52671 + Interview Questions in Chemistry in Class 11

1.	For the reaction H_(2) (g) + I_(2) (g) hArr 2 HI (g) , the equilibrium constant K_(C) is ………
Answer» SOLUTION :`([HI]^(2))/([H_(2)][I_(2)])`

Discussion

2.	For the reaction : H_(2) (g) + CO_(2) hArr CO(g) + H_(2) O (g) , if the intial pressure of [H_(2)]=[CO_(2)] and x moles / litre of hydrogen is consumed at equilibrium , the correct expression of K_(p) is
Answer» <P>`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),hArr,CO (g),+,H_(2)O(g)), ("Intial ",1,,1,,0,,0), ("At . EQM.",1-x,,1-x,,x,,x):}` ` K_(c) = (x xx x)/((1-x)(1-x))=x^(2)/(1-x)^(2)`. `K_(p)= K_(c)"" ( :' Deltan_(g)= 0)`

Discussion

3.	For the reaction Fe_(2)S_(3)+5O_2 to 2FeSO_4 +SO_(2) the equivalent weight of Fe_(2)S_(3) is-
Answer» `M/4` `M/(16)` `(M)/(22)` `M/(20)` ANSWER :D

Discussion

4.	For the reaction , FeCO_(3(s)) rarr FeO_((s)) + CO_(2(g)), Delta H= 82.8kJ " at " 25^(@)C, what is (Delta E or Delta U) " at " 25^(@)C?
Answer» 82.8kJ 80.32kJ `-2394.77` kJ 85.28 kJ Solution :`Delta E= Delta H- Delta n_(g) RT` `= 82.8 - (1 XX 8.314 xx 298)/(1000) = 80.32KJ`

Discussion

5.	For the reaction : F_(2)+H_(2)O overset("Ice cold") underset("temperature") rarrHOF + HF which one is not correct ?
Answer» It is INTRAMOLECULAR redox reaction It is INTERMOLECULAR redox REACTON It is AUTO redox reaction It is a disproportionation reaction Solution :`OVERSET(0)(F_2)+H_2OrarrHOoverset(-1)F+Hoverset(-1)F` It is not intermolecular redox reaction .

Discussion

6.	For the reaction C(s) + CO_(2) (g) hArr 2CO(g). the partial pressures of CO_(2) and CO are 2.0 and 4.0 atm respectively at equilibrium. What is the value of K_(p)for this reaction?
Answer» <P>0.5 4 8 32 Solution :`K_(P)=(P_(CO)^(2))/(P_(CO_(2)))`

Discussion

7.	For the reaction C(s) + CO_(2) (g) to 2 CO (g) K_(p) = 63 " atm at " 100 K . If at equilibriump_(CO) = 10_(p_(CO_(2)) , then the toal pressure of the gases at equilibrium is
Answer» `63` atm `693` atm `063 `atm `0693` atm Solution :`C(s) +CO_(2) (g) hArr 2 CO (g)` `K_(p)=(p^(2) CO)/p_(CO_(2)) or 63 = ((10p_(CO_(2))))/(p_(CO_(2)))` or `p_(CO_(2)) = 63/100 = 0.63 " atm"` ` p_(CO)= 10p_(CO_(2))= 10 xx 0.63 " atm " =6.3 " atm"` ` :. p_("TOTAL")= p_(CO_(2)) + p_(CO) = 0.63 + 6.3 = 6.93 " atm"`

Discussion

8.	For the reaction, CO_((g))+Cl_(2(g))hArrCoCl_(2(g)), the value of K_(p)//K_(c) is equal to
Answer» 1 RT `SQRT(RT)` `(1)/(RT)` ANSWER :D

Discussion

9.	For the reaction CO(g) + Cl_(2) (g) hArr COCl_(2)(g) , K_(p)//K_(c) is equal to
Answer» <P>`sqrt(RT)` RT `1/(RT)` `1*0` Solution :` K_(p) = K_(c) (RT)^(Delta n )` For the given REACTIONS, ` Delta n = 1 - 2 = - 1` ` :. (K_(p))/(K_(c)) = (RT)^(-1) = 1/(RT)`

Discussion

10.	For the reaction CO_((g))+CI_(2(g)) harr COCI_(2(g)).The K_p // K_c is equal to
Answer» <P>`1/(RT)` 1 `SQRT(RT)` RT Solution :`(K_(P))/(K_(C))=(RT)^(Delta ng)`

