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.

The activation energy for a reaction at the temperature TK was found to be 2.303 RT J mol^(-1) . The ratio of the rate constant to Arrhenius factor is

Answer»

`10^(-1)`
`10^(-2)`
`2 XX 10^(-3)`
`2 xx 10^(-2)`

Solution :2.303 log `k= 2.303` log `A - (E_(a))/(RT) , E_(a) = 2.303 RT` (given)
Putting the values of `E_(a)` and dividing by 2.303 on both sides , we get , log k = log A - 1 , log k - log A = -1
log `((k)/(A)) = -1 implies ` TAKING antilog `((k)/(A)) = 10^(-1)` .
Discussion
2.

The activation energy for a reaction is 9.0 K cal/mol . The increase in the rate constant when its temperature is increased from 298 K to 308 K is

Answer»

0.63
0.5
1
0.1

Solution :2.303 log `(k_(2))/(k_(1)) = (E_(a))/(R) [(T_(2) - T_(1))/(T_(1)T_(2))]`
log `(k_(2))/(k_(1)) = (9.0 XX 10^(3))/(2.303 xx 2) [(308 - 298)/(308xx 298)]`
`(k_(2))/(k_(1)) = 1.63 , K_(2) = 1.63 K_(1) , (1.63 k_(1) - k_(1))/(k_(1))xx 100 = 63.0` %
Discussion
3.

The activation energy for a reaction is 9 kcal/mol. The increase in late (XI 'slant when its temperature is increased from 298K to 308 K is(R = 2 cal K^(-1)mot^(-1), 10^(0.213) = 1.63)

Answer»

0.75
0.25
1
0.63

Answer :D
Discussion
4.

The activation energy for a reaction is 9.0 Kcal//mol. The increase in the rate constant when its temperature is increased from 298 K to 308 K is:

Answer»

`10%`
`100%`
`50%`
`63%`

ANSWER :D
Discussion
5.

The activation energy and enthalpy of chemisorption of oxygen on a metal surface is 37.3kJ mol^(-1) and -72.1kJ mol^(-1) At a certain pressure, the rateconstant for chemisorption is 1.2xx10^(-3) at 3198 K . What will be the value of the rate constant at 308K ?

Answer»

`7.6 xx 10^(-4)s^(-1)`
`1.6xx10^(-6)s^(-1)`
`7.6xx10^(-2)s^(-1)`
`1.6xx10^(-5)s^(-1)`

Solution :`k(` RATE CONSTANT `)=Ae^(En//RT)`
`ln.(K_(2))/(K_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`or `lnk_(2)+(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`
Now,`T_(1)=318K,E_(a)=37.3kJmol^(-1)` and `T_(2)=308K`
`logk_(2)=log(1.2xx10^(-3))+(37300)/(8.31xx2.303)((1)/(318)-(1)/(308))=-2.9208-(37300xx10)/(8.31xx2.303xx318xx308)`
`=-2.9208-0.1989`
`=k_(2)=7.6xx10^(-4)s^(-1)`
Discussion
6.

The activation energyfor a reaction at the temperature T K was found to be 2.303 RT J mol^(-1) . The ratio of the rate constant to Arrhenius factor is :

Answer»

`10^(-1)`
`10^(-2)`
`2XX10^(-3)`
`2x10^(-2)`

SOLUTION :(A) According to Arrhenius equation
`k = Ae^(-E_(a)//RT) `
or log k = log `A-(Ea)/(2.303RT)`
`E_(a)= 2.303 RT`
`:. ` log k = log A `-(2.303RT)/(2.303RT)`
log k = log A-1
or log k - log A=-1
log `(k)/(A) =-1`
`(k)/(A) = 10^(-1)`
Discussion
7.

The activation energy for a first-order reaction is 104.5 kJ/mole and the factor A in the Arrhenius equation is 5 xx 10^15 s^(-1). Find the temperature at which the half-life period of the reaction is 1 minute.

Answer»

SOLUTION :`75^@C`
Discussion
8.

The activation energies for the two reactions are Ea and Ea^(') with E_(a)gtE_(a)^('). If the temperature of the reaction system is increased from T_(1) to T_(2). Predict which alternative is correct. (k^(')) are the rate constants at highest temperature.

