4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 16 InterviewSolution

751.	`{:(,"Column-I",,"Column-II"),((a),"A process carried out infinitesimally",(p),"Adiabatic"),((b),"A process inwhich no heat enters or leaves the system",(q),DeltaG=0),((c),"A process carried out at constant temperature",(r),"Sublimation"),((d),"A process in equilibrium",(s),DeltaE=0DeltaH=0),((e),A(s)rarrA(g),(t),"Reversible"),((f),"Cyclic process",(u),"Isotermal"):}`
Answer» Correct Answer - `(a-t);(b-p);(c-u);(d-q);(e-r);(f-q,s)`

Discussion

752.	Statement-1: Combustion of all organic compounds is an exothermic reaction . Statement-2: The enthalpy of formation of all elements in their standard state are zero.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true and R is not the correct explanation of AC. A is true but R is falseD. A is false but R is true
Answer» Correct Answer - B Combustion of organic compounds is always exothermic. Heat or enthalpy of formation of natural elements in their state is considered to be zero

Discussion

753.	Statement-1: Combustion of all organic compounds is an exothermic reaction . Statement-2: The enthalpy of formation of all elements in their standard state are zero.A. Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True ,Statement-2 is False.D. Statement-1 is False ,Statement-2 is True.
Answer» Correct Answer - b

Discussion

754.	Assertion (A) : Combustion of all organic compounds is an exotherinic reaction ……Reason (R) : The enthalpies of all elements in their standard state are zero. Which of the above statement isare not correct? (a) both A and R are true and R is the correct explanation of A (b) both A and R are true and R is not correct explanation of A (c) both A and R are false (d) A is false but R is true
Answer» (b) both A and R are true and R is not correct explanation of A

Discussion

755.	Assertion (A). Combustion of all organic compounds is an exothermic reaction. Reason (R). The enthalpies of all elements in their standard state are zeroA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not the correct explanation of AC. A is true but R is falseD. A is false but R is true
Answer» Correct Answer - B Both assertion and reason are true but reason is not the correct explanation of assertion. Correct explanation. In a combustion reaction, sum of enthalpies of reactants is greater than the sum of the enthalpies of products.

Discussion

756.	All natural processes proceed spontaneously in a direction which :A. increase entropyB. increase free energyC. decrease entropyD. decrease free energy
Answer» Correct Answer - A::D

Discussion

757.	Statement-1: All combustion reactions are exothermic. Statement-2: Enthalpies of products greater than enthalpies of reactants `(Sv_(p)D_(f)H(P) gt Sv_(R)D_(f)(R))`A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true
Answer» Correct Answer - C

Discussion

758.	Statement -1: The Standard free energy changes of all spontaneously occurring reactions are negative . Statement-2: The standard free energies of the elements in their standard states at 1 bar and 2985K aare taken aszero.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true
Answer» Correct Answer - B

Discussion

759.	Statement -1: The Standard free energy changes of all spontaneously occurring reactions are negative . Statement-2: The standard free energies of the elements in their standard states at 1 bar and 2985K aare taken aszero.A. Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True ,Statement-2 is False.D. Statement-1 is False ,Statement-2 is True.
Answer» Correct Answer - b

Discussion

760.	All the naturally occurring processes proceed spontaneously in a direction which leads toA. Increase in enthalpy systemB. Decreases in entropy of systemC. Increase in entropy of systemD. Increase in entropy of Universe
Answer» Correct Answer - D

Discussion

761.	In a reaction `DeltaG` and `DeltaH` both are positive. The reaction would not be spontaneous if :A. `DeltaHgtTDeltaS`B. `DeltaS=(DeltaH)/T`C. `DeltaH=TDeltaS`D. All of these
Answer» Correct Answer - D `DeltaG=+ve` or zero in each case. `DeltaG=DeltaH-TDeltaS`

Discussion

762.	If the total enthalpy of reactants and products is `H_(R) and H_(P)` respectively, then for exothermic reactionA. `H_(R) = H_(P)`B. `H_(R) lt H_(P)`C. `H_(R) gt H_(P)`D. `H_(P) ge H_(R)`
Answer» Correct Answer - C

