94420 + Interview Questions in Chemistry in Class 12

1.	For the reaction :2NO(g)+Cl_(2)hArr 2NOCl (g):which is true:
Answer» `K_(p)=K_(C )XX RT` `K_(p)=K_(c )(RT)^(2)` `K_(p)=(K_(c ))/(RT)` `K_(p)=(K_(c ))/((RT)^(2))` Answer :C

Discussion

2.	For the reaction 2NO_(2(g))hArr2NO_((g))+O_(2(g)) K_(c)=1.8xx10^(-6)at 185^(@)C. At 185^(@)C, the value of K_(c) for reaction NO_((g))+1/2O_(2(g))hArrNO_(2(g)) is
Answer» `0.9xx10^(6)` `7.5xx10^(2)` `1.95xx10^(-3)` `1.95xx10^(3)` SOLUTION :Reaction is REVERSED and halved. `K=sqrt(1//18xx10^(-6))=7.5xx10^(2)`

Discussion

3.	For the reaction 2NO_(2(g))hArr2NO_((g))+O_(2(g)) (K_(c)=1.8xx10^(-6)at 184^(@)C) (R=0.0831kJ//(mol.K)
Answer» <P>`K_(p)` is greater then `K_(c)` `K_(p)` is LESS than `K_(c)` `K_(p)=K_(e)` Wheter `K_(p)` is greater than, less than or equal to `K_(c)` depnds upon the total GAS pressure SOLUTION :`K_(p)=K_(c)(RT)^(Deltan)` `Deltan=3-2=1,k_(p)gtk_(c).`

Discussion

4.	For the reaction :2NO_(2)(g)hArr 2NO(g)+O_(2)(g) [K_(c )=1.8xx10^(-6)"at" 184^(@)C, R=0.0831 "kJ (mol, K)"]When K_(p) and K_(c ) are compared at 184^(@)C, it is found that :
Answer» `K_(p)` is greater than `K_(c )` `K_(p)` is less than `K_(c )` `K_(p)=K_(c )` whether `K_(p)` is greater than, less than or equal to `K_(c )` depend upon the TOTAL gas pressure. Solution :`K_(p)=K_(a)(RT)^(DELTA n)` `Delta n = 3-2=1` `Delta n = 3-2=1` `(K_(p))/(K_(C ))=(0.0831xx457). therefore K_(p)` is greater than `K_(c )`

Discussion

5.	For the reaction :2NO_(2)(g)hArr 2NO(g)+O_(2)(g) K_(c )=1.8xx10^(-6) at 185^(@)C, the value of K_(c ) for the reactionNO+(1)/(2)O_(2)=NO_(2)is:
Answer» `0.9xx10^(6)` `7.5xx10^(2)` `1.95xx10^(-3)` `1.95xx10^(3)` Solution :`2NO_(2)(g)HARR 2NO(g)+O_(2)(g)K_(C )=1.8xx10^(-6)` `2NO+O_(2)hArr 2NO_(2)K.=(1)/(1.8xx10^(-6))` `NO+(1)/(2)O_(2)hArr NO_(2)` `K..=SQRT((1)/(1.8xx10^(-6)))=7.45xx10^(2)`

Discussion

6.	For the reaction 2NO_2(g) hArr 2NO(g) +O_2(g),K_c=1.8xx 10^(-6) at 185^@C . At 185^@C, the value of k_c for the reaction, NO(g)+1/2O_2(g) hArr NO_2(g)is
Answer» `0.9 XX 10^6` `7.5 xx 10^2` `1.95 xx 10^(-3)` `1.95 xx 10^3` ANSWER :B

