94420 + Interview Questions in Chemistry in Class 12

1.	For the reaction, P_((g))+3Q_((g))iff4R_((g)) Initial concentration of P is equal to that of Q. The equilibrium concentration of P and R are equal. K_(c) is equal to
Answer» `0.08` `0.8` 8 `(1)/(8)` SOLUTION :Let initial concentration of P and Q are equal and at equilibrium x concentration of P reacts, then `{:(,P_((g)),+,3Q_((g)),iff,4R_((g))),("Initial conc.",C,,C,,0),(("moles "L^(-1)),,,,,),("At equilibrium",(C-x),,(C-3x),,4x),("But it is given that",(C-x)=4x,,,,):}` `therefore C=5x` `K_(c)=([R]^(4))/([P][Q]^(3))=((4x)^(4))/((4x)(5x-3x)^(3))=(64X^(3))/(8x^(3))=8`

Discussion

2.	For the reaction Pcl_(5(g))hArrPCl_(3(g))+Cl_(2(g))
Answer» `K_(p)=K_(C)` `K_(p)=K_(c)(RT)^(-1)` `K_(p)=K_(c)(RT)` `K_(p)=K_(c)(RT)^(2)` SOLUTION :`Deltan=2-1=1` `K_(p)=K_(c)(RT)`

Discussion

3.	For the reaction : PCl_(5)(g) rarr PCl_(3)(g) + Cl_(2)(g)
Answer» `DELTA H = Delta U` `Delta H gt Delta U` `Delta H lt Delta U` `Delta H = 0`. ANSWER :B

Discussion

4.	For the reactionPCl_(3(g))+Cl_(2(g))hArrPCl_(5(g))at 250^(@)C, then value of K_(c) is 26, then the value of K_(p) the same temperaturee will be
Answer» <P>`0.61` `0.57` `0.83` `0.46` Solution :`K_(p)=K_(c)(RT)^(DELTAN)=26(0.0821xx523)^(-1)=0.61.` `Deltan_(G)=1-2=-1`

Discussion

5.	For the reaction PCL_5 hArr PCL_3+CL_2, the forward reaction at constant temperatureis fovoured by :
Answer» introducing an inret gas at CONSTANT VOLUME introducing CHLORINE gas at constant volume introdusing an inert gas at constant pressure increasing the volume of the CONTAINER ANSWER :D

Discussion

6.	For the reaction, PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g), the position of equilibrium can be shifted to the right by:
Answer» Increasing the TEMPERATURE Doubling the volume Additional of `Cl_(2)` at constant volume Additional of equimolar quantities of `PCl_(3) and PCl_(5)` Solution :ACCORDING to Le-chatelier principle when CONCENTRATION of REACTANT increases, the equlibrium shift in favour of forward reaction.

Discussion

7.	For the reaction :PCl_(3)(g)+Cl_(2)(g)hArr PCl_(5)(g)the value of K_(c ) at 250^(@)Cis 26.The value of K_(p)at this temperature will be :(R = 0.082 L atmmol^(-1)K^(-1))
Answer» `0.61` `0.46` `0.92` `0.57` ANSWER :A

Discussion

8.	For the reaction PCI_5 hArr PCI_3(g) +CI_2(g) the forward reaction at constant temperature is favoured by
Answer» INTRODUCING an inert gas at CONSTANT volume introducing chlorine gas at constant volume introducing an inert gas at constant pressure increasing the volume of the container introducing `PCI_5` at constant volume ANSWER :C::D

Discussion

9.	For the reaction PCI_3 (g) +CI_2(g) hArr PCI_5(g), the value of K_p at 250^@C is 0.61 atm^(-1). The value of K_c at this temperature will be
Answer» `15 (mol//I)^(-1)` `26 (mol//I)^(-1)` `35 (mol//I)^(-1)` `52 (mol//I)^(-1)` Solution :`PCI_3(g)+CI_2(g) hArrPCI_5(g)` `DELTAN=-,K_p=0.61 ATM ^(-1)` `K_c=K_p(RT)^(-Deltan)` `=0.61(0.0821 xx 523)^(+1)=26 mol//I`.