Discussion

763.	The vapour pressure of phoshorus trichoride is100 mm Hg at `21.0^(@)` and its normal boiling point is `74.2^(@)` C What is the enthalpy of vaporization in KJ `.mol^(-1)`A. `0.493`B. `3.93`C. `23.0`D. `32.4`
Answer» Correct Answer - d

Discussion

764.	Statement -1: The Standard free energy changes of all spontaneously occurring reactions are negative . Statement-2: The standard free energies of the elements in their standard states at 1 bar and 2985K aare taken aszero.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-8B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-8C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - B

Discussion

765.	Why is entropy of substance taken as zero at absolute zero of temperature?
Answer» At absolute zero of temperature , there is complete orderly molecular in the crystalline substance. Therefore, there is no randomness at 0 K and entropy is taken to be zero.

Discussion

766.	Statement -1: Enthalpy and entropy of any elements substance in the standard states are taken as zero . Statement-2: At absolute zero , partiles of the perfectly crystalline substance become completely motioness.A. Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True ,Statement-2 is False.D. Statement-1 is False ,Statement-2 is True.
Answer» Correct Answer - d

Discussion

767.	Assertion (A): There is no reaction known for which `DeltaG` is positive, yet it is spontaneous. Reason (R ) : For photochemical reaction, `DeltaG` is negative.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false.
Answer» Correct Answer - D

Discussion

768.	Assertion :- When a rubber band is stretched entropy increases. Reason :- During expansion entropy increases.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false.
Answer» Correct Answer - D

Discussion

769.	For exthermic reactions, the enthalpy of products is`"…........"` than enthalpy of reactants.
Answer» Correct Answer - less

Discussion

770.	The enthalpy of combustionof 1 mole of `H_(2)` to form `H_(2)O(l)` is`"…........."` kJ whereas to form `H_(2)O(g), ` it is `"…....................."` kJ.
Answer» Correct Answer - `285.8 kJ,241.8 kJ`

Discussion

771.	Statement -1: Enthalpy and entropy of any elements substance in the standard states are taken as zero . Statement-2: At absolute zero , partiles of the perfectly crystalline substance become completely motioness.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-9B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-9C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - D

Discussion

772.	Assertion :- Absolute value of enthalpy can not be determined. Reason :- Enthalpy is defined as H = E + PV, and value of internal energy can not be determined absolutely therefore absolute value of enthalpy can not be determined.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false.
Answer» Correct Answer - A

Discussion

773.	Out of different state parameters like `E, H, G, A` and S, only entropy(s) is the parameter whose absolute value can be determined, by using third law of thermodynamics. While perfect crystals hace zero entropy at `0K`, non-perfect crystals have some residual entropy at `0K`. From this info and the following data chart, answer the questions that follow : What will be molar entropy of liquid R at `300 K`?A. `150 J//K` moleB. `145 J//K` moleC. `45 J//K` moleD. `50 J//K` mole
Answer» Correct Answer - a

Discussion

774.	Out of different state parameters like `E, H, G, A` and S, only entropy(s) is the parameter whose absolute value can be determined, by using third law of thermodynamics. While perfect crystals hace zero entropy at `0K`, non-perfect crystals have some residual entropy at `0K`. From this info and the following data chart, answer the questions that follow : Which of the substances will have residual entropy at `0 K`A. Only RB. Only PC. Q, R and S all threeD. Q and P only
Answer» Correct Answer - c

Discussion

775.	Which out of the following can be determined ? Absolute internal energy, absolute enthalpy, absolute entropy.
Answer» The entropy possessed by a substance at absolute zero is called residual entropy.

Discussion

776.	A heating coil is immersed in a 100 g sample of `H_(2)O` (l) at a 1 atm and `100^(@)` C in a closed vessel. In this heating process , `60%` of the liquid is converted to the gaseous form at constant pressure of 1 atm . The densities of liquid and gas under these conditions are 1000 `kg//m^(3)` and 0.60 `kg//m^(3)` respectively . Magnitude of the work done forthe process is : (Take : 1L-atm= 100J)`A. 4997JB. 4970JC. 9994 JD. None of these
Answer» Correct Answer - C `w=-P_(ext)(V_(f)-V_(i))` `=-10^(5)((60xx10^(-3))/(0.60)+(40xx10^(-3))/(1000)-(100xx10^(-3))/(1000))` `=-10^(5)(100xx10^(-3)+0.04xx10^(-3)-0.1xx10^(-3))` `\|w\|=9994J`