Discussion

11.	For the reaction CO_((g))+2H_(2(g))hArrCH_(3)OH_((g)). If active mass of CO is kept constant and active mass of H_(2) is tripied, the rate of forward reaction will become
Answer» Three TIMES Six times Eight time NINE times ANSWER :D

Discussion

12.	For the reaction, CaCO_(3(g))hArrCaO_((s))+CO_(2(g))K_(p) is equal to
Answer» `K_(c)` `K_(c)RT` `K_(c)(RT)^(2)` `K_(c)(RT)^(-)` Answer :B

Discussion

13.	For the reaction C_(3) H_(8(g)) + 5O_(2(g)) to 3CI_(2(g)) + 4H_(2) O_((l)) , Delta H - Delta K=….....
Answer» `RT` `-3RT` `3RT` `-RT` ANSWER :B

Discussion

14.	For the reaction, C_2H_6(g) hArr C_2H_6(g) + H_2(g).K_p is 0.05 at 900 K. If an initial mixture comprising 20 mol of C_2H_6 and 80 mol of inert gas is passed over a dehydrogenation catalyst at 900 K, what is the equilibrium mole percentage of C_2H_6 in the gas mixture ? The total pressure is kept at 0.5 bar :
Answer» 4.3 9.67 8.76 72.5 Answer :C

Discussion

15.	For the reaction between KMnO_(4) " and " H_(2)O_(2) the number of electrons transferred per mole of H_(2)O_(2) is :
Answer» ONE two three four Solution :N//A

Discussion

16.	For the reaction at 300 K A_((g)) harr V_((g))+ S_((g) . Delta_(t) H^(@) = - 30 "KJ/mol" Delta_(t)S^(@) = - 0.1 K.J. K^(-1)."mole"^(-1) What Is the value of equilibrium constant ?
Answer» 0 1 10 0.1 Solution :`DELTA G^(@)=DELTAH^(@)-TDELTAS^(@)=-RTlnk`

Discussion

17.	For the reaction at 298K, A+ B rarr C, DeltaH =400 kJ mol^(-1) and DeltaS = 0.2 kJ K^(-1) mol^(-1) . At what temperature will be thereaction become spontaneous considerting DeltaHand DeltaS to be constant over the temperature range.
Answer» Solution :`DeltaG = DeltaH - T DELTAS` For the reaction to be spontaneous , `DeltaG` should be negative,i.e.,` Delta H - T DeltaSlt0` or `DeltaH lt T Delta S` or `T Delta S gt DeltaH` or `T gt (DeltaH)/( DeltaS) ` or`T gt ( 400 kJ MOL^(-1))/( 0.2 kJ K^(-1)mol^(-1))` or `T gt 2000K` This method can be applied only when `DeltaH ` and `DeltaS` both are `+ve`. ALTERNATIVELY, for equilibrium , `DeltaG= 0 `. Hence, `T DeltaS = DeltH` or `T = ( DeltaH )/( DeltaS = 0.2 = ( 400KJ mol^(-1))/( 0.2 kJ K^(-1) mol^(-1))= 2000 K `. For SPONTANEITY, `DeltaG = -ve`. This can be so only when `T gt 2000K`. ( so that `TDeltaS gt DeltaH`is magnitude).

Discussion

18.	For the reaction at 298 K, 2A + Bto C Delta H = 400 "kJ mol"^(-1) and Delta S = 0.2 "kJ K"^(-1) "mol"^(-1) At what temperature will the reaction become spontaneous considering Delta H and Delta S to beconstant over the temperature range.
Answer» Solution :For balanced REACTION `Delta G =0` `therefore Delta G = Delta H - T Delta S` `O = Delta H - T Delta S` `therefore T = (Delta H )/ (Delta S ) = (400)/(0.2) = 2000 K` `therefore` At 2000 K temperature of the reaction is equilibrium. So the reaction will be spontaneous over the RANGE of the temperature 2000 K.