Discussion

7.	For the reaction : 2NO_(2(g)) hArr 2NO_(g)+O_(2)(g), (K_(e )=1.8xx10^(-6) " at " 184^(@)C) (R=0.0831 kJ//("mol" K)) When K_(P) and K_(e ) are compared at 184^(@)C, it is found that
Answer» WHEATER `K_(p)` is greater than, less than or equal to `K_(E )` depends upon the total GAS pressure `K_(p)=K_(e )` `K_(p)` is less than `K_(e )` `K_(p) ` is greaterthan `K_(e )` Solution :For the reaction `2NO_(2)(g) hArr 2NO (g)+O_(2)(g)` given `K_(e )=1.8xx10^(-6) at 184^(@)C` R=0.0831 Kj/mol K `K_(p)=1.8xx10^(-6)xx0.0831xx457=6.836xx10^(-6)` `[because 184^(@)C=(273+184)=457k, Delta n =(2+1,-1)=1]` Hence it is clear that `K_(p) gt K_(e )`

Discussion

8.	For the reaction: 2NO_(2)(g) + F_(2)(g) to NO_(2)F +F (slow) ii) NO_(2) + F to NO_(2)F + F (fast) What is the predicted rate law?
Answer» SOLUTION :RATE = `K[NO_(2)][F_(2)]`

Discussion

9.	For the reaction 2NO_2(g) hArr 2NO(g) + O_2(g), K_c = 1.8 xx 10^-6 at 185^Oc. At 185^Oc, What is the value of K_c for NO(g) + 1/2O_2(g) hArr NO_2(g) ?
Answer» `0.9xx10^(-6)` `7.5xx10^2` `1.95xx10^(-3) `1.95xx10^(3) ANSWER :B

Discussion

10.	For the reaction , 2NO_(2) rarr 2NO+ O_(2) rate is expressed as :
Answer» `-(1)/(2)(d[NO_(2)])/(dt)=-(1)/(2)(d[NO])/(dt)= (d[O_(2)])/(dt)` `-(2D[NO_(2)])/(dt)= (2d[NO])/(dt)=(d[O_(2)])/(dt)` `-(1)/(2)(d[NO_(2)])/(dt)=(1)/(2)(d[NO])/(dt)=(d[O_(2]])/(dt)` `-(d[NO_(2)])/(dt)=d([NO])/(dt)=(2d[O_2])/(dt)` Answer :C

Discussion

11.	For the reaction, 2NO_(2) (g) hArr2NO (g) + O_(2)(g), K_(C) = 1.8 x× 10^(–6) at 185^(@)C. At 185^(@)C, the value of K_(C) for the reaction - NO(g) + 1/2 O_(2)(g)hArrNO_(2)(g) is-
Answer» `0.9xx10^(6)` 7.5 ×x 10^(2)` `1.95 x× 10^(-3)` `1.95 x× 10^(3)` ANSWER :B

Discussion

12.	For the reaction 2NO_(2)+F_(2) rarr 2NO_(2)F, the experimental rate law is r=K[NO_(2)][F_(2)]. Propose the mechanism of reaction.
Answer» SOLUTION :N//A

Discussion

13.	For the reaction 2NO_(2) + F_(2) rarr 2NO_(2) F which is proposed to occur as : NO_(2) + F_(2) rarr NO_(2) F+F (slow) NO_(2) +Frarr NO_(2) F (fast ) the rate law can be written as :
Answer» RATE `=K[NO_(2)][F]^(2)` rate `=k[NO_(2)][F_(2)]` rate `=k[NO_(2)][F_(2)]` rate=`k[NO_(2)F]` Answer :B

Discussion

14.	For the reaction : 2NO_(2)+F_(2) to 2NO_(2)F, the experimental rate law is given as : Rate =K[NO_(2)][F_(2)], propose the mechanism.
Answer» SOLUTION :The Rate `=K[NO_(2)][F_(2)]` implies that SLOW step of reaction uses only one molecule of each reactant to FORM product, thus we can write the mechanism as : `ul{:(NO_(2)+F_(2)toNO_(2)F+F,,("slow")),(NO_(2)+FtoNO_(2)F,,("FAST")):}` Overall reaction, `2NO_(2)+F_(2)to2NO_(2)F`

