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.

If DeltaG^@ for the reaction given below is 1.7KJ : The equilibrium constant of the reaction 2HI(g) hArr H_2(g)+ I_2(g) at 25^@C IS :

Answer»

24
3.9
2
0.5

Answer :D
Discussion
2.

If DeltaG^(@) for the reaction given below is 1.7 kJ, the equilibrium constant of the reaction, 2HI(g) hArr H_(2)(g)+I_(2)(g) at 25^(@)C, is

Answer»

0.5
2
3.9
24

Solution :`DELTAG^(@)=-2.303RTlogK`
GIVEN, `DeltaG^(@)=1.7kJ`
`T=25^(@)C=25+273=298K`
`therefore1.7=-2.303xx8.314xx10^(-3)xx298xxlogK`
`LOGK=-(1.7)/(2.303xx8.314xx10^(-3)xx298)`
`thereforeK=0.5`
Discussion
3.

If DeltaG = -177K cal for ""(1)2Fe(s)+(3)/(2)O_(2)(g) rarr Fe_(2)O_(3)(s) and DeltaG =- 19 K cal for ""(2)4Fe_(2)O_(3)(s)+ Fe(s) rarr 3Fe_(3)O_(4)(s) What is the Gibbs free energy of formation fo Fe_(3)O_(4)(s)?

Answer»

`+229.6` Kcal/mol
`-242.3` Kcal/mol
`-727` Kcal/mol
`-229.6` Kcal/mol

Solution :`DeltaG^(@)` for `3Fe(s) + 2O_(2)(g) rarr Fe_(3)O_(4)(s)` can be obtained by taking
`[(2) + 4 xx (1)] xx (1)/(3)`
Hence , we GET `DeltaG_(f) =[-19 + 4 xx (-177)] xx (1)/(3)=-242.3 K " cal for " 1 mol Fe_(3)O_(4)`
Discussion
4.

If DeltaE is the heat of reaction for C_(2)H_(5)OH(l)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(l) at constant volume, the DeltaH (heat of reaction at constant pressure), at constant temperature is

Answer»

`DeltaH=DeltaE+RT`
`DeltaH=DeltaE-RT`
`DeltaH=DeltaE-2RT`
`DeltaH=DeltaE+2RT`

Solution :We KNOW that, `DeltaH=DeltaE+DeltanRT`
where, `DELTAN`=number of MOLES of gaeous products-number of moles of GASEOUS reactants=2-3=-1
so, `DeltaH=DeltaE-RT`
Discussion
5.

If Delta_fH°(C_2H_4)and Delta_fH°(C_2H_6)| are in x_1| and x_2 kcalmol ^(-1),then heat of hydrogenation of C_2H_4 will be

Answer»

`x_1+x_2`
`x_1-x_2`
`x_2-x_1`
`x_1+2x_2`

ANSWER :C
Discussion
6.

If Delta_(f)G^(@) for NH_(3)(g) is - 16.4 kJ mol^(-1), then Delta G^(@) for the reaction : N_(2)(g) + 3 H_(2)(g) rarr 2NH_(3)(g) is

Answer»

`32.8 KJ MOL^(-1)`
`16.4 kJ mol^(-1)`
`- 16.4 kJ mol^(-1)`
`- 32.8 kJ mol^(-1)`

ANSWER :D
Discussion
7.

If Delta_(0)ltP then electronic arrangement of metal atom//ion in an octahederal complex with d^(4) configuration is t_(2g)^(x)e_(g)^(y). What is the value of x

Answer»


Solution :`(t_(2G))^(3) (e_(G))^(1)`
Discussion
8.

If Delta_(0) represents CFSE in an octahedral field, e_(g) orbitals lie ….. The bari centre and t_(2g) orbitals lie …… the bari centre.

Answer»

SOLUTION :0.6 `Delta_(0)` above, `0.4 Delta_(0)` below
Discussion
9.

If Delta_0 lt P , the correct electronic configuration for d^4 system will be

Answer»

`t_(2g)^4 e_g^0`
`t_(2g)^3 e_g^1`
`t_(2g)^0 e_g^4`
`t_(2g)^2 e_g^2`

ANSWER :A
Discussion
10.

