540 + Interview Questions in CHEMICAL-THERMODYNAMICS IN GENERAL AWARENESS Page 2 InterviewSolution

51.	The gas with the highest heat of combustion is `:`A. MethaneB. EthaneC. EtheneD. Ethyne
Answer» Correct Answer - B Gas with highest number of `C` atoms and having all single bonds should have highest value of `DeltaH_(comb)`.

Discussion

52.	One gram sample of `NH_(4)NO_(3)` is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by `6.12K` . The heat capacity of the system is `1.23KJ//g//deg` . What is the molar heat of decomposition for `NH_(4)NO_(3)` ?A. `-7.53KJ//mol`B. `-398.1KJ//mol`C. `-16.1KJ//mol`D. `-602KJ//mol`
Answer» Correct Answer - B Molecular weight of `NH_(4)NO_(3)=80` Heat evolved `=1.23xx6.12` `:.` Molar heat capacity `=1.23xx6.12xxC`

Discussion

53.	the sing of `DeltaG` for the process of melting of ice at 273 K and 1 atm pressurme isA. positiveB. negaitiveC. neither negative nor positiveD. either negative or positive
Answer» Correct Answer - c At 273 K water and ice are in equillibrium , hence` DeltaG=0`

Discussion

54.	The sign of `DeltaG` for the process of melting of ice at `273K` and `1` atm pressure isA. positiveB. negativeC. neither positive nor negativeD. either positive nor negative
Answer» Correct Answer - C Melting of ice is spontaneous above `273 K` at one atm pressure. At `273 K` and `1` atm, there is equilibrium `H_(2)O(s)hArrH_(2)O(l)` Hence, `DeltaG=0`

Discussion

55.	`DeltaG`, in process of melting of ice at `-15^(@)C`,isA. `DeltaG=-ve`B. `DeltaG=+ve`C. `DeltaG=0`D. all of these
Answer» Correct Answer - b becouse melting of ice at `-15^(@)C` is a non - spontaneous process.

Discussion

56.	For a certain reaction `X`rarr`Y` the value of `DeltaH` and `DeltaS` are `50.50KJ mol^(-1)` and `100.03JK^(-1)` respectively. The temperature at which `DeltaG=0` isA. `505^(@)C`B. `232^(@)C`C. `252^(@)C`D. `450^(@)C`
Answer» Correct Answer - A Given, `DeltaH=50.50KJ mol^(-1)` `DeltaS=100JK^(-1)=100xx10^(-3)=0.1KJK^(-1)` `:.` From `DeltaG=DeltaH-TDeltaS` `implies 0=DeltaH-TDeltaS` `implies T=(DeltaH)/(DeltaS)=(50.50)/(0.1)=505K` `:. T=505-273=232^(@)C`

Discussion

57.	Which of the following processes process towards more disordered state? `I` Stretching the rubberII. Sublimation of dry ice `III` Crystallisation of salt from solution `IV` . Dissolution of sugar from solutionA. `I,II,IV`B. `I,III`C. `III,IV`D. `II,IV`
Answer» Correct Answer - B In `I` and `IV` , entropy increaes due to increases in randomness.

Discussion

58.	A reaction has `DeltaH=-33KJ` and `DeltaS=-58(J)/(H)` . This reaction would be:A. spontanceous at all temperaturesB. non-spontaneous at all temperaturesC. spontaneous above a certain temperatureD. spontaneous below a certain temperature
Answer» Correct Answer - B `DeltaG=underset(-ve)undersetdarr((DeltaH))-Tunderset(-ve)undersetdarr((DeltaS))` Since both are `-ve` , the reaction would have a `-veDeltaG` below a temperature of `(33000)/(58)K(=569K)` .

Discussion

59.	From the following bond energies `H-H` bond energy `431.37 kJ mol^(-1)` `C=C` bond energy `606.10 kJ mol^(-1)` `C-C` bond energy `336.49 kJ mol^(-1)` `C-H` bond energy `410.5 kJ mol^(-1)` Enthalpy for the reaction `overset(H)overset(\|)underset(H)underset(\|)(C )=overset(H)overset(\|)underset(H)underset(\|)(C )+H-HtoH-overset(H)overset(\|)underset(H)underset(\|)(C )-overset(H)overset(\|)underset(H)underset(\|)(C )-H` will beA. `553.6 kJ mol^(-1)`B. `1573.6 kJ mol^(-1)`C. `-243.6 kJ mol^(-1)`D. `-120.0 kJ mol^(-1)`
Answer» Correct Answer - D `DeltaH-"reaction"=sum"Bond energies of reactants"-sum "Bond energies of products"` `=[BE(C=C+4xxBE(C-H)+BE(H-H)]-[BE(C-C)+6xxBE(C-H)]` `:. [606.10+1642.00+431.37]-[336.49+2463.49]` `=2679.47-2799.40=-120 kJ mol^(-1)`

Discussion

60.	From the following bond energies: `H--H` bond energy: `431.37KJmol^(-1)` `C=C` bond energy: `606.10KJmol^(-1)` `C--C` bond energy: `336.49KJmol^(-1)` `C--H` bond energy: `410.50KJmol^(-1)` Enthalpy for the reaction will be: `overset(H)overset(\|)underset(H)underset(\|)(C)=overset(H)overset(\|)underset(H)underset(\|)(C)+H-H to Hoverset(H)overset(\|)underset(H)underset(\|)(C)-overset(H)overset(\|)underset(H)underset(\|)(C)-H`A. `553.0KJmol^(-1)`B. `1523.6KJmol^(-1)`C. `-243.6KJmol^(-1)`D. `-120.0KJmol^(-1)`
Answer» Correct Answer - B For reaction `DeltaH_(r)=-[4xxBE_(C-H)+1BE_(H=-H)]` `-[6xxBE_(C-H)+1xxBE_(C-C)]` `=(4xx410.50+1xx606.10+1xx431.37)` `-[(6xx410.50)+(1xx336.49)]` `=-120.0KJmol^(-1)`

