InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
The molar heat of formation of `NH_(4)NO_(3(s))` is `-367.54kJ` and those of `N_(2)O_((g)), H_(2)O_((l))` are `81.46` and `-285.8kJ` respectively at `25^(@)C` and 1 atmosphere pressure. Calculate `DeltaH` and `DeltaU` of the reaction `NH_(4)NO_(3(s))rarr N_(2)O_((g))+2H_(2)O_((l))`. |
|
Answer» `NH_(4)NO_(3(s))rarr N_(2)O_((g))+2H_(2)O_((l)), DeltaH=?` `DeltaH_(Reaction)=DeltaH_(Products)-DeltaH_(Rea c t a nts)` `=DeltaH_(N_(2)O)+DeltaH_(H_(2)O)xx2-DeltaH_(NH_(4)NO_(3))` Given, `DeltaH_(N_(2)O)=+81.46kJ` ltBRgt `DeltaH_(H_(2)O)=-285.8kJ` `DeltaH_(NH_(4)NO_(3))=-367.54kJ` `:. DeltaH_(Reaction)=+81.46+2(-285.8)-(-367.54)` `DeltaH=-122.6kJ` Further `DeltaH=DeltaU+DeltanRT ` `(Deltan=1-0=1,R=8.314J,T=298K)` `:. -122.6xx10^(3)=DeltaU+1xx8.314xx298` `:. DeltaU=-125077j o u l e=-125.077kJ` |
|
| 52. |
For the reaction `N_(2)H_(4) (g) rarr N_(2)H_(2)(g) +H_(2)(g)" "Delta_(r)H^(@)=109 KJ//mol` Calculate the bond enthalpy of `N=N`. Given : `B.E. (N-N) =163 KJ//mol, B.E. (N-H) =391 KJ//mol, B.E. (H-H)=436 KJ//mol` |
|
Answer» The equation: `DeltaH=109 = in_(N-N)+4 in_(N-H)-in_(H-H)-2in_(N-H)-in_(N=N)` `in_(N equivN)=163+2xx391-436-109=400" kJ/mole"` |
|
| 53. |
If the bond dissociation energies of `XY,X_(2)` and `Y_(2)(` all diatomic molecules `)` are in the ratio `1:1:0.5` and `Delta_(f)H` of `XY` is `-200 kJ mol^(-1)`. The bond dissociation energy of `X_(2)` will be `:`A. `800kJ mol^(-1)`B. `200kJ mol^(-1)`C. `300 kJ mol^(-1)`D. `400kJ mol^(-1)` |
|
Answer» Correct Answer - A Given `(1)/(2)X_(2)+(1)/(2)Y_(2) rarr XY,DeltaH=-200kJ mol^(-1)` Let bond dissociation energy of `X_(2),Y_(2)` and `XY` be `a:(a)/(2)a(` the ratio given `)` `X_(2) rarr 2X, DeltaH=a ….(1)` `Y_(2) rarr 2Y, DeltaH=(a)/(2) …(2)` `XY rarr X+Y, DeltaH=a .....(3)` By `eq. (1)+(2)-(3)xx2` `X_(2)+Y_(2) rarr 2XY` `Delta H=a+(a)/(2)-2a` or `(1)/(2)X_(2)+(1)/(2)Y_(2) rarr XY,` `DeltaH=(a)/(2)+(a)/(2)-a=-(a)/(4)` `:. -(a)/(4)=-200` `:. a=800kJ mol^(-1)` |
|
| 54. |
Statement `:` Heat of neutralisation can be given as `:` `H^(+)+OH^(-) rarr H_(2)O, DeltaH=-13.6 kcal.` Explanation `:` Heat of neutralisation can be alternatively defined as heat of formation of water. |
|
Answer» Correct Answer - a Heat of formation of water is `:` `H_(2)+(1)/(2)O_(2) rarr H_(2)O,Delta_(f o r)`. |
|
| 55. |
The standard enthalpy of formation of ammonia gas is Given: `N_(2)H_(4) (g) + H_(2) (g) rarr 2NH_(3) (g), Delta H_(r)^(@) = - 40 kJ//mol` `DeltaH_(f)^(@) [N_(2)H_(4) (g)] = - 120 kJ//mol`A. `-60`B. `-180`C. 40D. `-80` |
|
Answer» Correct Answer - D `Delta_(r)H^(@)= sum_(i) a_(i) Delta_(f) H_("(products)")^(@)-sum_(i) b_(i) Delta_(f) H_("(reactants)")^(@)` `-40=2Delta_(f) H^(@)+120` `Delta_(f)H^(@)=-80` |
|
| 56. |
From the given table answer the following question: `{:(,CO(g),CO_(2)(g),H_(2)O(g),H_(2)(g)),(DeltaH_(298)^(@)(-"KCal"//"mole"),-26.42,-94.05,-57.8,0),(DeltaH_(298)^(@)(-"KCal"//"mole"),-32.79,-94.24,-54.64,0),(S_(298)^(@)(-"Cal"//"k mol"),47.3,51.1,?,31.2):}` Reaction:`H_(2)O(g) + CO(g) hArr H_(2)(g)+CO_(2)(g)` Calculate `S_(298)^(@) [H_(2)O(g)]` |
|