Answer»

`(k_(1)^('))/(k+(1)) = (k_(2)^('))/(k_(2))`
`k_(1) lt k_(2)` and `k_(2)^(') lt k_(1)^(')`
`k_(1) lt k_(2)`
`(k_(1)^(')/k_(1)) lt (2k_(2)^('))/(k_(2))`

SOLUTION :b) We know that
`k=Ae^((-Ea)/(RT))`
This means that more is the activation energy, lesser is the RATE constant.
Since `E_(a) gt E_(a)^(')`
`THEREFORE k_(1) lt k_(2)` and `k_(1)^(')` and `k_(2)^(')`
Discussion
9.

The activation energies of two reactions are E_1" and "E_2 (E_1 gt E_2). If the temperature of the system is increased from T_1" to "T_2, the rate constant changes from k_1" to "k'_1 in the first reaction and k_2" to "k'_2 in the second reaction. Predict which of the following expressions is/are incorrect?

Answer»

`(k'_1)/(k_1)=(k'_2)/(k_2)`
`(k'_1)/(k_1)GT(k'_2)/(k_2)`
`(k'_1)/(k_1)lt(k'_2)/(k_2)`
`(k'_1)/(k_1)=(k'_2)/(k_2)=1`

SOLUTION : `LOG(k_1.)/(k_1)=(E_1)/(2.303R)((T_2-T_1)/(T_1T_2))`…(i)
`[thereforeT_2 gtT_1]`
`log(k_2.)(k_2)=(E_2)/(2.303R)((T_2-T_1)/(T_1T_2))`…(ii)
As, `E_1 gt E_2`
From (i) and (ii), we GET
`log(k_1.)/(k_1) gt log (k_2.)/(k_2) implies (k_1.)/(k_1) gt (k_2.)/(k_2)`
Discussion
10.

The activation energies of two reactions are E_(1) and E_(2) (E_(1) gt E_(2)) . If the temperature of the system is increased from T_(1) to T_(2) , the rate constant of the reaction changes from k_(1) to k_(1) in the first reaction and k_(2) to k_(2) in second reaction predict which of the following expression is correct ?

Answer»

`(k_(1))/(k_(1))= (k_(2))/(k_(2))`
`(k_(1))/(k_(1)) GT (k_(2))/(k_(2))`
`(k_(1))/(k_(1)) lt (k_(2))/(k_(2))`
`(k_(1))/(k_(1))= (k_(2))/(k_(2))`=0

Solution :[B ] log `(k_(1))/(k_(1)) = (E_(1))/(2.303R) [(T_(2)-T_(1))/(T_(1)T_(2))]`
log`(k_(2))/(k_(2))=(E_(2))/(2.303R)[(T_(2)-T_(1))/(T_(1)-T_(2))]`
SINCE `E_(1) gt E_(2)`
`("log"(k_(1))/(k_(1)))/("log"k_(2)/(k_(2)))gt1`
or ` (k_(1))/(k_(1))gt (k_(2))/(k_(2))`
Discussion
11.

The activation energies of the forward and backward reactions in the case of a chemical reaction are 37.5 and 51.3 kJ/mole respectively The reaction is

Answer»

ENDOTHERMIC
EXOTHERMIC
Neither exothermic nor endothermic
INDEPENDENT of TEMPERATURE

ANSWER :B
Discussion
12.

The activation energies of two reactions are E_a andE_awithE_agt E_a if the temperature of the reacting systems is increased from T_1toT_2, predict which alternative is correct k are rate constants at higher temperature. Assume A being same for both the reactions:

Answer»

`k_1/k_2=k_2/k_2`
`k_1ltk_2` and `k_1ltk_2`
`k_1gtk_2` and `k_1gtk_2`
`k_1/k_2=(2k_2)/k_2`

ANSWER :B
Discussion
13.

The action of nitrous acid on an aliphatic primary amine gives

Answer»

ALCOHOL
ALKYL nitrile
Secondaryamine
NITROALKANE

SOLUTION :N
Discussion
14.

The action of sodium on alkyl halide to form an alkane is called

Answer»

GRIGNARD reaction
Wurtz COUPLING reaction
isocyanide reaction
halogenation reaction

Answer :A::B::C
Discussion
15.

The action that does not form an amine complex with excess of ammonia is:

Answer»

`AL^(3+)`
`AG^+`
`Cu^(2+)`
`CD^(2+)`

Answer :A
Discussion
16.

The action of nitrous acid on ehtyl amine gives

Answer»

Nitroethane
Ethyl nitrite
Ethane
Ethyl alcohol

Answer :D
Discussion
17.

The action of nitrous acid on a primary amine gives :

Answer»

NITRO alkane
Alkyl nitrite
Alcohol
SECONDARY AMINE

ANSWER :C
Discussion
18.