Discussion

777.	Enthalpy of polymerisation of ehtylene, as represented by the reaction , `nCH_(2)=CH_(2)rarr(-CH_(2)-CH_(2)-)` is `-100 kJ` per mole of ehtylene. Given bond enthalpy of C=C bond is `600 kJ mol^(-1)`, enthalpy of C-C bond ( in kJ mol) will be:A. 116.7B. 350C. 700D. indeterminate
Answer» Correct Answer - B

Discussion

778.	The polymerisation of ethylene to linear polyethylene is represented by the reaction, `nCH_(2)= CH_(2) rarr (- CH_(2) - CH_(2) -)` where n has a large integral value. Given that the average enthalpies of bond dissociation for`C=C` and`C-C` at298 K are`+590` and`+331kJ mol^(-1)` respectively, calculate the enthalpy of polymerisation per mole of ethyleneat298 K.
Answer» In the given polymerisation reaction,one molecule of ethylene involves breaking of C`=` C double bond and formation of threeC-C single bonds.However,in the complete polymer chain,the number of single C-C bonds formedin two per C=C double bond broken Energy required in breaking of one moleofC=C doublebonds `= 590 kJ` Energy released in the formation of two moles of`C-C` single bonds `= 2 xx 331 = 662kJ` `:.`Net energy released per moleof ethylene `= 662 - 590 = 72 kJ , i.e.,DeltaH =- 72kJ mol^(-10`

Discussion

779.	In a fuel celll, methanol is used as a fuel and `O_(2)` is used as oxidizer. The standard enthalpy of combustion of methanol is -726 kJ `mol^(-1)`. The standard free energies of formation of `CH_(3)OH(I), CO_(2)(g) and H_(2)O(I)` are -166.3, -394.4 and -237.1 kJ `mol^(-1)` respectively. The standard free energy change of the reaction will beA. `-727.24` kJ `mol^(-1)`B. `-724.76` kJ `mol^(-1)`C. `-728.48` kJ `mol^(-1)`D. `-723.42` kJ `mol^(-1)`
Answer» Correct Answer - B

Discussion

780.	In a fuel celll, methanol is used as a fuel and `O_(2)` is used as oxidizer. The standard enthalpy of combustion of methanol is -726 kJ `mol^(-1)`. The standard free energies of formation of `CH_(3)OH(I), CO_(2)(g) and H_(2)O(I)` are -166.3, -394.4 and -237.1 kJ `mol^(-1)` respectively. The standard free energy change of the reaction will beA. `-597.8` kJ `mol^(-1)`B. `-298.9` kJ `mol^(-1)`C. `-465.2` kJ `mol^(-1)`D. `-702.3` kJ `mol^(-1)`
Answer» Correct Answer - D

Discussion

781.	The standard molar heats of formation of ethane, carbon dioxide, and liquid water are `-21.1, -94.1`, and `-68.3kcal`, respectively. Calculate the standard molar heat of combustion of ethane.A. `-372 kcal`B. `-240 kcal`C. `162 kcal`D. `183.5 kcal`
Answer» `C_(2)H_(6)(g) +(7)/(2) O_(2)(g) rarr 2CO_(2)(g) +3H_(2)O(g)` `DeltaH^(Theta) = 2Delta_(f)H_(CO_(2))^(Theta)+3Delta_(f)H_(H_(2)O)^(Theta)-Delta_(f)H_(C_(2)H_(6))^(Theta)`. `= 2(-94.1) +3(-68.3)-(-21.1) =- 372 kcal`