Discussion

19.	For the reaction at 298 K : 2A +B to C DeltaH="400 KJ mol"^(-1) , DeltaS ="0.2 JK"^(-1) "mol"^(-1) Determine the temperature at which the reaction would be spontaneous .
Answer» Solution :Given: T=298 K `DeltaH=400 "J mol"^(-1) =400 "J mol"^(-1)` `DeltaS=0.2 JK^(-1) mol^(-1)` `DELTAG=DeltaH-TDeltaS` If T=2000 K `DeltaG`=400 -(0.2 x 2000) =0 if T > 2000 K `DeltaG` will be negative The reaction WOULD be spontaneus only BEYOND 2000 K .

Discussion

20.	For the reaction at 25^(@)C,X_(2)O_(4(l))to2XO_(2(g)),DeltaH=2.1kcal and DeltaS=20" cal "K^(-1). The reaction would be
Answer» spontaneous non-spontaneous at equilibrium unpredictable Answer :A

Discussion

21.	For the reaction at 298 K, 2A+BtoC DeltaH=400 kJ mol^(-1) and DeltaS=0.2 kJ K^(-1)mol^(-1). At what temperature will the reaction becomes spontaneous considering DeltaH and DeltaS to be constant over the temperature range?
Answer» SOLUTION :`GT 200 K`

Discussion

22.	For the reaction Ag(CN)bar(2) hArr Ag^(+) 2 CN^(-), K_(c)" at "25^(@)C" is "4 xx 10^(-19) . Calculate [Ag^(+)] in solution which was originally 01 M in KCN and 003M in AgNO_(3).
Answer» SOLUTION : Orignally on mixing KCN and `AgNO_(3)` , the reaction is `{:(,2 KCN,+,AgNO_(3),to,"Ag"(CN)_(2)^(-),+,KNO_(3),+,K^(+)),("Intial amounts",01 M,,003 M,,0,,0,,0),("Amounts after reaction ",(01-006),,0,,003 M,,003M,,003),(,=004 M,,,,,,,,):}` Thus , in the solution , now we have `[Ag (CN)_(2^(-)) ] = 03 M , [CN^(-)] = 0 04 M .`SUPPOSE x is the amount of `Ag(CN)_(2)^(-)`dissociated at equilibrium . Then ` {:(,AG(CN)_(2)^(-),hArr,AG^(+),+,2CN^(-)),("Intial amounts",003,,0,,004),("Amounts at eqm.",(003 -x),,x,+,004 + 2x):}` ` K_(c) = [ AG^(+) [CN^(-)]^(2))/([Ag(CN)_(2)^(-)])= (x (004 + 2x)^(2))/(003 - x) = 4 xx 10^(-19) ("Given")` As `K_(c)`is very small , DISSOCIATION of `Ag(CN)_(2)^(-)` is very smalli.e., x is very small . Hence , `004 + 2x approxand 03 - x approx003 .` ` :. (x( 004)^(2))/(003 )= 4 xx 10^(-19)or x = 75 xx 10^(-18)` Thus at equilibrium , `[Ag^(+)] = 7* 5 xx 10^(-18)M`

Discussion

23.	For the reaction Ag_(2)O(s)rarr 2Ag(s) + (1)/(2) O_(2)(g) , DeltaHis 30.56kJ mol^(-1) and Delta Sis 66JK^(-1) mol^(-1) at one atmosphere pressure. Calculate the temperature at which Delta G for it will be zero. What will be the direction of the reaction at this temperatureand at temperature above or below this temperature and why ?
Answer» Solution :When `DeltaG = 0, T = (DELTAH )/(DeltaS) = ( 30560J mol^(-1))/( 66JK^(-1)mol^(-1))= 463.0K` `DeltaG = DELTA H - T DeltaS `. Above 463K `, DeltaG = -ve` ( because `DeltaH ` and `Delta S` both are `+ve`) . Below 463K, `DeltaG = + ve`.

Discussion

24.	DeltaH and DeltaS for the reaction Ag_(2)O_((s))rarr2Ag_((s))+1/2O_(2_((g))) " are "30.56 kJ mol^(-1) and 66.0 Jk^(-1) mol^(-1)respectively. Calculate the temperature at which the free energy for this reaction will be zero.What will be the direction of reaction at this temperature and at temperature below this and why ? Given: DeltaH=30.56kJ mol^(-1)=30560 J mol^(-1) DeltaS=66.0JK^(-1)mol^(-1) DeltaG=0
Answer» Solution :Given: `DELTAH=30.56 "kJ mol"^(-1)` `=30560 J mol^(-1)` `DeltaS=6.66 xx 10^(-3) kJ K^(-1) "mol"^(-1)` T=? At which `DeltaG=0` `DeltaG=DeltaH-TDeltaS` `0=DeltaH-TDeltaS` `T=(DeltaH)/(DeltaS)` `T=("30.56 kJ mol"^(-1))/(6.66xx10^(-3) kJ K^(-1) "mol"^(-1))` T=4589 K (i) At 4589 K , `DeltaG=0` , the reaction is in equilibrium . (ii)At temperature below 4598 K, `DeltaH > TDeltaS` `DeltaG=DeltaH-T DeltaS > 0`, the reaction in the FORWARD direction, is non-spontaneous. In other words the reaction occurs in the backward direction.