If Delta_(0) is crystal field splitting energy and P is mean pairing energy per pair then stablizing energy of complex [FeF_(6)]^(4-) is :

Answer»

<P>`-2.4 Delta_(0)+3P`
`-2.0 Delta_(0)+3P`
`-1.6 Delta_(0)+P`
`-0.4 Delta_(0)`

Solution :
Stabilizing ENERGY
`=-0.4 Delta_(0)xx4+0.6 Delta_(0)xx2`
`=-0.4 Delta_(0)`
Discussion
11.

In octahedral complexes having co-ordination number 6, the degeneracy of the d-orbitals of central atom is removed due to ligand electron metal electron repulsions. In the octahedral complex three orbitals have lower energy, t_(2g) set and two orbitals have higher energy, eg set. This phenomenon is formed as crystal field splitting and the energy seperation is denoted by Delta_(0). Thus the energy of the two eg orbitals will increase by (3//5)Delta_(0) and that of the three t_(2g) will decrease by (2//5)Delta_(0). The erystal field splitling, Delta_(0) depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong field and in these cases, the splitting will be large whereas other produce weak fields and consequently result in small splitting of d-orbitals. If Delta_(0)ltP, the correct electronic configuration of d^(4) system will be

Answer»

`t_(2G)^4 e_g^0`
`t_(2g)^3 e_g^1`
`t_(2g)^0 e_g^4`
`t_(2g)^4 e_g^2`

ANSWER :A
Discussion
12.

If Delta_0 gt P the correct electronic confuguration for d^4 system will be

Answer»

`t_(2G)^(4)e_g^0`
`t_(2g)^(3)e_g^1`
`t_(2g)^(0)e_g^4`
`t_(2g)^(2)e_g^2`

Answer :A
Discussion
13.

If Delta H is the change in enthalpy and Delta E is the change in internal energy accompanying a gaseous reaction then

Answer»

`DELTA H` is ALWAYS greater than `Delta U`
`Delta H gt Delta U` only the NUMBER of moles of the reactnts is greater than the number of MOLE of products
`Delta H` is always less than `Delta U`
`Delta H lt Delta U` only if the number of moles of products is less than the number of moles of reactants.

Answer :D
Discussion
14.

If Delta H and Delta U are the changes in the enthalpy and internal energy when a liquid is converted into its vapours at temperature T, then :

Answer»

`Delta H - Delta U = 0`
`Delta H - Delta U = 22.4 kJ//mol`
`Delta H - Delta U = RT`
`Delta U - Delta H = RT`

Answer :C
Discussion
15.

If Delta G_(298K) for the reaction 2H_(2(g. 1atm)) + O_(2(g. 1atm)) rarr 2H_(2)O_((g.1atm)) is -240kJ, what will be Delta G_(298K) for the reaction H_(2)O_((g.0.2 atm)) rarr H_(2(g.4atm)) + (1)/(2) O_(2(g.0.26atm))

Answer»

245.7 KJ
239.3 kJ
125.7 kJ
`-125.7 kJ`

ANSWER :C
Discussion
16.

IfDelta G = Delta H - T Delta S and Delta G = Delta H + T [ (d(Delta G))/(dT) ]_P then variation of EMF of cell , with temperature T, is given by

Answer»

`(DELTA S)/(nF)`
`-(Delta S)/(nF)`
`(DELTAH)/(nF)`
`(Delta G)/(nF)`

Solution :On comparison
` Delta S = - [ (d (Delta G) )/(dT)] , Delta S = - (d (-NFE))/(dT) = nF ((dE)/(dT))`
` THEREFORE ((dE)/(dT)) = (Delta S)/(nF)`
Discussion
17.

If decomposition of Al_(2)O_(3) at 500 ""^(@)C temperature gives following reaction and gives Gibb's free energy : (2)/(3)Al_(2)O_(3) to (4)/(3) Al+O_(2). . . DeltaG=966kJ" "mol^(-1) then what is the difference of minimum required electrical energy for the reduction of Al_(2)O_(3) by electrolysis ?

