94420 + Interview Questions in Chemistry in Class 12

1.	For the reaction N_2+3H_2 to 2NH_3, if (Delta[NH_3])/(Delta t)=2 times 10^-4 mol l^-1 s^-1, then value of -(Delta[NH_3])/(Delta t) will be:
Answer» `1 times 10^-4 mol L^-1 s^-1` `3 times 10^-4 mol L^-1 s^-1` `4 times 10^-4 mol L^-1 s^-1` `6 times 10^-4 mol L^-1s^-1` ANSWER :B

Discussion

2.	For the reaction N_2+3H_2 to 2NH_3, rate of reaction with respect to hydrogen may be expressed as (-d[H_2])/(dt). In it, the expression as -(d[H_2])/(dt) represents
Answer» a negative RATE of reaction amount of hydrogen left unreated DECREASE in concentration of `H_2` in unit time decrease in the rate of CHANGE in concentration of hydrogen Solution :The MINUS sign in the expression `- d[H_(2)]//dt` indicates a decrease in the concentration of hydrogen in unit time .

Discussion

3.	For the reaction , N_(2)+3H_(2) to 2NH_(3), "if" (d[NH_(3)])/(dt)=2xx10^(-4)"mol"L^(-1)s^(-1),the value of(-d[H_(2)])/(dt) would be
Answer» `1XX10^(-4)"mol"L^(-1)s^(-1)` `3xx10^(-4)"mol"L^(-1)s^(-1)` `4xx10^(-4)"mol"L^(-1)s^(-1)` `6xx10^(-4)"mol"L^(-1)s^(-1)` Answer :B

Discussion

4.	For the reaction, N_2 + 3H_2
Answer» `1.25 xx 10^(-3) MOL L^(-1)s^(-1)` `2.5 xx 10^(-3) mol L^(-1)s^(-1)` `3.75 xx 10^(-3) mol L^(-1)s^(-1)` `6.25 xx 10^(-5) mol L^(-1)s^(-1)` Answer :B

Discussion

5.	For the reaction N_(2)+2O_(2)to2NO_(2) Given : at 1 atm, 300 K S_(N_(2))=180J//mol//K C_(p)(N_(2))=30J//mol//K S_(O_(2))=220J//mol//K C_(p)(O_(2))=30J//mol//K S_(NO_(2))=240J//mol//K C_(p)(NO_(2))=40J//mol//K Calculate (i) DeltaS_(300K,1"atm") (ii) DeltaS_(400K,1"atm") (iii) DeltaS_(300K,5"atm") (iv) DeltaS_(400K,5"atm")
Answer» Solution :(i) `(DeltaS_(R))_(300)=2S_(NO_(2))-2S_(O_(2))-S_(N_(2))` `=2xx240-2xx220-180` = `-140Jmol^(-1)K^(-1)` `(DeltaCp)_(r)=2Cp(NO_(2))-2Cp(O_(2))-Cp(N_(2))` = `2xx40-2xx30-30` `=-10Jmol^(-1)k^(-1)` (ii) `(DeltaS_(r))_(400)=(DeltaS_(r))_(300)+(DeltaC_(p))_(r)//nT_(2)/T_(1)` = `-140-10//n4/3` `=-142.88Jmol^(-1)k^(-1)` (III) `(DeltaS_(r))_(300k,5"atm")=(DeltaS_(r))_(300k,1"atm")+Deltan_(g)R//n` `p_(1)/p_(2)=-140+(-1)R//n1/5=-140+R//n5` = `-140+8.314//n5` = `-126.62 J mol^(-1)k^(-1)` (iv) `(DeltaS_(r))_(400k,5"atm")=(DeltaS_(r))_(400k,"1atm")-R//n1/5` = `142.88+R//n5` = `-129.5Jmol^(-1)k^(-1)`

Discussion

6.	For the reaction N_(2) + 2O_(2) to 2 NO_(2) Given : at 1 atm, 300 K {:(, S_(N_(2)) = 180 J//mol//K"",C_(P)(N_(2))= 30 J//mol//K), (,S_(O_(2))= 220 J//mol//K"",C_(P)(O_(2))= 30 J//mol//K),(,S_(NO_(2))= 240 J // mol //K"",C_(P)(NO_(2)) = 40J//mol//K):} Calculate(i)DeltaS_(300)K, 5 atm""(i)DeltaS_(400)K, 5 atm
Answer» SOLUTION :N//A

