1 K-1 3) 8.640 J g-1 K-1 4) 4.184 J g-1 K-1 10) The amount of heat necessary to raise the temperature of 60.0 g of aluminium from 15oC to 55oC is. (Molar heat capacity of Al is 24 J mol–1K–1) 1) 2.133 J 2) 1.066 J 3) 2.133 kJ 4) 2.4 kJ Formula: q=n.C m .T Where n = no. of moles Cm = molar heat capacity T = raise or lowering of temperature
11) The heat capacity of methyl alcohol (MW = 32.05 g/mol) is 80.3 J mol-1 K-1. The quantity of heat that will be evolved when the temperature of 2610 g of methyl alcohol falls from 22oC to 2oC 1) 1.5 x 102 kJ 2) 1.3 x 102 kJ 3) 1.7 x 102 kJ 4) 7 kJ 12) The amount of heat absorbed by one mole of an ideal gas in an isochoric process to raise the temperature from 1.1oC to 11.1oC is 120 kJ/mol. The Cv and Cp values of the gas in kJ mol-1K-1 will be 1) 12 kJ & 3.7 kJ 2) 10 kJ & 12 kJ 3) 12 kJ & 20.3 kJ 4) 20.3 kJ & 12 kJ
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value of work done on the system during compression is greater than the amount of heat lost.
Thermodynamics Part-I: Prepared by V. Aditya vardhan Use the formulae: Cv
Cp
U T
T
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qv
T
v
H
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qp T
p
C p C v nR
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13) The difference in Cp and Cv values for liquids and solids will be equal to 1) R 2) nR 3) 2R 4) 0 14) Choose the incorrect statement 1) The molar heat capacities of the metallic elements are almost identical. This is the basis of the Law of Dulong and Petit, which served as an important tool for estimating the atomic weights of some elements. 2) The intermolecular hydrogen bonding in water and alcohols results in anomalously high heat capacities for these liquids; the same is true for ice, compared to other solids. 3) The heat capacity values for graphite and diamond are very high as the solids that are more “ordered” tend to have larger heat capacities. 4) None. Note: The molar heat capacities of metals are almost equal to 3R.
1)
3R 2
2)
5R 2
3)
7R 2
4) R
Note: In case of noble gases (which are mono-atomic) only three translational degrees of freedom are possible and each of these contribute
1
2
3R
R to heat capacity and hence the Cv =
5
. The Cp value will be given by Cv + R =
2
R
2
16) The theoretical molar heat capacities of diatomic molecules at constant volume and at fairly high temperatures is almost equal to 1)
3R 2
2) R
3)
7R 2
4)
5R 2
Note: Diatomic molecules have 3 translational degrees of freedom (contribution of 3R/2), 2 rotational degrees of freedom (contribution of 1R/2 from each) and 1 vibrational degree of freedom ( contribution of R). Hence Cv =
7R 2
(at high temp.)
But at low temperatures (eg., room temperature), the vibrational degree of freedom can be neglected and hence Cv =
5R 2
In the same way, for poly-atomic molecules, the contributions are as follows From three translational degrees of freedom --
1
R
2
From three rotational degrees of freedom --
1 2
From 3N-6 vibrational modes -- (3N-6)R
R
1
R
2
1 2
R
1
R
2
1
R
3
R
2
3
R 2 2 (where N = number of atoms in poly-atomic molecule)
17) Calculate the enthalpy change on freezing of 1.0 mol of water at10.0oC to ice at –10.0oC. Given fusH = 6.03 kJ mol–1 at 0°C; Cp [H2O(l)] = 75.3 J mol–1 K–1 ; Cp [H2O(s)] = 36.8 J mol–1 K–1 1) 7.151 kJ 2) -6.03 kJ 3) 3.63 kJ 4) -7.151 kJ 18) If 1000 calories are added to 20 g of ice at -10oC, the final temperature will be (Specific heat of ice = 0.48 cal/g-K and Heat of fusion of ice = 80 cal/g.) 1) 380 K 2) 0 K 3) 273 K 4) 283.2 K Note: As the heat supplied is not sufficient to melt the ice completely, finally there will be a mixture of ice and water at 0oC