Answer»

5.0V
4.5V
2.5V
3.0V

Solution :`(2)/(3)Al_(2)O_(3)""THEREFORE DeltaG=966kJ`
`therefore Al_(2)O_(3)""DeltaG=(966xx3)/(2)=1449kJ`
`2Al^(3+)+6e^(-) to 2Al`
`DeltaG=nFE_(CELL)`, `therefore E_(cell)=(DeltaG)/(nF)|"Where, "n=6," "DeltaG=1449xx10^(3)J," "E=96500`
`=(1449xx1000)/(6xx96500)`
`=2.5V`
Discussion
18.

If degree of dissociation of HI is 0.1 then K_(P) for reaction is: 2HI (g) Harr H_(2) (g) + I_(2) (g)

Answer»

`(1)/(9)`
`(1)/(324)`
`(10)/(36)`
can't be determined

Solution :`{:(,2HI,HARR,H_(2),+,I_(2),,),("moles",a_(0) - a_(0) alpha,,(a_(0) alpha)/(2),,(a_(0) alpha)/(2),alpha = 0.1 "(given)",),(" at" eq^(m),,,,,,,):}`
`K_(p) = K_(C )` as `(DELTA n)_(g) = 0`
`K_(P) = (alpha^(2))/(4 (1 - alpha)^(2)) = ((0.1)^(2))/(4(1 - 0.1)^(2)) = ((0.1)^(2))/(4 xx (0.9)^(2))`
`= (1)/(81 xx 4)`
`K_(P) = (1)/(324)`
Discussion
19.

If degree of dissociation of 2M CH_3 COOH is 10% then degree of dissociation of this acetic acid in 3 Molar CH_3 COONa solution will be

Answer»

`=10%`
`LT 10%`
`gt10%`
Cannot be DETERMINE

ANSWER :B
Discussion
20.

If D-Glucaric acid is allowed to undergo intramolecular esterification the compound formed is/are

Answer»



It can't FORM a lactone

Answer :A::B
Discussion
21.

If current of one ampere is passed fro one second, then the mass deposited at the electrode is equal to

Answer»

ONE GRAM atomic
one gram MOLECULAR
one gram equivalent
one gram electro chemical equivalent

ANSWER :D
Discussion
22.

If Cu^(2+) and Cd^(2+) both are present , it is difficult to outline a scheme to analyes in a mixture

Answer»

Solution :`KCN` formcomples with `Cu^(2+)` and `Cu^(2+)`
`Cu^(2+) + KCN rarr K_(2) UNDERSET("Stable")([Cu(CN)_(4)])`
`Cd^(2+) + KCN rarr K_(2)underset("Unstable")([Cu(CN)_(4)])`
when `H_(2)S` gas pased anstable complex of `Cd^(2+)` given yellow ppt
`[Cu(CN)_(4)]^(2) = Cd^(2+) ,Cd^(2+) + S^(2-) rarr underset("Yellow") (CDS) darr`
Discussion
23.

If crystal field theory is completely followed, then which of the following statements must be true ?

Answer»

ANIONIC ligands such as `OH^(-)` should exert greater splitting effect than netural ligands such as `H_(2)O`
Complexes will have PARTIAL covalent bonds between ligands and central atom
Colour of complexes cannont be EXPLAINED
Complexes will have PURE covalent bonds between ligands and central atom

Answer :A
Discussion
24.

If C(s)+O_2(g)rarrCO_2(g), triangleH=X and CO(g)+1/2O_2(g)rarrCO_2(g), triangleH=Y then the heat of formation of CO is

Answer»

X+Y
X-Y
Y-X
XY

Answer :B
Discussion
25.

If, C(s)+O_2(g)rarrCO_2(g), triangle H=R and CO(g)+1/2O(g)rarrCO_2(g), triangleH=S, then the heat of formation of CO is:

Answer»

R+S
R-S
S-R
RXS

Answer :B
Discussion
26.

If coupling of reduction and oxidation results in…………..value of triangleG for overall reaction, the final reaction becomes feasible.

Answer»

SOLUTION :NEGATIVE
Discussion
27.