Discussion

7.	For the reaction N_2 + 3H_2 hArr 2NH_3 in a vessel after the addition of equal number of mole of N_2 and H_2 equilibrium state is formed.Which of the following is correct ?
Answer» `[H_2]=[N_2]` `[H_2]ltN_2]` `[H_2]gtN_2]` `[H_2]GT[NH_3]` ANSWER :B

Discussion

8.	For the reaction, M^(2+) + MnO_(4)^(-) rarr MO_(3)^(-) + Mn^(+2) + ( 1)/( 2) O_(2) if one mole of MnO_(4)^(-) oxidizes 1.67 mole of M^(+x) to MO_(3)^(-) , then the value of x in the metal ion is
Answer» 5 3 2 1 Answer :C

Discussion

9.	For the reaction: M^(n+)+ ne^(-) rarr M(s). The correct representation of Nernst equation is:
Answer» `E_(M^(N+)//M)=E _(M^(n+)//M)^(o) + (0.0591)/n log [MN^(+)]` `E_(M^(n+)//M)=E _(M^(n+)//M)^(o) -(0.0591)/n log [Mn^(+)]` `E_(M^(n+)//M)=E _(M^(n+)//M)^(o) -n/(0.0591) log [Mn^(+)]` NONE of these ANSWER :A

Discussion

10.	Forthe reaction M^X+MnO_(4) ^(- )to MO_3^(-) +Mn^(2+)+1//@O_2ifonemoleof MnO_(4)^(-)oxidises1.67molesof M^+ to MO_3^(-)thenthe valueof xin thereactionis
Answer» 5 3 2 1 Solution :`overset(+7 )(MnO_(4)^(-)+5E^(-) toMn^(2+)` since1 moleof `MnO_4^(-)`accepts5 MOLESOF electronstherefore, 5 moles of electrons are lostby 1.67molesof`M^(X+)` ` therefore ` 1 moleof`M^(x+)`will loseelectrons `= 5 //1.67= 3 ` moles( approx ) since ` M^(x+)` changesto `MO_(3)^(-)` ( whereO.N of`M= +5 )`byaccepting 3electrons ` thereforex= +5 -3 =+2`

Discussion

11.	For the reaction in aqueous solution Zn^(2+) + X^(-) hArr ZnX^(+), the K_(eq) is greatest when X is
Answer» `F^(-)` `NO_(3)^(-)` `ClO_(4)^(-)` `I^(-)` Solution :Because of `F^(-)` is a highly electronegative. So it is easily LOSE the electron and reaction OCCUR RAPIDLY.

Discussion

12.	For the reaction : I_(2)(g)hArr 2I(g), K_(c )=37.6xx10^(-6)at 1000 K. If 1.0 mole of I_(2) is introduced into a 1.0 litre flask at 1000 K at equilibrium, then
Answer» CONC. Of `I_(2)(g)` is LESS than that of I(g) Conc. of `I_(2)(g)`is much LARGER than that of I(g) `[I_(2)]=[I]` `[I_(2)=1//2[I]` Answer :B

Discussion

13.	For the reaction (i) A overset(k_(1))to P (ii) B overset(k_(II))to Q, following observation is made. Calculate (k_(I))/(k_(II)), where k_(I) and k_(II) are rate constant for the respective reaction.
Answer» 2.303 1 0.36 0.693 Solution :`k_(I)=(0.693)/(30)` `k_(II)=(a)/(2t_(1 // 2))=(2)/(2 XX t_(1 // 2))=(1)/(t_(1 // 2))=(1)/(30)` `(k_(I))/(k_(II))=(0.693)/(30 xx 1)=0.693`

Discussion

14.	For the reaction H_(2)O_((s))iff H_(2)O_((l))" at "0^(@)C and normal pressure
Answer» `DeltaH gt TDELTAS` `DeltaH = TDeltaS` `DeltaH=DeltaG` `DeltaH lt TDeltaS` Solution :`H_(2)O_((g))subH_(2)O_((l))` we know`DeltaG=DeltaH-TDeltaS` at EQUILIBRIUM `DeltaG=0` THEREFORE `DeltaH=TDeltaS`

Discussion

15.	For the reaction H_(2)O_((s)) hArr H_(2)O_((l)) at ""^(@)C and normal pressure
Answer» `DELTAH = Delta G` `Delta H lt T Delta S` `Delta H GT T Delta S` `Delta H= T Delta S` Solution :For the reaction to be spontaneous, `Delta G = Delta H - T Delta S` should be negative. As in the given reaction solid phase changes to the liquid phase entropy increases for `Delta G` MUST be negative. `Delta H lt T Delta S`

Discussion

16.	For the reaction, H_(2)(g)+I_(2)(g)hArrHI(g) the equlibrium constant K_(p) changes with
Answer» TOTAL PRESSURE Catalyst The amounts of `H_(2)and I_(2)` PRESENT TEMPERATURE Solution :Only temperature affect the EQULIBRIUM constant. Since here `DeltaH=2-2=0,` so there is no dependence on pressure.

