52671 + Interview Questions in Chemistry in Class 11

1.	Calculate the equilibrium constants of each of the indicated species necessary to reduce an initial 0.2M Zn^(2+) solution to 1.0 xx 10^(-4)Zn^(2+). a. Nh_(3) and Zn(NH_(3))_(4)^(2+) (assume no partial complexation) b. overset(Theta)OH in equilibrium with Zn(OH)_(2)(s). c. overset(Theta)OH and Zn(OH)_(4)^(2-). d. Calculate [overset(Theta)OH] which would be produced by each equilibrium concentration of NH_(3) in part (a). Predict whether Zn(OH)_(2) or Zn(OH)_(4)^(2-) would form in preference to Zn(NH_(3))_(4)^(2+) upon addition of suficient NH_(3) to produce the equilibrium concentration calculated in part(a). e. Explain what would be observeed if concentrated NH_(3) solution were added slowely to 0.2M solution of Zn^(2+). Given. K_(f)Zn(NH_(3))_(4)^(2+) = 5 xx 10^(8). K_(sp)ZN(OH)_(2) = 1.8 xx 10^(-14). K_(f)Zn(OH)_(4)^(2-) = 5 xx 10^(14). K_(b) NH_(4)OH = 1.8 xx 10^(-5).
Answer» Solution :d. `[Zn^(2+)]` final `=1 xx 10^(-4)M` `[Zn(NH_(3))_(4)^(2+)]aq = 0.2 - 10^(-4) ~~ 0.2` `Zn^(2+) + 4NH_(3) HARR Zn(NH_(3))_(4)^(2+)` `K_(f) = ([Zn(NH_(3))_(4)^(2+)])/([Zn^(2+)][NH_(3)]^(4))` `5 xx 10^(8) =(0.2)/((10^(-4))[NH_(3)]^(4))` `(NH_(3))^(4) = 4 xx 10^(-6) = 400 xx 10^(-8)` `(NH_(3)) = 4.5 xx 10^(-2)M` b. `Zn(OH)_(2) rarr Zn^(2+) +2 overset(Theta)OH` `K_(sp) = (1.8 xx 10^(-14))/(10^(-4)) = 1.8 xx 10^(-10)` `[overset(Theta)OH] = 1.3 xx 10^(-5)M` C. `Zn^(2+) + 4overset(Theta)OH rarr Zn(OH)_(4)^(2-)` `K_(f) - ([Zn(OH)_(4)^(2-)])/([Zn^(2+)][overset(Theta)OH]^(4))` `5 xx 10^(4) = (0.2)/((10^(-4))[overset(Theta)OH]^(4))` `rArr [overset(Theta)OH]^(4) = 4 xx 10^(-12), [overset(Theta(O)H] = 1.4 xx 10^(-3)` d. The concentration of `overset(Theta)OH` in equilibrium with the `NH_(3)` concentration of part (a) is given by, `NH_(3) + H_(2)O hArr overset(o+)NH_(4) + overset(Theta)OH`, (LET `x = [overset(o+)NH_(4)] = [overset(Theta)OH]` `K_(b) = ([overset(o+)NH_(4)][overset(Theta)OH])/([NH_(3)])` `1.8 xx 10^(-5) = (x^(2))/(4.5 xx 10^(-2))` `rArr x^(2) = 8.1 xx 10^(-7) rArr x = [overset(Theta)OH] = 9.0 xx 10^(-4)M`. `[overset(Theta)OH] (9.0 xx 10^(-4)M)` from the `4.5 xx 10^(-2)M NH_(3)` is sufficient precipitate `Zn(OH)_(2)` [part (b)] but not sufficient to from `Zn(OH_(4)^(2-))` [part (c)] e. On addition of `Nh_(3)` solution to `Zn^(2+)` solution, a precipitate of `Zn(OH)_(2)` will form, which will later dissolve to yield a clear solution containing `[Zn(NH_(3))_(4)]^(2+)`

Discussion

2.	Calculate the equilibrium pressure (in Pascal) for the conversion of graphite to diamong at 25^(@)C. The densities of graphite and diamond may be takes to be 2.20 and 3.40 g//cc respectively independent of pressure. (Express your answer in scientific notation x xx10gamma and write the value of y.) [Given : DeltaG_(298)^(@)(C_("graphite") rarr C _("diamond"))=2900J//mol]
Answer» ANSWER :9

Discussion

3.	Calculate the equilibrium constant, K for the following reaction at 400 K ? 2NOCl(g) hArr 2NO(g) + Cl_(2)(g) Given thatDelta_(r) H^(@) = 80.0 kJ mol^(_1) and Delta_(r)S^(@)=120 J K^(-1) mol^(-1) at 400 K.
Answer» SOLUTION :`Delta_(r) G^(@) = Delta_(r)H^(@) - TDelta_(r)S^(@) = 80000J - 400 xx 120 J = 32000 J` Now, `Delta_(r) G^(@) = -2.303 RT log K ` `:. 32000 = - 2.303 xx 8.314 xx 400 xx LOGK `or `log K = - 4.1782 = BAR(5).8218` `:.K = ` Antilog `bar(5).8218 = 6.634 xx 10^(-5)`

Discussion

4.	Calculate the equilibrium constant (in multiples of 10^(-4)) for the reaction PCl_(5(g)) rarr PCl_(3(g)) + Cl_(2(g)) at 400K, if Delta H^(0) = 77.2KJ "mole"^(-1) and Delta S^(0) = 122JK^(-1) "mole"^(-1)
Answer» SOLUTION :`Delta G^(0) = Delta H^(0) - T Delta S^(0)` `= 77.2 - (400 xx 122)/(1000) = 28.4 KJ` `Delta G^(0) = -RT LN K rArr K = 2 xx 10^(-4)`