Answer» Correct Answer - (i) -9.83 kCal; (ii) -6.81 kCal, (iii) -10.13 Cal/K, (iv) -9.83 kCal, (v) +45.13 Cal/K (i) `DeltaH=0-94.05-[-57.8-26.42]=-9.83` kCal (ii) `DeltaG=0+(-94.24) -[-54.64-32.79]=-6.81` kCal (iii) `DeltaS=(DeltaH-DeltaG)/(T)=-10.13` Cal (iv) `DeltaH=DeltaU+Deltan_(g)RT" "rArr Deltan_(g)=0` `DeltaU=-9.83` kCal (v) `-10.13=31.2+51.1-[x+47.3]` `x=45.13" Cal/K"` |
|
| 57. |
At temperature above 85 K, decarboxylation of acetic acid becomes a spontaneous process under standard state conditions. What is the standard entropy change (in J/K-mol) of the reaction. `CH_(3)COOH (aq) rarr CH_(4)(g) +CO_(2)(g)` `{:("Given :",DeltaH_(f)^(@)[CH_(3)COOH (aq)],=-484" kJ/mole"),(,DeltaH_(f)^(@)[CO_(2)(g)],=-392" kJ/mole"),(,DeltaH_(f)^(@)[CH_(4)(g)],=-75" kJ/mole"):}` |
|
Answer» Correct Answer - 200 J/K mole `DeltaS=(DeltaH)/(T) =(-392-75-(-484))/(85) =200 J//K` mole |
|
| 58. |
At temperature above 85 K, decarboxylation of acetic becomes a spontaneous process under standard state conditions. What is the standard entropy change (in J/K-mol) of the reaction. `CH_(3)COOH (aq) rarr CH_(4)(g)+CO_(2)(g)` `{:("Given :",DeltaH_(f)^(@)[CH_(3)COOH (aq)],=-484" kJ/mole"),(,DeltaH_(f)^(@)[CO_(2)(g)],=-392" kJ/mole"),(,DeltaH_(f)^(@) [CH_(4)(g)],=-75" kJ/mole"):}` |
|
Answer» At `85 K` the process must be at equilibrium under standard state condition: `DeltaG^(@)=0` `DeltaH^(@)=TDeltaS^(@)" "rArr" "DeltaS^(@)=(DeltaH^(@))/(T)=([-392-75-(-484)]xx10^(3))/(85)=2.00xx10^(2) J//K`-mol |
|
| 59. |
Calculate `|Delta_(f)G^(@)|` for `(NH_(4)Cl, s)` at `350 K`. Given : `Delta_(f)H^(@) (NH_(4)Cl, s)=-314.5" kJ/mol"` `{:(S_(N_(2)(g))^(@)=192 JK^(-1) mol^(-1),,S_(H_(2)(g))^(@)=130.5 JK^(-1) mol^(-1),),(S_(Cl_(2)(g))^(@)=223 JK^(-1) mol^(-1),,S_(NH_(4)Cl(s))^(@)=94.5 JK^(-1) mol^(-1),,"All given data at 300 K".),(Delta_(r)C_(P)=-20" J/mol-K",,ln(350/300)=0.15,):}` |
|
Answer» `Delta_(f)S^(@)(NH_(4)Cl, s)` at `300 K=S_(NH_(4)Cl(s))^(@)-[1/2 S_(N_(2))^(@)+2S_(H_(2))+1/2 S_(Cl_(2))^(@)]` `=94.5-(1/2xx192+2xx130.5+1/2xx223)` `=94.5-(96+261+111.5)` `Delta_(f)S_(300)^(@)" "=-374 JK^(-1) mol^(-1)` `Delta_(f)S_(310)^(@)" "=Delta_(f)S_(300)^(@)+Delta_(r)C_(p)ln (350/300)` `=-374-20xx(0.15)=-377 JK^(-1) mol^(-1)` `Delta_(f)H_(310)^(@)" "=Delta_(f)H_(300)^(@)+Delta_(r)C_(p) [350-300]` `=-314.5-(20xx50)/(1000)=-315.5` `Delta_(f) G_(350)^(@)" "=Delta_(f)H^(@)-T. Delta_(f) S^(@)` `=-315.5-(350xx(-377))/(1000) " "rArr-183.55" kJ/mol"` `|Delta_(f) G_(350)^(@)|" "=183.55" kJ/mol "rArr" "183" kJ/mol"` |
|
| 60. |
From the data of `DeltaH` of the following reactions `C(s)+1//2O_(2)(g) rarr CO(g), DeltaH=-110 kJ` and `C(s)+H_(2)O(g) rarr CO(g)+H_(2)(g), DeltaH=132 kJ` Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping the reaction temperature constant. |
|
Answer» Correct Answer - mole `% O_(2)(g)=37.5, H_(2)O(g)=62.5` Let x moles of `O_(2)` and y moles of `H_(2)O (g)` are present `:. 220 x=yxx132 rArr y/x=1.67` Calculate `%` moles of `O_(2)` and `H_(2)O (g)` |
|
| 61. |
The enthalpy changes of the following reactions at `27^(@)C` are `{:(Na(s)+1/2 Cl_(2) (g) rarr NaCl (s),Delta_(r)H=-411" kJ/mol",),(H_(2)(g)+S(s)+2O_(2)(g) rarr H_(2)SO_(4)(l),Delta_(r)H=-811" kJ/mol",),(2Na(s)+S(s)+2O_(2)(g) rarr Na_(2)SO_(4)(s),Delta_(r)H=-1382" kJ/mol",),(1/2 H_(2)(g)+1/2 Cl_(2)(g) rarr HCl(g),Delta_(r)H=-92" kJ/mol,",R=8.3" J/K-mol"):}` from these data, the heat change of reaction at constant volume (in kJ/mol) at `27^(@)C` for the process `2NaCl (s)+H_(2)SO_(4) (l) rarr Na_(2)SO_(4) (s)+2 HCl (g)` isA. 67B. 62.02C. 71.98D. None |