Discussion

61.	What percentage `T_(1)` is of `T_(2)` for a `10%` efficiency of a heat engine? `T_(1)` is the temperature of sink and `T_(2)` is the temperature of heat reservoir.A. `T_(1)=90%` of `T_(2)`B. `T_(1)=T_(2)`C. `T_(2)=90%` of `T_(1)`D. `T_(1)=50%` of `T_(2)`
Answer» Correct Answer - C Efficiency, `eta=(T_(2)-T_(1))/(T_(2)=1-(T_(1))/(T_(2))=0.1` or `T_(1)=0.9T_(2)`

Discussion

62.	The difference between `DeltaH` and `Delta E` at constant voluem is equal toA. `R`B. `pDeltaV`C. `Vdeltap`D. `(3)/(2)R`.
Answer» Correct Answer - C `H=E+PV` Taking `Delta` on both the sides `DeltaH=DeltaE+Delta(PV)=DeltaE+PDeltaV+VDeltaP` At constant volume, `DeltaV=0` `:. DeltaH=DeltaE+VDeltaP` or `DeltaH-DeltaE=VDeltaP`

Discussion

63.	When ammonium chloride is dissolved in water, the solution becomes cold. The change isA. EndothermicB. ExothermicC. SupercoolingD. None of the above
Answer» Correct Answer - A Endothermic because heat is absorbed.

Discussion

64.	Warming ammonium chloride with sodium hydroxide in a test tube is an example of :A. Closed systemB. Isolated systemC. Open systemD. None of these
Answer» Correct Answer - D An open system exchanges the mass and energy both.

Discussion

65.	Heat of formation, `Delta H_(f)^(@)` of an explosive compound like `NCl_(3)` is -A. positiveB. negativeC. zeroD. positive or negative
Answer» Correct Answer - A Since `DeltaH_(f)^(@)` is positive, it has greater heat content and is explosive.

Discussion

66.	when ammonium chloride is dissoved in water , the sloution becomes cold. The change isA. endothermicB. exothermicC. supercoolingD. none of these
Answer» Correct Answer - a the process is endothermic and takes up takes up heat from solution so that the solution becomes cold.

Discussion

67.	`S_("rhombic")+O_(2)(g)toSO_(2)(g), DeltaH = -297 .5 kcal` `S_("monoclinic")+ O_(2)(g) to SO_(2)(g) , DeltaH = -300 kJ`A. rhombic sulphur is yellow in colorB. monoclinic sulphur has metallic lustreC. monoclinic sulphur is more stableD. `DeltaH_("translition")of S_(R) of S_(M)` is endothermic
Answer» Correct Answer - d subtracting second equation from first , we get `S_((R))toS_((M))DeltaH=+2.5kJ`, endothermic

Discussion

68.	At `0^(@)C` , ice and water are in reuilibrium and `DeltaS` and `DeltaG` for the conversion of ice to liquid water isA. `0,21.98JK^(-1)mol^(-1)`B. `-ve` , zeroC. `-ve,21.98JK^(-1)mol^(-1)`D. Zero, zero
Answer» Correct Answer - C Since the given process is in equilibrium, i.e., `DeltaG=0` `DeltaG=DeltaH-TDeltaS` or, `0=Delta-TDeltaS` or `DeltaH=TDeltaS` or, `DeltaS=(DeltaH)/(T)=(6)/(273)xx10^(3)JK^(-1)K^(-1)` `=21.98JK^(-1)mol^(-1)`

Discussion

69.	Combustion of sucrose is used by aerobic oranisms for providing energy for the life sustaining processes. If all the capturing of energy from reaction is done through electrical process (non `P-V` work), then calculate maximum available energy which can be captured by combustion of `34.2g` of sucrose. Given, `DeltaH_("combustion")("surrose") =-6000KJ mol^(-1)` `DeltaS_("combustion")=180J//K-mol` and body temperatuire is `300K`A. `600KJ`B. `594.6KJ`C. `5.4KJ`D. `605.4KJ`
Answer» Correct Answer - D No. of moles of sucrose `=(34.2)/(342)=0.1` `-(DeltaG)T,P` =useful work done by the system `-DeltaG=-DeltaH+TDeltaS=+(6000xx0.1)+(180xx0.1xx300)/(1000)` `=605.4KJ`

Discussion

70.	At `0^(@)C` , ice and water are in equilibrium and `DeltaS` and `DeltaG` for the conversion of ice to liquid water isA. `0,21.98JK^(-1)mol^(-1)`B. `-ve` , zeroC. `+ve,21.98JKk^(-1)mol^(-1)`D. All of these
Answer» Correct Answer - D Since the given process is in equilibrium, i.e., `DeltaG=0` `DeltaG=DeltaH-TDeltaS` or, `0=DeltaH-TDeltaS` or `DeltaH=TDeltaS` or, `Delta=(DeltaH)/(T)=(6)/(273)xx10^(3)` `JK^(-1)mol^(-1)=21.98JK^(-1)mol^(-1)`