Discussion

10.	For the reaction PCI_5 (g) hArr PCI_3 (g) +CI_2(g). The forward reaction at constant temperature is favoured by
Answer» introducing an INERT GAS at constant volume introducing an inert gas at constant pressure increasing the volume of the container introducing `PCI_5` at constant volume Answer :B::C::D

Discussion

11.	For the reaction PCI_3(g) +CI_2(g) hArr PCI_5(g) , the value of K_p at 250^@C is 0.61 "atm"^(-1). The value of K_cat this temperature will be
Answer» `15 mol I^(-1)` `26 mol I^(-1)` `35 mol I^(-1)` `52mol I^(-1)` Answer :B

Discussion

12.	for the reaction P+Q hArr R + 2S, initially the concentration of P is equal to that of Q (1 molar) but at equilibrium the concentration of R will be twice of that of P, then the equilibrium constant of the reaction is
Answer» `4/3` `32/3` `3/10` `1/10` ANSWER :B

Discussion

13.	For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, DeltaU and w correspond to PCl_(5)(g) rarr PCl_(3) (g) + Cl_(2)(g)
Answer» `DeltaH = DELTAE` `DeltaH gt DeltaE` `DeltaH LT DeltaE` None of these Solution :`PCl_(5)(g)rarrPCl_(3)(g)+Cl_(2)(g)` For this reaction `Deltan_(g)=2-1=1` `Deltan_(g)` is positive, i.e., there is an increase in the number of GASEOUS moles then `DeltaHgtDeltaE`

Discussion

14.	For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, DeltaU and w correspond to
Answer» `DeltaU lt 0, w=0` `DeltaU=0, wlt0` `DeltaUgt0, w=0` `DeltaUlt0, wgt0` Solution :Bomb calorimeter is commonly used to find the heat of combustion of ORGANIC substances which consists of a scaled combustion chamber, called a bomb. If a process is RUN in a sealed container then no expansion or COMPRESSION is allowed, so w = 0 and `DeltaU = q`. `DeltaUlt0, w=0`

Discussion

15.	For the reaction NO_(3)^(-)toNO_(2) (acidic medium), E^(@)=0.790V NO_(3)^(-)toNH_(3)OH (acidic medium), E^(@)=0.731V Calculate the pH at which at the above two half reactions will have same E values (Assume the concentrations of all the species to be unity).
Answer» Solution :`NO_(3)^(-)+2H^(+)+E^(-)toNO_(2)+H_(2)O,""E^(@)=0.790V` `NO_(3)^(-)+7H^(+)+6e^(-)toNH_(2)OH+2H_(2)O,E^(@)=0.731V` Since E VALUES for both the reactions are same `E_(NO_(3)^(-)//NO_(2))=E_(NO_(3)^(-)//NH_(3)OH)` `thereforeE_(NO_(3)^(@)//NO_(2))+(0.059)/(1)"LOG"([H^(+)][NO_(3)^(-)])/([NO_(2)])=E_(NO_(3)^(-)//NH_(2)OH)^(@)+(0.059)/(6)"log"([H^(+)]^(7)[NO_(3)^(-)])/([NH_(2)OH])` or `0.790+0.059log[H^(+)]^(2)=0.731+(0.059)/(6)log[H^(+)]` (as concentrations of all SPECIES =1, given) or `0.790+0.118log[H^(+)]=0.731+0.0688log[H^(+)]` or `(0.118-0.0688)log[H^(+)]=0.731-0.790` or `0.0492log[H^(+)]=0.059` or `-log[H^(+)]=(0.059)/(0.0492)=1.1992` or `pH=1.1992`

Discussion

16.	For the reaction NOBr (g) hArrNO(g)+1/2Br_(2)(g),K_(p)=0.15 atm at 90°C. If NOBr, NO and Br_(2) are mixed at this temperature having partial pressures 0.5 atm,0.4 atm & 2.0 respectively, will Br_(2) be consumed or formed?
Answer» Solution :`Q_(P)=([P_(Br_(2))]^(1//2)[P_(NO)])/([P_(NOBR)])=([0.20^(1//2)[0.4]])/([0.50])=0.36` `K_(p)=0.15` HENCE, reaction will shift in backward direction, Therefor `Br_(2)` will be consumed.