Discussion

5.	Calculate the equilibriumconstant for the following reaction at 298 K and 1 atm pressure : NO(g) + (1)/(2) O_(2)(g) hArr NO_(2)(g) Given : Delta_(f) H^(@) at 298K For NO(g) = 90.4 kJ mol^(-1),ForNO_(2)(g) = 33.8 kJ mol^(-1) Delta S^(@)at 298 K for the reaction= - 70.8 J K^(-1)mol^(-1) Gas constant, R= 8.31 JK^(-1) mol^(-1)
Answer» Solution :`Delta_(r)H^(@) = Delta_(f) H^(@) ( NO_(2)) -[Delta_(f) H^(@) (NO)+(1)/(2) Delta_(f) H^(@) (O_(2))]= 33.8 -[90.4 + 0] = - 56.6kJ mol^(-1)` `= - 56600 J mol^(-1) - 298 K ( - 70.8 JK^(-1) mol^(-1)) = - 35501.6 J mol^(-1)` `:. - 35501.6 =- 2.303 xx 8.31 xx298 log K ` or `log K = 6.2250` or `K = 1.679 xx 10^(6)`

Discussion

6.	Calculate the entropy change when one kg of water is heated from 27^(@)C to 200^(@)C forming super heated steam under constant pressure. Specific heat of water and systems are respectively 4180 and 1670+0.49J/(kg)-K.
Answer» ANSWER :16.37 KJ

Discussion

7.	Calculate the entropy change when 1 mole of ethanol is evaporated at 351 K. The molar heat of vapourisation of ethanol is 39.84 "kJ mol"^(-1)
Answer» Solution :Given : `T_b`=351 K `DeltaH_"VAP"=39840 "J mol"^(-1)` `DeltaS_V`=? `DeltaS_V=(DeltaH_"vap")/T_b` `DeltaS_V=39840/351` `DeltaS_V=113.5 J K^(-1) "mol"^(-1)`

Discussion

8.	Calculate the entropy change when 1 kg of water is heated from 27^(@)C to 200^(@)C forming super heated steam under constant pressure. Given : Specific heat ofwater = 4180 J //kg .K Specific heat of steam = 1670 + 0.49 T J // kg. K Latent heat of vaporisation = 23 xx 10^(5) J // kg
Answer» Solution :Entropy change for heatgin 1 kg of water from `27^(@)C` to `100^(@)C` `Delta S = 2.303 n C_(p) log. ( T_(2))/( T_(1))` `=2.303xx ( 1000)/( 18) "moles"XX ( 4180xx18)/( 1000) J mol^(-1) xx log. ( 373K)/( 300K) ` `= 910.55J` Entropy change for heating 1 kg of water at `100^(@)C` to 1 kg of steam at `100^(@)C` `Delta S = ( DeltaH_(v))/( T) = ( 23 xx 10^(5)J)/( 373 ) = 6166.21J` Entropy change for heating 1kg steam from `100^(@)C ( 373K) ` to `200^(@)(473K) ` `DeltaS=int_(373)^(473)(nC_(p) dT)/(T) = int_373^473((1670 +0.49T))/(T) = dT` `=int_(373)^(473)((1670)/(T)+0.49)dT` `=[1670ln T + 0.49 T]_(373)^(473)` `= 1670 xx 2.303 ( log473- log 373) + 0.49( 473- 373)` `=1670 xx 2.303 ( 0. 1032 ) + 49` `= 396.91 +49 =445. 9 J` `:. `Total entropychange `= 910.55 + 6166.21 +45.9 J = 7522.6 J`

Discussion

9.	Calculate the entropy change when 1 mole of ethanol is evaporated at 351 K. The molar heat of vaporization of ethanol is 39.84 kJ "mol"^(-1)
Answer» Solution :Given : `DeltaH_"VAP"=39840 "J MOL"^(-1) "" DeltaS_v=(DeltaH_"vap")/T_b` `T_b=351 K"" DeltaS_v=39840/351` `DeltaS_v=113.5 J K^(-1) "mol"^(-1)`

Discussion

10.	Calculate the entropy change of a process possessing DeltaH_(t)=2090 J mol^(-1).
Answer» Solution :GIVEN : `S_(n(alpha, 13^@C)) oversettolarr S_(n(BETA, 13^@C))` `DeltaH_t="2090 J mol"^(-1) "" DeltaS_t=(DeltaH_t)/T_t` `T_t` =13+273=286 K `DeltaS_t=2090/286` `DeltaS_t=7.307 J K^(-1) "mol"^(-1)`

Discussion

11.	Calculate the entropy change of a process, possessing DeltaH_(t)=2090Jmol^(-1) S_(n)(alpha,10^(@)C)hArrS_(n)(beta,13^(@)C)
Answer» SOLUTION :GIVEN: `S_(N(alpha,13^(@)C))HARR S_(n(beta,13^(@)C))`

Discussion

12.	Calculate the entropy change of a process H_2O_((l)) to H_2O_((g)) at 373K. Enthalpy of vaporization of water is 40850 J "Mole"^(-1)
Answer» `120 J K^(-1) "mol"^(-1)` `9.1xx10^(-3) J K^(-1) "mol"^(-1)` `9.1xx10^(-4) J K^(-1) "mol"^(-1)` `109.52 J K^(-1) "mol"^(-1)` Solution :ENTHALPY of vaporizationof WATER is=40850 J `"mol"^(-1) ` Boiling point =373 K = `T_b` `DELTAS=(DeltaH)/T_b=40850/373` = 109.517 `=109.52 J K^(-1) "mol"^(-1)`