Discussion

71.	The standard heat of formation of `U_(3)O_(8)` is -853.5 kcl//mol and the standard heat of the reaction `3UO_(2) +O_(2) rarr U_(3)O_(8)` is -76.01 Kcal. The standard heat of formation of `UO_(2)` isA. `-1083KJmol^(-1)`B. `-108.3KJmol^(-1)`C. `-10.83KJmol^(-1)`D. `-1.083KJmol^(-1)`
Answer» Correct Answer - C `3U O_(2) + O_(2) rarr U_(3)O_(8)` `Delta H^(@) = Delta H_(f)^(@) (U_(3)O_(8)) - Delta H_(f)^(@) (O_(2)) - 3 Delta H_(f)^(@) (U O_(2))` `implies 3DeltaH_(f)^(@) (UO_(2))=-DeltaH^(@) + DeltaH_(f)^(@)(U_(3)O_(8)) - DeltaH_(f)^(@)(O_(2))` or `Delta H_(f)^(@) (UO_(2)) = (-(-76.0) + (-583.5)-0)/(3)` `=-259.17` Kcal `=-1083.33` KJ/mol

Discussion

72.	Heat of neutralisation is least whenA. `NaOH` is neutralised by `CH_(3)COOH`B. `NaOH` is neutralised by `HNO_(3)`C. `NaOH` is neutralised by `HCL`D. `NaOH` is neutralised by `H_(2)SO_(4)`
Answer» Correct Answer - A `Delta H` is least when weak acid or base in neutralized by weak base or acid, ww

Discussion

73.	For the raction `B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(2)(s)+3H_(2)O(l)` `DeltaE=-2143.2KJ` Calculate `DeltaH` for the reaction at `25^(@)C`A. `-2148.2KJmol^(-1)`B. `-2138.6KJmol^(-1)`C. `-2133.2KJmol^(-1)`D. `-2143.2KJmol^(-1)`
Answer» Correct Answer - C For `B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(l)` `DeltaE=-2143.2KJ` `Deltan=0-2=-2` Now from `DeltaH=DeltaE+DeltanRT` `-2143.2+(-2)xx8.314xx10^(-3)xx298` `=-2148.2KJ mol^(-1)`

Discussion

74.	Standard enthalpy and standard entropy changes for the oxidation of ammonia at `298 K` are `-382.64 kJ mol^(-1)` and `-145.6 jK^(-1)mol^(-1)` respectively. Standard Gibbs energy change for the same reaction at `298 K` isA. `-523.2 kJ mol^(-1)`B. `-22.1 kJ mol^(-1)`C. `-339.3 kJ mol^(-1)`D. `-439.3 kJ mol^(-1)`
Answer» Correct Answer - C `NH_(3)(g)+(3)/(4)O_(2)(g)to(1)/(2)N_(2)(g)+(3)/(2)H_(2)O(l)` `T=298 K` `DeltaH=-382.64 kJ mol^(-1)` `DeltaS=-145.6 JK^(-1) mol^(-1)` `=-0.1456 kJ K^(-1) mol^(-1)` `DeltaG=DeltaH-TDeltaS` `=-382.64 kJ mol^(-1)-(298 K)xx(0.1456 k JK^(-1)mol^(-1))` `=-382.64 kJ mol^(-1)+43.3888 kJ mol^(-1)` `=-339.25 kJ mol^(-1)`

Discussion

75.	Standard enthalpy and standard entropy change for the oxidation of `NH_(3)` at `298K` are `-382.64KJ mol^(-1)` and `145.6Jmol^(-1)` respectively. Standard free energy change for the same reaction at `298K` isA. `-221.1KJmol^(-1)`B. `-339.3KJmol^(-1)`C. `-439.3KJmol^(-1)`D. `-523.2KJmol^(-1)`
Answer» Correct Answer - A `DeltaH^(@)=-382.64KJmol^(-1)` Given that, `DeltaS^(@)=-145.6JK^(-1)mol^(-1)` `=-145.6xx10^(-3)KJK^(-1)` or `DeltaG^(@)=-382.64-(298xx1-145.6xx10^(3))` `=-339.3KJmol^(-1)` .

Discussion

76.	For the reaction, `X_(2)O_(4)(l)rarr2XO_(2)(g),DeltaE=2.1Kcal` , `DeltaS=20cal//K` at `300K` . Hence `DeltaG` isA. `2.7Kcal`B. `-2.7Kcal`C. `9.3Kcal`D. `-9.3Kcal`
Answer» Correct Answer - B From `DeltaH=DeltaE+DeltanRT` `DeltaH=2.1+2xx(2xx10^(-3))xx300` `3.3` Now from `DeltaG=DeltaH-TDeltaS` `=3.3-300xx(20xx10^(-3))` `=-2.7Kcal`

Discussion

77.	Molar heat capacity of water in equilibrium with ice at constant pressure isA. ZeroB. Infinity `(oo)`C. `40.45KJK^(-1)mol^(-1)`D. `75.48JK^(-1)mol^(-1)`
Answer» Correct Answer - C By definition `C_(p,m)=(dq_(p))/(dT)` For `h_(2)O(l)` Temperature does not change if some heat is given to the system. Hence, `C_(p,m)=(+ve)/(zero)=oo`