Discussion

25.	For the reaction, Ag_(2)O_((s)) hArr 2Ag_((s)) + (1)/(2) O_(2(g)) Delta H, Delta S and T are 40.63 kJ mol^(-1), 108.8 JK^(-1) mol^(-1) and 373K respectively. Free energy change Delta G of the reaction will be _____
Answer» Solution :`Delta G= Delta H- T Delta S` `= 40.43 - (373 xx 108.8)/(1000) = 0.476 KJ ~~0`

Discussion

26.	For the reaction A(g) + B(s) hArr C (g)+ D (g), K_(c) = 49 mol dm^(-3) at 127^(@)C. Calculate K_(p).
Answer» Solution :`DELTA n = n_(p) - n_(r) = 2-1 = 1, K_(p) = K_(c) (RT)^(Delta n) = (49 mol DM^(-3)) (0.0821 dm^(3) "atm" K^(-1) mol^(-1) xx 400 K)^(1) = 1.61 xx 10^(3)` atm.

Discussion

27.	For the reaction A_((g)) + 3B_((g)) hArr 2C_((g))at 27^(@)C.2 mole of A, 4 moles of B and 6 moles of C are present in 2 lit vessel. If K_(c) for the reaction is 1.2, the reaction will proceed in
Answer» FORWARD direction backward direction neither direction none of these Solution :`Q_(C) lt K_(c)` forward is favoured

Discussion

28.	For the reaction A_((g))+B_((g)) harr C_((g))+D_((g)) K is 0.63 at 700^(@)C and 1.66 at 1000^(@)C. Then DeltaH^(@) is x KCal. What is x?
Answer» SOLUTION :`LOG(K_(2))/(K_(1))=(DeltaH^(@))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `log((1.66)/(0.63))=(DeltaH^(@))/(2.303R)[(1)/(973)-(1)/(1273)]` `DeltaH^(@)=8000` cal= 8 kcal

Discussion

29.	For the reaction, A_((g))+2B_((g))hArr 3C_((g))+3_((g)),K_(p)=0.05" atm at "1000K. The value of K_(c) is represented by
Answer» `5XX10^(-4)R` `(5xx10^(-4))/(R)` `5xx10^(-5)R` `(5xx10^(-5))/(R)` ANSWER :D

Discussion

30.	For the reaction, ADP+"phosphate" hArr ATP, DeltaG^circ=30.50kJmol^-1. What is the value of a equilibrium constant, K for this process under physiological conditions of 37.5^circC?.
Answer» `4.5xx10^-6` `7.4xx10^-6` `1.3xx10^5` `2.2xx10^5` ANSWER :B

Discussion

31.	For the reaction AB(g)hArrA(g)+B(g), at equilibrium AB is 20% dissociated at a total pressure of P, The equilibrium constant K_(p) is related to tha total pressure by the expression
Answer» <P>`P=24K_(p)` `P=8K_(p)` `24P=K_(p)` NONE of these ANSWER :A

Discussion

32.	For the reaction, AB(g) hArr A(g)+ B(g), at equilibrium, AB is 20% dissociated at a total pressure of P. The equilibrium constant K_p is related to the total pressure by the expression……………… .
Answer» `P = 24 K_P` `P = 8 K_P` `24 P = K_P` NONE of these Solution : Total no. of moles at equilibrium`= 80+ 20 + 20 + 120` `K_P = (P_a.P_B)/(P_(AB)) = ((20/120xxP)(20/120 xx P))/((80/120xxP)) = P /24` `24 K_P = P`