Discussion

782.	In a fuel cell (device used for producing electricity directly from a chemical reaction )methanol is used as a fuel and oxygen gas is used asan oxidizer. The standard enthalpy of combustion of methanol is `-721kJ mol^(-1)` . The standard free energiesof formation of `CH_(3)OH(l, CO_(2)(g)` and `H_(2)O(l)` are- 166.3 , -394.4 and `- 237 .1 kJ mol^(-1)` respectively. The standard free energy change of the reaction will beA. `-597.8 kJ mol^(-1)`B. `-298 kJmol^(-1)`C. `- 465.2 kJ mol^(-1)`D. `-702.3kJ mol^(-1)`
Answer» Correct Answer - d The reaction is `CH_(3)OH(l) + (3)/(2) O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) ` `Delta_(r)G^(@) = SigmaDelta_(f)G^(@)(` Products) - `SigmaDelta_(f)G^(@)(`Reactants) `= [ Delta_(f)G^(@) ( CO_(2) +2Delta_(f) G^(@) (H_(2)O)] -[Delta_(f)G^(@) (CH_(3)OH)+(3)/(2)Delta_(f)G^(@) (O_(2))]` `= [( - 394.4) + 2 ( - 237.1) ] - [ - 166.3 + (3)/(2) (0)]` `=- 702 . 3 kJ mol^(-1)`

Discussion

783.	Find the heat of formation of ethyl alcohol for following data `C(s) +O_(2)(g) rarr CO_(2)(g) DeltaH =- 94 kcal` `H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l), DeltaH =- 68 kcal` `C_(2)H_(5)OH(l)+3O_(2)(g) rarr 2CO_(2)(g) +3H_(2)O(l) DeltaH =- 327 kcal`
Answer» `2C(s) +3H_(2)(g)+(1)/(2)O_(2)(s) rarr C_(2)H_(5)OH(l) DeltaH = ?` Given: `C(s) +O_(2)(g) rarr CO_(2)(g) ……..(i)` `DeltaH_(1) =- 94 kcal` `H_(2)(s) +(1)/(2)O_(2)(g) rarr H_(2)O(l) ……..(ii)` `DeltaH_(2) =- 68 kcal` `C_(2)H_(5)OH (l)+3O_(2)(g) rarr 2CO_(2)(g) +3H_(2)O(l)......(iii)` `DeltaH_(3) =- 372 kcal` Operate: `2(i) +3(ii) `-(iii) to get `Delta_(f)H = 2(DeltaH_(1))+3(DeltaH_(2)) - DeltaH_(3)` `=(-94) +3(-68) -(-327) = 65 kcal mol^(-1)`

Discussion

784.	Staement -1: The heat absorbed during the isothermal expansion of an ideal gas againt vacuum is zero . Statement -2: The volume occupied by the molecules of an ideal gas is zero.A. statement -1 is a true, statement -2 is true, statement -2 is a correct explanation for statements -1B. statement-1 is true, statement-2 is true, statement-2 Not a correct explanation for statement-1.C. statement-1 is true, statement-2 is falseD. statement-1 is false, statement-2 is true
Answer» Correct Answer - B

Discussion

785.	A cyclic process `ABCD` is shown is shown in the following `P-V` diagram. Which of the following curves represent the same process ? A. B. C. D.
Answer» Correct Answer - B From given `P-V` diagram`ArarrB` is isobaric and `CrarrD` is ischoric so option b is correct.

Discussion

786.	A cyclic process ABCD is shown in the figure P - V diagram. Which of the following curves represent the same process A. B. C. D.
Answer» Correct Answer - A

Discussion

787.	A cyclic process ABCD is shown in the p-V diagram. Which of the following curves represent the same process?
Answer» Correct Answer - C

Discussion

788.	The standed free energy of fromation of NO(g) is 86.6 kj/ mol at 298 K what is the standed free energy of fromation of ` NO_(2) g` at 298 k? `K _(p)= 1.6 xx 10^(12)`A. `R(298) ln (1.6xx10^(12))-86600`B. `86600+R(298) ln (1.6xx10^(12))`C. `86600-(ln (1.6xx10^(12)))/(R(298))`D. `0.5[2xx86, 600-R(298) ln (1.6xx10^(12))`
Answer» Correct Answer - D `DeltaG_(r)^(@)=Sigma DeltaG_(g)^(@)("Products")-Sigma DeltaG_(f)^(@)("reactants")=-RT ln K_(P)` `2DeltaG_(f(NO_(2)))^(@)-[2DeltaG_(f(NO))^(@)+DeltaG_(f(O_(2)))^(@)]` `=-RT ln K_(P)` `2 DeltaG_(f(NO_(2)))^(@)-[2xx86,600+0]` `=-RT ln K_(P)` `DeltaG_(f(NO_(2)))^(@)= 1/2[2xx86600-Rxx298xxln(1.6xx10^(12))]`