Discussion

78.	The molar heat capacity of water at constant pressure, `C_(P)` , is `75 JK^(-1) mol^(-1)`. When `1.0 kJ` of heat is supplied to `100 g` of water which is free to expand, the increase in temperature of water isA. `6.6 K`B. `1.2 K`C. `2.4 K`D. `4.8 K`
Answer» Correct Answer - C `Q=1 kJ=1000 J` `m=100 g=(100g)/(18g mol^(-1))=5.55 mol` `Q=mC.Delta t` `Delta t=(Q)/(m.C.)=(1000J)/(5.55molxx75 JK^(-1)mol^(-1))` `=2.4 K`

Discussion

79.	If for an ideal gas, the ratio of pressure and volume is constant and is equal to 1 atm `L^(-1)` , the molar heat capacity at constant pressure would beA. `(3)/(2)R`B. `2R`C. `(5)/(2)R`D. zero
Answer» Correct Answer - C By definition, `H=E+PV` `((dH)/(dT))_(P)=((dE)/(dT))_(P)+P((dV)/(dT))_(p)=C_(p),m` For the given ideal gas, we will have `PV=RT` or `V^(2)=RT` or `2V((dV)/(dT))_(P)=R` or `((dV)/(dT))_(P)=(R)/(2V)` `E=(3RT)/(2)` or `((dE)/(dT))_(P)=(3)/(2)R=C_(v),m` `C_(P),m=(3)/(2)R+Pxx(R)/(2V)=((3)/(2)+(1)/(2))R=2R`

Discussion

80.	Heat exchanged in a chemical reaction at constant temperature and pressure is calledA. entropy changeB. enthalpy changeC. internal energy changeD. free energy change
Answer» Correct Answer - B Enthalpy change is the heat exchanged at constant tempertaure and pressure.

Discussion

81.	The law formulated by Nernst isA. First law of thermodynamicsB. Second law of thermodynamicsC. Third law of thermodynamicsD. Both `A` and `B`
Answer» Correct Answer - C Third law of thermodynamics was put forward by Nernst.

Discussion

82.	For a melting of a solid at `25^(@)C`, the fusion process requires energy equivalent to` 2906`joules o be added to system considering the process to be reversible at fusion point, the entropy change of the process isA. `9.75 JK^(-1)`B. `11.272 JK^(-1)`C. `2.33 JK^(-1)`D. insufficient data
Answer» Correct Answer - A `DeltaS=(2906)/(298)=9.75jK^(-1)`

Discussion

83.	A natural gas may be assumed to be a mixture of methane and ethane only. On complete combustion of `10L` of gas at `STP` the heat evolved was `474.6 kJ`. Assuming `Delta_(comb) H^(Theta) CH_(4)(g) =- 894 kJ mol^(-1)` and `Delta_(comb) H^(Theta) C_(2)H_(6)(g) =- 1500 kJ mol^(-1)` . So, find composition of the mixture by volume.
Answer» x litre `toCH_(4)," mole of "CH_(4)=x//22.4` `(10-x)litre to C_(2)H_(6)," mole of "C_(2)H_(6)=(10-x)//22.4` Heat evolved `=(x)/(22.4)xx894+((10-x))/(22.4)xx1500` `474.6=(x)/(22.4)xx894+((10-x))/(22.4)xx1500` `x=0.745,%CH_(4)=74.5%`

Discussion

84.	The difference between `Delta H` and `Delta U` for the combustion of methane at `27^(@)C` will be (in `J mol^(-1)`)A. `8.314xx27xx(-3)`B. `8.314xx300xx(-3)`C. `8.314xx(300)xx(-2)`D. `8.314xx300xx1`
Answer» Correct Answer - C `CH_(4)(g)+2O_(2)toCO_(2)(g)+2H_(2)O(l)` `Deltan=1-3=(-2)` ` :. DeltaH-DeltaE=Deltan_(g)RT=(-2)(8.314)(300)`

Discussion

85.	A process is spantaneous at a given temperature ifA. `DeltaH=0`, `DeltaS lt 0`B. `DeltaHgt0`, `DeltaS lt 0`C. `DeltaHlt0`, `DeltaS gt 0`D. `DeltaHgt0`, `DeltaS = 0`
Answer» Correct Answer - C `(A)` Energy has no rule, entropy decreases non spontaneous `(B)` Energy factor opposes, entropy factor also opposes- non spontaneous `(C )` Energy factor favours, entropy factor also opposes-spontaneous `(D)` Energy factor opposes, entropy factor has no role- non spontaneous

Discussion

86.	Which relations among the following is/are correct ?A. `G=H-TS`B. `PV=H-V`C. `w=DeltaW-q`D. `qv-qp=-DeltanRT`
Answer» Correct Answer - A::C::D `(A)`, `(C )`, `(D)` represent correct expressions.

Discussion

87.	The bond dissociation energies for `Cl_(2)`, `I_(2)` and `IC l` are `242.3`, `151.0` and `211.3 kJ//"mole"` respectively. The enthalpy of sublimation of iodine is `62.8 kJ //"mole"`. What is the standard enthalpy of formation of `ICI(g)` nearly equal toA. `-211.3 kJ//"mole"`B. `-14.6 kJ//"mole"`C. `16.8 kJ//"mole"`D. `33.5 kJ//"mole"`
Answer» Correct Answer - C We aim at `(1)/(2)I_(2)(s)+(1)/(2)Cl_(2)(g)toICl(g)` We are given `(i) Cl_(2)(g)to2Cl(g)`, `DeltaH=242.3 kJ` `(ii) I_(2)(g)to2I(g)`, `DeltaH=151 kJ` `(iii) Icl(g)toI(g)+Cl(g)`, `DeltaH=211.3 kJ` `(iv) I_(2)(s)toI_(2)(g)`, `DeltaH=62.8kJ` `(1)/(2)(ii)+(1)/(2)(i)-(iii)+(1)/(2)(iv)` gives the required equation `DeltaH=(1)/(2)(151)+(1)(2)(242.3)-211.3+(1)(2)(62.8)` `=16.75 kJ`