Discussion

15.	For the reaction, 2NH_3rarrN_2+3H_2, if (-d[NH_3])/(dt)=k_1[NH_3], (d[N_2])/(dt)=k_2[NH_3], (d[H_2])/(dt)= k_3[NH_3]then the relation between k_1,k_2 and k_3 is .....
Answer» `k_1 = k_2 = k_3` `k_1 = 3 k_2=2k_3` `1.5k_1=3k_2=k_3` `2k_1=k_2=3k_3` SOLUTION :`(-1/2)(d[NH_3])/(DT)=(d[N_2])/(dt)=(1/3)(d[H_2])/((dt))` `(1/2)k_1[NH_3]=k_2[NH_3]=(1/3)k_3[NH_3]` `(3/2)k_1=3k_2=k_3` `1.5k_1=3k_2=k_3`

Discussion

16.	For the reaction : 2NH_(3)(g) underset("Rate"=k)overset(Pt)to N_(2)(g)+3H_(2)(g) (i) Write the order and molecularity of the reaction. (ii) Write the unit of k.
Answer» SOLUTION :(i) ZERO ORDER, BIMOLECULAR (ii) mol `L^(-1)s^(-1)`

Discussion

17.	For the reaction 2NH_3 to N_2 +3H_2, if -(d[NH_3])/(dt)=k_1[NH_3], (d[N_2])/(dt)=k_2[NH_3], (d[H_2])/(dt)=k_3[NH_3] then the relation between k_1,k_2 and k_3 is
Answer» `k_1= k_2=k_3` `k_1=3 k_2=2k_3` `1.5k_1= 3k_2 =k_3` `2k_1=k_2=3k_3` SOLUTION :`2 NH_3 to N_2 + 3H_2` `RATE = - 1/2(d[NH_3])/(DT)= (d[N_2])/(dt) = 1/3(d[H_2])/(dt)` `1.5k_1=3k_2= k_3`

Discussion

18.	For the reaction 2N_(2) + 3H_(2)(g)what is the order?
Answer» SOLUTION :ZERO ORDER.

Discussion

19.	For the reaction, 2N_2O_5rarr4NO_2+O_2 , rate of reaction in terms of O_2 is d[O_2]/dt . In term of N_2O_5 will be:
Answer» `-d[N_2O_5]/DT` `+d[N_2O_5]/dt` `-1/2D[[N_2O_5]]/dt` `-2d[[N_2O_5]]/dt Answer :C

Discussion

20.	For the reaction 2N_(2)O_(5)(g)to4NO_(2)(g)+O_(2)(g), the following results have been obtained : {:("S. No.",,,[N_(2)O_(5)]" mol L"^(-1),,,"Rate of disappearance of"),(,,,,,,N_(2)O_(5)","" mol L"^(-1)min^(-1)),(1,,,1.13xx10^(-2),,,34xx10^(-5)),(2,,,0.84xx10^(-2),,,25xx10^(-5)),(3,,,0.62xx10^(-2),,,18xx10^(-5)):} (a) Calculate the order of reaction (b) Write rate law (c) Calculate rate constant of the reaction.
Answer» Solution :Rate `=k[N_(2)O_(5)]^(n). :.(R_(1))/(R_(2))={([N_(2)O_(5)]_(1))/([N_(2)O_(5)]_(2))}^(n)`. Substitute given values and calculate n. Rate of reaction `=(1)/(2)XX" Rate of disappearance of "N_(2)O_(5)` WHE `[N_(2)O_(5)]=1.13xx10^(-2)" mol L"^(-1)"","""Rate"=(1)/(2)xx34xx10^(-5)=17xx10^(-5)" mol L"^(-1)min^(-1)` `k=("Rate")/([N_(2)O_(5)])=(17xx10^(-5)"mol L"^(-1)min^(-1))/(1.13xx10^(-2)"mol L"^(-1))=1.5xx10^(-2)min^(-1)`