Discussion

17.	For the reaction H_(2)O_((i))hArrH_(2)O_((g)) at 373 K and 1 atmospheric pressure
Answer» `DeltaE=0` `DeltaH=TDeltaS` `DeltaH=DeltaE` `Delta=0` Solution :In the given reaction, liquid water is in EQUILIBRIUM with water VAPOURS at 373 K and 1 atm PRESSURE, THUS `DeltaG=0` `:' DeltaG=DeltaH-Tdelta S :.DeltaH=T DeltaS`

Discussion

18.	For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g), the standard free energy is Delta G^(@)gt 0. The equilibrium constant (K) would be
Answer» K = 0 `K GT 1` K = 1 `K lt 1` ANSWER :D

Discussion

19.	For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g) at 712K the value of equlibrium constant (K_(c)) is 50. When the equlibrium concentration of both is 0.5M, the value of K_(p) under the same conditions will be
Answer» <P>`0.002` `0.2` `50.0` `50//RT` Solution :For the reaction `H_(2)+I_(2)hArr2HI Deltan=0` So `K_(p)=K_(C)THEREFORE 50.0`

Discussion

20.	For the reaction H_2(g)+I_2(g) hArr 2HI(g) at 720 K the value of equlibrium constant is 50. When equilibrium concentration of H_2 and I_2 IS 0.5 M.K_p under the same conditions will be :
Answer» 0.02 0.2 50 50RT Answer :C

Discussion

21.	For the reaction H_2(g) +I_2(g) hArr 2HI(g), the equilibriumconstant K_p Changes with
Answer» TOTAL pressure catalyst the AMOUNT of `H_2 and I_2`PRESENT temperature Answer :D

Discussion

22.	For the reaction , H_(2(g))+Hg_(2)Cl_(2) to 2Hg_((l))+2HCl_((aq)) electrochemical cell will be:
Answer» `Pt,H_(2)\|HCl\|\|`Calomel electrode `HG\|Hg(NO_(3))_(2)\|\|`Calomel electrode Calomel electrode`\|\|HCl\|H_(2)_((G)), Pt` `Hg\|Hg_(2)Cl_(2),KCl\|\|Hg(NO_(3))_(2)\|Hg` Solution :Large difference in SOP gives spontaneous cell REACTION.

Discussion

23.	For the reaction :H_(2(g))+CO_(2(g))hArr CO_((g))+H_(2)O_((g)), if the initial concentraion of [H_(2)]=[CO_(2)]and x moles/litre of hydrogenis consumed at equlibrium, the correct expression of K_(p) is
Answer» `(x^(2))/((1-x)^(2))` `(1+x)^(2)/((1-x)^(2))` `(x^(2))/((2+x)^(2))` `(x^(2))/(1-x^(2))` Solution :`H_(2(G))+CO_(2(g))hArrCO_((g))+H_(2)O_((g))` `{:("Initial conc",1,1,0,0):}` `{:("At eqm",(1-x),(1-x),x,x):}` `K_(p)=(P_(CO).P_(H_(2)O))/(P_(H_(2)).P_(CO_(2)))=(x.x)/((1-x)(1-x))=(x^(2))/((1-x)^(2))`

Discussion

24.	For the reaction ,H_2(g)+Br(g)=2HBr(g),the reaction rate=K[H2][Br2]^1/2. Which statement is true about this reaction:
Answer» The REACTION is of second ORDER Molecularity of the reaction is 3/2 The unit of K is `sec^-1` Molecularity of the reaction is 2 Answer :D

Discussion

25.	For the reaction H_(2)(g)+Br_(2)(g)to2HBr(g), the reaction rate =K[H_(2)][Br_(2)]^(1//2), which statement is true about this reaction?
Answer» The REACTION is of second order Molecularity of the reaction is 3/2 Order of the reaction is 3/2 Molecularity of the reaction is 2 Solution :For `H_(2)+Br_(2)hArr2HBr=r=k[H_(2)]^(1)[Br_(2)]^(1/2)`.