The action of HCI on silver nitrate produces:

Answer»

A PRECIPITATE of AgCl
Chlorine gas
No visible CHANGE
A black stain of SILVER

ANSWER :C
Discussion
19.

The action of boiling water on actinoids gives

Answer»

OXIDES and hydroxides
oxides and hydrides
hydroxides and hydrides
only oxides

Answer :B
Discussion
20.

The action of enzyme is inhibited during the fermentation once the concentration of alcohol exceeds …….

Answer»

14 PERCENT
15 percent
20 percent
10 percent

Solution :14 percent
Discussion
21.

The action of Bleaching powder on ethyl alcohol gives

Answer»

`C CCl_(4)`
`CHCl_(3)`
`COCl_(2)`
Ether

Solution :`CaOCl_(2)+2H_(2)O to Ca(OH)_(2)+3HCl+2(O)`
`C_(2)H_(5)OH+(O) to CH_(3)CHO+H_(2)O`
`CH_(3)CHO+3Cl_(2) to C Cl_(3) CHO+3HCl`
`2C Cl_(3)CHO+CA(OH)_(2)) to Ca(HCOO)_(2)+2CHCl_(3)`
Discussion
22.

The action of halogen acids on an ether, has the following order of reactivity:

Answer»

`HCl GT HBr gt HI`
`HI gt HCl gt HBr`
`HI gt HBr gt HCl`
`HCl gt HI gt HBr`

ANSWER :C
Discussion
23.

The action of dil. HNO_3 on magnesium gives:

Answer»

NO
`H_2`
`NO_2`
`NH_4NO_3`

ANSWER :B
Discussion
24.

The actinoids exihibits more number of oxidation states in general than the lanthanoids.This is because

Answer»

the 5f ORBITALS extend farther from the nucleus than the 4f orbitals
the 5f orbitals are more buried than the 4f orbitals
there is a similarly between 4f and 5f orbitals in their angular part of the wave function
the actinoids are more reactive than the lanthanoids

Solution :As 5f orbitals extend farther from the nucleus than 4f orbitals, the attraction on the 5f electrons in actinoid series decreases. Hence, they can easily PARTICIPATE in bonding and, therefore,exhibit more number of oxidation states.
Discussion
25.

The action of aniline with benzoyl chloride in the presence of aqueous NaOH gives :

Answer»

BENZAMIDE
N-Phenylbenzamide
p-Aminobenzophenone
N-Phenylacetamide.

Answer :B
Discussion
26.

The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because

Answer»

the 5f orbitals extend further from the NUCLEUS than the 4f orbitals
the 5f orbitals are more BURIED than the 4f orbitals
there is a similarity between 4fand 5f orbitals in their angular part of the wave function
the ACTINOIDS are more REACTIVE than the lanthanoids.

Solution :More the disatance between nucleus and outer orbitals, lesser will be force of attration on them . Distance between 4f orbital and nucleus.
So actinoieds EXHIBIT more number of oxidation states in gerneral than the lanthnoids.
Discussion
27.

The actinoid elements which show the highest oxidation state of +7 are

Answer»

Np, PU Am
U, Fm, TH
U, Th, Md
Es, No, LT

Solution :Np, Pu Am
Discussion
28.

Theactinoidelements whichshow thehighestoxidationstate of +7are

Answer»

`NP, Pu, An`
`U , FM , TH`
`U, Th , Md`
`Es , No, LR`

Answer :A
Discussion
29.

The actinide element which can exhibit +7 oxidation state in its compounds

Answer»

THORIUM
Curium
NEPTUNIUM
Nobelium

ANSWER :C
Discussion
30.

The actinide contraction differs from lanthanide contraction? Explain.

Answer»

Solution :Lanthanide CONTRACTION refers to the gradual DECREASE in the size of the lanthanoids and their trivalent ions, `( Z = 58 ` to `71)` whereas actinoid contraction refers to the gradual decrease in the size of the actinoids or their ions `( M^(3+)` or `M^(4+)) ( Z = 90 ` to `103 )`. The differ in the FACT that int he actinoid series, the contraction is greater from element to element DUE to poor shielding by 5f electrons thanby 4f electrons.
Discussion
31.

The acidity of phenol is due to the ____of its anion.

Answer»

SOLUTION :RESONANCE STABILIZATION
Discussion
32.