Discussion

13.	Calculate the entropy change of process H_(2)O_((l))rarrH_(2)O_((g)) at 373K. Enthalpy of vaporization of water is 40850"J Mole"^(-1)
Answer» SOLUTION :`H_2O_((l)) to H_2O_((g))` TEMPERATURE T=373 K Enthalpy of vapourisation of WATER =`DeltaH_"vap"=40580 "J mol"^(-1)` `DeltaS`=entropy CHANGE =`(DeltaH_"vap")/T_b=40850/373` `DeltaS` (entropy change ) =`109.51 "J mol"^(-1) K^(-1)`

Discussion

14.	Calculate the entropy change of a process H_2 O_((l)) to H_2 O_((g))at 373K. Enthalpy of vapourization of water is 40850 J" Mole"^(-1).
Answer» `120 JK^(-1) "mol"^(-1)` `9.1 XX 10^(-3) JK^(-1) mol^(-1)` `9.1 xx 10^(-4) "mol"^(-1)` `109.52 J K^(-1) "mol"^(-1)` SOLUTION :`109.52 JK^(-1) "mol"^(-1)` Enthalpy of vaporization of water is `=DeltaH = 40850 J "mol"^(-1)` Boiling point = 373 K `= T_b` `DeltaS = (DeltaH)/(T_b) = (40850)/(373) = 109.517` `=109.52 JK^(-1) "mol"^(-1)`

Discussion

15.	Calculate the entropy change of a process H_2 O_((f)) toH_2 O_((g)) at 373K. Enthalpy of vaporization of water is 40850 J Mole""^(-1)
Answer» SOLUTION :`H_2 O_((l))toH_2 O` temperature`T =373 K` enthalpyof vapourisationof water =` DELTA H_(vap )= 40580jmol^(-1)` `DeltaS ="entropychange"` ` =( Delta H_(vap))/(T_(b)) = ( 40850 )/( 373 )` ` DeltaS` ( entropychange)` = 109.51 J MOL ^(-1)K^(-1) `

Discussion

16.	Calculate the entropy change involved in the converion of one mole of liquid water at 373 K to vapour at the same temperature ( latenet heat of vaporization of water Delta_(vap) H = 2.257 kJ //g)
Answer» Solution : For the conversion of water `rarr` vapour , the entropy change is given by `Delta_(vap) S = (Delta_(vap) H)/( T_(b))` Here, `Delta_(vap) H=2.257 kJ //g = 2.257 xx18 kJ //mol =40.626 kJ //mol, T_(b) = 373 K` `:. Delta _(vap ) S = (40.626k J mol^(-1))/( 373K) = 0.1089K J K^(-1) mol^(-1) = 108.9 J K^(-1) mol^(-1)`

Discussion

17.	Calculate the entropy changeinvolved in conversion of one mole ( 18g) of solid ice at 273K to liquid water at the same temperature ( latent heat of fusion =6025J mol^(-1))
Answer» Solution :Entropy changefor ice `RARR` WATER is GIVEN by `Delta_(f) S = (Delta_(f)H)/(T_(f))` Here, `Delta_(f) H = 6025 J mol^(-1) , T_(f) = 273 K:.Delta_(f) S = (6025J K^(-1) mol^(-1))/( 273K ) = 22.1 J K^(-1) mol^(-1)`

Discussion

18.	Calculate the entropy change involved in convertion of 1 mole of water at 373 K to vapours at the same temperature. Latent heat of vaporisation of water = 2.257 kJ g^(-1).
Answer» Solution :`k = 1.8 xx 10^(-7)` `DELTA H = 2.257 xx 18 KJ = 40.626 kJ mol^(-1)` `Delta S_("vap") = (Delta H_(vap))/("B.pt. in kelvin") = (40.626 kJ mol^(-1))/(373 K) = 0.1089 kJ mol^(-1) = 108.9 kJ^(-1) mol^(-1)`.

Discussion

19.	Calculate the entropy change in the system and surroundings and the total entropy change in the Universe during a process in which 245 J of heat flow out of the system at 77^@C to the surrounding at 33^@C.
Answer» Solution :Given: `T_("SYS") = 77^(@)C = (77+273) = 350K` `T_("surr") = 33^(@)C = (33+273) = 306K` q=245J `DeltaS_("sys") = (q)/(T_("sys")) = (-245)/(350) = -0.7 JK^(-1)` `DeltaS_("UNIV") = (q)/(T_(sys")) = (+245)/(350) = +0.8JK^(-1)` `DeltaS_("univ") = DeltaS_("sys")+DeltaS_("surr")` `DeltaS_("univ") = -0.7JK^(-1)+0.8JK^(-1) = 0.1 JK^(-1)`

Discussion

20.	Calculate the entropy change in the system and surroundings, and the total entropy change in the universe during a process in which 245 J of heat flow out of the system at 77^@C to the surrounding at 33^@C.
Answer» Solution :Given `T_"SYS"=77^@C =(77+273)`=350 K `T_"surr"=33^@C` =(33+273) =306 K q=245 J `DeltaS_"sys" =q/T_"sys" =(-245)/350=-0.7 JK^(-1)` `DeltaS_"UNIV"=q/T_"sys" =(+245)/350=+0.8 JK^(-1)` `DeltaS_"univ"=DeltaS_"sys" +DeltaS_"surr"` `DeltaS_"univ"=-0.7 JK^(-1) + 0.8 JK^(-1)` `DeltaS_"univ" =0.1 JK^(-1)`