Discussion

21.	For the reaction, 2N_(2)O_(5)to4NO_(2)+O_(2) rate and rate constant are 1.02xx10^(-4) M sec^(-1) and 3.4xx10^(-5)sec^(-1) respectively, the concentration of N_(2)O_(5), at that time will be
Answer» 1.732 M 3 M `3.4xx10^(5)M` `1.02xx10^(-1)M` ANSWER :B

Discussion

22.	For the reaction: 2N_(2)O_(5)(g) to 4NO_(2)(g)+O_(2)(g), the rate of formation of NO_(2)(g) is 2.8xx10^(-3)Ms^(-1). Calculate the rate of disappearance of N_(2)O_(5)(g).
Answer» Solution :`2N_(2)O_(5)(g) to 4NO_(2)(g)+O_(2)(g)` Rate of formation of `NO_(2)(g) =2.8xx10^(-3)M s^(-1)` Rate of disappearance of `N_(2)O_(5)(g)=2.8xx10^(-3)XX(2)/(4)=1.4xx10^(-3)MS^(-1)`

Discussion

23.	For the reaction, 2N_(2)O_(5)rarr 4 NO_(2)+O_(2), select the correct statement.
Answer» Rate of FORMATION of `O_(2)` is same as rate of formation of `NO_(2)` Rate of disappearance of `N_(2)O_(5)` is two times the rate of formation of `O_(2)`. Rate of formation of `O_(2)` is 0.5 times rate of disappearance of `N_(2)O_(5)`, Rate of formation of `NO_(2)` is EQUAL to rate of disappearance of `N_(2)O_(5)` ANSWER :C

Discussion

24.	For the reaction 2N_(2)O_(5(g))rarr 4NO_(2(g))+O_(2(g)) in liquid bromine, whichof the following rate equation is'incorrect' ?
Answer» `-(1)/(2)(d[N_(2)O_(5)])/(dt)` `-(1)/(4)(d[NO_(2)])/(dt)` `(d[O_(2)])/(dt)` `1/4 (d[NO_(2)])/(dt)` Answer :A::B::D

Discussion

25.	The rate of reaction. 2N_(2)O_(5) to 4NO_(2) + O_(2) can be written in three ways. (-d[N_(2)O_(5)])/(dt) = k[N_(2)O_(5)] (d[N_(2)O_(5)])/(dt) =( k^(')[N_(2)O_(5)]) (d[O_(2)])/(dt) = (k^(')[N_(2)O_(5)]) The relation between k and k^(') are:
Answer» k=k 2k=k k=2k' k=4k' Solution :Rate of DISAPPEARANCE of reactant = Rate of appearance of PRODUCTS `(1)/(2)(d[N_(2)O_(5)])/(DT)=(1)/(4)(d[NO_(2)])/(dt)` `(1)/(2)k[N_(2)O_(5)]=(1)/(4) k.[N_(2)O_(5)]` `(k)/(2)=(k.)/(4), :. k.=2k`

Discussion

26.	For the reaction 2N_(2)O_(5)to4NO_(2)+O_(2), rate and rate constant are 1.02xx10^(-4)M"sec"^(-1) and 3.4xx10^(-5)"sec"^(-1) respectively then concentration of N_(2)O_(5) at that time will be (in moles /lit)
Answer» 3 M 4 M 1 M 1.5 M Answer :A

Discussion

27.	For the reaction 2N_2O_2(g) to 4NO_2 (g)+O_2(g). the rate of fromation of NO_2(g) is 2.8 times 10^-3 MS^-1 , Calculate the rate of disappearance of N_2O_3(g).
Answer» SOLUTION :RATE=`1/4 (DELTA (NO_2))/(Delta (t))=-1/2 (Delta (N_2O_s))/(Delta(t))` `1/4(2.8 TIMES 10^-3)=-1/2 (Delta (N_2O_5))/(Delta (t))` Rate of DISAPPEARANCE of `N_2O_3 (- (Delta (N_2O_5))/(Delta (t)))=1.4 times 10^-3 M//S`