Discussion

33.	For the reaction AB (g) hArr A(g) + B(g), ABis 33 % dissociated at a totalpressure of P . Therefore, P is related to K_(p)by one of the following options
Answer» `P=K_(p)` `P=3 K_(p)` ` P=4K_(p)` ` P=8 K_(p)` SOLUTION :`{:(,AB (g),hArr,A(g),+,B(g)),("Intial moles:",1,,0,,0),("Moles at eqm:",1-0.33,,0.33,,0.33):}` `0.67` Total moles at eqm. =1.33 `p_(A) =(0.33/1.33)P, p_(B)=(0.33/1.33) P,p_(AB)=(0.67/1.33)P` `K_(p) = (p_(A) xxp_(B))/p_(AB) = (0.33 xx 0.33 xx P)/(0.67 xx 1.33)` `P/K_(p)=8 or P =8 K_(p).`

Discussion

34.	For the reaction AB_(2(g)) hArr AB_((g)) + B_((g)) if alpha is negligiable w.rt 1 then degree of dissociation (alpha) of AB_(2) is proportional to
Answer» `1/p` `1/V` `1/(SQRT(p)` `sqrt(v)` Solution :`underset((1-alpha))(AB_(2)) HARR underset(alpha)(AB)+underset(alpha)(B)` Total moles = `1+alpha` `K_(P)=(Palpha^(2))/(1-alpha^(2)), alpha LT lt IMPLIES K_(P)=P alpha^(2) implies alpha(1)/(sqrt(P))` & `P alpha(1)/(V)`

Discussion

35.	For the reaction: A_2(g)+ B_2(g) hArr 2AB(g): DeltaH is -ve.
Answer» Solution :The FOLLOWING MOLECULAR scrnes represent different reaction mixture (A - DARK grey, B - light grey) (i) Calculate the equilibrium constant `K_P and K_C`. (ii) For the reaction mixture represented by scene (X), (y) the reaction proced in which direactions ? (iii) What is the effectof increase in pressure for the mixture at equilibrium.

Discussion

36.	For the reaction A to B , Delta E = 85 kJ mol^(-1) , if the system proceeds from A to B by a reversible path and returns to A by an irreversible path , the net change in internal energy (in KJ) is
Answer» 170 zero 42.5 85 Answer :B

Discussion

37.	For the reaction (A to B+C), the energy profile diagram is given in the figure .
Answer» z z - p y - z z - x Answer :B

Discussion

38.	For the reaction A rarr B,DeltaH = + 24kJ //mole For the reaction B rarrC , DeltaH= -18 kJ // mole . The decreasing order of enthalpy of A,B,C follows the order :
Answer» A,B,C B,C,A C,B,A C,A,B Solution :`A rarr B, Delta H = +ve` means that `H (A)lt H(B)` `B rarr C, DeltaH = -ve` means that`H(B) gt H(C )` or `H(C ) lt H(B)` Addingthe given equations,we get `A rarrC, DeltaH = + 6 kJ mol^(-1)`. This means that `H(A)lt H(C )` COMBINING the above results. `H(A) lt H (C ) lt H(B)` or `H( B) gt H( C) gt H(A)` . Hence, decreasing ORDER of ENTHALPY is B,C,A.

Discussion

39.	For the reaction A+BhArrC+D, the equilibrium constant is 0.05 at 300K. Calculate the equilibrium constant for C+DhArrA+B at the same temperature.
Answer» SOLUTION :`A+BhArrC+D""K=([C][D])/([A][B])` at EQUILIBRIUM For `C+DhArrA+B,""K'=([A][B])/([C][D])` at equilibrium `thereforeK'=1/K=1/0.05=20`.

Discussion

40.	For the reaction, A+BhArr2C K_(C)=1. If the initial concentration of A,B and C are 1m, 1m and 2m respectively then, at equilibrium.
Answer» `[A]=[B]=[C]` `[A]=(4)/(3)M` `[B]=(2)/(3)M` `[A]=(1)/(2)[C]` ANSWER :A::B

Discussion

41.	For the reactionA + B hArr C+ D , the concentrations of A and B are equal . The equilibrium concentration of C is twice that of A . K_(C) of the reaction is
Answer» `1//4` `4` `1//9` 9 Answer :B

Discussion

42.	For the reaction A+ 2B to C: 5 moles of A and 8 moles of B will produce
Answer» 5 MOLES of C 4 moles of C 8 moles of C 13 moles of C Answer :B