Discussion

21.	Calculate the entropy change in the system, and in the surroundings and the total entropy change in the universe when during a process 75 J of heat flow out of the system at 55^(@)C to the surrounding at 20^(@)C.
Answer» Solution :HEAT FLOW (q) = 75 J Entropy change = `DeltaS`= ? Temperature of the system =`55^@C+273` = 328 K Temperature of the surroundings =`20^@C`+273 =293 K `DeltaS_"system"=q/T_"system"=75/328=0.2286 "J K"^(-1)` `DeltaS_"surroundings"=q/T_"surroundings"=75/293=0.2559 J K^(-1)` `DeltaS_"UNIVERSE"`=Entropy change =`DeltaS_"system"+DeltaS_"surroundings"` =0.2286+0.2559 `DeltaS_"universe"=0.4845 JK^(-1)`

Discussion

22.	Calculate the entropy change in the surrounding when 1.0 mol of H_(2)O(l) is found under standard conditions : Delta_(f)H^(@) = -286 kJ mol^(-1)
Answer» Solution :`H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l), Delta_(f)H^(@) = - 286kJ mol^(-1)` This means that when 1 mol of `H_(2)O(l)`, is formed , 286kJof HEAT is released . This heat is absorbed by the surroundings, i.e., `q_(surr)= +286 KJ mol^(-1)` `:. DeltaS =( q_(surr))/(T) = ( 286kJ mol^(-1))/( 298K) = 0.9597kJ K^(-1) mol^(-1)= 959.7 JK^(-1) mol^(-1)`

Discussion

23.	Calculate the entropy change in surroundings when 1.00 mol of H_(2) O_((l)) is formed under standard conditions. Delta_(f)H^( Theta )= -286 "kJ mol"^(-1).
Answer» SOLUTION :`H_(2(G)) + (1)/(2) to H_(2) O_((l)) ""Delta_(f) H= -286 "kJ/mol" -286 "kJ"` heat is released when ONE mole `H_2 O` formed. So heat is absorbed by surrounding. `q_("surr") = +286 "kJ/mol"` `therefore Delta S= (q_("surr"))/( T) = (286)/( 286)` `= 0.9597 "kJ/mol"= 259.7 "J/K mol"`

Discussion

24.	Calculate the entropy change in surroundings when 1.00 mole of H_(2)O_((l)) is formed under standard conditions Delta_(f)H^(theta)=-286 kJmol^(-1).
Answer» SOLUTION :`959.7 JK ^(-1) "MOLE"^(-1)`

Discussion

25.	Calculate the entropy change for the rusting of iron according to the reaction : 4Fe(s) + 3O_(2)(g) rarr Fe_(2)O_(2)(s) , DeltaH^(@) =- 1648 kJ mol^(-1) Given that the standard entropies of Fe,O_(2)and Fe_(2)O_(3) are 27.3,205.0 and 87.4 JK^(-1) mol^(-1) respectively.Will the reaction be spontandeous at room temperature ( 25^(@)C) ?Justify your answer with appropriate calculations.
Answer» SOLUTION :`Delta_(r) S^(@)= SigmaS^(@) `( PRODUCTS ) `- SigmaS^(@) ` ( Reactants) `=2S^(@) (Fe_(2)O_(3)) -[4 S^(@) ( Fe) + 3S^(@) ( O_(2))]` `= 2xx 87.4- [4 xx 27.3 + 3 xx 205.0 ]JK^(-1) mol^(-1) =- 549 JK^(-1) mol^(-1)` This is the entropy change for the reaction, i.e., system `( Delta S_("system"))` Now, `Delta_(r) G^(@) = Delta_(r) H^(@) - TDelta_(r) S^(@)= - 1648000 J mol^(_1) -298 K xx ( - 549 .4 J K^(-1) mol^(-1))` `= - 1648000 +163721 J K^(-1) mol^(_1) = - 1484279 J K^(-1) mol^(-1)` As `Delta G^(@)`is - ve , the reaction is spontaneous. Alsternatively, as the reaction is exothermic, heat given out by THEREACTION is absorbed by the surroundings at room temperature `(25^(@)C)`. Hence, entropy ofthe surroundings increases. `DeltaS_("surroundings")= (1648xx10^(3)JK^(-1) mol^(-1))/( 298 K ) = 5530 JK^(-1) mol^(_1)` `:. Delta S_("total") = DeltaS_("system") + DeltaS_("surroundings") = -549.4 + 5530 JK^(-1) mol^(-1) = + 4980. 6JK^(-1) mol^(-1)` As `Delta S_("total") ` is `+ ve` ,therefore, the reaction is spontaneous.

Discussion

26.	Calculate the entropy change in surroundings when 1 mol of H_2O_((l)) is formed under standard conditions. Given DeltaH^Ө=-286 "kJ mol"^(-1)
Answer» Solution :`q_"rev"=(-Delta_fH^Ө)=-286 "kJ mol"^(-1) =286000 "J mol"^(-1)` `DeltaS_"(Surroundings)"=q_"rev"/T=("286000 J mol"^(-1))/"298 K"=959 J K^(-1) "mol"^(-1)`

Discussion

27.	Calculate the entropy change during the melting of one mole of ice into water at 0^@C Enthalpy of fustion of ice is 6008 J mol^(-1) (ii) Write the balanced chemical equation for K_(c) = ([CaO_((s))][CO_(2(g))])/([CaCO_(3(s)))
Answer» Solution :Given `DeltaS_("FUSION")=6008 J mol^(-1)` `T_(R) =0^@C = 273 K` `H_(2)O(S) overset(273K) rarr H_(2) O(l)` `Delta S_("fusion")= (Delta H_("fusion"))/(T_f)` `Delta S_("fusion")= (6008)/(273)` `Delta S_("funsion")=22.007 JK^(-1) mol ^(-1)` (ii) `CaCO_(3(s)) hArr CaO_((s))+CO_(2(g))`

