Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

For the reaction 2NH_(3) hArrN_(2) + 3H_(2) equilibrium constant at 25^(@)C and 400^(@)C and 3.5 and 35 respcctively. This means that the forward reaction is:

Answer»

EXOTHERMIC
ENDOTHERMIC
Exorgic
Onpredic table

Solution :as `T uarr`, K is INCREASING `IMPLIES` endothermic reaction
2.

For the reaction 2KClO_(3) rarr 2KCl +3O_2which statments (s) is (are ) correct ?

Answer»

It is disproportionation
It is intramolecular redoxchange
CL atoms are reduced
OXYGEN atoms are OXIDIZED

SOLUTION :
3.

For the reaction: 2H_(2(g)) + O_(2(g)) rarr 2H_2O_((g)) , DeltaH = -571 kJ bond energy of (H-H) = 435 kJ, of (O = O) = 498kJ , then the average bond energy of O-H bond using the above data

Answer»

`484kJ`
`-84kJ `
`271kJ`
`-271kJ`

ANSWER :A
4.

For the reaction 2H_(2(g)) + o_(2(g)) rarr 2H_(2)O_((g)), Delta H^(@) = - 573.2 kJ. The heat of decomposition of water per mol is:

Answer»

286.6kJ
573.2 KJ
`-28.66` k Cal
68.56 kCal

Solution :For DECOMPSITION `+573.2 rarr 2` mol
? `rarr` 1MOL
`Delta = (573.2)/(2)` = 286.6kJ/mol = 68.56 Kcal/mol
5.

For the reaction, 2Fe^(3+) + Sn^(2+)rarr2Fe^(2+) + Sn^(4+) the normality of SnCl, (mol. wt. = 189.7) solution prepared by dissolving 47.5g in acid solution and diluting with H_(2)Oto a total of 2.25 litre is

Answer»

0.222N
0.111 N
0.333 N
0.444 N

Solution :`VN=W/Eimplies 2.25 XXN = ((475)/(189.7))/2 impliesN= 0.222` N
6.

For the reaction, 2Co +O_(2)rarr 2CO_(2), DeltaH = - 560 kJ . Two moles of CO andone mole ofO_(2) are takenin a container of volume 1 L. They completely reacts to form two moles of CO_(2). The gases deviate appreciably from ideal behaviour. Ifthe pressure in the vessel changesfrom 70 to 40 atm,find the magnitude ( absolute value)of DeltaU at 500 K ( 1 L atm = 0.1 kJ ).

Answer»

Solution :`H =U +PV`
`DELTAH+ DeltaU + Delta(PV) ` or ` DeltaH = DeltaU + VDeltaP`
or `DeltaU = DeltaH -VDeltaP = - 560kJ - 1L( 40 atm-70 atm)`
`=- 560 KJ + 30 Latm = - 560 kJ + 3-0 XX 0.1kJ = - 557 kJ `
Absolute value `=557 kJ . `
7.

For the reaction , 2Cl(g)rarrCl_(2)(g), what are thesigns ofDeltaHand DeltaS ?

Answer»

Solution :The GIVEN reaction repesents the formation of BONDS. Hence, energy is released , ie.,`DELTAH` is `-ve`. Further, 2 moles of atomshave greater randomness than1 mole of molecules. Hence, randomness DECREASES, i.e.,`DeltaS` is `-ve`.
8.

For the reaction 2Cl(g)rarrCl_(2)(g), the signs of DeltaHand DeltaS respectively are

Answer»

`+,-`
`+,+`
`-,-`
`-,+`

Solution :Whenatoms combine to form molecule,energy is released,i.e., `DeltaH ` is `-ve`. As TWO moles combine to give ONE mol, randomness ( ENTROPY) DECREASES,i.e., `eltaS` is `-ve`.
9.

For the reaction, 2Cl(g)rarr Cl_2(g) what are the signs of DeltaH and DeltaS?

Answer»

SOLUTION :Since the reaction involves formation of bonds, enery is released. So `DELTAH` is NEGATIVE. Since chlorine atoms combine together to form `Cl_2` molecules, RANDOMNESS decreases. i.e. `DeltaS` is negaive.
10.