Discussion

28.	For the reaction 2N_2O_5 (g) to 4NO_2(g) +O_2(g) which of the following graph would yield a straight line.
Answer» `logp_(N_2O_5)` vs timewith -ve slope `log p_(N_2O_5)` vs time the +ve slope `(p_(N_2O_5))^(-1)` vs time `p_(N_2O_5)` vs time Solution :As it is a REACTION of 1st order , log `[N_(2)O_(5)]` vs time GIVES a STRAIGHT line graph with -ve slope .

Discussion

29.	For the reaction 2HI_((g))hArrH_(2(g))+I_(2(g))-QkJ,, the equilibrium constant depends upon
Answer» catalyst volume temperature pressure Solution :In general , the temperature dependence of the equilibrium constant depends on the nature of the reaction whether it is ENDOTHERMIC or exothermic. The equilibrium constant for an exothermic reaction (negative `DeltaQ)` decreases as temperature INCREASES and for an endothermic reaction (POSITIVES `DeltaQ)` equilibrium constant increases as temperature increases.

Discussion

30.	For the reaction 2HItoH_(2)+I_(2) the expression -1/2(d[HI])/(dt) represents
Answer» The RATE of formation of HI The rate of disappearance of HI The instantaneous rate of the REACTION The AVERAGE rate of reaction Answer :C

Discussion

31.	For the reaction 2HIrarrH_(2)+I_(2) .Write its molecularity.
Answer» SOLUTION :TWO or 2.

Discussion

32.	For the reaction: 2H_(2(g))+O_(2(g))iff2H_(2)O_((g)) Which of the following fact holds good?
Answer» `K_(p)=K_(c )` `K_(p)gtK_(c )` `K_(p)ltK_(c )` `K_(p)andK_(c )` cannot be correlated SOLUTION :As we know, `K_(p)=K_(c)(RT)^(Deltan)` `2H_(2(g))+O_(2(g))iff2H_(2)O_((g))` `Deltan=2-(2+1)=-1` `K_(p)=K_(c)(RT)^(-1)=(K_(c))/(RT)` Hence, `K_(p)ltK_(c)`

Discussion

33.	For the reaction, 2H_(2)O(g) rarr 2H_(2)(g) + O_(2)(g), Delta H = 571.6 kJ Delta_(f) H^(theta) of water is:
Answer» 285.8kJ `-285.8kJ` 1143.2kJ `-1143.2kJ` SOLUTION :`Delta_(f) H (H_(2)O)` MAY be written as `H_(2)(G) + (1)/(2) O_(2)(g) rarr H_(2)O(g)` `Delta_(f)H (H_(2)O) = - 571.6/2 = - 285.8 KJ`

Discussion

34.	For the reaction 2H_(2)(g)+O_(2)(g) to 2H_(2)O(g), DeltaH^(ө)=-573.2kJ The heat of decomposition of water per mole is
Answer» 286.6 kJ 573.2 kJ `-28.66 kJ` zero Solution :HEAT of decomposition of water is `H_(2)O(G)to H_(2)(g)+(1)/(2)O_(2)(g),DeltaH=(+573.2)/(2)=286.6kJ//mol`

Discussion

35.	For the reaction, 2H_(2)(g)+2NO(g) rarr N_(20(g) +2H_(2)O(g) Rate =k[H_(2)][NO]^(2). This mechanism has been proposed: Step 1: H_(2)+NO rarr H_(2)O+N Step2 : N=NO rarr N_(2)+O Step 3 : O+H_(2) rarr H_(2)O Which statement about this rate law and mechanism is correct?
Answer» The MECHANISM is consistent with the RATE LAW if step 1 is the rate determineing step. This mechanism is consistent with the rate law if step 2 is the rate detemining step. The mechanism is consistent with the rate law if step 3 is the rate determining step. This mechanism can not be consistent with the rate law, REGARDLESS of which step is rate-determining. Answer :B