Discussion

26.	For the reaction ,H_2(g)+Br(g)=2HBr(g),the reaction rate=K[H_2][Br_2]^(1/2). Which statement is true about this reaction:
Answer» The reaction is of second order MOLECULARITY of the reaction is 3/2 The UNIT of K is `sec^-1` Molecularity of the reaction is 2 Answer :D

Discussion

27.	Forthe reaction H_(2(g))+Br_(2(g)), the rate lawis rate, =K[H_2][Br_2]^(1//2) . Which of the following statements is true about this rection ?
Answer» the RACTION is second order one MOLECULARITY of the raction is m+ n both a and b NONE of corrcet Solution :In the reaction `H_(2(G)) + Br_(2(g)) to 2 H Br_((g))` , molecules (one each of `H_(2)` and `Br_(2)` ) take part in the reaction . Hence , its molecularity is 1 + 1 = 2 .

Discussion

28.	For the reaction H_(2)(g) +Br_(2)(g)rarr 2HBr(g) the experimental data suggestion that r=k[H_(2)][Br_(2)]^(1//2). The molecularity and order of the reaction are respectively:
Answer» `2,(3)/(2)` `(3)/(2),(3)/(2)` not DEFINED , `(3)/(2)` `1,(1)/(2)` ANSWER :C

Discussion

29.	For the reaction H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l),DeltaH=-285.8kJ mol^(-1)DeltaS=-0.163 kJ mol^(-1) K^(-1). What is the value of free energy change at 27^(@)C for the reaction
Answer» `-236.9 KJ mol^(-1)` `-281.4 kJ mol^(-1)` `-334.7 kJ mol^(-1)` `+334.7 kJ mol^(-1)` Solution :`DELTAG=DeltaH-TDeltaS, T=27+273=300 K` `DeltaG=(-285.8)-(300)(-0.163)=-236.9 kJ mol^(-1)`

Discussion

30.	For the reaction H_2(g)+2AgCl(s)=2Ag(s)+2H^+ (aq)(0.1M)+2Cl^(-) (aq)(0.1M) (1 atm) Delta G^@=42927 joules at 25^@C.Calculate the emf of the cell in which the given reaction takes place .
Answer» SOLUTION :We know `DELTAG^@=-NFE^@` `therefore E^@=- (DeltaG^@)/(nF)=42927/(2 times 96500)=0.2224` volt Now, for the above cell REACTION `E_(cell),E^@-0.0591/2log""([H^+]^2[Cl^-]^2)/([H_2])` `=0.2224-0.0591/2log""((0.1)^2(0.1)^2)/((1))` `=0.3406` volt `therefore E_(Hg_2^(2+))^@=0.77+00.021=0.791` volt

Discussion

31.	For the reaction, H_(2(g)) + 1/2O_(2(g)) to H_2O_((l)) B.E._((H – H)) = x_1, B.E._((O=O)) = x_2 and B.E._((O-H))= x_3. If the latent heat of vaporisation of water liquid into water vapour = x_4, then Delta_fH (heat of formation of liquid water) is
Answer» `x_1+x_2/2 - x_3+x_4` `2x_3-x_1-x_2/2-x_4` `x_1+x_2/2- 2x_3-x_4` `x_1+x_2/2-2x_3+x_4` Solution :`Delta_fH=(B.E.)_"reactants"-(B.E.)_"PRODUCTS"` But all the species must be in gaseous state , so in product `[H_2O_((l)) to H_2O_((g))] DeltaH_(vap)` must be ADDED . HENCE for the reaction , `H_(2(g)) + 1/2O_(2(g)) to H_2O_((l))` `Delta_fH = [(B.E.)_(H-H)+1/2(B.E.)_(O=O)]-[DeltaH_(vap)+2(B.E.)_(O-H)]` `=x_1+x_2/2-[x_4+2x_3] rArr x_1+x_2/2-x_4-2x_3`

Discussion

32.	For the reaction, H_2(g)+1/2O_2(g)→H_2O(l),
Answer» `7.63xx (373-297)-68.3` `7.63xx 10^-3xx(373-298)-68.3` `7.63xx 10^-3xx(373-298)+68.3` `7.63xx(373-297)+68.3` ANSWER :B

Discussion

33.	For the reaction : H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l), Delta H = - 68 kcalmol^(-1) The heat change for the decomposition of 7.2 g of water is
Answer» `13.6 KCAL` `27.2 kcal` `54.4 kcal` `-34 kcal` Solution :HEAT change for the decomposition of 1 MOLE or 18 g water = 68 kcal Heat change for the decomposition of 7.2 g of water = `(68 XX 7.2)/(18) = 27.2 kcal`

Discussion

34.	For the reaction H_(2)+I_(2)hArr2Hl, the equilibrium concentration of H_(2),I_(2)and HI are 8.0. 3.0 and 28.0 mol per litre respectively, the equlibrium constant of the reaction is
Answer» `30.66` `32.66` `34.66` `36.66` SOLUTION :`K_(c)=([HI]^(2))/([H_(2)][I_(2)])=((28)^(2))/(8xx3)=32.66`