Discussion

17.	For the reaction, "NO"_(2)(g)+"CO "(g)to"CO"_(2)(g)+"NO"(g), the experimentally determined rate expression below 400 K is : rate =k["NO"_(2)]^(2). What mechanism can be proposed for this reaction ?
Answer» ANSWER :SEE EXAMPLE (III) on PAGE `4//20`

Discussion

18.	For the reaction NO_(2)+CO rarr CO_(2)+NO the experimental rate expression is (dc)/(dt)=k[NO_(2)]^(2) the number of molecules of CO involved in the slowest step will be:
Answer» 0 1 2 3 Answer :A

Discussion

19.	For the reaction : Ni(s)+2Ag^(+)(1 M) to Ni^(2+)(1 M) +2Ag(s) which species gets reduced ?
Answer» SOLUTION :`AG^(+)` IONS GET REDUCED.

Discussion

20.	For the reaction Ni^(2+)+4NH_(3) to [Ni(NH_(3))_(4)]^(2+), at equilibrium is 0.5M. Then the instability constant of the complex will be approximately equal to :
Answer» `1.0xx10^(-5)` `1.5xx10^(-16)` `1.0xx10^(7)` `1.5xx10^(-17)` Solution :`K_(3)=(100)/(1.6xx10^(-4)XX(0.5)^(4))` `K_("INS")=(1)/(K_(s))=1.6xx(0.5)^(4)xx10^(6)=10^(7)`

Discussion

21.	For the reaction Ni^(2+) +4NH_(3) iff [Ni(NH_(3))_(4)]^(2+), at equilibrium, if the solution contain 1.65xx10^(-4)% ofnickel in the free state, and the concentration of NH_(3) at equilibrium is 0.5 M. Then the instability constant of the complex will be approximately equal to :
Answer» `1.0xx10^(-5)` `1.5xx10^(-16)` `1.0xx10^(-7)` `1.5xx10^(-17)` SOLUTION :`NI^(2+)+4NH_(3) iff [Ni(NH_(3))_(4)]^(2+)"":. K=([Ni(NH_(3))_(4)]^(2+))/([Ni^(2+)[NH_(3)]^(4)])` But ` ([Ni^(2+)])/([Ni^(2+)]+[Ni(NH_(3))_(4)]^(2+))=1.6xx10^(-6)` `or (Ni^(2+))/([Ni(NH_(3))_(4)]^(2+))~~1.6xx10^(-6)"":. K=(10^(6))/(1.6xx(0.5)^(4))=10^(7)` Hence instability constant`=10^(-7)`

Discussion

22.	For the reaction : NH_3 + OCI^(-) rarr N_2H_4 + Cl^(-) in basic medium, the coefficients of NH_3, OCI^(-) and N_2H_4 for the balanced equation are respectively
Answer» 2,2,2 2,2, I 2, 1, I 4,4,2 Solution :(C) The BALANCED equation:`2NH_3 + OCI^(-) RARR N_2,H_4 + CL^(I) + H_2O`

Discussion

23.	For the reaction N_(2)O_(6)to2NO_(2)+1/2O_(2) the only correct combination is
Answer» (III)(II)(Q) (I)(iii)(P) (II)(i)(S) (IV)(i)(P) Solution :For a `1^(st)` ORDER reaction `k=2.303/t"log" a/(a-x)`