Discussion

28.	Calculate the entropy change for the conversion of2 moles of liquid water at 373K to vapours , if Delta_(vap) H is 37.3kJ mol^(-1).
Answer» ANSWER :`200 JK^(-1)`

Discussion

29.	Calculate the entropy change during the melting of one mole of ice into water at 0^@C and 1 atm pressure. Enthalpy of fusion of ice is 6008 J "Mole"^(-1)
Answer» `22.007 "J K"^(-1) "MOLE"^(-1)` 22.007 J K Mole `220.07 "J K"^(-1) "Mole"^(-1)` `2.2007 "J K"^(-1)` Mole Solution :Enthalpy of fusion of ice = 6008 J `"MOL"^(-1) =DeltaH` `DELTAS=(DeltaH)/T_m "" T_m =0^@C` = 273 K Entropy of fusion = `"Enthalpy of fusion of ice "/"Melting point (Kelvin)"` `DeltaS=6008/273 =22.007 J K^(-1) "mol"^(-1)`

Discussion

30.	Calculate the enthalypy of combustion of ethylene (gas) to form CO_(2) (gas ) and H_(2) O (gas) at 298 K and 1 atmospheric pressure. The enthalpies of formation ofCO_(2), H_(2)O and C_(2)H_(4) are - 393.7, -241.8 ,+52.3kJ per mole respectively.
Answer» Solution :We are given `:` (i) `C(s) +O_(2)(g) rarr CO_(2)(g), DELTAH^(@) = - 393.5kJ mol^(-1)` (ii)`H_(2)(g) + (1)/(2) O_(2)(g)rarr H_(2)O(g), Delta H^(@) = - 241.8 kJ mol^(-1)` (iii)`2C(s) + 2H_(2)(g) rarr C_(2)H_(4)(g), Delta H^(@)= + 52.3 kJ mol^(-1)` We aim at `:``C_(2)H_(4)(g) + 3O_(2)(g)rarr 2CO_(2)(g) + 2H_(2)O(g) , Delta _(c ) H^(@) = ?` ` 2 xx` Eqn.`(i) + 2 xx ` Eqn. (ii) - Eqn. (iii) gives `ul{(" "2C(s) ,+2O_(2)(g)rarr2CO_(2)(g)+O_(2)(g),+2H_(2)(g)+2H_(2)O(g)),(-2C(s),,-2H_(2)(g)-C_(2)H_(4)(g)):}` `3O_(2)(g)rarr 2CO_(2) +2H_(2)(g)-C_(2)H_(4)(g)` or `C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g)` `Delta_(c) H^(@)= 2 ( -393.5) + (-241.8) - (52.3) = -1322.9 kJ mol^(-1)` Alternative Method ` :` We aim at `:``C_(2)H_(4)(g)+3O_(2)(g) rarr2CO_(2)(g)+2H_(2)O(g)` We are given `:``Delta_(f) H_((CO_(2)))^(@) = - 393.5 kJ mol^(-1)` `Delta_(f) H_((H_(2)O))^(@)= - 241.8 kJ mol^(-1)` `Delta _(f) H_((C_(2)H_(4)))^(@)= + 52.3 kJ mol^(_1)` `Delta_(R)H^(@) = ("Sum of " Delta_(f)H^(@)" valuesof Products")-("Sum of" Delta_(f)H^(@)" values of Reactants")` `=[ 2 xx Delta_(f) H^(@) (CO_(2))+2xxDelta_(f)H^(@)(H_(2)O)]-[Delta_(f)H^(@)(C_(2)H_(4))+3xxDelta_(f)H^(@)(O_(2))]` `=[ 2 xx ( -393.5) + 2 xx ( -241.8) ] - [ ( 52.3) + 0]( :' Delta_(f) H^(@)` for elementary subtance `=0 )` `=[ -787 .0 -483.6 ] - 52.3 = - 1322.9 kJ mol^(-1)`

Discussion

31.	Calculate the entropy change during the melting of one mole of ice into water at 0^@C and 1 atm pressure. Enthalpy of fusion of ice is 6008 "J mol"^(-1) .
Answer» Solution :`H_(2)O(S)OVERSET(273K)to H_(2)O(l)` `DeltaS_("FUSION")=(DeltaH_("fusion"))/(T_(f))` `= (6008)/(273)` `DeltaS_("fusion")=22.007 J K^(-1)"mole"^(-1)`

Discussion

32.	Calculate the enthalpy of vapourisation of ethanol, given enthalpies of formation of liquid ethanol and gaseous ethanol as -277.6J and -235.4 kJ respectively.
Answer» <P> SOLUTION :`DeltaH_(vap)=H_(p)-H_(R)` `=-235.4kJ-(-277.6kJ)` `=42.2kJ`

Discussion

33.	Calculate the enthalpy of reaction forCO(g) + (1)/(2) O_(2)(g) rarr CO_(2)(g) GivenC(s) + O_(2) (g) rarr CO_(2)(g) , DeltaH = -393.5 kJ mol^(-1) C(s) + (1)/(2) O_(2)(g) rarr CO(g) Delta H = - 110 .5 kJ mol^(-1)
Answer» ANSWER :`-283kJ MOL^(-1)`

Discussion

34.	Calculate the enthalpy of the following reaction : H_(2)C = CH_(2)(g) + H_(2)(g) rarr H_(3)C - CH_(3)(g) Thebond energiesof C-H,C-C,C=C and H-H are 414,347,615 and 435 kJ mol^(-1) respectively
Answer» ANSWER :`-125kJ MOL^(-1)`