If conductivity of 0.020 M KCl solution at 298 K temperature is 0.0248 S cm^(-1), then find out its molar conductivity.

Answer»

`124Omega^(-1)cm^(2)MOL^(-1)`
`224Omega^(-1)cm^(2)mol^(-1)`
`24Omega^(-1)cm^(2)mol^(-1)`
`1.24Omega^(-1)cm^(2)mol^(-1)`

Solution :Here, `K=0.0248=2.48xx10^(-2)OMEGA^(-1)cm^(-1)`
`M=0.20" mol "L^(-1)`
`Lamda_(m)=(Kxx1000)/(M)=(1000xx2.48xx10^(-2))/(0.20)`
`=(1000xx248xx10^(-2))/(20xx100)XX100`
`Lamda_(m)=124Omega^(-1)cm^(2)mol^(-1)`.
Discussion
28.

If concentrations of two acids are same, their relative strengths can be compared by which of the following relations(i)(a_(1))/(a_(2) (ii)(K_(1))/(K_(2)) (iii)([H^(+)]_(1))/(H^(+)]_(2))(iv) sqrt((K_(1))/(K_(2)))

Answer»

(i),(II),(III)
(i),(iii),(IV)
(i),and(iii)
All

Answer :B
Discussion
29.

If concentration of reactants is made x times the rate constant k becomes

Answer»

`E^(k//x)`
k/x
unchanged
x/k

Answer :C
Discussion
30.

If concentration of reactant is increased by 'm' then k becomes:

Answer»

`ln (K//x)`
`K//x`
`K+x`
K

Solution :There is no EFFECT of CHANGE in concentration on EQULIBRIUM constant.
Discussion
31.

If concentration of reactants is increased by ‘X’, the rate of constant K becomes:

Answer»

`E^(K//X)`
`(K//X)`
K
`(X/K)`

ANSWER :C
Discussion
32.

If concentration of reactants is increased by ‘X’, the rateconstant K becomes:

Answer»

`E^(K//X)`
`(K//X)`
K
`(X/K)`

ANSWER :C
Discussion
33.

If concentration of reactants is increased by X the rate constant k becomes

Answer»

`E^(k//X)`
`k//X`
k
X/E

Solution : Rate constant does not DEPEND on CONCENTRATION of REACTANTS.
Discussion
34.

If concentration is expressed as mol L^(-), the equilibrium constant, K for the reaction :2N_(2)O_(5)(g)hArr 4NO_(2)(g)+O_(2)(g)has the units :

Answer»

`mol^(3)L^(-3)`
`mol L^(-1)`
`mol^(3)L^(-1)`
no units.

Solution :`K=([NO_(2)]^(4)[O_(2)])/([N_(2)O_(5)]^(2))=((mol//L)^(4)(mol//L))/((mol//L)^(2))`
`= mol^(3)L^(-3)`
Discussion
35.

If concentration are measured in moles/lit and time in minutes, the unit for the rate constant of a 3^(rd) order reaction are

Answer»

`MOL LIT^(-1)min^(-1)`
`lit^(2)mol^(-2)min^(-1)`
`lit mol^(-1)min^(-1)`
`min^(-1)`

ANSWER :B
Discussion
36.

If compound MX_(4)has vecmu=0 then most probable geometry will be:-

Answer»

TRIGONAL bipyramidal
Square bipyramidal
Pentagonal bipyramidal
distorted tetrahedral

Answer :B
Discussion
37.

If compound AX_(3)is a hypervalent compound then the group no of element A is :-

Answer»

III A
V A
VI A
VII A

ANSWER :D
Discussion
38.

If common salt is dissolved in water, the vapour pressure of the solution will

Answer»

Increase
Decrease
REMAIN UNCHANGED
Can not be PREDICTED

ANSWER :B
Discussion
39.

If CO_(2) gas is passed through 500 ml of 0.5 (M) Ca(OH)_(2), the amount of CaCO_(3) produced is

Answer»

10 G
20 g
50 g
25 g

Solution :By equating the equivalents
`W/100xx2=(500xx0.5xx2)/1000`
`W=25 g""(W="mass of "CaCO_(3))`
Discussion
40.