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15) The molar heat capacity values of noble gases at contant volume are almost equal to
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19) A coffee-cup calorimeter is calibrated by adding 1840 J of heat to the water in the calorimeter and measuring a 1.72oC rise in temperature. When some NH4Cl(s) is added to the same water in the calorimeter, the temperature falls by 1.04oC. The enthalpy change due to the dissolving of NH4Cl(s) is 1) -1112 J 2) +1250 J 3) +3040 J 4) +1112 J 20) Which of the following statements is true? 1) q = dH at constant P; q = dE at constant T 2) q = dH at constant T; q = dE at constant V 3) q = dH at constant V; q = dE at constant P 4) q = dH at constant P; q = dE at constant V 21) An ice cube at 0oC weighing 9.0 g is dropped into an insulated vessel containing 72 g of water at 50oC. What is the final temperature of water after the ice has melted and a constant temperature has been reached? The latent heat of fusion of ice is 6.01 kJ/mol and the molar heat capacity of H2O is 75.4 J mol-1K-1. 1) 36oC 2) 40oC 3) 44oC 4) 32oC o
22) What is U o when one mole of liquid water vaporises at 100 C if the heat of vaporisation vap H o -1
Note:
Cp Cv
5
3) 36.73 kJ.mol-1 3) 6 cal
4) -40.66 kJ.mol-1 4) 10 cal
1.67 cal
3
3R
3 cal 2 C p Cv R 5 cal
For mono atomic gases Cv
24) Temperature of one mole of Neon gas is increased by 1oC, hence, increase in internal energy is 1) 5 cal 2) 3 cal 3) 9 cal 4) 2 cal 25) Enthalpy change for a reaction does not depend upon 1) the physical states of reactants and products. 2) use of different reactants for the same product. 3) the nature of intermediate reaction steps. 4) the differences in initial or final temperatures of involved substances. 26) The specific heat of water is 4.18 J g-1 K-1 and that of stainless steel is 0.51 J g-1 K-1. The heat that must be supplied to a 750.0 g stainless steel vessel containing 800.0 g of water to raise its temperature from 20.0oC to the boiling point of water 1) 6.98 kJ 2) 29.8 kJ 3) 69.8 kJ 4) 298 kJ 27) The temperature of a substance represents 1) the total heat content of a the particles in a substance 2) the speed of the fastest particles in the substance 3) the speed of the slowest particles in the substance 4) the average kinetic energy of the particles in a system. 28) A bomb calorimeter was calibrated by burning a sample of benzoic acid (C6H5COOH) which has a known heat of reaction, H° = –3227 kJ/mol. When 1.22 g of benzoic acid is burned in the calorimeter, the temperature is increased by 0.75 °C. The heat capacity of the calorimeter and its contents will be 1) 32 kJ K-1 2) 4.3 kJ K-1 3) 83 kJ K-1 4) 43 kJ K-1 29) The heat of combustion of methyl alcohol, CH3OH, is -715 kJ mol-1. When 2.85 g of CH3OH
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o
of water at 100 C is 40.66 kJ.mol ? 1) 40.66 kJ.mol-1 2) 24.66 kJ.mol-1 23) The value of Helium gas is equal to 1) 1.67 cal 2) 1.4 cal
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was burned in a bomb calorimeter, the temperature of the calorimeter changed from 24.05 °C to 29.19 °C. The heat capacity of the calorimeter is 1) 12.4 kJ/°C 2) 124 kJ/°C 3) 12.4 J/°C 4) 1.24 kJ/°C
3)
4)
1) 208 kJ 5)
SiO2(s) + 3C(s) SiC(s) + 2CO(g) 2) 5.06 kJ 3) 10.4 kJ
4) 31.2 kJ
H of which of the following reactions is equal to the standard enthalpy of formation( f H o ) of
NH3?
1) 2NH3(g) 3H2(g) + N2(g)
2) NH3(g) 3/2H2(g) + 1/2N2(g) 3) 3H(g) + N(g) NH3(g)
4) 3/2H2(g) + 1/2N2(g) NH3(g)
o
Note: The standard enthalpy of formation, f H , is the amount of heat either liberated or absorbed during the formation of one mole of a compound from its elements under standard conditions.