Discussion

35.	Calculate the enthalpy of reaction (Delta H ^(@))when ammonia is oxidized : 4NH_(3) (g) + 5O_(2)(g) rarr 6H_(2)O(g) + 4NO(g) Standard enthalpies of formation (Delta_(f) H^(@)) at 25^(@)CforNH_(3)(g),H_(2)O(g)and NO(g) are - 46.2, -241.8 and + 90.4 kJ // mole respectively.
Answer» ANSWER :`-904.4 KJ MOL^(-1)`

Discussion

36.	Calculate the enthalpyof hydrogenation ofC_(2)H_(2)(g)to C_(2)H_(4)(g). ( Given bond energies:C-H= 414.0 kJ mol^(-1), C=C= 827.6 kJ mol^(-1), C=C= 606 .0 kJ mol^(-1), H-H = 430 .5 kJ mol^(-1))
Answer» Solution :`H-C-=C-H+H-Hrarr underset(H)underset(\|) overset(H) overset(\|) (C) = underset(H) underset(\|)overset(H) overset(\|) (C) ` `Delta_(r)H^(@)=`B.E. ( REACTANTS) - B.E. ( PRODUCTS ) `=[ B.E.(C-=C) + 2 XX B.E. ( C-H) + B.E. (H-H) ] - [ B.E.(C=C) + 4B.E.(C-H)]` `[827.6+2 xx414.0+ 430 .5]-[606.0 + 4 xx 414.0] = - 175.9 kJ MOL^(-1)`

Discussion

37.	Calculate the enthalpy of isomerization of ethanol to dimethy ether: C_(2)H_(5)OH(l)toCH_(3)OCH_(3)(g) Given : Enthalpy of vaporisation of equal =41 KJ/mole bond enthalpies C-C=348KJ/mole C-H=415KJ/mole C-o=352KJ/mole O-h=463KJ/mole
Answer» 65KJ/mole 25KJ/mole 125KJ/mole 85KJ/mole Answer :d

Discussion

38.	Calculate the enthalpy of hydrogenation of ethylene, given that the enthalpy of combustion of ethylene, hydrogen and ethane are -1410.0,-286.2 and -1560.6 kJmol^(-1) respectively at 298 K
Answer» Solution :We aregiven (i)`C_(2)H_(4)(G) +3O_(2)(g)rarr 2CO_(2)(g) +2H_(2)O(l), DeltaH= - 1410kJ mol^(-1)` (II) `H_(2)(g) +(1)/(2) O_(2)(g) rarr H_(2)O(l), DeltaH = - 286.2 kJ mol^(-1)` (iii)`C_(2)H_(6)(g) + 3(1)/(2) O (g) rarr 2CO_(2)(g) + 3H_(2)O(l), Delta H = - 1560.6 kJ mol^(_1)` We aim at `:``C_(2)H_(4)+ H_(2)(g) rarr C_(2)H_(6)(g), DeltaH = ?` Equation (i) `+` Equation (ii) - Equation (iii) gives `C_(2)H_(4)(g) +H_(2)(g) rarr C_92)H_(6)(g), Delta H = - 1410.0 + ( - 286.2) - ( 1560.6)` `= - 135.6 kJ mol^(-1)`

Discussion

39.	Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of C-H , C-C , C=C and H-H are 414,347,618 and 435 kJ mol^(-1) .
Answer» Solution :Given : `E_(C-H)=414 "kJ mol"^(-1)` `E_(C-C)=347 "kJ mol"^(-1)` `E_(C=C)=618 "kJ mol"^(-1)` `E_(H-H)=435 "kJ mol"^(-1)` `DeltaH_r=SUM "(Bond energy)"_r -sum "(Bond energy)"_p` `DeltaH_r=(E_(C=C)+4E_(C-H)+E_(H-H))-(E_(C-C) +6E_(C-H))` `DeltaH_r` = (618 + ( 4 x 414 ) + 435 ) -(347 +(6 x 414)) `DeltaH_r`=2709-2831 `DeltaH_r=-122 "kJ mol"^(-1)`

Discussion

40.	Calculate the enthalpy of hydration of anhydrous copper sulphate (CuSO_(4).5H_(2)O). Given that the enthalpies of solutions of anhydrous copper sulphate and hydrated copper sulphate are -66.5 and +11.7 KJ mol^(-1) respectively
Answer» Solution :We are given (i) `(CuSO_(4).5H_(2)O)` `Delta_("sol")H=-66.5KJmol^(-1)` (II) `CuSO_(4).5H_(2)O_((s))+aqrarrCuSO_(4)(aq)` `Delta_("sol")H=-66.5KJmol^(-1)` (ii) `CuSO_(4).5H_(2)O_((s))+aqrarrCuSO_(4)(aq)` `Delta_(sol)H=+11.7KJmol^(-1)` We need `CuSO_(4(s))+5H_(2)O_((l))rarrCuSO_(4).5H_(2)O_((s)),Delta_(hyd)H=?` Substracting equation (ii) from equation (i). we get `CuSO_(4(s))+5H_(2)O_((l))rarrCuSO_(4).5H_(2)O_((s))` `Delta_(hyd)H=-78.2KJmol^(-1)`