|
Answer» Correct Answer - B `{:(2NaCl(s)rarr 2Na(s)+Cl_(2)(g)" "Delta_(r)H=411xx2" kJ/mole"),(H_(2)SO_(4)(l) rarr H_(2)(g)+S(s)+2O_(2)(g)" "Delta_(r)H=811" kJ/mole"),(2Na(s) +S(s)+2O_(2)(g)rarrNa_(2)SO_(4)(s)" "Delta_(r)H=-1382" kJ/mole"),(H_(2)(g)+Cl_(2)(g) rarr 2HCl(g)" "Delta_(r)H=-184" kJ/mole"),(bar(2NaCl(s)+H_(2)SO_(4)(l) rarr 2HCl(g)+Na_(2)SO_(4)(s)" "Delta_(r)H=67"kJ/mole")):}` `67=Delta_(r)U+(2xx8.3xx300)/(1000), Delta_(r)U=62.0" kJ/mole"` |
|
| 62. |
For which of the following reactions heat of formation `(Delta_(f)H)` is equal to `DeltaH//2` ?A. `N_(2)+O_(2)rarr 2NO,DeltaH=X_(1)`B. `CO+O_(2)rarrCO_(2),DeltaH=X_(2)`C. `PCl_(3)+Cl_(2) rarr PCl_(5),DeltaH=X_(3)`D. `C+O_(2) rarrCO_(2),DeltaH=X_(4)` |
|
Answer» Correct Answer - A Two moles of `NO` are formed in the reaction. |
|
| 63. |
From the following data at `25^(@)C` `{:("Reaction",Delta_(r)H^(@)" kJ/mol"),(1/2 H_(2)(g)+1/2 O_(2)(g) rarr OH(g),42),(H_(2)(g)+1/2 O_(2)(g) rarr H_(2)O(g),-242),(H_(2)(g) rarr 2H(g),436),(O_(2)(g) rarr 2O(g),495):}` Which of the following statement (s) is/are correct:A. `Delta_(r)H^(@)` for the reaction `H_(2)O(g) rarr 2H(g)+O(g)` is 925.5 kJ/molB. `Delta_(r)H^(@)` for the reaction `OH(g) rarr H(g)+O(g)` is 502 kJ/molC. Enthalpy of formation of `H(g)` is `-218` kJ/molD. Enthalpy of formation of `OH(g)` is `42` kJ/mol |
|
Answer» Correct Answer - A::D (A) `H_(2)O rarr 2H+O` `DeltaH=925.5` `B rarr 507.5` `C rarr 432/2 =218` `D rarr 42` |
|
| 64. |
If `Delta_(f)H^(@)``_(C_(3)H_(8)(g))=-85kJ moll^(-1),` `Delta_(f)H^(@)``_(C_(3)H_(8)(g))=-104kJ mol^(-1),` `DeltaH` for `C_((s))rarr C_((g))` is `718kJ mol^(-1)` and heat of formation of `H-` atom is `218kJ mol^(-1)`, then `:`A. `e_(C-H)=414kJ`B. `e_(C-C)=345kJ`C. `e_(H-H)=218kJ`D. `e_(H-H)=436kJ` |
|
Answer» Correct Answer - A::B::D |
|
| 65. |
During the melting of an ice slab at `273K` at atmospheric pressure `:`A. Positive work is doen on the ice`-` water system by the atmosphere.B. Positive work is done by the ice`-` water system on the atmosphereC. The internal energy of the ice`-` water system decreasesD. The internal energy of the ice`-` water system increases. |
|
Answer» Correct Answer - A::D |
|
| 66. |
The standard heat of combustion of solid boron is equal to `:`A. `Delta_(f)H^(@)(B_(2)O_(3))`B. `(1)/(2)Delta_(f)H^(@)(B_(2)O_(3))`C. `2Delta_(f)H^(@)(B_(2)O_(3))`D. `-(1)/(2)Delta_(f)H^(@)(B_(2)O_(3))` |
|
Answer» Correct Answer - B `2B+(3)/(2)O_(2)rarr B_(2)O_(3),Delta_(f)H^(@)of B_(2)O_(3)` `Delta_(comb.)H^(@)of bo ro n=(1)/(2)Delta_(f)H^(@)ofB_(2)O_(3)` |
|
| 67. |
The data given below are for vapour phase reactions at constant pressure. `C_(2)H_(6)rarroverset(*)(C_(2))overset(*)(H_(5))+overset(*)(H), DeltaH=-4200kJmol^(-1)` `overset(*)(C_(2))H_(6) rarr C_(2)H_(4)+overset(*)(H), DeltaH=168kJ mol^(-1)` The enthalpy change for the reaction `,` `2overset(*)(C_(2))H_(5) rarr C_(2)H_(6)+C_(2)H_(4)` is `:`A. `+250kJ mol^(-1)`B. `+588kJ mol^(-1)`C. `-252kJ mol^(-1)`D. `-588kJ mol^(-1)` |
|
Answer» Correct Answer - C `overset(*)(C_(2))H_(5)+overset(*)(H)rarrC_(2)H_(6),DeltaH=-420kJ mol^(-1)` `ul( overset(*)(C_(2))H_(5)rarr C_(2)H_(4)+overset(*)(H),DeltaH=168kJ mol^(-1) )` `:. 2C_(2)H_(5) rarrC_(2)H_(6)+C_(2)H_(4),DeltaH=-252kJ mol^(-1)` |