Discussion

43.	For the reaction: 5A(g)+3B(g)+7C(g) to 5D(g)+4E(l) DeltaH=-56kcal//"mole" Find heat exchanged when 2 moles of A, 1.5 moles of B and 2.1 moles of C react in a closed rigid container to 300K:
Answer» `-62 KCAL` `-15 kcal` `-18.6 kcal` `-50 kcal` ANSWER :B

Discussion

44.	For the reaction 2SO_(2)(g)+O_(2)(g) to 2SO_(3)(g). What is the realation between DeltaH and DeltaE ?
Answer» SOLUTION :`DeltaH=Delta E+DeltanRT` `Deltan=n_(P)-n_(R)=2-3=-1` `DeltaH=DELTAE-RT" (or) "DeltaH +RT =DeltaE` Therefore, the ralationship Between `DeltaH" and" DeltaE` is `DeltaH lt DeltaE`

Discussion

45.	For the reaction : 2SO_(2)(g)+O_(2)(g) hArr2SO_(3)(g),DeltaH=ve. An increase in temperature shows:
Answer» more dissociation of `SO_(3)` and a DECREASES in `K_(c)` less dissociation of `SO_(3)` and an INCREASES in `K_(c)` more dissociation of `SO_(3)` and an increase in `K_(c)` less dissociation of `SO_(3)` and a decrease in `K_(c)` Answer :A

Discussion

46.	For the reaction, 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)""DeltaH^(@)lt0 Which change(s) will increase the fraction of SO_(3)(g) in the equilibrium mixture? (P) Increasing the pressure (Q) Increasing the temperature (R) Adding a catalyst
Answer» <P>P only R only P and R only P, Q and R Answer :A

Discussion

47.	For the reaction : 2NO(g) + O_(2)(g) rarr 2NO_(2)(g), the enthalpy and entropy changes are - 113.0 kJ mol^(-1) and -145 JK^(-1) mol^(-1) respectively. Find the temperature below which this reaction is spontaneous.
Answer» Solution :`DELTAG = DELTAH - T DeltaS`. When reaction is in equilibrium , `DeltaG =0.``:. T = (DeltaH)/(DeltaS) = ( -113000J mol^(-1))/( - 145 JK^(-1)mol^(-1)) = 779.3K`. For reaction to be SPONTANEOUS , `Delta G ` should be -ve which can be so if temperature is below ` 779.3K`

Discussion

48.	For the reaction, 2NOCl(g)hArr2NO(g)+Cl_(2)(g) the value of K_(c)=3.75xx10^(-6) at 1069K. Calculate K_(p).
Answer» Solution :Since unit is not mentioned, it is to be assumed that the concentration is EXPRESSED in mol `L^(-)`. HENCE, R = 0.0831 L BAR `K^(-1)mol^(-1).DELTAN` for the REACTION is 3 - 2 = 1. `K_(p)=K_(c)(RT)^(Deltan)` becomes `K_(p)=K_(c)RT` `K_(p)=(3.75xx10^(-6))0.0831xx1069=3.33xx10^(-4)`.

Discussion

49.	For the reaction 2NOCl_((g)) hArr 2NO_((g)) + Cl_(2(g)) the value of equilibrium constant K_p is 0.033 bar at 1060 K temp. then calculate value of K_c.
Answer» SOLUTION :`K_c=3.7xx10^(-4)`

Discussion

50.	For the reaction, 2NO_(2)(g) hArr N_(2)O_(4)(g) at 300 K The value of K_(P) is 2 atm^(-1). The total pressure at equilibrium is 10 atm. If volum of cantainer become two times of its original volume, what will be its equilibrium pressure at 300 K?
Answer» `6.4` ATM `4.51` atm `6.0` atm `5.19` atm Answer :D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For the reaction H_(2) (g) + I_(2) (g) hArr 2 HI (g) , the equilibrium constant K_(C) is ………

For the reaction : H_(2) (g) + CO_(2) hArr CO(g) + H_(2) O (g) , if the intial pressure of [H_(2)]=[CO_(2)] and x moles / litre of hydrogen is consumed at equilibrium , the correct expression of K_(p) is

For the reaction Fe_(2)S_(3)+5O_2 to 2FeSO_4 +SO_(2) the equivalent weight of Fe_(2)S_(3) is-

For the reaction , FeCO_(3(s)) rarr FeO_((s)) + CO_(2(g)), Delta H= 82.8kJ " at " 25^(@)C, what is (Delta E or Delta U) " at " 25^(@)C?