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7) 8)
o Given: H2(g) + Br2(l) 2 HBr(g) ; r H = –72.8 kJ Calculate the amount of energy absorbed or released when 15.0 g of HBr (g) is formed. 1) 6.75 kJ released 2) 13.5 kJ released 3) 4.85 kJ absorbed 4) 607 kJ absorbed Which of the following has non zero standard enthalpy of formation at 25oC? 1) Cl2(g) 2)O2(g) 3) Na(g) 4) F2(g) The standard heat of formation of NH3 is
(Given N2(g) + 3H2 (g) 2NH (g) ; r H o = -91 kJ ) 3 1) +91 kJ 2) + 45.5 kJ 3) -19 kJ 4) -45.5 kJ 9) The enthalpy change for the reaction of 50 mL of ethylene with 50 mL of H2 at 1.5 atm pressure is H = -0.31 kJ. The change in internal energy will be 1) 0.66 kJ.mol-1 2) -0.3024 kJ.mol-1 3) +0.3024 kJ.mol-1 4) -0.66 kJ.mol-1 10) In a constant-volume bomb calorimeter an unknown compound reacted with excess oxygen to give carbon dioxide and water. The temperature of 2 kg of water in the calorimeter rose from 12.72oC to 20.72oC. The heat capacity of the calorimeter is 2.02 kJ.K-1 and the specific heat of water is 4.184 J/goC. The heat given off by the combustion reaction under these conditions is: 1) 41,600 J 2) -41.6 kJ 3) -83.2 kJ 4) -33.5 kJ Note: The heat given off indicates the internal energy change, as the reaction is occuring at constant volume, and
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THERMOCHEMISTRY Which statement is incorrect about endothermic reactions? 1) the system absorbs energy from its surroundings 2) the enthalpy of products is lower than the enthalpy of the reactants 3) the thermal kinetic energy of the surroundings will decrease 4) the enthalpy change will have a positive value Which of the following is the standard state of carbon at STP? 1) C as CO2 (g) 2) C as graphite (s) 3) C as CH4 (g) 4) C as diamond (s) Which of the following statements is/are true about an exothermic reaction? I) the energy absorbed in bond breaking is more than the energy released in bond formation II) the system absorbs energy III) the potential energy of the reactants is less than the potential energy of the products IV) the thermal kinetic energy of the surroundings will increase 1) I and III only 2) II and III only 3) II and IV only 4) IV only How much heat is absorbed when 3.00 grams of SiO2 react with excess carbon according to the reaction below? r H o for the reaction is +624.7 kJ.
Thermodynamics Part-I: Prepared by V. Aditya vardhan not is not equal to the enthalpy change.
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11) In a constant - volume bomb calorimeter, 4g of methane is burned in excess of oxygen. The temperature of 0.5 Kg of water in the calorimeter rose from 12oC to22oC The heat capacity of calorim-1 eter is 20.1 kJ.K . The enthalpy of combustion of methane under standard conditions will be 1) 223.1 kJ.mol-1 2) 891.6 kJ.mol-1 3) 888 kJ.mol-1 4) 20.92 kJ.mol-1 Note: Enthalpy of combustion is the amount of heat liberated when one mole of substance is completely burned in excess of oxygen at standard conditions (constant temperature and pressure). In this case the amount of heat liberated is equal to U . This value should be converted to H by using the formula. H U n g RT
12) B5H9 burns in air according to the following reaction.
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ; r H o = -890.4 kJ At 1.0 atm and 273 K , how much work is involved per mole of CH4(g) with the volume change that occurs upon reaction. 1) -4.5 kJ 2) -2.2 kJ 3) 2.2 kJ 4) 4.54 kJ Note: The work is done on the system and is equal to 4.54 kJ. Here n g =-ve and hence, H E i.e., -890.4 kJ< - 885.8 kJ. But if the absolute values are considered, the decrease in internal energy is less than heat liberated. This is because Some amount of decrease in internal energy, due to loss of heat, is compensated by compression work done on the system.