|
| 68. |
If `x_(1),x_(2) and x_(3)` are enthalpies of H-H , O=O and O-H bonds respectively , and `x_(4)` is the enthalpy of vaporisation of water , estimate the standard enthalpy of combustion of hydrogen:A. `x_(1) + (x_(2))/(2) - 2x_(3) + x_(4)`B. `X_(1) + (x_(2))/(2) - 2x_(3) - x_(4)`C. `x_(1) + (x_(2))/(2) - x_(3) + x_(4)`D. `2x_(3) - x_(1) - (x_(2))/(2) - x_(4)` |
|
Answer» Correct Answer - B `H_(2)+1/2 O_(2) rarr H_(2)O(l)` `DeltaH=x_(1)+x_(2)/2-2x_(3)` `H_(2)O(l) rarr H_(2)O(g)` `DeltaH=x_(1)+x/2-2x_(3)-x_(4)` |
|
| 69. |
Reactions involving gold have been of particular interest to a chemist. Consider the following reactions, `{:(Au (OH)_(3) + 4 HCl rarr HAuCl_(4) + 3 H_(2) O",",Delta H = - 28 k Cal,),(Au (OH)_(3) + HBr rarr HAuBr_(4) + 3 H_(2)O",",Delta H = -36.8 k Cal,):}` In an experiment there was an absorption of 0.44 k Cal when one mole of `HAuBr_(4)` was mixed with 4 moles of `HCl`. what is the percentage conversion of `HAuBr_(4)` into `HAuCl_(4)` ?A. 0.005B. 0.006C. 0.05D. 0.5 |
|
Answer» Correct Answer - C Target reaction `HAuBr_(4)+4HCl rarr HAuCl_(4)+4HBr` `DeltaH=8.8` `%` conversion `=100/8.8 xx0.44=5%` |
|
| 70. |
Heat of reaction is the change in enthalpy or internal energy as represented by a balanced thermochemical equation . The amount of energy released during a chemical change depends upon the state of reactants and products, the conditions of pressure and volume at which reaction is carried out, and temperature. The variation of heat of reaction `(DeltaH_(1) or DeltaE)` with temperature is given as `DeltaH_(2)-DeltaH_(1)=DeltaC_(P)[T_(2)-T_(1)] or DeltaE_(2)-DeltaE_(1)=DeltaC_(v)(T_(2)-T_(1).` Standard heat enthalpy of elements in their most stable state is assumed to be zero whereas standard heat enthalpy of compound is referred as heat of formation of that compound at `1 atm` pressure and `25^(@)C`. Oxidation of `N_(2)` to `N_(2)O,NO,NO_(2)` shows absorption of energy whereas heat of combustion of `N_(2)` is exothermic like other heat of combustion. Heat of combustion of carbon in diamond and amorphous form ar `-94.3 ` and `-97.6 kcal//mol` . The heat required to convert `6g` carbon from diamond to amorphous form is `:`A. `-1.65kcal`B. `+1.65kcal`C. `-3.3kcal`D. `+3.3kcal` |
|
Answer» Correct Answer - a `C_(D)rarr C_(G), DeltaH=-94.3+97.6=+3.3kcal` `12g C_(D rarr A)-=-3.3kcal` `:. 6gC_(D rarr A)-=-(3.3xx6)/(12)=-1.65kcal` |
|
| 71. |
Calculate the amount of heat released when heat of neutralization is -57.0 kJ : (a) 0.5 mole of HNO3 is mixed with 0.3 mole of NaOH in aqueous solution. (b) 200 mL of 0.1 MH2SO4 is mixed with 150 mL of 0.2 MKOH. |
|
Answer» (a) 1 mole of HNO3 and 1 mole of NaOH give heat = 57.0 kJ 0.3 mole of HNO3 and 0.3 mole of NaOH give heat = 57 x 0.3 = 17.1 kJ (b) Meq. of H2SO4 = 200 x 0.1 x 2 = 40; Meq. of KOH = 150 x 0.2 x 1 = 30 1000 Meq. of H2S04 and 1000 Meq. of KOH on mixing produce heat = 57.0 kJ 30 Meq. of H2SO4 and 30 Meq. of KOH on mixing produce heat = 57x30 =1000 =1.71 kJ |
|
| 72. |
The incorrect expression among the following is `:`A. in isothermal process, `w_(reversibl e)=-nRTln``(V_(f))/(V_(i))`B. `lnK=(DeltaH^(@)-TDeltaS^(@))/(RT)`C. `K=e^(-DeltaG//RT)`D. `(DeltaG_(system))/(DeltaS_(t otal))=-T` |