Discussion

88.	The sublimation energy of `I_(2)` (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole. The enthalpy of vapourisation of `I_(2)` isA. `41.8 kJ //mol`B. `-41.8 kJ //mol`C. `72.8 kJ //mol`D. `-72.8 kJ //mol`
Answer» Correct Answer - A `I_(2)(s)overset(Delta_("sub")H)(to)I_(2)(g)` `:. Delta_("sub")H=Delta_(f)H+Delta_(vap)H` `:. 57.3=15.5+Delta_(vap)H` `Delta_(vap)H=41.8 kJ//mol`

Discussion

89.	For vaporization of water at 1 atmospheric pressure the values of `DeltaH` and `DeltaS` are `40.63KJmol^(-1)` and `108JK^(-1)mol^(-1)` , respectively. The temperature when Gibbs energy change `(DeltaG)` for this transformation will be zero isA. `273.4K`B. `393.4K`C. `373.4K`D. `293.4K`
Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` `Delta=0` , `:. DeltaH=TDeltaS` `T=(DeltaH)/(DeltaS)=(40.63xx10^(3))/(108.8)=373.4K`

Discussion

90.	The molar heat of formation of `NH_(4)NO_(3)(s)` is `-367.5kJ ` and those of `N_(2)O(g)` and `H_(2)O(l)` are `+81.46kJ and -285.78kJ` respectively at `25^(@)C` and 1 atmospheric pressure. Calculate the `DeltaH` and `DeltaU` for the reaction, `NH_(4)NO_(3)(s)toN_(2)O(g)+2H_(2)O(l)`
Answer» `DeltaH^(@)=DeltaH_(f("products"))^(@)-DeltaH_(f("reactants"))^(@)` `=[DeltaH_(f(N_(2)O))^(@)+2xxDeltaH_(f(H_(2)O))^(@)]-[DeltaH_(f(NH_(4)NO_(3)))^(@)]` `=81.46+2xx(-285.78)-(-367.5)` `=81.46-571.56+367.5` `=-122.56kJ` We know that `DeltaH=DeltaU+DeltanRT` or `DeltaU=DeltaH-DeltanRT` `Deltan=1,R=8.314xx10^(-3)kJ" "mol^(-1)K^(-1),T=298K` `DeltaU=-122.56-(1)(8.314xx10^(-3))(298)` `=-122.56-2.477` `=-125.037kJ`

Discussion

91.	Calculate the enthalpy of formation of `Delta_(f)H` for `C_(2)H_(5)OH` from tabulated data and its heat of combustion as represented by the following equaitons: i. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH^(Theta) =- 241.8 kJ mol^(-1)` ii. `C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393.5kJ mol^(-1)` iii. `C_(2)H_(5)OH (l) +3O_(2)(g) rarr 3H_(2)O(g) + 2CO_(2)(g), DeltaH^(Theta) =- 1234.7kJ mol^(-1)` a. `-2747.1 kJ mol^(-1)` b. `-277.7 kJ mol^(-1)` c. `277.7 kJ mol^(-1)` d. `2747.1 kJ mol^(-1)`A. `-2747.1kJ" "mol^(-1)`B. `-277.7kJ" "mol^(-1)`C. `277.7kJ" "mol^(-1)`D. `2747.1kJ" "mol^(-1)`
Answer» Correct Answer - B Required equation: `2C(s)+3H_(2)(g)+1//2O_(2)(g)toC_(2)H_(5)OH(l)` `2C(s)+2O_(2)(g)to2CO_(2)(g),DeltaH^(@)=-2xx393.5kJ` `3H_(2)(g)+(3)/(2)O_(2)(g)to3H_(2)O(g),DeltaH^(@)=3xx241.8kJ` `underline(3H_(2)O(l)+2CO_(2)(g)toC_(2)H_(5)OH(l)+3O_(2)(g)," "DeltaH=1234.7kJ)` On adding `underline(2C(s)+3H_(2)(g)+(1)/(2)O_(2)(g)toC_(2)H_(5)OH(l)," "DeltaH^(@)=-277.7kJ" "mol^(-1))`

Discussion

92.	What would be the heat released when an aqueous solution containing `0.5 mol` if `HNO_(3)` is mixed with `0.3` mol of `OH^(-1)` (enthalpy of neutralisation is `-57.1 kJ`)A. `28.5 kJ`B. `17.1 kJ`C. `45.7 kJ`D. `1.7 kJ`
Answer» Correct Answer - B `0.5 mol HNO_(3)-=0.5 molH^(+)` `H^(+)+OH^(-)toH_(2)O`, `DeltaH=-57.1 kJ` Here `OH^(-)=0.3 mol` Thus, `OH^(-)` is the limiting reactant When `1` mol of `OH^(-)` ions are neutralised heat released` = 57.1 kJ` `0.3` mol of `OH^(-)` ions are neutralised heat released `=57.1xx0.3=17.13 kJ`

Discussion

93.	What would be the heat released when an aqueous solution containing `0.5 mol` if `HNO_(3)` is mixed with `0.3` mol of `OH^(-1)` (enthalpy of neutralisation is `-57.1 kJ`)A. 28.5 kJB. 17.1 kJC. 45.7 kJD. 1.7 kJ
Answer» Correct Answer - B 0.3 mole `OH^(-)` ion will be completely neutralised, `thereforeDeltaH=-57.1xx0.3=-17.13kJ`