14) When burned in oxygen, 10.0 g of phosphorus generated enough heat to raise the temperature of 2950 g of water from 8.0oC to 28.0oC. The heat of formation of P4O10 from P4(s) and O2(g) is 1) -30.6 kJ mol-1 2) -306 kJ mol-1 3) -3060 kJ mol-1 4)-6120 kJ mol-1 15) Using the following data: I) N2(g) + 3O2(g) + H2(g) 2HNO3(aq) ; H = -414.8 kJ II) N2O5(g) + H2O(g) 2HNO3(aq) ; H = 218.4 kJ
III) 2H2O(g) 2H2(g) + O2(g) ; H = 483.6 kJ What is the H for the reaction: 2N2O5(g) 2N2(g) + 5O2(g) 1) 90.8 kJ 2) -876.4 kJ 3) 782.8 kJ 16.
4)1750 kJ
Which thermochemical equation corresponds to the enthalpy diagram shown above? 1) 2 H2 (g) + O2 (g) 2 H2O (g) + 486.3 kJ 2) 2 H2 (g) + O2 (g) + 486.3 kJ 2 H2O (g) 3) 2 H2 (g) + O2 (g) 2 H2O (g) – 486.3 kJ 4) H2 (g) + 1/2 O2 (g) – 486.3 kJ H2O (g) 17. The standard heat of combustion (Hº ) of ethanol is – 1372 kJ/mol ethanol. How much heat is released when a 20.0 g sample burns? 1) 68.6 kJ 2) 29.8 kJ 3) 3.16 x 103 kJ 4) 595 kJ 18. Which is the correct enthalpy level diagram for the following reaction: CaO(s) + 3C(s) + 462.3 kJ CaC2(s) + CO(g)
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2B5H9(g) + 12O2(g) 5B2O3(s) + 9H2O(l) What is the molar heat of reaction for the combustion of B5H9 if the reaction between 0.1 g of B5H9 and excess oxygen in a bomb calorimeter raises the temperature of the 852 g of water surrounding the calorimeter by 1.57oC? [At wts: B = 10.81 amu; H = 1.008 amu; O = 16.00 amu; the heat capacity of water is 4.184 J / goC] 1) 5.60 x 103 J/mol 2) 9.14 x 102 kJ/mol 3) 3.54 x 103 kJ/mol 4) 4.46 x 103 kJ/mol 13) In the reaction,
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19. Given the following thermochemical equation 2 C2H3Br (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (g) + 2 HBr (g) Hº = 1150 kJ What is the change in energy accompanying the production of 1.5 moles of CO2(g)? 1) 1725 kJ 2) 766.7 kJ 3) 431.3 kJ 4) 9.799 kJ 20. When 1.75 g of CaCl2 dissolves in 125 g of water in a coffee-cup calorimeter, the temperature increases by 2.44ºC. What is the heat change per mole of CaCl2 dissolved? Assume that all the heat is absorbed by the water. 1) 11.3 kJ/mol of CaCl2 2) 1.13 kJ/mol of CaCl2 3) 729 kJ/mol of CaCl2 4) 80.9 kJ/mol of CaCl2 21. 5.5 g of sodium hydroxide is dissolved in 175 mL of water. Using a coffee-cup calorimeter, the temperature change of the water is measured to be +2.10C. The specific heat capacity of water is 4.184 J/gC. What is the thermochemical equation for this process? 1) NaOH(s) NaOH(aq) 1.54 kJ 2) NaOH(s) 1.54 kJ NaOH(aq) 3) NaOH(s) NaOH(aq) 11.2 kJ 4) NaOH(s) 11.2 kJ NaOH(aq) 22. A mass of 100.0 g of dilute hydrochloric acid is placed in a coffee cup calorimeter. The temperature of the solution is recorded to be 17.5C. A piece of magnesium ribbon of mass 0.601 g is placed in the solution and the final temperature is recorded to be 39.6C. Calculate the molar enthalpy change for this reaction : Mg(s) 2 HCl (aq) MgCl2(aq) H2 (g) 1) + 15.4 kJ/mol of Mg 2) - 263 kJ/mol of Mg 3) - 374 kJ/mol of Mg 4) + 5.56 kJ/mol of Mg 23. Which process is not endothermic? 1) H2O(s) H2O(l) 2) 2H2O(g) 2H2(g) + O2(g) 3) H2O(g) H2O(l) 4) Al2O3 + 2Fe(l) 2Al + Fe2O3 o 24. Calculate the H for the following reaction: 2ClF3(g) + 2NH3(g) N2(g) + 6HF(g) + Cl2(g) Given this information : ClF3(g) f Ho -261.0 kJ/mol HF(g) f Ho -271.1 kJ/mol NH3(g) f Ho -46.11 kJ/mol 1) -2241 kJ 2) - 1013 kJ 3) -578.2 kJ 4) 578.2 kJ