|
Answer» Correct Answer - B `:. DeltaG^(@)=-RTlnK` and `DeltaG^(@)=DeltaH^(@)-T DeltaS^(@)` `DeltaH^(@)-TDeltaS^(@)=-RTlnK` `lnK=-((DeltaH^(@)-TDeltaS^(@))/(RT))` |
|
| 73. |
Calculate the enthalpy change when infinitely dilute solutions of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed. `Delta_(f)H^(@)` for `Ca_((aq.))^(2+),CO_(3(aq.))^(2-)` and `CaCO_(3(s))` are `-129.80,-161.65` and `288.46kcal//mol` respectively. |
|
Answer» Correct Answer - 3 The given reaction on two solution mixing is `CaCl_(2)+Na_(2)CO_(3) rarr CaCO_(3)+2NaCl` Each` 100%` dissociated at infinite dilution, thus `Ca_((aq.))^(2+)+2Cl_(aq.)^(-)+2Na_((aq.))^(+)+CO_(3(aq.))^(2-)rarr 2Na_((aq.))^(+)+2Cl_((aq.))^(-)+CaCO_(3(s))` or `CO_(3(aq.))^(2-)+Ca_((aq.))^(2+)rarr CaCO_(3(s))` `:. Delta_(r)H=Sigma_(Pr od uct)^(@)-Sigma_(reactant)^(@)` `=Delta_(r)H^(@)Sigma_(CaCO_(3))^(@)-[Delta_(f)H^(@)Sigma_(Ca^(2+))+Delta_(f)H^(@)Sigma_(CO_(3)^(2-))]` `=-288.46-(-129.80-161.65)` `=2.99kcal~~3kcal` |
|
| 74. |
The dissolution of `1 mol e` of `NaOH_((s))` in `100 mol e` of `H_(2)O_((l))` give rise to evolution of heat equal to `-42.34kJ` . However if `1 mol e` of `NaOH_((s))` is dissolved in `1000 mol e `of `H_(2)O_((l))`, the heat given out is `42.76kJ`. What would be enthalpy change when 900 mole of `H_(2)O_((l))` are added to a solution containing 1 mole of `NaOH_((s))` in 100 mole of `H_(2)O` ? |
|
Answer» Correct Answer - `-0.42kJ;` |
|
| 75. |
1.00 `l` sample of a mixture of `CH_(4) (g)` & `O_(2)(g)` measured at `25^(@)C` & `740` torr was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of 1260 Cal/K. The Complete combustion of methane to `CO_(2)` & `H_(2)O` caused a temperature rise, in the Calorimeter, of 0.667 K. What was the mole percent of `CH_(4)` in the original mixture ? `DeltaH_("comb")^(@) (CH_(4))=-215" k Cal mol"^(-1)`. |
|
Answer» Correct Answer - `9.82" mol"% CH_(4)` Total energy released `=1260xx0.667=840.42 cal` Moles of `CH_(4) =(840.42)/(215.10^(3)) =3.9xx10^(-13)` Calculate V of `CH_(4)` By applying `PV=nRT` `V=0.098246 L` `%` moles of `CH_(4)=(0.098246)/(1) xx100=9.82 %` |
|
| 76. |
Two solutions initially at `25^(@)C` were mixed in an insulated bottle. One contains `400mL` of `0.2N` weak monoprotic acid solution. The other contains `100mL` of `0.8N NaOH` solution. After mixing the temperature rises to `26.17^(@)C` . Calculate the heat of neutralisation of weak acid with `NaOH`. Assume density of final solution `1.0 g cm^(-3)`. and speicific heat of final solution `4.2J g^(-1)K^(-1)`. |
|
Answer» Correct Answer - `30.71kJ eq uival ent^(-1);` |
|
| 77. |
A swimmer coming out from a pool is covered with a film of water weighing about 80 g. How much heat must be supplied to evaporate this water. If latent heat of evaporation for H2O is 40.79 kJ mol-1 at 100°C. |
|
Answer» Amount of water present on the body of swimmer = 80 g = 80/18 mole Thus heat required to evaporate 80/18 mole H2O = 40.79 x 80/18 = 181.2 kJ |
|
| 78. |
How much heat is produced when 4.50 g methane gas is burnt in a constant pressure system.Given : CH4 + 2O2 → CO2 + 2H2O: ΔH = -802 kJ |
|
Answer» 16 g CH4 on burning produces heat = - 802 kJ ∴ 4.5 g CH4 on burning produces heat = -802 x4.5/16 = -225.6 kJ |