Discussion

94.	The standard enthalpy of formation `(Delta_(f)H^(@))` at `298K ` for methane `(CH_(4(g)))` is `-74.8kJ mol^(-1)`. The additional information required to determine the average energy for `C-H` bond formation would be `:`A. The dissociation energy of hydrogen molecule, `H_(2)`B. The dissociation energy of `H_(2)` and enthlpy of sublimation of carbonC. Latent heat vaporisation of methaneD. The first four ionisation energies of carbon and electron gain enthalpy of hydrogen
Answer» Correct Answer - B `C+2H_(2)rarrCH_(4)` `Delta_(f)H^(@)=[e_(C(Srarrg))+2xxe_(H-H)-4xxe_(C-H)]`

Discussion

95.	If a `298K` the bond energies of `C-H,C-C,C=C` and `H-H` bonds are respectivly `414, 347, 615KJmol^(-1)` , the vlaue of enthalpy change for the reaction `H_(2)C=CH_(2)(g)+H_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` at `298K` will beA. `+250 kJ`B. `-250 kJ`C. `+125 kJ`D. `-125 kJ`
Answer» Correct Answer - D `DeltaH=sumB.E."of reactants"-sumB.E."of products"` `=B.E.(C=C)+4xxB.E.(C-H)+B.E.(H-H)-B.E.(C-C)-6xxB.E.(C-H)` `=B.E.(C=C)+B.E.(H-H)-B.E.(C-C)-2xxB.E.(C-H)` `=615+435-347-2xx414=1050-1175` `=-125 kJ`

Discussion

96.	If a `298K` the bond energies of `C-H,C-C,C=C` and `H-H` bonds are respectivly `414, 347, 615KJmol^(-1)` , the vlaue of enthalpy change for the reaction `H_(2)C=CH_(2)(g)+H_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` at `298K` will beA. `+250KJ`B. `-250KJ`C. `+125KJ`D. `-125KJ`
Answer» Correct Answer - B `CH_(2)=CH_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` `4E_(C-H)implies 414xx4=16566E_(C-H) implies 414xx6=2484` `1E_(C=C) implies 615xx1=61E_(C-C) implies 347xx1=347` `1E_(H-H) implies 435xx1=435` `4DeltaH_(C-H)+DeltaH_(C=C)+DeltaH_(H-H)=2706rarr` `6DeltaH_(C-H)+DeltaH_(C-C)=2831` `DeltaH=2706-2831=-125KJ`

Discussion

97.	The standard enthalpy of formation of `C_(2)H_(4)(l)`, `CO_(2)(g)` and `H_(2)O(l)` are `52`, `394` and `-286 kJ mol^(-1)` respectively. Then the amount of heat evolved by burning `7g` of `C_(2)H_(4)(g)` isA. `1412 kJ`B. `9884 kJ`C. `353 kJ`D. `706 kJ`
Answer» Correct Answer - C Desired equation `C_(2)H_(4)(g)+3O_(2)to2CO_(2)+2H_(2)O` , `DeltaH=?` `DeltaH=[2DeltaH_(f)^(@)(CO_(2))+2DeltaH_(f)^(@)(H_(2)O)-DeltaH_(f)^(@)(C_(2)H_(2))-3DeltaH_(f)^(@)(O_(2))]` `=2(-394)+2(-286)-52-0=1412 kJ mol^(-1)` or `1` mol (287) of `C_(2)H_(4)` will give heat `=1412 kJ` `:. 7 g` of `C_(2)H_(4)` will give `=(1412)/(28)xx7=353 kJ`

Discussion

98.	The enthalpy of formation for `C_(2)H_(4)(g)`, `CO_(2)(g)` and `H_(2)O(l)` at `25^(@)C` and `1` atm. Pressure be `52`, `-394` and `-286 kJ mol^(-1)` respectively. The enthalpy of combustion of `C_(2)H_(4)(g)` will beA. `+1412 kJ mol^(-1)`B. `-1412 kJ mol^(-1)`C. `+141.2 kJ mol^(-1)`D. `-141.2 kJ mol^(-1)`
Answer» Correct Answer - B `C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O` , `DeltaH=2(-286)+2(-394)-(52)=-1412kJ`

Discussion

99.	In the reaction equilibrium `N_(2)O_(4) hArr 2NO_(2)(g)` When `5` mol of each is taken and the temperature is kept at `298 K`, the total pressure was found to be `20` bar. Given : `Delta_(f)G_(n_(2)O_(4))^(ɵ)=100 kJ, Delta_(f)G_(NO_(2))^(ɵ)=50 KJ` a. Find `DeltaG` of the reaction at `298 K`. b. Find the direction of the reaction.
Answer» The reaction is: `N_(2)O_(4)(g)hArr2NO_(2)(g)` Since, number of moles of both `N_(2)O_(4)` and `NO_(2)` are same hence their partial pressure will also be same. `P_(N_(2)O_(4))=P_(NO_(2))=(20)/(2)=10` bar `Q_(P)=([P_(NO_(2))]^(2))/([P_(N_(2)O_(4))])=(10^(2))/(10)=10`bar `DeltaG_("reaction")^(@)=2DeltaG_(f)^(@)NO_(2)-DeltaG_(f)^(@)N_(2)O_(4)` `=2xx50-100=0` We know that, `DeltaG=DeltaG^(@)-2.303RT" "log" "Q` `=0-2.303xx8.314xx298" log "10` `=5705J` Since, `DeltaG` is negative hence reaction will be spontaneous in forward direction.