Discussion

35.	For the reaction. H_2+I_2hArr2HI,K=47.6. If the initial number of moles of each reactant product is 1 mol then at equilibrium-
Answer» [I2]=[H2],[I2]>[HI] [I2]=[H2],[I2]<[HI] [I2]<[H2],[I2]=[HI] [I2]>[H2],[I2]=[HI] Solution :`A+2B initial : 220 At equilibrium`2-x2-x2x `=1.5=1=1 As 1 mol of C is formed at equilibrium, `2x=1to=0.5` `:.` At equilibrium NUMBER of moles of A = 1.5 , B=1 and C=1 `:.` Equilibrium CONCENTRATION of A = 1.5/ 10 `K_c=([C]^2)/([A][B]^2)=([1//10]^2)/([1.5//10][1//10]^2)=6.67`

Discussion

36.	For the reaction, H_2+I_2 hArr 2HIthe K_p and K_care related as :
Answer» `K_C = 2K_P` `K_C GT K_P` `K_C =K_P` `K_C LT K_P` ANSWER :C

Discussion

37.	For the reaction H_(2 (g)) + Br_(2) (g) to 2HBr(g) experimental data suggest , rate = K[H_(2)][Br_(2)]^(1//2), molecularity and order of the reaction are respectively
Answer» ` 2 , (3)/(2)` `(3)/(2) , (3)/(2)` `1, 1` `1 , (1)/(2)` Solution :The RATE of REACTION is `(3)/(2)` and molecularity is 2.

Discussion

38.	For the reaction H_(2) + Cl_(2) overset("Sunlight")(rarr) 2HCl taking place on water, the order of reaction is
Answer» `1` `2` `3` `0` Solution :The RATE of this photochemical reaction is independent of the concentration, therefore, it is zeroorder reaction.

Discussion

39.	For the reaction H_(2) + Cl_(2) overset ("Sunlight")(to) 2HCl taking place on water , the order of reaction is
Answer» 1 2 3 0 Solution :The RATE of this photochemical reaction is independent of the concentration , THEREFORE , it is ZERO ORDER reaction .

Discussion

40.	For the reaction : H^()+""^()CH_(2)CH_(3)overset(K)(to)CH_(3).CH_(3), a chain termination step of the reaction: 2C_(2)H_(6)toCH_(2)=CH_(2)+H_(2), the activation energy and Arrhenius parameter can be given as
Answer» 0,K 0,Ea A,1 1,K Solution :`K=A.e^(-Ea//RT),K=A.e^((0)//RT,K=A=0,K`

Discussion

41.	For the reaction H_(2)+Br_(2)to2HBr, the rate expression is rate =K[H_(2)][Br_(2)]^(1//2) which statement is true about this reaction
Answer» The REACTION is of second order Order of the reaction is `3//2` The UNIT of K is `sec^(-1)` MOLECULARITY of the reaction is `2` ANSWER :B

Discussion

42.	For the reaction Fe_(2)O_(3)+3CO rarr 2Fe+3CO_(2), the volume of carbon monoxide required to reduce one mole of ferric oxide is
Answer» `67.2 dm^(2)` `11.2 dm^(2)` `22.4 dm^(3)` `44.8 dm^(3)` Solution :`{:(Fe_(2)O_(3),+,3CO,rarr,2Fe,+,3CO_(2)),("1 vol.",,"3 vol.",,"2 vol.",,"3 vol."),("1 mol.",,"3 mol.",,"2 mol.",,"3 mol."):}` `( :' "vol% = mol%")` One gram mol of any GAS occupies 22.4 litre at NTP. 1 mol of `Fe_(2)O_(3)` requires 3 mol of `CO` for its reduction i.e., 1 mol of `Fe_(2)O_(3)` requires `3xx22.4` litre or `67.2 dm^(3) CO` to GET itself REDUCED.

Discussion

43.	For the reaction Fe_(2)O_(3)+3COto2Fe+3CO_(2),the volume of carbon monoxide required to reduceone mole of ferricoxide is
Answer» `67.2dm^(3)` `11.2dm^(3)` `22.4dm^(3)` `44.8dm^(3)` SOLUTION :`{:(Fe_(2)O_(3)+3COto2Fe+3CO_(2),),(1vol." " 3vol " " 2vol . " " 3vol.,),(1mol " " 3 mol. " " 2 mol.""3MOL,):}` `(becausevol%=mol%)`. One GRAM mol of any gas occupies 22.4 litre at NTP. 1 mol of `Fe_(2)O_(3)` REQUIRES 3 mol of CO for its REDUCTION i.e., 1 mol of `Fe_(2)O_(3)` requires `3xx22.4` litre or 67.2 `dm^(3)` CO to get itself reduced .