For the reaction : F_(2)+H_(2)O overset("Ice cold") underset("temperature") rarrHOF + HF which one is not correct ?

For the reaction C(s) + CO_(2) (g) hArr 2CO(g). the partial pressures of CO_(2) and CO are 2.0 and 4.0 atm respectively at equilibrium. What is the value of K_(p)for this reaction?

For the reaction C(s) + CO_(2) (g) to 2 CO (g) K_(p) = 63 " atm at " 100 K . If at equilibriump_(CO) = 10_(p_(CO_(2)) , then the toal pressure of the gases at equilibrium is

For the reaction, CO_((g))+Cl_(2(g))hArrCoCl_(2(g)), the value of K_(p)//K_(c) is equal to

For the reaction CO(g) + Cl_(2) (g) hArr COCl_(2)(g) , K_(p)//K_(c) is equal to

For the reaction CO_((g))+CI_(2(g)) harr COCI_(2(g)).The K_p // K_c is equal to

For the reaction CO_((g))+2H_(2(g))hArrCH_(3)OH_((g)). If active mass of CO is kept constant and active mass of H_(2) is tripied, the rate of forward reaction will become

For the reaction, CaCO_(3(g))hArrCaO_((s))+CO_(2(g))K_(p) is equal to

For the reaction C_(3) H_(8(g)) + 5O_(2(g)) to 3CI_(2(g)) + 4H_(2) O_((l)) , Delta H - Delta K=….....

For the reaction between KMnO_(4) " and " H_(2)O_(2) the number of electrons transferred per mole of H_(2)O_(2) is :

For the reaction at 300 K A_((g)) harr V_((g))+ S_((g) . Delta_(t) H^(@) = - 30 "KJ/mol" Delta_(t)S^(@) = - 0.1 K.J. K^(-1)."mole"^(-1) What Is the value of equilibrium constant ?

For the reaction at 298K, A+ B rarr C, DeltaH =400 kJ mol^(-1) and DeltaS = 0.2 kJ K^(-1) mol^(-1) . At what temperature will be thereaction become spontaneous considerting DeltaHand DeltaS to be constant over the temperature range.

For the reaction at 298 K, 2A + Bto C Delta H = 400 "kJ mol"^(-1) and Delta S = 0.2 "kJ K"^(-1) "mol"^(-1) At what temperature will the reaction become spontaneous considering Delta H and Delta S to beconstant over the temperature range.

For the reaction at 298 K : 2A +B to C DeltaH="400 KJ mol"^(-1) , DeltaS ="0.2 JK"^(-1) "mol"^(-1) Determine the temperature at which the reaction would be spontaneous .

For the reaction at 25^(@)C,X_(2)O_(4(l))to2XO_(2(g)),DeltaH=2.1kcal and DeltaS=20" cal "K^(-1). The reaction would be

For the reaction at 298 K, 2A+BtoC DeltaH=400 kJ mol^(-1) and DeltaS=0.2 kJ K^(-1)mol^(-1). At what temperature will the reaction becomes spontaneous considering DeltaH and DeltaS to be constant over the temperature range?

For the reaction Ag(CN)bar(2) hArr Ag^(+) 2 CN^(-), K_(c)" at "25^(@)C" is "4 xx 10^(-19) . Calculate [Ag^(+)] in solution which was originally 0*1 M in KCN and 0*03M in AgNO_(3).

For the reaction, Ag_(2)O_((s)) hArr 2Ag_((s)) + (1)/(2) O_(2(g)) Delta H, Delta S and T are 40.63 kJ mol^(-1), 108.8 JK^(-1) mol^(-1) and 373K respectively. Free energy change Delta G of the reaction will be _____

For the reaction A(g) + B(s) hArr C (g)+ D (g), K_(c) = 49 mol dm^(-3) at 127^(@)C. Calculate K_(p).

For the reaction A_((g)) + 3B_((g)) hArr 2C_((g))at 27^(@)C.2 mole of A, 4 moles of B and 6 moles of C are present in 2 lit vessel. If K_(c) for the reaction is 1.2, the reaction will proceed in

For the reaction A_((g))+B_((g)) harr C_((g))+D_((g)) K is 0.63 at 700^(@)C and 1.66 at 1000^(@)C. Then DeltaH^(@) is x KCal. What is x?