Discussion

24.	For the reaction N_2O_(5(g))rarr2NO_(2_(g))+1/2O_(2(g)) , the value of rate of disappearance of N_2O_5 is given as 6.5xx10^-2) molL^(-1)s^(-1). The rate of formation of NO_2 and O_2 is given respectively as .......
Answer» `(3.25xx10^(-2) "MOL L"^(-1)s^(-1)) and (1.3xx10^(-2)"MOLL"^(-1)s^(-1))` `(1.3xx10^(-2)"molL"^(-1)s^(-1) and (3.25xx10^(-2) "mol L"^(-1)s^(-1)) ` `(1.3xx10^(-1)"molL"^(-1)s^(-1)and (3.25xx10^(-2) "mol L"^(-1)s^(-1))` none of these Solution :Rate `(d[N_2O_5])/(dt)=(1/2)(d[NO_2])/(dt)=(2D[O_2])/(dt)` Given that , `(d[N_2O_5])/(dt)=6.5xx10^(-2)"mol L"^(-1)s^(-1)` `(d[NO_2])/(dt)=2xx6.5xx10^(-2)=1.3xx10^(-1)"mol L"^(-1)s^(-1)` `(d[O_2])/(dt)=(6.5xx10^(-2))/2=3.25xx10^(-2)"mol L"^(-1)s^(-1)`

Discussion

25.	For the reaction :NH_(2)COONH_(4)(g) hArr 2NH_(3)(g)+CO_(2)(g) equilibrium pressure was found to be 3 atm at 1000 K. K_(p) is :
Answer» `(4)/(27)` 4 27 `(27)/(4)` ANSWER :B

Discussion

26.	For the reaction N_(2)O_(5) (g) to 2NO_(2) (g) + (1)/(2) O_(2) (g) the value of rate of disappearance of N_(2)O_(5) is given as 6.25 xx 10^(-3) mol L^(-1) S^(-1) . The rate of formation of NO_(2) and O_(2) is given respectively as
Answer» `1.25 xx 10^(-2)` mol `L^(-1) S^(-1)` and ` 6.25 xx 10^(-3)` mol `L^(-1) S^(-1)` `6.25 xx 10^(-3)` mol `L^(-1) S^(-1)` and ` 6.25 xx 10^(-3)` mol `L^(-1) S^(-1)` `1.25 xx 10^(-2)` mol `L^(-1) S^(-1)` and ` 3.125 xx 10^(-3)` mol `L^(-1) S^(-1)` `6.25 xx 10^(-2)` mol `L^(-1) S^(-1)` and ` 3.125 xx 10^(-3)` mol `L^(-1) S^(-1)` Solution :GIVEN `(-d[N_(2)O_(5)])/(dt) = 6.25 xx 10^(-3)` mol `L^(-1) S^(-1)` Fore the reaction , `N_(2)O_(5) to 2NO_(2) + (1)/(2) O_(2)` `(-d[N_(2)O_(5)])/(dt) = (1)/(2) (d[NO_(2)])/(dt) = (2d[O_(2)])/(dt)` `therefore (d[NO_(2)])/(dt) = -(2d[N_(2)O_(5)])/(dt) = 1.25 xx 10^(-2)` mol `L^(-1) S^(-1)` `therefore (d[O_(2)])/(dt) = - (1)/(2) (d[N_(2)O_(5)])/(dt) = 3.125 xx 10^(-3) ` mol `L^(-1) S^(-1)`.

Discussion

27.	For the reaction N_(2)O_(5)(g) to 2NO_(2)(g)+(1)/(2)O_(2)(g) the rate of disappearance of N_(2)O_(5) is 6.25xx10^(-3)"mol"L^(-1)s^(-1). The rate of formation of NO_(2) and O_(2) will be respectively
Answer» `6.25xx10^(-3)"mol"L^(-1)s^(-1)"and"` `6.25xx10^(-3)"mol"L^(-1)s^(-1)` `1.25xx10^(-2)"mol"L^(-1)s^(-1)"and"` `3.125xx10^(-3)"mol"L^(-1)s^(-1)` `6.25xx10^(-3)"mol"L^(-1)s^(-1)"and"` `3.125xx10^(-3)"mol"L^(-1)s^(-1) `1.25xx10^(-2)"mol"L^(-1)s^(-1)"and"` `6.25xx10^(-3)"mol"L^(-1)s^(-1)` ANSWER :B