The acidic strength of the given compounds follows the order I. CH_(3)-CH=CH-overset(O)overset(||)C-OH II. CH_(3)-overset(..)underset(..)O-CH=CH-overset(O)overset(||)C-OH III. CH_(3)-CH_(2)-overset(O)overset(||)C-OH

Answer»

`IIgtIIIgtI`
`IIIgtIIgtI`
`IIgtIgtIII`
`IgtIIgtIII`

ANSWER :D
Discussion
33.

The acidic strength of saturated aliphatic carboxylic acids depends mainly upon the inductive effect of the substituent and its position w.r.t. the -COOH group. Whereas electron-donating substituents tend to decrease, electron-withdrawing substituents tend to increase the acid strength. the acid weakening effect of electron donating substituents and acid-strengthening effect of the electron-withdrawing substituents is more pronounced at p-position than that at m-position. Due to ortho effect, o-substituted benzoic acids are usually stronger than benzoic acid regardless of the nature of substituent whetehr electron-donating or electron-withdrawing Q. The pK_(a) of acetylsalicylic acid (aspirin) is 3.5 The pH of gastric juice in human stomach is about 2-3 and pH in the small intestine is about 8. Aspirin will be

Answer»

Unionized in the SMALL intestine and in the stomach
Completely ionized in the stomach and ALMOST unionized in the small intestine
Ionised in the stomach and almost unionized in the small intestine.
Ionised in the small intestine and almost unionized in the stomach.

Solution :Since the pH in the small intestine is about 8 (basic), ASPIRIN is completely ionized here and further since the pH in the stomach is 2-3 (highly acidic) therefore, aspirin which is a weak acid is completely unionized here.
Discussion
34.

The acidic strength of saturated aliphatic carboxylic acids depends mainly upon the inductive effect of the substituent and its position w.r.t., the -COOH group. Whereas electron donating substituents tend to decrease, electron withdrawing substituents tend to increase the acid sttrength. the acidic strength of aromatic carboxylic acids, on the other hand, depends upon both the inductive and the resonance effect of the substituents Q. Which of the followig is obtained when 4-methylbenzenesulphonic acid is hydrolysed with excess of sodium acetate?

Answer»




Solution :4-Methylbenzenesulphonic ACID is a stronger acid tha acetic acid, therefore, it displaces acetic acid from sdoium ACETATE. THUS,
`4-Me-C_(6)H_(4)-SO_(3)H+CH_(3)COONa to 4-Me-C_(6)H_(4)SO_(3)^(-)Na^(+)+CH_(3)COOH`
Discussion
35.

The acidic strength of saturated aliphatic carboxylic acids depends mainly upon the inductive effect of the substituent and its position w.r.t., the -COOH group. Whereas electron donating substituents tend to decrease, electron withdrawing substituents tend to increase the acid sttrength. the acidic strength of aromatic carboxylic acids, on the other hand, depends upon both the inductive and the resonance effect of the substituents Q. Among the following, the strongest acid is

Answer»

`CH_(3)COOH`
`C_(6)H_(5)COOH`
`m-CH_(3)OC_(6)H_(4)COOH`
`p-CH_(3)OC_(6)H_(4)COOH`

Solution :Since in `C_(6)H_(5)COOH,-COOH` is attached to more electronegative `sp^(2)`-carbon of the `C_(6)H_(5)` ring while in `CH_(3)COOH`, it is attached to less electronegative `sp^(3)`-carbon of `CH_(3)` group, therefore, benzoic ACID is a STRONGER acid than ACETIC acid. since methoxy group acts as an electron-withdrawing group at m-position but electron-donating group at p-position, therefore, m-methoxybenzoic acid is even more acidic than benzoic acid. thus, option (c) is correct.
Discussion
36.

The acidic strength of oxyacids follow the order

Answer»

`HCIO_3gtHCIO_2gtHOCl`
`HCIO_3ltHCIO_2ltHOCI`
`HCIO_3gtHCIO_2ltHOCI`
`HCIO_3ltHCIO_2gtHOCI`

ANSWER :A
Discussion
37.

The acidic strength decreases in the order : HCl gt H_(2)S gt PH_(3). Explain. Or The acidic strength of compounds increases in the order : PH_(3) lt H_(2)S lt HCl.