For the reaction, A_((g))+2B_((g))hArr 3C_((g))+3_((g)),K_(p)=0.05" atm at "1000K. The value of K_(c) is represented by

For the reaction, ADP+"phosphate" hArr ATP, DeltaG^circ=30.50kJmol^-1. What is the value of a equilibrium constant, K for this process under physiological conditions of 37.5^circC?.

For the reaction AB(g)hArrA(g)+B(g), at equilibrium AB is 20% dissociated at a total pressure of P, The equilibrium constant K_(p) is related to tha total pressure by the expression

For the reaction, AB(g) hArr A(g)+ B(g), at equilibrium, AB is 20% dissociated at a total pressure of P. The equilibrium constant K_p is related to the total pressure by the expression……………… .

For the reaction AB (g) hArr A(g) + B(g), ABis 33 % dissociated at a totalpressure of P . Therefore, P is related to K_(p)by one of the following options

For the reaction AB_(2(g)) hArr AB_((g)) + B_((g)) if alpha is negligiable w.rt 1 then degree of dissociation (alpha) of AB_(2) is proportional to

For the reaction: A_2(g)+ B_2(g) hArr 2AB(g): DeltaH is -ve.

For the reaction A to B , Delta E = 85 kJ mol^(-1) , if the system proceeds from A to B by a reversible path and returns to A by an irreversible path , the net change in internal energy (in KJ) is

For the reaction (A to B+C), the energy profile diagram is given in the figure .

For the reaction A rarr B,DeltaH = + 24kJ //mole For the reaction B rarrC , DeltaH= -18 kJ // mole . The decreasing order of enthalpy of A,B,C follows the order :

For the reaction A+BhArrC+D, the equilibrium constant is 0.05 at 300K. Calculate the equilibrium constant for C+DhArrA+B at the same temperature.

For the reaction, A+BhArr2C K_(C)=1. If the initial concentration of A,B and C are 1m, 1m and 2m respectively then, at equilibrium.

For the reactionA + B hArr C+ D , the concentrations of A and B are equal . The equilibrium concentration of C is twice that of A . K_(C) of the reaction is

For the reaction A+ 2B to C: 5 moles of A and 8 moles of B will produce

For the reaction: 5A(g)+3B(g)+7C(g) to 5D(g)+4E(l) DeltaH=-56kcal//"mole" Find heat exchanged when 2 moles of A, 1.5 moles of B and 2.1 moles of C react in a closed rigid container to 300K:

For the reaction 2SO_(2)(g)+O_(2)(g) to 2SO_(3)(g). What is the realation between DeltaH and DeltaE ?

For the reaction : 2SO_(2)(g)+O_(2)(g) hArr2SO_(3)(g),DeltaH=ve. An increase in temperature shows:

For the reaction, 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)""DeltaH^(@)lt0 Which change(s) will increase the fraction of SO_(3)(g) in the equilibrium mixture? (P) Increasing the pressure (Q) Increasing the temperature (R) Adding a catalyst

For the reaction : 2NO(g) + O_(2)(g) rarr 2NO_(2)(g), the enthalpy and entropy changes are - 113.0 kJ mol^(-1) and -145 JK^(-1) mol^(-1) respectively. Find the temperature below which this reaction is spontaneous.

For the reaction, 2NOCl(g)hArr2NO(g)+Cl_(2)(g) the value of K_(c)=3.75xx10^(-6) at 1069K. Calculate K_(p).

For the reaction 2NOCl_((g)) hArr 2NO_((g)) + Cl_(2(g)) the value of equilibrium constant K_p is 0.033 bar at 1060 K temp. then calculate value of K_c.

For the reaction, 2NO_(2)(g) hArr N_(2)O_(4)(g) at 300 K The value of K_(P) is 2 atm^(-1). The total pressure at equilibrium is 10 atm. If volum of cantainer become two times of its original volume, what will be its equilibrium pressure at 300 K?

For the reaction Ag(CN)bar(2) hArr Ag^(+) 2 CN^(-), K_(c)" at "25^(@)C" is "4 xx 10^(-19) . Calculate [Ag^(+)] in solution which was originally 01 M in KCN and 003M in AgNO_(3).