For the reaction 2CI_((s) ) to CI_(2(s)) the value of Delta H and Delta S are respectively .......... .

Answer»

`+,-`
`+,+`
`-,-`
`-,+`

ANSWER :C
11.

For the reaction 2BrCl hArr Br_(2)(g) + Cl_(2)(g) K_(c)" is " 32 " at " 500 K. If initially pure BrCl is present at a concentration of 3.30 xx 10^(-3), M, what is its molar concentration in the mixture at equilibrium?

Answer»


ANSWER :`3.0 XX 10^(-4) M`
12.

For the reaction, 2CI_((g)) to CI_(2(g)) what are the signs of Delta H and Delta S ?

Answer»

SOLUTION :`2CI_((G)) to CI_(2(g))`shows the reaction of formation of bond. So heat is released.
`THEREFORE Delta H -ve`
One mole `CI_(2)` molecule is formed from 2 mol of Cl ATOMS. So RANDOMNESS is decreased
`therefore Delta S = -ve`
13.

for the reaction 2A_((g)) hArr B_((g)) + 3C_((g)), at a given temperature, K_(c) = 16. What must be the volume of the falsk, If a mixture of 2 mole cach A, B and C exist in equilibrium ?

Answer»

`(1)/(4)`
`(1)/(2)`
1
2

Solution :`K_(C)=(n_(B)n_(C)^(2))/(n_(A)^(2)) xx (1)/(V^(2)) IMPLIES 16=(2 xx 2^(3))/(2^(2) xx V^(2)) implies V=(1)/(2)`
14.

For the reaction 2A(g)+B(g)rarr2D(g), DeltaU^theta=-10.5 kJ and DeltaS^theta=-44.1Jk^-1 Calculate DeltaG^theta for the reaction and predict whether the reaction may occur spontaneously.

Answer»

Solution :`Deltan` for the reaction=2-(2+1)=-1
`therefore DeltaH^theta=DeltaU^theta+DeltanRT=-10.5+(-1)xx8.314xx10^-3kJxx298=-12.98`
DeltaG^theta=DeltaH^theta-TDeltaS=-12.98-298(-44.1xx10^-3kJk^-1)=+0.16kJ`
Since `DeltaG^theta is POSITIVE, the reaction will not occur SPONTANEOUSLY at 298K.
15.

For the reaction 2A_((g)) + B_((g)) to 2D_((g)) DeltaU^( Theta ) = -10.5 "kJ" and S^( Theta ) =-44.1 "JK"^(-1) Calculate Delta G^( Theta ) for the reaction, and predict whether the reaction may occur spontaneously.

Answer»

Solution :`Deltan_((g)) = 2-3 = -1`
`Delta H = Delta U + Deltan_((g)) "RT"`
`= - 10.5 + (-1) (8.314 XX 10^(-3) ) (298)`
`= -10.5 - 2.48 = - 12.98 "kJ"`
`Delta G = Delta H- T Delta S`
`= -12.98 - 298 (-44.1 xx 10^(-3) )`
`= - 12.98 "kJ" + 13.14 "kJ" = 0.16 "kJ"`
`Delta G` is positive, so the reaction may not occur SPONTANEOUSLY.
16.

For the reaction, , 2A(g)+B(g) rarr 2D(g), DeltaU^(@) = - 10.5 kJ andDeltaS^(@) = - 44.10JK^(-1) Calculate DeltaG^(@) for the reaction and predictwhether the reaction mayoccur spontaneously.

Answer»

Solution :For the GIVEN reaction,`Deltan_(G) =2- ( 3)= - 1`
`:.DeltaH^(@) =DELTAU^(@) + Deltan_(g) RT =- 10.5 kJ + ( -1) ( 8.314 xx 10^(-3) kJ ) ( 298) = - 10.5-2.48= - 12.98 kJ`
`DeltaG^(@) = DeltaH^(@)- T DeltaS=-12.98 kJ -298 ( -44.1xx 10^(-3) kJ)=-12.98 kJ+13.14 kJ =0.16 kJ `
As`DeltaG^(@) `COMES out to be `+ve`,the reaction will not occur spontaneously.
17.