|
| 79. |
An stoichiometric mixture of hydrogen gas and the air at `25^(@)C` and a total pressure of 1 atm, is exploded in a closed rigid vessel. If the process occurs under adiabatic condition then using the given data answer the question that follow: Given : (i) `C_(P)=8.3" Cal deg"^(-1)" mol"^(-1), (ii) C_(P)=11.3" Cal deg"^(-1)" mol"^(-1), DeltaH_(f) [H_(2)O(g)]= -57.8 kCal` [Take air as `80% N_(2), 20 % O_(2)` by volume] What will be the maximum temperature attained if the process occurs in adiabatic container ?A. ` cong 2937 K`B. `cong 2665 K`C. `cong 1900 K`D. `cong 298 K` |
|
Answer» Correct Answer - A `C_(V_(N_(2)))=C_(P)-R=8.3-2=6.3` `C_(V_(H_(2)O))=C_(P)-R=11.3=9.3` `DeltaU=nC_(V_(N_(2))) (T_(2)-T_(1))+nC_(V_(H_(2)O)) (T_(2)-T_(1))` `T_(2)=2937 K` |
|
| 80. |
An stoichiometric mixture of hydrogen gas and the air at `25^(@)C` and a total pressure of 1 atm, is exploded in a closed rigid vessel. If the process occurs under adiabatic condition then using the given data answer the question that follow: Given : (i) `C_(P)=8.3" Cal deg"^(-1)" mol"^(-1), (ii) C_(P)=11.3" Cal deg"^(-1)" mol"^(-1), DeltaH_(f) [H_(2)O(g)]= -57.8 kCal` [Take air as `80% N_(2), 20 % O_(2)` by volume] What will be final pressure in atm ?A. `cong 8.5`B. `cong 7.6`C. `cong 5.46`D. `cong 0.85` |
|
Answer» Correct Answer - A `P_(1)/(n_(1)T_(1))=P_(2)/(n_(2)T_(2))rArr P_(2) cong 8.5` |
|
| 81. |
An stoichiometric mixture of hydrogen gas and the air at `25^(@)C` and a total pressure of 1 atm, is exploded in a closed rigid vessel. If the process occurs under adiabatic condition then using the given data answer the question that follow: Given : (i) `C_(P)=8.3" Cal deg"^(-1)" mol"^(-1), (ii) C_(P)=11.3" Cal deg"^(-1)" mol"^(-1), DeltaH_(f) [H_(2)O(g)]= -57.8 kCal` [Take air as `80% N_(2), 20 % O_(2)` by volume] The value of `C_(P)` of `N_(2)` & `H_(2)O` will be (in Cal.`"deg."^(-1)" mol"^(-1)`)A. 8.3, 8.3B. 8.3, 11.3C. 113, 11.3D. 11.3, 8.3 |
|
Answer» Correct Answer - B `C_(P)` is directly proportional to no. of atoms |
|
| 82. |
For which change `DeltaH cancel(=)DeltaU` ?A. `H_(2(g))+(I)_(2(g))hArr2HI_((g))`B. `HCl_((aq.))+NaOH_((aq.))rarr NaCl_((aq.))+H_(2)O_((l))`C. `C_((s))+O_(2(g))hArrCO_(2(g))`D. `N_(2(g))+3H_(2(g)) rarr 2NH_(3(g))` |
|
Answer» Correct Answer - D `DeltaH=DeltaU+DeltanRT` for `(a), (b), (c )` Deltan=0` and for `(d) Deltan=-2` |
|
| 83. |
Molar heat capacity of water in equilibrium with the ice at constant pressure is `:`A. zeroB. infinity `(oo)`C. `40.45.45kJ K^(-1)mol^(-1)`D. `75.48JK^(-1)mol^(-1)` |
|
Answer» Correct Answer - B `C_(p)=((deltaH)/(deltaT))_(p),` At equilibrium `T` is constant, `i.e., deltaT=0,` Thus `C_(p)=oo` |
|
| 84. |
The dissociation energy of `CH_(4)` and `C_(2)H_(6)` are respectively `360` and `620 kcal//mol.` The bond energy of `C-C` bond is `:`A. `260kcal//mol`B. `180kcal//mol`C. `130kcal//mol`D. `80kcal//mol` |
|
Answer» Correct Answer - D `CH_(4) rarr C+4H,DeltaH=360kcal` `e_(C-H)=90kcal` `C_(2)H_(6)rarr2C+6H,DeltaH=620` `:. 620=e_(C-C)+6e_(C-H)` `:. E_(C-C)=620-540=80kcal.` |
|
| 85. |
If, combustion of `4g` of `CH_(4)` liberates `2.5kcal` of heat, the heat of combustion of `CH_(4)` is `:`A. `-20kcal`B. `-10kcal`C. `2.5kcal`D. `-5kcal` |