Discussion

36.	For the reaction 2H_(2)+O_(2)rarr2H_(2)O, DeltaH=-571.Bond energy of H-H = 435, O=O = 498, then calculate the average bond energy of O-H bond using the above data
Answer» 484 `- 484` 271 `- 271` Solution :`2H_(2)+O_(2)rarr2H_(2)O, DeltaH=-571` `DeltaH_("REATION")=2xxB.E(H_(2))+B.E(O_(2))-2xxB.E.(H_(2)O)` `THEREFORE -571=(2xx435)+498-2xxB.E. (H_(2)O)` `or, 2xxB.E(H_(2)O)=870+498+571` `B.E(H_(2)O)=(1939)/(2)=(969.5)` `B.E(O-H)=(969.5)/(2)=484.75`.

Discussion

37.	For the reaction, 2H_(2)+O_(2) to 2H_(2)O,DeltaH=571. bond energy of H-H=435, O=O=498, then calculate the average bond energy of O-H bond using the above data
Answer» 484 `-484` `271` `-271` SOLUTION :`DeltaH_(("reaction"))=SUM` bond energies of reactants `-sum`bond energies of produts. Given, `2H_(2)+O_(2)to2H_(2)O,DeltaH=-571` `H-H=435,O=O` is 498, O-H is? `DeltaH=(2xx"B E of "H_(2)+" B E of "O_(2))-2("B E of "H_(2)O)` or `-571=(2xx435+498)-2xx`B E of `H_(2)O` or `-571=870+498-2xx`Be of `H_(2)O` or `2BE` of `H_(2)O =870+498+571=1939` `therefore`BE of 1 `H_(2)O` molecule`=(1939)/(2)=969.5` `therefore `BE of 1 O-H bond `=(969.5)/(2)` (`because` `1H_(2)O` molecule has 2O-H bonds)=484.75

Discussion

38.	For the reaction. 2CO(g)+O_(2)(g)to2CO_(2)(g),DeltaH=-560kJmol^(-1) In one litre vessel at 500 K the initial pressure is 70 atm and after the reaction it becomes 40 atm at constant volume of one litre. Calculate change in internal energy. All the above gases show significant deviation from ideal behaviour. (1 L atm = 0.1 kJ)
Answer»

Discussion

39.	For the reaction, 2CO+O_(2)rarr2CO_(2), DeltaH=560 kJ. Two moles of CO and one mole of O_(2) are taken in a container of volume 1L. They completely form two moles of CO_(2), the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm, find the magnitude (absolute value) of DeltaU at 500 K (1 L atm = 0.1 kJ)
Answer» 563 575 585 595 Solution :`DeltaH=DeltaU+Delta(PV)` `DeltaH=DeltaU+VDeltaP` `DeltaU=DeltaH-VDeltaP=-560-1xx30xx0.1` ABSOLUTE VALUE = 563 KJ.

Discussion

40.	For the reaction, 2Cl(g) to Cl_(2)(g), what are the signs of DeltaH and DeltaS?
Answer» `+,+` `+,-` `-,-` `-,+` Solution :In the GIVEN REACTION, a MOLECULE of `Cl_(2)` is formed from its two gaseous atoms and the energy is relased witth the formation of BOND. Hence, `DeltaH` is negative. In this reaction randomness (entropy) ALSO decreases because 2 atoms of Cl have more randomness than one molecule of chlorine. hence, `DeltaS` is negative.