For the reaction, 2A(g)+B(g)to2D(g) DeltaU^(theta)=-10.5 kJ and DeltaS^(theta)=-44.1 JK^(-1) Calculate DeltaG^(theta) for the reaction, and predict whether the reaction can occur spontaneously or not.

Answer»

SOLUTION :`DELTA G ^(@) = + 161.8 kJ MOL ^(-1) , `NON spontaneous
18.

For the reaction 2A+BrarrC at 298 K, DeltaH=400kJ mol^-1 and DeltaS=0.2 kJk^-1mol^_1. At what temperature will be reaction become spontaneous considering DeltaH and DeltaS to be constant over the temperature range.

Answer»

Solution :At equilibrium, `DeltaG`=0`therefore` DELTAH-TDELTAS`=0 or `DeltaH=TDeltaS` `therefore` T=`DeltaH/DeltaS=400/0.2=2000K`
i.e the reaction will be at equilibrium at 2000K. Since `DeltaH`and `DeltaS` are positive`DeltaG` will be negative when `Tdeltas>DeltaH`. This takes PLACE at higher temperature. Therefore, the reaction will be sopntaneous at temperature above 2000K.
19.

For the reaction 2 NO_(2) (g) hArr 2NO (g) + O_(2) (g) , K_(C) = 1.8 xx 10^(-6) at 185^(@) C . At the same temperature the value of K_(C) for the reaction . NO (g) + (1)/(2) O_(2) 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)`

Solution :The reaction is REVERSED and halved.
`K_(C) = (1)/(sqrt(1.8 xx 10^(-6)) = 7.5 xx 10^(2)`
20.

For the reaction 298 K A_((g)) + B_((g)) to C_((g)) at 298K, Delta U = -5 "cal" and Delta S= -10 "cal" K^(-1). Delta G = …..... "cal".

Answer»

`2612`
`2379`
`261.2`
`-2612`

ANSWER :B
21.

For the reaction , 2 NO (g) + Cl_(2) (g) hArr 2 NOCl (g) " if partial pressure " p_(NOCl)= 0*32 " atm, " p_(No) = 0*22 " atm" and p_(Cl_(2))= 0*11 " atm, then find "K_(p).

Answer»


Solution :`K_(p) = p_(NOCI)^(2)/ (p_(NO)^(2) xx p_(Cl_(2)))= (0*32 " atm")^(2)/(0*22 " atm"^(2)xx 0*11 "atm")=19*23 " atm"^(-1)`
22.

For the reaction 2 HI (g) hArr H_(2) (g) + I_(2) (g)at 720 K , the equilibrium constant value is 50 . The equilibrium constant for the reaction H_(2) (g) + I_(2) (g) hArr 2 HI (g) at the same temperature will be ………..

Answer»

Solution :`0.02`
Forward reaction EQUILIBRIUM constant `K_1= 50`
Reverse reaction equilibrium constant `K_(2) = ?`
`K_(2) = (1)/(K_(1)) = (1)/(50) = 0.02`
23.

For the reaction 2 H_(2) + O_(2) rarr 2H_(2)O, DeltaH= - 571 . Bond energy of H-H =435 and O=O=498 . Then the average bond energy of O-H bond will be

Answer»

484
`-484`
271
`-271`

Solution :For the reaction
`2H_(2)+O_(2)RARR 2H_(2)O(2H_(2)O(2H-O-H)`
`Delta_(r)H= 2xx BE(H_(2)) +BE(O_(2)) - 4 xx BE(O-H)-571= 2 xx 435 +498 -4 xxBE (O-H) ` or `BE(O-H) =484`
24.

For the raction: PCI_5(g)hArrPCI_3(g)+CI_2(g) The forward reaction at constant temperature is favouraved by:

Answer»

introduncing chlorine gas at constant VOLUME
introducing an inert gas at constant PRESSURE
INCREASING the volune of the CONATAINER
introducing `PCI_5` at constant volume

ANSWER :B::C::D
25.