Answer»

Solution :The strength of an acid depends upon the stability of the anion (i.e., conjugate base) it gives after release of a proton. Now the disssociation of `PH_(3), H_(2)S` and HCl in aqueous solution occurs as FOLLOWS :
`{:(PH_(3)(g)+aq rarr H^(+) (aq) + H_(2)P^(-) (aq), K_(a) = 1.6 xx 10^(-29)),(H_(2)S(g) + aq rarr H^(+) (aq) + HS^(-)(aq), K_(a)=1.6 xx 10^(-7)),(HCl(g)+aq rarr H^(+) (aq) + Cl^(-)(aq), K_(a) = 1 xx 10^(10)):}`
Since the ELECTRONEGATIVITY decreases in the order : `Cl gt S gt P`, therefore, stability of the conjugate bases decreases in the same order, i.e., `Cl^(-) gt HS^(-) gt H_(2)P^(-)`, In other words, acidity of the corresponding acids decreases in the same order, i.e., `HCl gt H_(2)S gt PH_(3)`.
Discussion
38.

The acidic organic compound, which does not have carboxylic function group is …

Answer»

ASCORBIC ACID
Vinegar
Oxalic acid
PICRIC acid

Solution :Picric acid
Discussion
39.

The acidic strength of HF,HCl, HBr, HI depends on

Answer»

ELECTRO negativity difference
extent of hydration
electron affinity of HALOGEN
depends on bond pair

Solution :Acidic STRENGTH depends on (1)EN (2) EA (3) Extent of hydration. 2
Discussion
40.

The acidic nature of carboxylic acids is due to :

Answer»

High degree of ionisation of the ACID
GREATER resonance
Greater resonance STABILISATION of carboxylate ions
None of these

Answer :C
Discussion
41.

The acidic nature of carboxylic acid is due to

Answer»

HIGH DEGREE of ionisation of acid
greater resonance stabilization of the acid
greater resonance stabilization of its anion
all of the above

Answer :C
Discussion
42.

The acidic dehydration method is not souitble for converting a tertiary alcohol into ether. Justify.

Answer»

SOLUTION :TERTIARY alcohols are dehydrated to alkenes and ETHERS are formed in negligible AMOUNTS.
Discussion
43.

The acidic hydroysis of ether (X)shown below is fastest when :

Answer»

one PHENYL group is REPLACED by a methyl group
one phenyl group is replacedby a paramethoxyphenyl group
TWO phenyl groups are replaced by two paramethoxyphyenyl groups
no structural CHANGE is made to X

Solution :N//A
Discussion
44.

The acidic nature of carboxylic acid is due to :

Answer»

Higher DEGREE of ionisation of theacid
GREATER resonance stabilisation of the aicd
Greater resonance stabilisation of carboxylate ion
None of the above STATEMENT is correct

Answer :C
Discussion
45.

The acidiccharacterof phenol is due to __________ .

Answer»

greaterresonance STABILIZATION of PHENOXIDEION overphenol.
greaterresosnancestabilization of phenol over phenoxide ION.
tautomerism occurring in phenol
the FACT that oxygen s moreelectronegative than hydrogen

Answer :A
Discussion
46.

The acidic character of phenol is due to

Answer»

GREATER RESONANCE stabilization of phenoxide ion over phenol
Greater resonance stabilization of phenol over phenoxide ion
Because of tautomerism OCCURRING in phenol
Because OXYGEN is more ELECTRONEGATIVE than hydrogen

Answer :A
Discussion
47.

The acidic character of glycine is due to ..................group and the basic character is due to ..........group.

Answer»

SOLUTION :`-OVERSET(N)H_(3),-COO^(-)`
Discussion
48.

The acidic character of fatty acids…… with an increase in molecular weight:

Answer»

INCREASES
Decreases
Sometimes increases or decreases
None

ANSWER :B
Discussion
49.

The acidic aqueous solution of ferrous ion forms a brown complex in the presence of NO_(3)^(-), by the following two steps : [Fe(H_(2)O)_(6)]^(2+) + NO_(3)^(-)+H^(+) rarr .... + [Fe(H_(2)O)_(6)]^(3+) [Fe(H_(2)O)_(6)]^(2+) + .... rarr ..... + H_(2)O Complete and balance the equation.

Answer»
Discussion
50.

The acidic character of dioxides of members of oxygen family decreases in the order

Answer»

`SeO_(2) gt SO_(2) gt TeO_(2) gt PoO_(2)`
`SO_(2) gt SeO_(2) gt TeO_(2) gt PoO_(2)`
`PoO_(2) gt TeO_(2) gt SeO_(2) gt SO_(2)`
`TeO_(2) gt PoO_(2) gt SeO_(2) gt SO_(2)`

ANSWER :B
Discussion