Discussion

41.	For the reaction : 2Cl(g)-.Cl_2(g).
Answer» `/_\H =+ve and /_\S =-ve` `/_\H=-ve and/_\S=+ve` `/_\H= -ve and/_\ S= -ve `/_\H= -ve and /_\S = +ve Answer :C

Discussion

42.	For the reaction : 2Cl_((aq))^(-) + 2H_2O_((l)) to 2OH_((aq))^(-) + H_2 + Cl_(2(l)) The free energy change is equal to ..........
Answer» `-242 KJ ` `+242 kJ` `+422 kJ ` `-422 kJ ` ANSWER :C

Discussion

43.	For the reaction: 2AgCl(s) + H_(2)(g) (1 atm) to 2Ag (s) + 2H^(+) (0.1 M) + 2Cl^(-) (0.1 M), DeltaG^(@) =-43600 J at 25^(@) C Calculate the emf of the cell. [log 10^(-n) =-n] (b) Define fuel cell and write its two advantages.
Answer» Solution :(a) `DeltaG^(@) =-nFE_("cell")^(@)` `=-43600 J =-2 xx 96487 C mol^(-1) xx E_("cell")^(@)` or `E_("cell")^(@) =(-43600)/(-2 xx 96487) = 0.2259 V` `E_("cell") =E_("cell")^(@)- 0.059/2 LOG ([0.1]^(2) [0.1]^(2))/1 =0.2259 -0.059/2 log [0.1]^(4)` `=0.2259 + 0.118 = 0.3439 V` (b) Galvanic cells that are DESIGNED to convert the energy of combustion of fuels like hydrogen, methane, methanol etc., directly into electrical energy are called fuel cells. Advantages : 1. Fuel cells PRODUCE electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%. 2. Fuel cells are pollution free

Discussion

44.	For the reaction: 2A(g) + nB(g) rarr 3C(g)IfK_(p) "and" K_(c) are 0.0105 and 0.45 at 250^(@)C. The value of n is ________.
Answer» SOLUTION :2A(g) + nB(g) rarr 3C(g) `K_(p)/K_(c) = (RT)^(Deltan)` `(0.0105/0.45) = (0.0821 xx 523)^(Delta n)` `Delta n = -1` `Delta n = 3-(2+n) = -1` n = 2

Discussion

45.	For the reaction : 2A(g) rarr 3B(g) + C(l), the rate law is : r=-(1)/(2).(dP_(A))/(dt)=K.P_(A) The reaction is performed at constant volume and temperature, starting with only pure A(g). If P_(T) and P_(oo) are the total pressure of system at t = (oo), and P_(0) is the vapour pressure of C(l), then the rate constant of reaction may be expressed as :
Answer» `k=(1)/(T_(min))."ln"(P_(OO))/(P_(oo)-P_(T))` `k=(1)/(2T_(min))."ln"(P_(oo))/(P_(oo)-P_(T))` `k=(1)/(2T_(min))."ln"(P_(oo)-P_(0))/(3(P_(oo)-P_(T)))` `k=(1)/(2T_(min))."ln"(P_(oo)-P_(0))/(P_(oo)-P_(T))` Answer :C

Discussion

46.	For the reaction: 2A(g) ltimplies 3B(g) + C(g) , K_(p) = 2.7 xx 10^(-9) atm^(2) at 298 K. The degree of dissociation of A(g) at 298 K and 40 atm is-
Answer» 0.1 0.001 0.001 0.0001 Answer :C