Discussion

100.	Calculate the internal energy change for the process in which 1.0 kcal of heat is added to1 .2 litre of `O_(2)` gas in a cyclinder at constant pressure of 1.0 atm and the volume changes to 1.5 litre.
Answer» Correct Answer - 0.993kcal

Discussion

Explore topic-wise InterviewSolutions in .

The gas with the highest heat of combustion is `:`A. MethaneB. EthaneC. EtheneD. Ethyne

the sing of `DeltaG` for the process of melting of ice at 273 K and 1 atm pressurme isA. positiveB. negaitiveC. neither negative nor positiveD. either negative or positive

The sign of `DeltaG` for the process of melting of ice at `273K` and `1` atm pressure isA. positiveB. negativeC. neither positive nor negativeD. either positive nor negative

`DeltaG`, in process of melting of ice at `-15^(@)C`,isA. `DeltaG=-ve`B. `DeltaG=+ve`C. `DeltaG=0`D. all of these

For a certain reaction `X`rarr`Y` the value of `DeltaH` and `DeltaS` are `50.50KJ mol^(-1)` and `100.03JK^(-1)` respectively. The temperature at which `DeltaG=0` isA. `505^(@)C`B. `232^(@)C`C. `252^(@)C`D. `450^(@)C`

Which of the following processes process towards more disordered state? `I` Stretching the rubberII. Sublimation of dry ice `III` Crystallisation of salt from solution `IV` . Dissolution of sugar from solutionA. `I,II,IV`B. `I,III`C. `III,IV`D. `II,IV`

A reaction has `DeltaH=-33KJ` and `DeltaS=-58(J)/(H)` . This reaction would be:A. spontanceous at all temperaturesB. non-spontaneous at all temperaturesC. spontaneous above a certain temperatureD. spontaneous below a certain temperature

What percentage `T_(1)` is of `T_(2)` for a `10%` efficiency of a heat engine? `T_(1)` is the temperature of sink and `T_(2)` is the temperature of heat reservoir.A. `T_(1)=90%` of `T_(2)`B. `T_(1)=T_(2)`C. `T_(2)=90%` of `T_(1)`D. `T_(1)=50%` of `T_(2)`

The difference between `DeltaH` and `Delta E` at constant voluem is equal toA. `R`B. `pDeltaV`C. `Vdeltap`D. `(3)/(2)R`.

When ammonium chloride is dissolved in water, the solution becomes cold. The change isA. EndothermicB. ExothermicC. SupercoolingD. None of the above

Warming ammonium chloride with sodium hydroxide in a test tube is an example of :A. Closed systemB. Isolated systemC. Open systemD. None of these

Heat of formation, `Delta H_(f)^(@)` of an explosive compound like `NCl_(3)` is -A. positiveB. negativeC. zeroD. positive or negative

when ammonium chloride is dissoved in water , the sloution becomes cold. The change isA. endothermicB. exothermicC. supercoolingD. none of these

`S_("rhombic")+O_(2)(g)toSO_(2)(g), DeltaH = -297 .5 kcal` `S_("monoclinic")+ O_(2)(g) to SO_(2)(g) , DeltaH = -300 kJ`A. rhombic sulphur is yellow in colorB. monoclinic sulphur has metallic lustreC. monoclinic sulphur is more stableD. `DeltaH_("translition")of S_(R) of S_(M)` is endothermic

At `0^(@)C` , ice and water are in reuilibrium and `DeltaS` and `DeltaG` for the conversion of ice to liquid water isA. `0,21.98JK^(-1)mol^(-1)`B. `-ve` , zeroC. `-ve,21.98JK^(-1)mol^(-1)`D. Zero, zero

At `0^(@)C` , ice and water are in equilibrium and `DeltaS` and `DeltaG` for the conversion of ice to liquid water isA. `0,21.98JK^(-1)mol^(-1)`B. `-ve` , zeroC. `+ve,21.98JKk^(-1)mol^(-1)`D. All of these

The standard heat of formation of `U_(3)O_(8)` is -853.5 kcl//mol and the standard heat of the reaction `3UO_(2) +O_(2) rarr U_(3)O_(8)` is -76.01 Kcal. The standard heat of formation of `UO_(2)` isA. `-1083KJmol^(-1)`B. `-108.3KJmol^(-1)`C. `-10.83KJmol^(-1)`D. `-1.083KJmol^(-1)`

Heat of neutralisation is least whenA. `NaOH` is neutralised by `CH_(3)COOH`B. `NaOH` is neutralised by `HNO_(3)`C. `NaOH` is neutralised by `HCL`D. `NaOH` is neutralised by `H_(2)SO_(4)`

For the raction `B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(2)(s)+3H_(2)O(l)` `DeltaE=-2143.2KJ` Calculate `DeltaH` for the reaction at `25^(@)C`A. `-2148.2KJmol^(-1)`B. `-2138.6KJmol^(-1)`C. `-2133.2KJmol^(-1)`D. `-2143.2KJmol^(-1)`

Standard enthalpy and standard entropy change for the oxidation of `NH_(3)` at `298K` are `-382.64KJ mol^(-1)` and `145.6Jmol^(-1)` respectively. Standard free energy change for the same reaction at `298K` isA. `-221.1KJmol^(-1)`B. `-339.3KJmol^(-1)`C. `-439.3KJmol^(-1)`D. `-523.2KJmol^(-1)`