Discussion

41.	Calculate the enthalpy of hydration of anhydrouscopper sulphate ( CuSO_(4)) into hydrated copper sulphate (CuSO_(4). 5H_(2)O). Given that the enthalpies of solutionsof anhydrous copper sulphate and hydrated copper sulphate are - 66.5 and + 11.7 kJ mol^(-1) respectively.
Answer» Solution :We are given (i) `CuSO_(4)(s) + aq rarr CuSO_(4)(aq), Delta_(sol)H= - 66.5 kJ mol^(-1)` (ii) `CuSO_(4) . 5H_(2)O(s) +aq rarr CuSO_(4)(aq), Delta_(sol) H= + 11.7 kJ mol^(-1)` we aim at`CuSO_(4) (s) + 5H_(2)O(l) rarr CuSO_(4).5H_(2)O(s), Delta _(HYD) H= ?` Eqaution (i) can be written in two steps as`:` `(ii) CuSO_(4) (s) +5H_(2)O(l) rarr CuSO_(4).5H_(2)O(s) , Delta H= Delta H_(1) kJ mol^(-1)` (iv)`CuSO_(4). 5H_(2)O(s)+ aq rarr CuSO_(4)(aq) , Delta H = DeltaH_(2) kJ mol^(-1)` According to Hess's law, `Delta H_(1) + Delta H_92) = - 66.5 kJ mol^(-1)` Further, equations (ii) and (iv) are same `:. Delta H_(2) = + 11.7 k J mol^(-1)` Putting this value above, we get `Delta H_(1) + 11.7 = - 66.5 ` or `Delta H_(1) = - 66.5- 11.7 kJ = - 78.2 kJ mol^(-1)` Thus, equation (iii) may be written as `CuSO_(4)(s) + 5H_(2)O(l) rarr CuSO_(4). 5H_(2)O(s), Delta _(hyd) H- 78.2kJ mol^(-1)` This is what we AIMED at . Hence, the required value of the ENTHALPY of hydration is`Delta _(hyd) H = - 78.2k J mol^(-1)`.

Discussion

42.	Calculate the enthalpyof formation of water, given that the bond energies of H-H, O=O and O-H bond are 433 kJ mol^(-1), 492 kJ mol^(-1) and 464kJ mol^(-1) respectively.
Answer» Solution :Aim `: H_(2)+(1)/(2)O_(2) rarr H_(2)O,Delta_(f) H^(@) = ? `or `H-H+(1)/(2) O = O rarr H-O-H ,DeltaH = ?` `433 + (1)/(2) XX 492 ` `2 xx 464` `ubrace("=679kJ" )_("Energy ABSORBED")``ubrace("=928kJ" )_("Energy released")` `:.`Netreleased`= 928- 679 = 249 kJ MOL^(-1) , i.e., Delta _(f) H^(@) = - 249 kJ mol^(-1)` Alternatively , `DeltaH = B.E. (H_(2))+ (1)/(2) B.E. (O_(2))- 2B.E. ( O-H)` `=433 + (1)/(2) xx 492 - 2 xx 464 = - 249 kJ mol^(-1)`

Discussion

43.	Calculate the enthalpy of formationn of carbon disulphidegiven that the enthalpy of combustion of carbon disulphide is110.2 kJ mol^(-1) and thoseof sulphur and carbon are 297.4 kJ and 394.5 kJ // g atom respectively.
Answer» Solution :Aim `: C(s)+ 2S(s) rarr CS_(2)(l), DeltaH = ?` GIVEN `: (i) CS_(2)(l) + 3O_(2)(G) rarr CO_(2)(g) + 2SO_(2)(g), DELTA H- 110.2 kJ mol^(-1)` (ii)`S(s) + O_(2)(g) rarr SO_(2)(g) , Delta H= - 297.4 kJ mol^(-1)` (iii) `C(s)+O_(2)(g) rarrCO_(2)(g), Delta H =- 394.5 kJ mol^(-1)` Eqn. (iii) ` +2 xx ` Eqn. (ii) - Eqn. (i) gives the required result.

Discussion

44.	Calculate the enthalpy of formation of methane, given that the enthalpiedof combustion of methane, graphite and hydrogen are 890.2 kJ, 393.4 kJ and 285.7 kJ mol^(-1) respectively.
Answer» SOLUTION :We are given `:(i) CH_(4) + 2O_(2) rarr CO_(2)+ 2H_(2)O,DeltaH== -890.2 kJ mol^(-1)` (II) `C+ O_(2) rarr CO_(2), Delta H = - 393.4 kJ mol^(-1)``(iii) H_(2)+(1)/(2)O_(2) rarr H_(2)O , Delta H = - 285.7 kJ mol^(-1)` We aim at `: C+2 H_(2)rarr CH_(4), Delta H = ?` In order to get thisthermochemical equation, multiplyeqn. (iii) by 2 and it to eqn. (ii) and then subtract eqn. (i) from their sum. We get `:` `C+2H_(2) rarr CH_(4), Delta H = -393.4+2(-285.7) - (-890.2) kJ mol^(-1) = -74.6 kJ mol^(-1)` Hence, the heat of formation of methane is `:Delta _(F) H= -74.6 kJ mol^(-1)`

Discussion

45.	Calculate the enthalpy of formation of methanol from the following data, CH_(3)OH(l) + (3)/(2)O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) ….(i)""Delta H^(@) = -726.4 kJ mol^(-1) C"(graphite)" + O_(2)(g) rarr CO_(2)(g) ….(iii) ""Delta H^(@) = -393.5 kJ mol^(-1) H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l) ....(iii) ""Delta H^(@) = -285.8 kJ mol^(-1).
Answer» Solution :Required equation, `C "(graphite)" + 2H_(2)(g) + (1)/(2)O_(2)(g) rarr CH_(3)OH(l)` Reverse Eq. (1) MULTIPLY Eq. (3) by 2 and add to Eq. (2) `{:(CO_(2)(g)+2H_(2)O(l) rarr CH_(3)OH(l)+ (3)/(2)O_(2)(g) "" DELTA H^(@) = +726.4 kJ MOL^(-1)),(2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l) ""Delta H^(@) = -571.6 kJ mol^(-1)),("C(graphite)" + O_(2)(g) rarr CO_(2)(g) ""Delta H^(@) = -393.5 kJ mol^(-1)):}/("C(graphite)" + 2H_(2)(g) + (1)/(2)O_(2)(g) rarr CH_(3)OH(l)Delta H^(@) = -238.5 kJ mol^(-1))` `Delta_(f)H^(@)` of `CH_(3)OH = -238.5 kJ mol^(1)`.