Discussion

789.	An ideal gas is expanded form `(P_(1) V_(1) T_(1))` to `(p_(2),V_(2) T_(2)` used different conditions. The correct statement (s) among the following is (are)A. The work done by the gas is less when it is expanded revesibly from `V_(1),(to) V_(2)` under abiabitc conditions as compared to that when expanded reversibly form `v_(1) (to)V_(2)` under isothemal conditions.B. The change in internal energy of the gas is (i) zero , if it is expanded reversibly with `T_(1)= T_(2) ` and (ii) positive if it is expaned reversiby under abiabatic conditions with `T_(1) ne T_(2)`C. If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabaticD. The work done on the gas is maximum when it is compressed irrversibly from `(p_(2)V_(2)) " to " (P_(1), V_(1))` against constant pressen `P_(1)`
Answer» Correct Answer - a,c,d

Discussion

790.	An ideal gas in thermally insulated vessel at internal `(pressure)=P_(1), (volume)=V_(1)` and absolute `temperature = T_(1)` expands irreversiby against zero external, pressure , as shown in the diagram, The final internal pressure, volume and absolute temperature of the gas are `p_(2), V_(2) and T_(2)`, respectively . For this expansion A. `q=0`B. `T_(2)=T_(1)`C. `p_(2)V_(2)= p_(1)V_(1)`D. `P_(2)V_(2)^(gamma) =P_(1)V_(2)^(gamma)`
Answer» Correct Answer - a,b,c

Discussion

791.	An ideal gas in thermally insulated vessel at internal `(pressure)=P_(1), (volume)=V_(1)` and absolute `temperature = T_(1)` expands irreversiby against zero external, pressure , as shown in the diagram, The final internal pressure, volume and absolute temperature of the gas are `p_(2), V_(2) and T_(2)`, respectively . For this expansion A. `q = 0`B. `T_(2) = T_(1)`C. `p_(2)V_(2) = p_(1)V_(1)`D. `p_(2)V_(2)^(gamma) = p_(1)V_(1)^(gamma)`
Answer» Correct Answer - A::B::C

Discussion

792.	A thermally isolated vessel contains `100g` of water at `0^(@)C`. When air above the water is pumped out, some of the water freezes and some evaporates at `0^(@)C` itself. Calculate the mass of the ice formed such that no water is left in the vessel. Latent heat of vaporization of water at `0^(@)C=2.10xx10^(6)J//kg` and latent heat of fusion of ice `=3.36xx10^(5)J//kg`.
Answer» Total mass of the water `=M=100g` Latent heat of vaporization of water at `0^(@)C` `=L_(1)=21.0xx10^(5)J//Kg` & Latent heat of fusion of ice `=L_(2)=3.36xx10^(5)J//kg` Suppose, the mass of the ice formed `=m` Then, the mass of water evaporated `=M-m` Heat lost by the water in freezing`=` Heat taken by water in evaporation. Thus, `mL_(2)=(M-m)L_(1) or m-86 g`

Discussion

793.	Internal enregy change during a reversible is isothermal expansion of an ideal is :-A. Always negativeB. Always positiveC. ZeroD. May be positive or negative
Answer» Correct Answer - C In isothermal process for ideal gas `Delta E = 0`

Discussion

794.	Calculate `Delta_(r)G^(@)` for `(NH_(4)Cl,s)` at 310K. Given :`Delta_(r)H^(@)(NH_(4)Cl,s)` =-314 kj/mol,`Delta_(r)C_(p)=0` `S_(N_(2)(g))^(@)=192 JK^(-1mol^(-1)),S_(H_(2)(g))^(@)=130.5JK^(-1)mol^(-1),` `S_(Cl_(2)(g))^(@)=233JK mol^(-1), S_(NH_(4)Cl(s))^(@)=99.5JK^(-1)mol^(-1)` All given data at 300KA. `-198.56 KJ//"mol"`B. `-426.7 KJ//"mol"`C. `-202.3 KJ//"mol"`D. None of these
Answer» Correct Answer - A `Delta_(f)S^(@) (NH_(4)Cl,s) "at" 300k` `= S_(NH_(4)Cl(s))^(@) - [(1)/(2) S_(N_(2))^(@) + 2S_(H_(2))^(@) + (1)/(2)S_(Cl_(2))^(@)] =- 374 Jk^(-1) mol^(-1)` `Delta_(f)C_(P) =0` `therefore Delta_(f)S_(310)^(@)=Delta_(f)S_(300)^(@)` `=-374JK^(-1) mol^(-1)` `Delta_(f)H_(300)^(@) =Delta_(f)H_(300)^(@) =- 314.5` `Delta_(f)G_(310)^(@) = Delta_(f)H^(@) - 310 DeltaS^(@) =- (310(-374))/(1000) =- 198.56 KJ//"mol"` .