For the reaction, `X_(2)O_(4)(l)rarr2XO_(2)(g),DeltaE=2.1Kcal` , `DeltaS=20cal//K` at `300K` . Hence `DeltaG` isA. `2.7Kcal`B. `-2.7Kcal`C. `9.3Kcal`D. `-9.3Kcal`

Molar heat capacity of water in equilibrium with ice at constant pressure isA. ZeroB. Infinity `(oo)`C. `40.45KJK^(-1)mol^(-1)`D. `75.48JK^(-1)mol^(-1)`

The molar heat capacity of water at constant pressure, `C_(P)` , is `75 JK^(-1) mol^(-1)`. When `1.0 kJ` of heat is supplied to `100 g` of water which is free to expand, the increase in temperature of water isA. `6.6 K`B. `1.2 K`C. `2.4 K`D. `4.8 K`

If for an ideal gas, the ratio of pressure and volume is constant and is equal to 1 atm `L^(-1)` , the molar heat capacity at constant pressure would beA. `(3)/(2)R`B. `2R`C. `(5)/(2)R`D. zero

Heat exchanged in a chemical reaction at constant temperature and pressure is calledA. entropy changeB. enthalpy changeC. internal energy changeD. free energy change

The law formulated by Nernst isA. First law of thermodynamicsB. Second law of thermodynamicsC. Third law of thermodynamicsD. Both `A` and `B`

For a melting of a solid at `25^(@)C`, the fusion process requires energy equivalent to` 2906`joules o be added to system considering the process to be reversible at fusion point, the entropy change of the process isA. `9.75 JK^(-1)`B. `11.272 JK^(-1)`C. `2.33 JK^(-1)`D. insufficient data

The difference between `Delta H` and `Delta U` for the combustion of methane at `27^(@)C` will be (in `J mol^(-1)`)A. `8.314xx27xx(-3)`B. `8.314xx300xx(-3)`C. `8.314xx(300)xx(-2)`D. `8.314xx300xx1`

A process is spantaneous at a given temperature ifA. `DeltaH=0`, `DeltaS lt 0`B. `DeltaHgt0`, `DeltaS lt 0`C. `DeltaHlt0`, `DeltaS gt 0`D. `DeltaHgt0`, `DeltaS = 0`

Which relations among the following is/are correct ?A. `G=H-TS`B. `PV=H-V`C. `w=DeltaW-q`D. `qv-qp=-DeltanRT`

The sublimation energy of `I_(2)` (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole. The enthalpy of vapourisation of `I_(2)` isA. `41.8 kJ //mol`B. `-41.8 kJ //mol`C. `72.8 kJ //mol`D. `-72.8 kJ //mol`

For vaporization of water at 1 atmospheric pressure the values of `DeltaH` and `DeltaS` are `40.63KJmol^(-1)` and `108JK^(-1)mol^(-1)` , respectively. The temperature when Gibbs energy change `(DeltaG)` for this transformation will be zero isA. `273.4K`B. `393.4K`C. `373.4K`D. `293.4K`

The molar heat of formation of `NH_(4)NO_(3)(s)` is `-367.5kJ ` and those of `N_(2)O(g)` and `H_(2)O(l)` are `+81.46kJ and -285.78kJ` respectively at `25^(@)C` and 1 atmospheric pressure. Calculate the `DeltaH` and `DeltaU` for the reaction, `NH_(4)NO_(3)(s)toN_(2)O(g)+2H_(2)O(l)`

What would be the heat released when an aqueous solution containing `0.5 mol` if `HNO_(3)` is mixed with `0.3` mol of `OH^(-1)` (enthalpy of neutralisation is `-57.1 kJ`)A. `28.5 kJ`B. `17.1 kJ`C. `45.7 kJ`D. `1.7 kJ`

What would be the heat released when an aqueous solution containing `0.5 mol` if `HNO_(3)` is mixed with `0.3` mol of `OH^(-1)` (enthalpy of neutralisation is `-57.1 kJ`)A. 28.5 kJB. 17.1 kJC. 45.7 kJD. 1.7 kJ

If a `298K` the bond energies of `C-H,C-C,C=C` and `H-H` bonds are respectivly `414, 347, 615KJmol^(-1)` , the vlaue of enthalpy change for the reaction `H_(2)C=CH_(2)(g)+H_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` at `298K` will beA. `+250 kJ`B. `-250 kJ`C. `+125 kJ`D. `-125 kJ`

If a `298K` the bond energies of `C-H,C-C,C=C` and `H-H` bonds are respectivly `414, 347, 615KJmol^(-1)` , the vlaue of enthalpy change for the reaction `H_(2)C=CH_(2)(g)+H_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` at `298K` will beA. `+250KJ`B. `-250KJ`C. `+125KJ`D. `-125KJ`

The standard enthalpy of formation of `C_(2)H_(4)(l)`, `CO_(2)(g)` and `H_(2)O(l)` are `52`, `394` and `-286 kJ mol^(-1)` respectively. Then the amount of heat evolved by burning `7g` of `C_(2)H_(4)(g)` isA. `1412 kJ`B. `9884 kJ`C. `353 kJ`D. `706 kJ`

Calculate the internal energy change for the process in which 1.0 kcal of heat is added to1 .2 litre of `O_(2)` gas in a cyclinder at constant pressure of 1.0 atm and the volume changes to 1.5 litre.