If CO ligands are substituted by NO in respective neutral carbonyl compounds then which of the following will not be correct formula?

Answer»

`Cr(CO)_(3)(NO)_(2)`
`Fe(CO)_(2)(NO)_(2)`
`Cr(NO)_(4)`
`Ni(CO)_(2)(NO)_(2)`

Solution :LIGAND NO is `3E^(-)` DONAR hence three CO ligands can be SUBSTITUTED by two NO ligands.
Discussion
41.

If Cl_2 gas is passed into aqueoussolution of KI containing some C Cl_4 and the mixture is shaken then

Answer»

Upper layer BECOMES VIOLET
Lower layer becomes violet
Homogenous violet layer is formed
None of these

Answer :B
Discussion
42.

If CI_2 is passed through hot aqueous NaOH, the products formed have CI in different oxidation states. These are indicated as

Answer»

`-1 and + 1`
`-1 and + 5`
`+ 1 and + 5`
`-1 and + 3`

Solution :With hot and concentrated aqueous NaOH (THOUGH .concentrated. was not MENTIONED in the question PAPER, HOWEVER, it should have been), `Cl_2` reacts as below :
`underset(("hot, conc."))(3overset(0)(Cl_2) + 6NaOH) to underset(("Sodium chloride"))(5Na overset(-1)(Cl)) + underset(("sodium chlorate"))(Naoverset(+5)(Cl)O_3) +3H_2O`
Discussion
43.

If Cl_(2) gas is passed into aqueous solution of KI containing some C Cl_(4) and the mixtures is shaken

Answer»

UPPER LAYER BECOME VIOLET
homogeneous violet layer is formed
organge COLOUR appears
lower layer beomes violet

Answer :D
Discussion
44.

If chorine gas is passed through hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ......and...... .

Answer»

0 to + 5
0 to + 3
0 to -1
0 to + 1

Solution :`3overset(0)(Cl)_(2)+6NaOH OVERSET(Delta)(rarr) 5 Na overset(-1)(Cl)+Na overset(+5)(Cl)O_(3)+3H_(2)O`
Options (a) and (c) are correct.
Discussion
45.

If chloroform is left open in air in the presence of sunlight, it gives

Answer»

CARBON tetrachloride
carbonyl chloride
mustard GAS lewisie

Answer :B
Discussion
46.

If Chloroform is left open in air in presence of sun-rays:

Answer»

EXPLOSION takes place
Poisonous phosgene gas is formed
Polymerisation takes place
No reaction takes place

Answer :B
Discussion
47.

If CH_(4) is known as methane, then C_(9)H_(20) is known as

Answer»

HEXANE
Nonane
Octane
BUTANE

ANSWER :B
Discussion
48.

If chlorine gas is passed through hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ........... and ...........

Answer»

0 to (+5)
0 to (+3)
0 to (-1)
0 to (+1)

SOLUTION :`6NaOH + underset(0)(3Cl_(2)) to underset(-1)(5NaCl) + underset(+5)(NaClO_(3)) + 3H_(2)O`
Discussion
49.

If CH_(3)COOH + OH^(-) rarr CH_(3)COO^(-) + H_(2)O + x kJ H^(+) + OH^(-) rarr H_(2)O + y kJ The enthalpy change for the reaction : CH_(3)COOH rarr CH_(3)COO^(-) + H^(+) is

Answer»

x + y
x - y
y - x
`x - y//2`

Solution :(i) `CH_(3)COOH + OH^(-) rarr CH_(3)COO^(-) + H_(2)O, Delta H = -x`
(ii) `H^(+) + OH^(-) rarr H_(2)O, Delta H = -y` SUBTRACTING (ii) from (i),
`CH_(3)COOH rarr CH_(3)COO^(-) + H^(+), Delta H = -x + y "or" = y - x`
Discussion
50.

If CFSE (Delta_(0)) is less than pairing energy (P), the ligand is a …… ligand and the complex formed is a …… complex.

Answer»

SOLUTION :WEAK FIELD, HIGH SPIN.
Discussion