Discussion

44.	For the reaction, following data is given AtoB:K_(1)=10^(15) exp ((-2000)/T)""CtoD,K_(2)=10^(14) exp ((-1000)/T) The temperature at whilch K_(1)=K_(2) is
Answer» 1000 K 2000 K 868.4 K 434.22 K Solution :`K_(1)=K_(2),10^(15).e^(((-2000)/T))=10^(14).e^(((-1000)/T)),10.e^(((-2000)/T))=e^(((-1000)/T)),10=(e^((1000/T)))/(e^(((-200)/T))),10=e^(((-1000)/T))XXE^(((-2000)/T))` `10=e^((-1000-2000)/T),10=e^((1000/T)),10=1000/T,2.303log_(10)^(10)=1000/T,T=1000/2.303=434.2K`

Discussion

45.	For the reaction equlibrium N_(2)O_(4)hArr2NO_(2(g))the concentration of N_(2)O_(4) and NO_(2) at equlibrium are 4.8xx10^(-2)and 1.2xx10^(-2)"mol lite"^(-1) respectively. The value of K_(c) for the reaction is
Answer» `3.3xx10^(2)"mol litre"^(-1)` `3xx10^(-1)"mol litre"^(-1)` `3xx10^(-3)"mol litre"^(-1)` `3xx10^(3)"mol litre"^(-1)` Solution :`K=([NO_(2)]^(2))/([N_(2)O_(4)])=([1.2xx10^(-2)])/([4.8xx10^(-2)])=0.3xx10^(-2)=3xx10^(-3)`

Discussion

46.	For the reaction [Cu(NH_(3))_(4)]^(2+) + H_(2) O [ Cu ( NH_(3))_(3) H_(2) O ] ^(2+ ) + NH_(3) Net rate is ((dx)/(dt)) = 2.0 xx 10^(-4) s^(-1) [[ Cu (NH_(3))_(4) ]^(2+) ] - 3.0 xx 10^(5) L mol^(-1) s^(_1) [[ Cu ( NH_(3))_(3) H_(2) O ] (NH_(3))] Ratio of rate constant of the forward and backward reactions is
Answer» `0.66 XX 10^(-9) mol L^(-1)` `1.5 xx 10^(-9) mol L^(-1)` `2.0 xx 10^(-4) s^(-1)` `3.0 xx 10^(5) L mol^(-1) s^(-1)` ANSWER :A

Discussion

47.	For the reaction : F_(2)(g) + 2HCl(g) hArr 2HF(g) + Cl_(2)(g) Delta H^(theta) at 25^(@)C is =-84.4 kcalmol^(-1) Delta_(f) H^(theta)(HF) = -64.2kcalmol^(-1) Delta_(f) H^(theta) for HCl(g) per gram is :
Answer» `- 0.603 kcalg^(-1)` `0.603 kcalg^(-1)` `0.0603 kcalg^(-1)` `6.03 kcalg^(-1)` Solution :`Delta H = 2 Delta H_(f) [HF(g)] + Delta H_(f)(Cl_(2)) - 2 Delta H_(f) [HCL(g)] - Delta H_(f)(F_(2))` `- 84.4 = 2 xx (-64.2) - 0 - 2 Delta H_(f) [HCl(g)] - 0` `2 Delta H_(f)[HCl(g)] = - 128.4 + 84.4 = - 44.0 KCAL` `Delta H_(f) [HCl(g)] = - 22.0 kcal` `Delta H_(f)^(@)` for HCl per GRAM `= (-22.0)/(36.5)` `= - 0.603 kcal`

Discussion

48.	For the reaction, CuSO_4.5H_2O(s) hArr CuSO_4.3H_2O(s) + 2H_2O(v). Which one is correct representation ?
Answer» <P>`K_p=(p_(H_20)^2` `K_c=[H_2O]^2` `K_P = K_c (RT)^-2` All of the above Answer :D

Discussion

49.	For the reaction C(s)+O_(2)(g) to CO_(2)(g) DeltaH^(0)=-393.51 kJ//"mole" and DeltaS^(0)=2.86 J//"mole" . K at 25^(@)C. Does the reaction become more or less favourable as the temperature increases ?
Answer» ANSWER :( more FAVOURABLE)