|
Answer» Correct Answer - B `DeltaH=(2.5xx16)/(4)=-10kcal mol^(-1)` |
|
| 86. |
A calorimeter will heat capacity equivalent to having `13.3` moles of water is used to measured the heat of combustion from `0.303 g` of sugar `(C_(12)H_(22)O_(11))`. The temperature increase was found to be `5.0 K`. Calculate the heat released, the amount of heat released by `1.0 g`, and `1.0` mole of sugar. |
|
Answer» Heat released, `q_(v)`, `q_(v)=13.3xx75.2xx5.0 K` `=5000 J` The amount of heat released by `1.0g` would be, `5000 J//0.303 g=16.5" kJ/g"` since the molecular weight of sugar is 342 h/mol, the amount of hear released by `1.0` mole would be `16.5xx342=56431" kJ/mol"`. |
|
| 87. |
A person requires `2870kcal` of energy to lead normal daily life. If heat of combustion of cane sugar is `-1349kcal`, then his daily comsumption of sugar is `:`A. `728g`B. `0.728 g`C. `342g`D. `0.342g` |
|
Answer» Correct Answer - A Amount of sugar needed `=(2870xx342)/(1349)=727.6g` |
|
| 88. |
The commercial production of water gas utilizes the reaction under standard conditions `:C+H_(2)O_((g)) rarr H_(2)+CO`. The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to `CO_(20`. How many gram of carbon must be burnt to `CO_(2)` to provide enough heat for the water gas conversion of `100g` carbon ? Neglect all heat losses to the environment. Also `DeltaH_(f)^(@)` of `CO,H_(2)O_((g))` and `CO_(2)` are `-110.53, -241.81` and `-393.51kJ //mol` respectively. |
|
Answer» Correct Answer - `33.36g ` carbon; |
|
| 89. |
From the following data of `DeltaH`, of the following reaction, `C_((s))+(1)/(2)O_(2(g))rarr CO_((g)),DeltaH=-110kJ` `C_((g))+H_(2)O_((g)) rarr +H_(2(g)),DeltaH=132kJ` Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at `1273K`, keeping temperature constant. |
|
Answer» Correct Answer - `(0.6)/(1);` |
|
| 90. |
A person inhales `640g ` of `O_(20` per day. If all the `O_(2)` is used for converting sugar into `CO_(2)` and `H_(2)O`, how much sucrose `(C_(12)H_(22)O_(11))` is consumed in the body in one day and what is the heat evolved ? `(DeltaH_("combination of sucrose ")=-5645kJ mol^(-1))` |
|
Answer» Correct Answer - `570g,9408.33kJ;` |
|
| 91. |
The heat of formation of carbon dioxide from graphite at `15^(@)C` and constant volume is `97400 cal`, and that of carbon monoxide under the same conditions is `25400 cal`. What heat should be evolved when `100 ` litre of carbon monoxide measured at `N.T.P. ` is burnt in an excess of oxygen, both reactants and products being at `15^(@)C` ? |
|
Answer» Correct Answer - `-321408cal;` |
|
| 92. |
Ethanol can undergo decompostion to form two sets of products. If the molar ratio of `C_(4)H_(4) "to" CH_(3)CHO` is 8:1 in a set of product gases, then the energy involved in the decomposition of 1 mole of ethanol is: A. 65.98 kJB. 48.137 kJC. 48.46 kJD. 48.46 kJ |
|
Answer» Correct Answer - B Energy released `=(8/9 xx45:54)+(1/9xx68.91)=48.137 kJ` |
|
| 93. |
Calculate `DeltaH_(f)^(@)` for chloride ion from the following data `:` `(1)/(2)H_(2(g))+(1)/(2)Cl_(2(g))rarr HCl_((g)),DeltaH_(f)^(@)=-92.4kJ` `HCl_((g))+nH_(2)O rarr H_((aq.))^(+)+Cl_((aq.))^(-),DeltaH^(@)=-74.8kJ` `DeltaH_(f)^(@) H_((aq.))^(+)=0.0kJ` |
|