Discussion

46.	Calculate the enthalpy of formation of methane from the following data. C_((s))+O_(2_((g)))toCO_(2)(g)DeltaH^(@)=-393.5kJ…………..1 H_(2_((g)))+1/2O_(2(g))toH_(2)O_((l))DeltaH^(@)=-285.8kJ…………2 CH_(4_((g)))+2O_(2_((g))) rarr CO_(2)(g)+2H_(2)O_((l))DeltaH^(@)=-890kJ......................3
Answer» SOLUTION :required thermochemical equation is `C(S)+2H_(2)(g)toCH_(4)(g)DeltaH^((-))=?, C_((S))+2H_(2(g))toCH_(4(g))` Eqn 1 as it is `C_((s))+O_(2(g))toCO_(2(g))DeltaH^((-))=-393//5kJ` Eqn 2 `X^(2)` eqn Reverse Eqn 3 `2H_(2(g))+2/2O_(2)(g)to2H_(2((l)))DeltaH^((-))=571.6kJ` `CO_(2(g))+2H_(2)O_((l))toCH_(4(g))+2O_(2(g))=-57kJ` `DeltaH^((-1))=+890kJ` `C_((s))+2H_(2(g)toCH_(4(g))DeltaH^((-))=-75kJ`

Discussion

47.	Calculate the enthalpyof formationof sucrose(C_(12)H_(22)O_(11)) from the following data : (i)C_(12)H_(22)O_(11)+12O_(2) rarr12 CO_(2) + 11H_(2)O,Delta H = - 5200.7 kJ mol^(-1) (ii) C+ O_(2) rarr CO_(2), Delta H = - 395.5 kJ mol^(-1)(iii)H_(2)+ (1)/(2) O_(2)rarr h_(2)O , Delta H = - 285.8 kJ mol^(-1)
Answer» SOLUTION :Aim `: 12 C(s) + 11 H_(2) (g) + (11)/(2) O_(2) (g) rarr C_(12) H_(12) O_(11) (s), DeltaH = ?`

Discussion

48.	Calculate the enthalpy of formation of anhydrous Al_(2)Cl_(6) from the following data : (i) 2Al(s) + 6HCl(aq) rarr Al_(2)Cl_(2) + 3H_(2)(g) + 1004.2kJ .kJ mol^(-1) (ii)H_(2)(g) +Cl_(2)(g) rarr 2HCl(g) + 184.1 kJ mol^(-1) (iii) HCl(g) + aq rarr HCl(aq) + 73.2kJ mol^(-1) (iv)Al_(2)Cl_(6)(s) +aq rarr Al__(2)Cl_(6)(aq) +643.1 kJ mol^(-1)
Answer» Solution :AIM `: 2 Al(s) 3Cl_(2)(g) rarr Al_(2)Cl_(6)(s), DELTAH = ?` Eqn. `(I ) + 3 xx `Eqn. (ii) - Eqn. (iv) `+ 6 xx` Eqn. (iii) gives the required result.

Discussion

49.	Calculate the enthalpy of formation of carbon monoxide (CO) from the following data: (i) C(s) + O(g) rarr CO_(2)(g), Delta_(r)H^(@) = - 393.5 kJ mol^(-1) (ii) CO(g) + (1)/(2) O_(2)(g) rarrCO_(2)(g) , Delta _(r) H^(@) = - 283. 0 kJ mol^(-1)
Answer» Solution :We AIM at` :C(s) +(1)/(2) O_(2)(g) rarr CO(g) , Delta_(f) H^(@) = ?` Subtracting eqn. (ii) fromeqn. (i), we get `C(s) +(1)/(2) O_(2)(g)- CO(g) rarr0, Delta_(R) H^(@) = - 393.5 - (- 283.0) = - 110.5 KJ mol^(-1)` or `C(s) +(1)/(2) O_(2)(g) rarr CO (g) , Delta _(r) H^(@) = - 110.5 kJ` `:. ` Heat of FORMATION of CO is`: Delta _(f) H^(@) = - 110.5 kJ mol^(-1)`

Discussion

50.	Calculate the enthalpy of formation of CO from the given information. C_((s)) + O_(2(g)) to CO_(2(g)) , DeltaH^(@) = -x " kJ" 2CO_((g)) + O_(2(g)) to 2CO_(2(g)) , Delta H^(@) = - y " kJ"
Answer» `(2x-y)/( 2)` `(y-2x)/( 2)` `2x-y` `y-2x` ANSWER :B

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

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Calculate the enthalpy of hydrogenation of ethylene, given that the enthalpy of combustion of ethylene, hydrogen and ethane are -1410.0,-286.2 and -1560.6 kJmol^(-1) respectively at 298 K

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Calculate the enthalpy of hydration of anhydrous copper sulphate (CuSO_(4).5H_(2)O). Given that the enthalpies of solutions of anhydrous copper sulphate and hydrated copper sulphate are -66.5 and +11.7 KJ mol^(-1) respectively

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Calculate the enthalpy of formation of methane, given that the enthalpiedof combustion of methane, graphite and hydrogen are 890.2 kJ, 393.4 kJ and 285.7 kJ mol^(-1) respectively.

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Calculate the enthalpy of formation of CO from the given information. C_((s)) + O_(2(g)) to CO_(2(g)) , DeltaH^(@) = -x " kJ" 2CO_((g)) + O_(2(g)) to 2CO_(2(g)) , Delta H^(@) = - y " kJ"