Discussion

47.	For the reaction :2A+BtoA_(2)B the rate =k[A][B]^(2) with k=2.0xx10^(-6) mol^(-2) L^(2)s^(-1) Calculate the initial rate of the reaction when [A]=0.1 mol L^(-1),[B] =0.2 mol L^(-1).Calculate the rate of reaction after [A] is reduced to 0.06 mol L^(-1)
Answer» SOLUTION :Given reaction :`2A+BtoA_(2)B` Rate EQUATION r=`k[A][B]^(2)` Where k=`2.0xx10^(-6) mol^(-2) L^(-2)s^(-1)` (i)Initial rate =`r_(1)` at initial [A]=0.1 mol `L^(-1)` [B]=0.2 mol `L^(-1)` `THEREFORE r_(1)(2.0xx10^(-6) mol^(-2) L^(2) s^(-1))` `(0.1 mol L^(-1))(0.2 mol L^(-1))^(2)` `=8xx10^(-9) mol L^(-1)s^(-1)` ..........(i) (ii)When [A]=0.06 mol `L^(-1)` than rate `r_(2)`: Where ,`r_(2)=k[A][B]^(2)` Reaction :`2A+BtoA_(2)B` Initial concentration 0.1M0.2M CHANGE in reaction -2x-x Concentration at NEW 0.06M(0.2-x)M equilibirum =(0.1-2x)M=0.18 M Decrease in concentration of A 0.06=(0.1-2x)M `therefore` (2x)=(0.1-0.06)M `therefore` 2x=0.04M `therefore` x=0.02M `therefore r_(2)=k[A][B]^(2)` (0.06 mol `L^(-1))xx(0.18 mol L^(-1))^(2)` `=3.888xx10^(-9) mol L^(-1)s^(-1)`

Discussion

48.	For the reaction :2A(g)+B(g)hArr 3C(g)+D(g)Two moles each of A and B are taken in a 2L flask. The following must always be true at equilibrium :
Answer» `[A] = [B]` `[A] LT [B]` `[B] - [D]` `[A] GT [B]` ANSWER :B

Discussion

49.	For the reaction 2A+Bto3C+D which of the following does not express the reaction rate:
Answer» `(d[D])/(dt)` `-(d[A])/(2dt)` `-(d[C])/(3dt)` `-(d[B])/(dt)` SOLUTION :RATE of reaction `=13xx(d[C])/(dt)`

Discussion

50.	For the reaction 2A+Bto product ,(d[A])/(dt)=K[A]^(2)[B]. What will be rate equation for -(d[B])/(dt) ?
Answer» `K[2A]^(2)[B]` `(1)/(2)K[A]^(2)[B]` `K[A][B]^(2)` `K[A][B]^((1)/(2))` ANSWER :B

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For the reaction :2NO(g)+Cl_(2)hArr 2NOCl (g):which is true:

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For the reaction :2NO_(2)(g)hArr 2NO(g)+O_(2)(g) [K_(c )=1.8xx10^(-6)"at" 184^(@)C, R=0.0831 "kJ (mol, K)"]When K_(p) and K_(c ) are compared at 184^(@)C, it is found that :

For the reaction :2NO_(2)(g)hArr 2NO(g)+O_(2)(g) K_(c )=1.8xx10^(-6) at 185^(@)C, the value of K_(c ) for the reactionNO+(1)/(2)O_(2)=NO_(2)is:

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For the reaction , 2NO_(2) rarr 2NO+ O_(2) rate is expressed as :

For the reaction, 2NO_(2) (g) hArr2NO (g) + O_(2)(g), K_(C) = 1.8 x× 10^(–6) at 185^(@)C. At 185^(@)C, the value of K_(C) for the reaction - NO(g) + 1/2 O_(2)(g)hArrNO_(2)(g) is-

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For the reaction 2NO_(2) + F_(2) rarr 2NO_(2) F which is proposed to occur as : NO_(2) + F_(2) rarr NO_(2) F+F (slow) NO_(2) +Frarr NO_(2) F (fast ) the rate law can be written as :

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For the reaction, 2NH_3rarrN_2+3H_2, if (-d[NH_3])/(dt)=k_1[NH_3], (d[N_2])/(dt)=k_2[NH_3], (d[H_2])/(dt)= k_3[NH_3]then the relation between k_1,k_2 and k_3 is .....

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For the reaction :2A(g)+B(g)hArr 3C(g)+D(g)Two moles each of A and B are taken in a 2L flask. The following must always be true at equilibrium :

For the reaction 2A+Bto3C+D which of the following does not express the reaction rate:

For the reaction 2A+Bto product ,(d[A])/(dt)=K[A]^(2)[B]. What will be rate equation for -(d[B])/(dt) ?