For the p_(z) orbital, conventionally m is

Answer»

`-2`
`+2`
0
Any of these

Answer :C
26.

For the properties mentioned, the correct trend for the different species is in

Answer»

Strength as LEWIS acid`-BCl_(3)gtAlCl_(3)gtGaCl_(3)`
Inert pair effect-`AlgtGagtIn`
Oxidising property-`Al^(3+)gtIn^(3+)gtTi^(3+)`
FIRST ionisation ENTHALPY-`BgtAlgtTI`

Solution :(a) The tendency to behave as Lewisacid DECREASES with the increases in the size down the group. `BCl_(3)` easily accepts a lone pair of ELECTRONS from ammonia to form `BCl_(3)cdotNH_(3)`. Hence, strength as Lewis acid
`BCL_(3)gtAlCl_(3)gtGaCl_(3)`
27.

For the process X_((g)) + e^(-) rarr X_((g))^(-), Delta H =x and X_(g)^(-) rarr X_(g) + e^(-) , Delta H = y Select correct alternat :

Answer»

IONISATION ENERGY of `X_((G))^- is Y`
electron affinity of `X_((g)) is X`
electron affinity of `X_((g)) is -Y`
all the correct statements.

Answer :D
28.

For the process to occue under adiabatic conditions, the correct conditions is : (i) Delta T = 0 (ii) Delta p =0(iii) q=0(iv) w=0

Answer»

SOLUTION :`DELTAP= 0`
29.

For the process to occur under adiabatic conditions, the correct condition is :

Answer»

<P>`DELTA T = 0`
`Delta p = 0`
`q=0`
`w=0`

ANSWER :C
30.

Forthe process H_(2)O(l) rarr H_(2)O(g) at T=100 ^(@)C and 1 atmosphere pressure, the correct choice is

Answer»

`DeltaS_("system") gt 0` and `DELTAS _("surroundings") gt 0`
`DeltaS_("system") gt 0` and `DeltaS _("surroundings") lt 0`
`DeltaS_("system") lt 0` and `DeltaS _("surroundings") gt 0`
`DeltaS_("system") lt 0` and `DeltaS _("surroundings") lt 0`

Solution :At `T= 100^(@)C` and 1 atmosphere presssure (i.e.,boilingpoint ) ,thereexists the EQUILIBRIM
`H_(2)O(l) hArr H_( 2)O (g)`,
For this conditions,`DeltaS_("total") = 0`
Hence,`DeltaS_("system") + Delta_("surroundings") = 0`
or`DeltaS_("system") = - DeltaS_("surroundings")`
But as `H_(2)O(l)` changes into `H_(2)O(g)`,
`DeltaS_("system") = + ve , i.e.,gt 0`
`:. DeltaS_("surroundings")` willbe`lt 0`.
31.

H_(2) O(l) hArr H_(2) O_((s)) , Delta H = - qapplication of pressure on this equilibrium

Answer»

`G _((ICE )) = G _(("WATER")) = 0`
`G _((ice) = G _(("water")) NE 0`
`G _((ice)) GT G _(("water")) `
`G _("ice") lt G _(("water"))`

ANSWER :A
32.

For the process Dry ice rarr CO_(2(g))

Answer»

`Delta H` is positive and `Delta S` is negative
Both `Delta H and Delta S` are negative
Both `Delta H and Delta S` are positive
`Delta H` is negative and `Delta S` is positive

Solution :GASES have more entropy than SOLIDS sublimation is ENDOTHERMIC
33.

For the reaction 2H_(2)O_((l)) rarr H_(3)O_((aq))^(+) + OH_((aq))^(-), the value of Delta H is

Answer»

`+348 KJ`
`-348KJ`
`-174KJ`
`+87KJ`

ANSWER :A
34.

For the process, 3/2A to B, at 298 K , DeltaG^@ is kJ mol^(-1). The composition of the reaction mixture is [B] =1 and [A] = 10000. Predict the direction of the reaction and the relation between reaction quotient (Q) and the equilibrium constant (K).