Discussion

795.	Calculate `Delta_(r)G^(@)` for `(NH_(4)Cl,s)` at 310K. Given :`Delta_(r)H^(@)(NH_(4)Cl,s)` =-314 kj/mol,`Delta_(r)C_(p)=0` `S_(N_(2)(g))^(@)=192 JK^(-1mol^(-1)),S_(H_(2)(g))^(@)=130.5JK^(-1)mol^(-1),` `S_(Cl_(2)(g))^(@)=233JK mol^(-1), S_(NH_(4)Cl(s))^(@)=99.5JK^(-1)mol^(-1)` All given data at 300KA. `-198.56kJ//mol`B. `-426.7KJ//mol`C. `-202.3KJ//mol`D. none of these
Answer» Correct Answer - a

Discussion

796.	Show that the reaction `CO(g) +(1//2)O_(2)(g) rarr CO_(2)(g)` at `300K` is spontaneous and exothermic, when the standard entropy change is `-0.094 k J mol^(-1) K^(-1)`. The standard Gibbs free energies of formation for `CO_(2)`and `CO` are `-394.4`and `-137.2kJ mol^(-1)`, respectively.
Answer» Correct Answer - `(-285.4 kJ)`

Discussion

797.	Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` isA. `93 KJ "mol"^(-1)`B. `-245 KJ "mol"^(-1)`C. `-93 KJ "mol"^(-1)`D. `245 KJ "mol"^(-1)`
Answer» Correct Answer - C Given, `DeltaH_(H-H)=434` KJ/mol `DeltaH_(Cl-Cl)=242` KJ/mol `DeltaH_(H-Cl) = 431` KJ/mol `(1)/(2)H_(2) + (1)/(2) Cl_(2) to HCl, DeltaH_(r)=?` `DeltaH_(r)=(1)/(2)xxDeltaH_(H-H)+(1)/(2)xxDeltaH_(Cl-Cl)-DeltaH_(H-Cl)` `=(1)/(2)xx434 + (1)/(2)xx 242 - 431` `=217xx121 - 431 =- 93` KJ/mol

Discussion

798.	Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` isA. `93 kJ mol^(-1)`B. `- 245 kJ mol^(-1)`C. `- 93 kJ mol^(-1)`D. `245 kJ mol^(-1)`
Answer» Correct Answer - C The enthalpy of formation of `HCl` is the enthalpy change for the following reaction: `(1)/(2) H_(2) (g) + (1)/(2) Cl_(2) (g) rarr HCl (g),` `Delta_(r) H^(@) = ?` That is, one mole of `HCl` is synthesized in its standard state from tis elements in their standard states. According to Eq., we have `Delta_(r) H^(@) = sum "Bond enthalpies"_("reactants") - sum "Bond enthalies"_("products")` `((1)/(2) Delta_(H - H) H^(@) + (1)/(2) Delta_(Cl - Cl) H^(@)) - Delta_(H - Cl) H^(@)` `= (1)/(2) (434) + (1)/(2) (242) - (431)` `= (271) + (121) - (431) = - 93 kJ mol^(-1)`