Answer» Given, `(1)/(2)H_(2(g))+aq. rarr H_((aq.))^(+)+E, DeltaH^(@)=0 ....(1)` `(1)/(2)H_(2(g))+(1)/(2)Cl_(2(g))rarr HCl_((g)), DeltaH^(@)=-92.4kJ ....(2)` `CHl_((g))+nH_(2)O_((l)) rarr H_((aq.))^(+)+Cl_((aq.))^(-), DeltaH^(@)=-74.8kJ ...(3)` Add `eq. (2)` and `(3)` `(1)/(2) H_(2(g))+(1)/(2)Cl_(2(g))+nH_(2)Orarr H_((aq.))^(+)+Cl_((aq.))^(-), DeltaH^(@)=-167.2kJ ....(4)` Subtract `eq. (2)` from `(4)` `ul({:((1)/(2)H_(2(g)),+aq.,rarr,H_((aq.))^(+),+e,,DeltaH^(@)=0),(-,-,,-,-,-):})` `:. e+(1)/(2)Cl_(2(g))+aq. rarr Cl_((aq.))^(-), DeltaH=-167.2kJ` Heat of formation for `Cl_((aq.))^(-)=-167.2kJ` |
|
| 94. |
Calculate the enthalpy change when `6.80g` of `NH_(3)` is passed over heated `CuO`. The standard heat enthalpies of `NH_(3(g)),CuO_((s))` and `H_(2)O_((l))` are `-46.0,-155.0` and `-285.0kJ mol^(-1)` respectively and the change is `NH_(3)+(3)/(2)CuO rarr (1)/(2) N_(2(g))+(3)/(2) H_(2)O_((l))+(3)/(2)Cu_((s))` . |
|
Answer» Correct Answer - `-59.6kJ;` |
|
| 95. |
`1.00litre` sample of a mixture of `CH_(4(g))` and `O_(2(g))` measured at `25^(@)C` and `740 t or r` was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of `1260cal//K`. The complete combustion of the methane to`CO_(2)` and `H_(2)O` caused a temperature rise in the calorimeter of `0.667K`. What was the mole per cent of `CH_(4)` in the original mixture ? `(DeltaH_(comb.)^(@)(CH_(4))=-215kcal mol^(-1))` |
|
Answer» Correct Answer - `9.75%;` |
|
| 96. |
Which of the following is `(are)` true ?A. The evaporation of water is an endothermic changeB. The conversion of white phosphours to red phosphorus is an exothermix reactionC. The heat of neutralization of a strong acid with a strong base is always the sameD. `DeltaH` is negative for endothermic reactions |
|
Answer» Correct Answer - A::B::C |
|
| 97. |
Which of the following are applicable for a thermochemical equation ? It tells `:`A. about the physical state of reactants and productsB. about the allotropic form `(` if any `)` of the reactantsC. whether the reaction is exothermic or endothermicD. whether a particular reaction is spontaneous or not |
|
Answer» Correct Answer - A::B::C |
|
| 98. |
The enthalpy changes of formation of the gaseous oxides of nitrogen `(N_(2)O` and `NO)` are positive because of `:`A. the high bond energy of the nitrogen moleculeB. the hight electrone affinity of oxygen atomsC. the high electron affinity of nitrogen atomsD. the tendency of oxygen to form `O^(2-)` |
|
Answer» Correct Answer - A Due to high bond energy of `N-=N`, more heat is absorbed to break up `N_(3)` molecule. |
|
| 99. |
`DeltaC_(p)` for change `,N_(2(g))+3H_(2(g))=2NH_(3(g))` is `:`A. `C_(pNH_(3))-(C_(pN_(2)))`B. `2C_(pNH_(3))-(C_(pN_(2))+3C_(pH_(2)))`C. `2C_(pNH_(3))-(C_(pH_(2)))`D. `2C_(pNH_(3))+(C_(pN_(2))+3C_(pH_(2)))` |
|
Answer» Correct Answer - B `DeltaC_(p)=C_(p)` of products `-C_(p)`of reactants. |
|
| 100. |
Identify the correct statement regarding a spontaneous process `:`A. For a spontaneous process in an isolated system, the change in entropy is positiveB. Endothermic processes are never spontaneousC. Exothermic processes are always spontaneousD. Lowering of energy in the reaction process is the only criterion for spontanity |
|
Answer» Correct Answer - A For a spontaneous process in an isolated system, the change in entropy is positive. |
|