Answer»

FORWARD DIRECTION because QgtK
Reverse direction because QgtK
Forward direction because QltK
Reverse direction becauseltK

Answer :B
35.

For the preparation of alkanes, aqueous solution of sodium or potassium salt of carboxylic acid is subjected to

Answer»

HYDROGENATION
oxidation
ELECTROLYSIS
hydrolysis

Solution :`2CH_3COOK+2H_2Ooverset"Electrolysis"to ubrace(CH_3-CH_3+2CO_2)_"ANODE"+ubrace(2KOH+H_2)_"CATHODE"`
36.

For the preparation of a litre of N//10 solution of H_(2)SO_(4), we need :

Answer»

9.8 g
4.9 g
10 g
98 g

Solution :N//A
37.

For the precipitation reaction of Ag ^(+) ions with NaCl, which of the following statements is correct ?

Answer»

`Delta H` for the REACTION is zero
`Delta G` for the reaction is zero
`Delta G` for the reaction is negative
`Delta G` is EQUAL to `Delta H`

ANSWER :C
38.

For the precipitation reaction of Ag^(+) ions with NaCl, which of the following statements is correct?

Answer»

`DELTAH` for the reaction is ZERO 
`DELTAG` for the reaction is zero 
`DeltaG` for the reaction is NEGATIVE
`DeltaG = DeltaH`

ANSWER :C
39.

For the physical equilibrium ice hArr water, the forward reaction is not favoured by

Answer»

INCREASING pressure
Increasing temperature
Keeping in CONTACT with HOT water
TAKING more ICE

Solution :Taking more ice
40.

For the nomenclature of alkene, give the ruler of giving the names of structure ?

Answer»

Solution :(i) For alkene nomenclature, select the carbon atom chain having double bond.
(ii) Double bonded carbon having less numbers of carbon or double carbon should gets less number. Give the chain side where double bond must GET less number.
(iii) Seeing the total number of alkene, (ANE) and (ENE) suffix are used after the main carbon.
(iv) If there are more than 1 double bond then according to double bond diene. triene, tetraene prefix are used.
(v) For the SUBSTITUTED carbon, see the RULES which you have studied earlier.
41.

For the n=2 energy level, how many orbitals of all kinds are possible

Answer»

2
3
4
5

Solution :For the n=2, ENERGY LEVEL ORBITALS of all KINDS are possible `n^2 , 2^2 =4`
42.

For the isothermal reversible expansion of an ideal gas fromvolume V_(1) to volume V_(2), the work done is given by w_(rev)="…................." whereas for the irreversible expansion , it is given by w_("irrev") = "….........".

Answer»

<P>

Solution :`w_("rev") = -2.303 n RT log. (V_(2))/( V_(1)), w_("irrev") =-P( V_(2) -V_(1)) = - P DELTAV`
43.

For the ideal gas , the work of reversible expansion under isothermal conditions can be calculated byusing the expressionw=-nRT"In"(v_(f))/(v_(I)) A samplecontaining 1.0 mol ofan ideal gasis expanded isothermally and reversiblyto ten times of its original is carried put at 300 K and at 600 Krespectvely . Choose the correct option .

Answer»

Workdone at 600 K is20 timesthe work done at 300 K.
Workdone at 300 K istwicethework done at 600 K.
Workdone at 600 K istwicethework done at 300 K.
`DeltaU=0 ` in both cases.

Answer :c,d
44.

For the hypothetical reactions, the equilibrium constant (K) values are given : A hArr B , K_(1) = 2*0 B hArr C , K_(2) = 4*0 C hArr D , K_(3) = 3*0 The equilibrium constant (K) for the reaction A hArr D is

Answer»

48
6
12
24

Answer :A::B::C::D
45.

For the hypothetical reactions, the equilibrium constant (K) values are given A harr B, K_1=2.0 B harr C, K_2=4.0 C harr D, K_3=3.0 The equilibrium constant for the reaction A harr D is

Answer»

48
6
24
12

Solution :`K_(C)=K_(1) XX K_(2) xx K_(3)`
46.