Discussion

28.	For the reaction, N_(2)O_(4(g))iff2NO_(2(g)), the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is/are correct?
Answer» The reaction is endothermic The reaction is exothermic If `NO_(2(g))andN_(2)O_(4(g))` are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more `N_(2)O_(4(g))` will be FORMED The entropy of the SYSTEM remains constant Solution :For an endothermic reaction, K increases with increase in temperature. At 400 K, `Q=((p_(NO_(2)))^(2))/((p_(N_(2)O_(4))))=((20)^(2))/(2)=200` As `QgtK`, hence, reaction proceeds in backward DIRECTION.

Discussion

29.	For the reaction, N_(2)O_(4(g))hArr2NO_(2(g)) DeltaU=2.1kCal,DeltaS=20" Cal "K^(-1) at 300K, then DeltaG is:-
Answer» `+9.3` KCAL `-9.3` kCal `-3.3` kCal `+3.3` kCal Solution :`DeltaH=DeltaU+Deltan_(G)RT` `=2.1+(1xx2xx300)/(1000)` `=2.1+0.6` `=2.7` kCal `DeltaG=DeltaH-TDeltaS` `=2.7-(300xx20)/(1000)` `=2.7-6` `=-3.3` kCal

Discussion

30.	For the reaction, N_2O_5 to 2NO_2 + O_2 , Given(-d)/(dt)[N_2O_5]=K_1[N_2O_5] d/(dt)[NO_2]=K_2[N_2O_5] d/(dt)[O_2]=K_3[N_2O_5], the relation in between of K_1K_2K_3 is
Answer» `2K_1=K_2=4K_3` `K_1=K_2=K_3` `2K_1=4K_2=K_3` `2K_1=2K_2=3K_3` ANSWER :A

Discussion

31.	For the reaction:N_(2)O_(4)(g)hArr 2NO_(2)(g)the concentration of the equilibrium mixture at 300 K are [N_(2)O_(4)]=4.8xx10^(-2)mol L^(-1) and [NO_(2)]=1.2xx10^(-2)mol L^(-1).K_(c ) for the reaction is :
Answer» `0.25` `3xx10^(-1)MOL L^(-1)` `3xx10^(3)mol L^(-1)` `3xx10^(-3)mol L^(-1)` Solution :`K_(C )=([NO_(2)]^(2))/([N_(2)O_(4)])` `=((1.2xx10^(-2))^(2))/((4.8xx10^(-2)))` `=3xx10^(-3)mol L^(-1)`

Discussion

32.	For the reaction, N_2O_4(g)0 mention two factors whose change at equilibrium of the reaction will increase the yield of NO_2(g)
Answer» SOLUTION :DECREASE in temperature, INCREASE in pressure at CONSTANT temperature

Discussion

33.	For the reaction, N_2O_3(g)iff 2NO_2(g) +0.5 O_2(g) , calculate the mole fraction of N_2O_5(g)decomposed at a constant volume and temperature, if the initial pressure is 600 mmHg and the pressure at any time is 960 mmHg. Assume ideal gas behaviour.
Answer» SOLUTION : If .p. MM of `N_2O_5` DECOMPOSES then 600 - p + 2p +p/2 = 960 0.25

Discussion

34.	For the reaction : N_(2)O_(3(g))hArrNO_((g)) + NO_(2(g)) , total pressure = P, degree of dissociation = 50%. Then Kp would be –
Answer» 3P 2P `P/3` `P/2` ANSWER :C

Discussion

35.	For the reactionN_(2(g))+O_(2(g))hArr2NO_((g)), the equlibrium constant is K_(1),The equilibrium constant is K_(2) for the reaction 2NO_((g))+2_(2(g))hArr2NO_(2(g)). What is K for the reaction NO_(2(g))hArr1/2N_(2(g))+O_(2(g))
Answer» `(1)/((K_(1)K_(2)))` `(1)/((2K_(1)K_(2)))` `(1)/((4K_(1)K_(2)))` `[(1)/(K_(1)K_(2))]^(1//2)` Solution :`N_(2(g))+O_(2(g))hArr2NO_((g)),K_(1)` `(2NO_((g))+O_(2(g))hArr2NO_(2(g)),K_(2))/(N_(2(g))+2O_(2(g))hArr2NO_(2(g)),K=K_(1)xxK_(2))` `THEREFORE"For"NO_(2(g))hArr1/2N_(2(g))+O_(2(g)),K'[(1)/(K_(1)K_(2))]^(1/2).`