Discussion

50.	For the reaction : C(s) + O_(2)(g) rarr CO_(2)(g)
Answer» `DELTA H LT Delta U` `Delta H GT Delta U` `Delta H = Delta U` `Delta H = 0` Answer :C

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For the reaction N_2+3H_2 to 2NH_3, if (Delta[NH_3])/(Delta t)=2 times 10^-4 mol l^-1 s^-1, then value of -(Delta[NH_3])/(Delta t) will be:

For the reaction N_2+3H_2 to 2NH_3, rate of reaction with respect to hydrogen may be expressed as (-d[H_2])/(dt). In it, the expression as -(d[H_2])/(dt) represents

For the reaction , N_(2)+3H_(2) to 2NH_(3), "if" (d[NH_(3)])/(dt)=2xx10^(-4)"mol"L^(-1)s^(-1),the value of(-d[H_(2)])/(dt) would be

For the reaction, N_2 + 3H_2

For the reaction N_2 + 3H_2 hArr 2NH_3 in a vessel after the addition of equal number of mole of N_2 and H_2 equilibrium state is formed.Which of the following is correct ?

For the reaction, M^(2+) + MnO_(4)^(-) rarr MO_(3)^(-) + Mn^(+2) + ( 1)/( 2) O_(2) if one mole of MnO_(4)^(-) oxidizes 1.67 mole of M^(+x) to MO_(3)^(-) , then the value of x in the metal ion is

For the reaction: M^(n+)+ ne^(-) rarr M(s). The correct representation of Nernst equation is:

Forthe reaction M^X+MnO_(4) ^(- )to MO_3^(-) +Mn^(2+)+1//@O_2ifonemoleof MnO_(4)^(-)oxidises1.67molesof M^+ to MO_3^(-)thenthe valueof xin thereactionis

For the reaction in aqueous solution Zn^(2+) + X^(-) hArr ZnX^(+), the K_(eq) is greatest when X is

For the reaction : I_(2)(g)hArr 2I(g), K_(c )=37.6xx10^(-6)at 1000 K. If 1.0 mole of I_(2) is introduced into a 1.0 litre flask at 1000 K at equilibrium, then

For the reaction﻿ (i)﻿ A overset(k_(1))to P (ii) B overset(k_(II))to Q, following observation is made. Calculate (k_(I))/(k_(II)), where k_(I) and k_(II) are rate constant for the respective reaction.

For the reaction H_(2)O_((s))iff H_(2)O_((l))" at "0^(@)C and normal pressure

For the reaction H_(2)O_((s)) hArr H_(2)O_((l)) at ""^(@)C and normal pressure

For the reaction, H_(2)(g)+I_(2)(g)hArrHI(g) the equlibrium constant K_(p) changes with

For the reaction H_(2)O_((i))hArrH_(2)O_((g)) at 373 K and 1 atmospheric pressure

For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g), the standard free energy is Delta G^(@)gt 0. The equilibrium constant (K) would be

For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g) at 712K the value of equlibrium constant (K_(c)) is 50. When the equlibrium concentration of both is 0.5M, the value of K_(p) under the same conditions will be

For the reaction H_2(g)+I_2(g) hArr 2HI(g) at 720 K the value of equlibrium constant is 50. When equilibrium concentration of H_2 and I_2 IS 0.5 M.K_p under the same conditions will be :

For the reaction H_2(g) +I_2(g) hArr 2HI(g), the equilibriumconstant K_p Changes with

For the reaction , H_(2(g))+Hg_(2)Cl_(2) to 2Hg_((l))+2HCl_((aq)) electrochemical cell will be:

For the reaction :H_(2(g))+CO_(2(g))hArr CO_((g))+H_(2)O_((g)), if the initial concentraion of [H_(2)]=[CO_(2)]and x moles/litre of hydrogenis consumed at equlibrium, the correct expression of K_(p) is

For the reaction ,H_2(g)+Br(g)=2HBr(g),the reaction rate=K[H2][Br2]^1/2. Which statement is true about this reaction:

For the reaction H_(2)(g)+Br_(2)(g)to2HBr(g), the reaction rate =K[H_(2)][Br_(2)]^(1//2), which statement is true about this reaction?

For the reaction ,H_2(g)+Br(g)=2HBr(g),the reaction rate=K[H_2][Br_2]^(1/2). Which statement is true about this reaction:

Forthe reaction H_(2(g))+Br_(2(g)), the rate lawis rate, =K[H_2][Br_2]^(1//2) . Which of the following statements is true about this rection ?