Discussion

799.	The initial state of one mole of monoatomic ideal gas is P = 8 atm, T = 400 K. Calculate the change of entropy ifa. Gas expands isothermally and reversibly to a pressure of 1 atm.b. Gas decreases its pressure at constant volume to a pressure of 5 atm.C. Gas expands adiabatically and irreversibly to the pressure of 2 atm while doing work of 400 joules.
Answer» We have given, initial pressure of monoatomic gas = 8 atm initial temperature T = 400k (a) ΔS= Cp ln T₂/T₁ - Rln P₂/P₁ for isothermal process \(\triangle\)T = 0 i.e T₂= T₁ \(\therefore\)\(\triangle\)s = -Rln P₂/P₁ \(\triangle\)s = -8.314 ln 1/8 \(\triangle\)s = -8.314 x (-2.1) \(\triangle\)s = 17.3 Jk^-1 (b) when, gas decreases its pressure at constant volume to a pressure of 5 atm, there will be also change in temperature. \(\therefore\) final temperature T₂ = \(\frac{P_2\times T_1}{P_1}\) = \(\frac{5atm\times400k}8\) = 250 k \(\therefore\) \(\triangle\)s = C_p ln T₂/T₁ - Rln P₂/P₁ = 5R/2 ln 250/400 - Rln(5/8) (for monoatomic gas C_p = 5R/2) \(\triangle\)s = 5/2 x 8.314 x ln(250/400) - 8.314 ln(5/8) = -9.77 jk^-1- (-3.91Jk^-1) \(\triangle\)s = -5.86 Jk^-1 (c) \(\triangle\)u = q + w for adiabatre process - q = 0 \(\triangle\)u = w (w = -ve, expression) C_v x (T₂ - T₁) = -400 3/2 R(T₂ - 400k) = -400 (T₂ - 400k) = \(\frac{-400\times2}{3\times8.314}\) T₂ = 368 k \(\triangle\)s = C_pln T₂/T₁ - Rln P₂/P₁ = \(\frac52\) 8.314 ln 368/400 - 8.314 ln2/8 = (-1.73) - (-11.520) \(\triangle\)s = 9.8 Jk^-1

799.

The initial state of one mole of monoatomic ideal gas is P = 8 atm, T = 400 K. Calculate the change of entropy ifa. Gas expands isothermally and reversibly to a pressure of 1 atm.b. Gas decreases its pressure at constant volume to a pressure of 5 atm.C. Gas expands adiabatically and irreversibly to the pressure of 2 atm while doing work of 400 joules.

Answer»

We have given,

initial pressure of monoatomic gas = 8 atm

initial temperature T = 400k

(a) ΔS= Cp ln T₂/T₁ - Rln P₂/P₁

for isothermal process \(\triangle\)T = 0 i.e T₂= T₁

\(\therefore\)\(\triangle\)s = -Rln P₂/P₁

\(\triangle\)s = -8.314 ln 1/8

\(\triangle\)s = -8.314 x (-2.1)

\(\triangle\)s = 17.3 Jk^-1

(b) when, gas decreases its pressure at constant volume to a pressure of 5 atm, there will be also change in temperature.

\(\therefore\) final temperature T₂ = \(\frac{P_2\times T_1}{P_1}\)

= \(\frac{5atm\times400k}8\) = 250 k

\(\therefore\) \(\triangle\)s = C_p ln T₂/T₁ - Rln P₂/P₁

= 5R/2 ln 250/400 - Rln(5/8)

(for monoatomic gas C_p = 5R/2)

\(\triangle\)s = 5/2 x 8.314 x ln(250/400) - 8.314 ln(5/8)

= -9.77 jk^-1- (-3.91Jk^-1)

\(\triangle\)s = -5.86 Jk^-1

(c) \(\triangle\)u = q + w

for adiabatre process - q = 0

\(\triangle\)u = w (w = -ve, expression)

C_v x (T₂ - T₁) = -400

3/2 R(T₂ - 400k) = -400

(T₂ - 400k) = \(\frac{-400\times2}{3\times8.314}\)

T₂ = 368 k

\(\triangle\)s = C_pln T₂/T₁ - Rln P₂/P₁

= \(\frac52\) 8.314 ln 368/400 - 8.314 ln2/8

= (-1.73) - (-11.520)

\(\triangle\)s = 9.8 Jk^-1

Discussion

800.	The efficiency of a Carnot engine working between TH = 400 K and TC = 300 K is (A) 75% (B) 25% (C) 1/3 (D) 4/7.
Answer» Correct option is (B) 25%

Discussion

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