Discussion

36.	For the reaction N_(2(g))+O_(2(g))hArr2NO_((g)), the value of K_(c) at 800^(@)C is 0.1. When the equilibrium concentrations of both the reactants is 0.5 mol, what is the value of K_(p) at the same temperatuer
Answer» `0.5` `0.1` `0.01` `0.025` Solution :`N_(2(G))+O_(2(g))hArr2NO_((g))` `K_(C)=0.1 K_(p)=K_(c)(RT)^(DELTAN)` `Deltan=0,K_(p)=K_(c)=0.1`

Discussion

37.	For the reactionN_(2(g))+O_(2(g))rArrNO_((g)),the value of K_(c)at800^(@) C is 0.1 . What is the value of K_(p) atthis temperature ?
Answer» `0.5` `0.01` `0.05` `0.1` ANSWER :D

Discussion

38.	For the reaction, N_(2(g))+3H_(2(g))hArr2NH_(3(g))at 400K, K_(p)=41 Find the vlaue of K_(p) for the following reaction 1/2N_(2(g))+3/2H_(2(g))hArrNH_(3(g))
Answer» `6.4` `0.02` 50 `4.6` Solution :`K_(p)=([p_(NH_(3))])/([P_(N_(2))][P_(H_(2))]^(3))=41` `impliesP_(NH_(3))=SQRT(K_(p)[P_(N_(2))][P_(H_(2))]^(3))""...(i)` `K'_(p)=([P_(NH_(3))])/([P_(N_(2))]^(1//2)[P_(H_(2))]^(3//2))""...(II)` Putting (i) in (ii), we get `K'_(p)=sqrt(K_(p)[P_(N_(2))][P_(H_(2))]^(3))=sqrt(41)=6.4.`

Discussion

39.	For the reaction N_(2(g)) + O_(2(g)) hArr 2NO_((g))the value of K, at 800^(@)C is 0.1. When the equilibrium concentrations of both the reactants is 0.5 mol, what is the value of K_(p) at the same temperature?
Answer» <P>0.5 0.1 0.01 0.025 Solution :`N_(2(g))+O_(2(g))hArr2NO(g)` `K_(c)=0.1,K_(p)=K_(c)(RT)^(/_\N)` `/_\n=0,K_(p)=K-(c)=0.1`

Discussion

40.	For the reaction N_(2(g))+3H_(2(g))rarr 2NH_(3(g)) the rate of the reaction in terms of ammonia is __________
Answer» `+(1)/(2)(d[NH_(3)])/(DT)` `-(1)/(2)(d[NH_(3)])/(dt)` `(-d[NH_(3)])/(dt)` `(+d[NH_(3)])/(dt)` ANSWER :A

Discussion

41.	For the reaction, N_2(g)+3H_2(g)rarr2NH_3(g), which is true:
Answer» `triangleH=triangleU` `triangleHlttriangleU` `triangleHgttriangleU` None Answer :B

Discussion

42.	For the reaction N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) at 298K, enthalpy and entropy changes are -92.4 kJ and -198.2 JK^(-1) respectively. Calculate the equilibrium constant of the reaction (R=8.314JK^(-1)mol^(-1)).
Answer» SOLUTION :`DELTAG=DeltaH-TDeltaS`. CALCULATE `DeltaG`.