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Thermodynamics Part-I: Prepared by V. Aditya vardhan
Thermodynamics Part-I: Prepared by V. Aditya vardhan
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25. Given the thermochemical equation 2Al(s) + 3/2O2(g) Al2O3(s) ; Ho = -95.6 kJ determine Ho for the following reaction 2Al2O3(s) 4Al(s) + 3O2(g) 1) -95.6 kJ 2) +95.6 kJ 3) -47.8 kJ 4) +191.2 kJ 26. For the reaction: NH4NO3(s) N2(g) + 2H2O(g) + 1/2O2(g) o H = -1.50 kJ/g NH4NO3 (molecular mass = 80.05 amu). If 0.105 g NH4NO3 decompose in a bomb calorimeter with a heat capacity of 5.510 J/oC initially at 21.00oC, the final temperature of the calorimeter and its contents will be numerically (in K): 1) 223 2) 281 3) 323 4) 381 27. Calculate the molar enthalpy of combustion of propylene, C3H6(g), when it reacts with O2(g) to
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give CO2(g) and H2O(l). [ f H o (C3H6(g)) = +20.4 kJ/mol, f H o (CO2(g)) = -393.5 kJ/mole,
28.
29.
30.
31.
32.
1) +699.7 kJ 2) -658.6 kJ 3) -1926.3 kJ 4) -2018 kJ Using the following data: N2(g) + 3O2(g) + H2(g) 2HNO3(aq) H = -414.8 kJ N2O5(g) + H2O(g) 2HNO3(aq) H = 218.4 kJ 2H2O(g) 2H2(g) + O2(g) H = 483.6 kJ Determine H for the reaction: 2N2O5(g) 2N2(g) + 5O2(g) 1)149.6 kJ 2) 90.8 kJ 3) -876.4 kJ 4) 782.8 kJ Calculate the standard molar enthalpy of formation of CO2(g) in the reaction: C(s) + O2(g) CO2(g) given the following standard enthalpy changes: 2C(s) + O2(g) 2CO(g) Ho = -221.0 kJ 2CO(g) + O2(g) 2CO2(g) Ho = -566.0 kJ 1) -393.5 kJ 2) +393.5 kJ 3) +787.0 kJ 4) -787.0 kJ o Calculate the H for the reaction: S(s) + O2(g) SO2(g) Given the following data: S(s) + 3/2O2(g) SO3(g) Ho = -395.2 kJ/mol 2SO2(g) + O2 2SO3(g) Ho = -198.2 kJ/mol 1) 592.2 kJ/mol 2) -197.0 kJ/mol 3) -296.1 kJ/mol 4) -593 kJ/mol Given the following data: 3/2O2(g) + 2B(s) B2O3(s) Ho = -1264 kJ/mol O3(g) + 2B(s) B2O3(s) Ho = -1406 kJ/mol The change in enthalpy for the reaction converting oxygen (O2(g)) to ozone (O3(g)) at 298 K and 1 atm will be 3/2O2(g) O3(g) 1) -1406 kJ 2) -1264 kJ 3) -2670 kJ 4) +142 kJ Given the following data N2(g) + O2(g) 2NO(g) H = 180.8 kJ 1/2N2(g) + O2(g) NO2(g) H = 33.9 kJ calculate the enthalpy change for the following reaction NO(g) + 1/2O2(g) NO2(g) 1) 214.7 kJ 2) 146.9 kJ 3) -56.5 kJ 4) 56.5 kJ
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f H o (H2O(l)) = -285.8 kJ/mole]
Thermodynamics Part-I: Prepared by V. Aditya vardhan
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33. Calculate the enthalpy change, Ho, for the combustion of benzene, C6H6, given the following C6H6(l) + 15/2O2(g) 6CO2(g) + 3H2O(l)
f H o values in kJ/mol are as follows: C6H6(l) = 49.0; CO2(g) = -393.5; H2O(l) = -285.8 1) -3169.4 kJ 2) 3267.4 kJ 3) -728.3 kJ 4) -3267.4 kJ o 34. Calculate H for the reaction: Na2O(s) + SO3 (g) Na2SO4(s) given the following: (1) Na(s) + H2O(l) NaOH(s) + 1/2 H2(g)
Ho = – 146 kJ
38.