For the reaction H_(2)(g) +Br_(2)(g)rarr 2HBr(g) the experimental data suggestion that r=k[H_(2)][Br_(2)]^(1//2). The molecularity and order of the reaction are respectively:

For the reaction H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l),DeltaH=-285.8kJ mol^(-1)DeltaS=-0.163 kJ mol^(-1) K^(-1). What is the value of free energy change at 27^(@)C for the reaction

For the reaction H_2(g)+2AgCl(s)=2Ag(s)+2H^+ (aq)(0.1M)+2Cl^(-) (aq)(0.1M) (1 atm) Delta G^@=42927 joules at 25^@C.Calculate the emf of the cell in which the given reaction takes place .

For the reaction, H_(2(g)) + 1/2O_(2(g)) to H_2O_((l)) B.E._((H – H)) = x_1, B.E._((O=O)) = x_2 and B.E._((O-H))= x_3. If the latent heat of vaporisation of water liquid into water vapour = x_4, then Delta_fH (heat of formation of liquid water) is

For the reaction, H_2(g)+1/2O_2(g)→H_2O(l),

For the reaction : H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l), Delta H = - 68 kcalmol^(-1) The heat change for the decomposition of 7.2 g of water is

For the reaction H_(2)+I_(2)hArr2Hl, the equilibrium concentration of H_(2),I_(2)and HI are 8.0. 3.0 and 28.0 mol per litre respectively, the equlibrium constant of the reaction is

For the reaction. H_2+I_2hArr2HI,K=47.6. If the initial number of moles of each reactant product is 1 mol then at equilibrium-

For the reaction, H_2+I_2 hArr 2HIthe K_p and K_care related as :

For the reaction H_(2 (g)) + Br_(2) (g) to 2HBr(g) experimental data suggest , rate = K[H_(2)][Br_(2)]^(1//2), molecularity and order of the reaction are respectively

For the reaction H_(2) + Cl_(2) overset("Sunlight")(rarr) 2HCl taking place on water, the order of reaction is

For the reaction H_(2) + Cl_(2) overset ("Sunlight")(to) 2HCl taking place on water , the order of reaction is

For the reaction : H^(*)+""^(*)CH_(2)CH_(3)overset(K)(to)CH_(3).CH_(3), a chain termination step of the reaction: 2C_(2)H_(6)toCH_(2)=CH_(2)+H_(2), the activation energy and Arrhenius parameter can be given as

For the reaction H_(2)+Br_(2)to2HBr, the rate expression is rate =K[H_(2)][Br_(2)]^(1//2) which statement is true about this reaction

For the reaction Fe_(2)O_(3)+3CO rarr 2Fe+3CO_(2), the volume of carbon monoxide required to reduce one mole of ferric oxide is

For the reaction Fe_(2)O_(3)+3COto2Fe+3CO_(2),the volume of carbon monoxide required to reduceone mole of ferricoxide is

For the reaction, following data is given AtoB:K_(1)=10^(15) exp ((-2000)/T)""CtoD,K_(2)=10^(14) exp ((-1000)/T) The temperature at whilch K_(1)=K_(2) is

For the reaction equlibrium N_(2)O_(4)hArr2NO_(2(g))the concentration of N_(2)O_(4) and NO_(2) at equlibrium are 4.8xx10^(-2)and 1.2xx10^(-2)"mol lite"^(-1) respectively. The value of K_(c) for the reaction is

For the reaction : F_(2)(g) + 2HCl(g) hArr 2HF(g) + Cl_(2)(g) Delta H^(theta) at 25^(@)C is =-84.4 kcalmol^(-1) Delta_(f) H^(theta)(HF) = -64.2kcalmol^(-1) Delta_(f) H^(theta) for HCl(g) per gram is :

For the reaction, CuSO_4.5H_2O(s) hArr CuSO_4.3H_2O(s) + 2H_2O(v). Which one is correct representation ?

For the reaction C(s)+O_(2)(g) to CO_(2)(g) DeltaH^(0)=-393.51 kJ//"mole" and DeltaS^(0)=2.86 J//"mole" . K at 25^(@)C. Does the reaction become more or less favourable as the temperature increases ?

For the reaction : C(s) + O_(2)(g) rarr CO_(2)(g)

For the reaction (i) A overset(k_(1))to P (ii) B overset(k_(II))to Q, following observation is made. Calculate (k_(I))/(k_(II)), where k_(I) and k_(II) are rate constant for the respective reaction.

For the reaction : H^()+""^()CH_(2)CH_(3)overset(K)(to)CH_(3).CH_(3), a chain termination step of the reaction: 2C_(2)H_(6)toCH_(2)=CH_(2)+H_(2), the activation energy and Arrhenius parameter can be given as