Discussion

43.	For the reaction, N_(2)(g) + 3H_(2)(g) to 2NH_(3), how are the rate of reaction expressin inter- related?
Answer» SOLUTION :Rate `= (-dN_(2))/(DT) =-1/3dH_(2)[Br_(2)]^(1//2), "ORDER" = 11/2`

Discussion

44.	For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(3) ( g) under certain conditions of temperature and partial pressure of the reactants, the rate of formation of NH_(3) is 0.001kgh^(-1). The rate of conversion of H_(2) under the same conditions is
Answer» `1.82 XX 10^(-4) kg //HR` `0.0015 kg //hr` `1.52 xx 10^(4) kg //hr` `1.82 xx 10^(-14) kg //hr` Answer :B

Discussion

45.	For the reaction N_(2)(g) + 2 H_(2) (g) to 2 NH_(3) (g) under certain conditions of temperature and partial pressure of the reactants , the rate of formation of NH_(3) is 0.001 kg h^(-1) . The rate of conversion of H_(2) under the same conditions is
Answer» `1.82 xx 10^(-4)`kg/hr `0.0015` kg/hr `1.52 xx 10^(4)` kg/hr `1.82 xx 10^(-14)` kg/hr Solution :`(-dN_(2))/(dt) = (-1)/(3) (dH_(2))/(dt) = (1)/(2) (dNH_(5))/(dt) implies (dH_(2))/(dt) = (3)/(2) xx 0.001 = 0.001 ` 5 kg 1`hr^(-1)`.

Discussion

46.	For the reaction N_(2)+3H_(2)to2NH_(3), the rate of disappearance of H_(2) is "0.01 mol lit"^(-1)"min"^(-1). The rate of appearance of NH_(3) would be
Answer» `"0.001 mol LIT"^(-1)"MIN"^(-1)` `"0.02 mol lit"^(-1)"min"^(-1)` `"0.007 mol lit"^(-1)"min"^(-1)` `"0.002 mol lit"^(-1)"min"^(-1)` ANSWER :C

Discussion

47.	For the reaction N_(2)+3H_(2)to2NH_(3) , the rate (d[NH_(3)])/(dt)=2xx10^(-4)Ms^(-1).Therfore the rate -(d[N_(2)])/(dt) is given
Answer» `10^(-4)M"sec"^(-1)` `10^(4)M"sec^(-1)` `10^(-2)M"sec"^(-1)` `4xx10^(-4)M"sec"^(-1)` Solution :`(r_(N_(2)))/1=(r_(H_(2)))/3=(r_(NH_(3)))/2,r_(N_(2))=(2xx10^(-4))/2=10^(-4)`

Discussion

48.	For the reaction N_(2)+3H_(2)iff2NH_(3),DeltaH=?
Answer» `DeltaE-RT` `DeltaE-2RT` `DeltaE+RT` `DeltaE+2RT` SOLUTION :`DeltaH=DeltaE+DeltanRT` SINCE `Deltan=-2` Than `DeltaH=DeltaE-2RT`.

Discussion

49.	For the reaction : N_(2)+3H_(2)hArr 2NH_(3), equal number of moles of N_(2) and H_(2) were taken in a 1L flask. Which of the following is correct at equilibrium ?
Answer» `[H_(2)]=[N_(2)]` `[H_(2)]GT [N_(2)]` `[H_(2)]LT [N_(2)]` `[H_(2)]` and `[N_(2)]=0` Answer :C

Discussion

50.	For the reaction N_2+ 3H_2 to 2NH_3 the rate of change of concentration for hydrogen is -0.3 xx 10^(-4) Ms^(-1). The rate of change of concentration of ammonia is
Answer» `-0.2 xx 10^(-4)` `0.2 xx 10^(-4)` `0.1 xx 10^(-4)` `0.3 xx 10^(-4)` SOLUTION :`(d[H_2])/(DT)=-0.3xx 10^(-4)Ms^(-1)` `"Rate"=-1/3(d[H_2])/(dt)=+1/2(d[NH_3])/(dt)` Hence `(d[NH_3])/(dt)=-2/3(d[H_2])/(dt)` `=-2/3 xx (-0.3xx 10^(-4))` `=0.2xx 10^(-4)Ms^(-1)`

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