39)
40)
Hint: When weak acids or bases participate in the neutralisation, some amount of heat is required for their complete dissociation. Hence the amount of heat liberated during neutralisation will be less than 53.7 kJ mol-1.
41) The heat of neutralisation of a strong acid and a strong alkali is 57.0 kJ mol-1. The heat released when 0.5 mole of HNO3 solution is mixed with 0.2 mole of KOH is 1) 57.0 kJ 2) 11.4 kJ 3) 28.5 kJ 4) 34.9 kJ 42) If the bond dissociation energies of XY, X2 and Y2 are in the ratio of 1:2:0.5 and standard heat of formation of XY is -100 kJ mol-1, then the bond dissociation energy of Y2 will be 1) 400 kJ mol-1 2) 100 kJ mol-1 3) 200 kJ mol-1 4) 800 kJ mol-1
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S(s) + O2(g) Hº = – 296.8 kJ/mol SO2(g) 2 SO3(g) Hº = + 197.8 kJ/mol 2 SO2(g) + O2(g) Determine the enthalpy change of the reaction: 2 S(s) + 3 O2(g) 2 SO3(g) 1) – 99 kJ 2) – 495 kJ 3) + 495 kJ 4) – 791.4 kJ Use the following reactions to calculate H° for: 4 HCl (g) + O2(g) 2 Cl2(g) + 2 H2O(g) H2(g) + Cl2 (g) 2 HCl(g) H° = – 185.0 kJ 2 H2(g) + O2(g) 2 H2O(g) H° = – 483.7 kJ 1) +113.7 kJ 2) –298.7 kJ 3) +298.7 kJ 4) –113.7 kJ Given equations (I), (II), and (III), calculate the standard enthalpy of formation of acetylene, C2H2, as shown in equation (IV). (I) C (s) + O2 (g) CO2 (g) Hº = – 393.5 kJ (II) H2 (g) + ½ O2 (g) H2O (l) Hº = – 285.8 kJ (III) 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (l) Hº = – 2598.8 kJ (IV) 2 C (s) + H2 (g) C2H2 (g) 1) 253.6 kJ 2) 226.6 kJ 3) 1801 kJ 4) -3278 kJ A volume of 50.0 mL of 0.5 M NaOH was added to 20.0 mL of 0.5 M H2SO4 in a calorimeter whose heat capacity is 39.0 J/K. The temperature of the resulting solution increased by 3.6 °C. The standard enthalpy of neutralization of H2SO4(aq) with NaOH(aq) will be 1) 13.7 kJ mol-1 2) 53.7 kJ mol-1 3) 5.37 kJ mol-1 4) 23.7 kJ mol-1 The standard enthalpy of neutralization of acetic acid with sodium hydroxide will be 1) =53.7 kJ mol-1 2) <53.7 kJ mol-1 3) >53.7 kJ mol-1 4) None
36. Given that :
37.
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(2) Na2SO4 (s) + H2O(l) Ho = + 418 kJ 2NaOH(s) + SO3(g) (3) 2Na2O (s) + 2H2(g) Ho = + 259 kJ 4Na(s) + 2H2O(l) 1) + 823 kJ 2) – 581 kJ 3) – 435 kJ 4) + 531 kJ 35. Given equations (1) and (2), calculate the enthalpy change for equation (3). (1) Pb(s) + PbO2(s) + 2 SO3(g) Ho = -775 kJ 2 PbSO4(s) (2) SO3(g) + H2O(l) H2SO4(aq) Ho = -133 kJ (3) Pb(s) PbO2(s) 2 H2SO4(aq) 2 PbSO4(s) 2 H2O(l) 1) – 908 kJ 2) – 642 kJ 3) – 509 kJ 4) + 642 kJ