For the hypothetical reaction A_(2(g)) + B_(2(g)) hArr 2AB_((g)) Delta_(r ) G^(@) and Delta_(r)S^(@) are 20 kJ/mol and -20 JK^(-1) mol^(-1) respectively at 200K. If Delta_(r)C_(P) is 20 JK^(-1) mol^(-1) then Delta_(r ) H^(@) at 400K is

Answer»

20kJ/mol
7.98kJ/mol
28kJ/mol
none of these

Solution :`Delta H_(200K)^(0) = Delta G^(0) + T Delta S^(0)`
`= 20 + ((-20) xx 200)/(1000) = 16` kJ/mol `Delta H_(400K)^(0) = (20 xx 200)/(1000) + 16 = 20`
47.

For the hypothetical reaction : A+B hArrC+D, the equilibrium constant,K, is less than 1.0 at 25^(circ)C and decreased by 35% on changing the temperature to 45^(circ)C. What must be ture according to this information?

Answer»

The `DeltaH^circ` for the REACTION is NEGATIVE .
The `DeltaS^circ` for the reaction is POSITIVE.
The `DeltaG^circ`for the reaction at `25^cricC` is negative.
The `DeltaG^circ` for the reaction at `45^circC` is ZERO.

ANSWER :A
48.

For the hypothetical reaction A +B hArr C+2D equilibrium constant at 400 K is 1.8 xx 10^(-6) mol L^(-1)

Answer»

`9.5 xx 10^(-5) mol L^(-1)`
`9.5 xx 10^(-4) mol L^(-1)`
`4.75 xx10^(-4) mol L^(-1)`
None of these

Solution :`{:(,A,+,B,hArr,C,+,2D),("Initial CONC." (mol L^(-1)),-,,-,,1.0,,1.0),("At EQUILIBRIUM" (mol L^(-1)),((1-0-x)/(2)),,((1.0-x)/(2)),,((1.0+x)/(2)),,x):}`
D used up `=(1.0 -x )` mol `L^(-1)`
C used up `=((1.0-x)/(2)) mol L^(-1)`
C at equilibrium `(1.0-(1.0-x)/(2))mol L^(-1)`
`=((1.0+x)/(2)) mol L^(-1)`
A ( or B ) produced at equilibrium `=C` used up
`=((1.0-x)/(2))mol L^(-1)`
A very small value of K indicates that at equilibrium CONCENTRATIONS of C and D are very small i.e., x is very small. As such at equilibrium
`[C]=((1.0+x)/(2))=0.5 mol L^(-1)`
`[D]=x mol L^(-1) , [A]=[B]`
`=((1.0-x)/(2)) mol L^(-1)`
`=0.50 mol L^(-1)`
`:. K=([C][D]^(2))/([A][B])`
`1.8 xx 10^(-6)=(0.50xx x^(2))/(0.50 xx 0.80)`
`x^(2)=9.0 xx 10^(-9)`
`implies x=9.5 xx10^(-4) mol L^(-1)`
`:. [D]=x=9.5 xx 10^(-4) mol L^(-1)`
49.

For the homogeneous are reaction at 600 K, 4NH_(3(g))+5O_(2(g))hArr4NO_((g))+6H_(2)O_((g)). The equilibrium K_(c) has the unit.

Answer»

`("mol dm"^(-3))^(-1)`
`("mol dm"^(-3))^(1)`
`("mol dm"^(-3))^(10)`
`("mol dm"^(-3))^(-9)`

SOLUTION :`4NH_(3(g))+5O_(2(g))hArr4NO_((g))+6H_(2)O_((g))`
`Deltan_(g)=10-9=1`
`"Unit of "K_(c)=("mol dm"^(-3))^(Deltan_(g))`
`:."Unit"=("mol dm"^(-3))^(1)`
50.

For the half cell reaction, 2BrO_(3)^(-) + 12H^(+)+10e to Br_(26H_(2)O the equivalent mass of sodium bromate is :

Answer»

EQUAL to its MOL. MASS
`1//3` of its mol. Mass
`1//6` of its mol. Mass
`1//5` of its mol. Mass

Solution :N//A