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86151.

According to Arrhenius equation rate constant k is equal to A e^(-E_(a)//RT).Which of the following options represents the graph of In k va (1)/(T)?

Answer»




Solution :K=`A e^(-E_(a)//RT)`
`therefore` In k=In `(A e^(-E_(a)//RT))`
`therefore` k=In `A-(E_(a))/(RT)`
`therefore`

k=`Ae^(-E_(a)//RT) therefore` In k=In `-(E_(a))/(R )`Here ,graph of In k vs `(1)/(T)` will be linear with negative SLOPE and INTERCEPT =In A,i.e (A)
Discussion
86152.

According to Arrhenius equation,the slope of log Kto(1)/(T) plot is………

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`(-E_(a))/(2.303)`
`(-E_(a))/(2.303R)`
`(E_(a))/(2.303RT)`
`(E_(a))/(2.303RT)`

ANSWER :B
Discussion
86153.

According to Arrhenius equation rate constant k is equal to Ae^(-E_(a)//RT). Which of the following option. Represents the graph of ln k us (1)/(T)?

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ANSWER :A
Discussion
86154.

According to Arrhenius equation, rate constant k is equal to "A e"^(-E_(a)//"RT"). Which of the following options represents the graph of ln k vs (1)/(T) ?

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Solution :`K="A e"^(-E_(a)//"RT"):.lnk=lnA-(E_(a))/(RT)`. Hence, graph of LN k vs `(1)/(T)` will be linear with negative slope and INTERCEPT = ln A, i.e., (a).
Discussion
86155.

According to Arrhenius equation,

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a high activation energy usually implies a FAST reaction
rate CONSTANT increases with increase in temperature. This is due to greater number of collisions whose energy exceeds the activation energy
higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant
the pre-exponential factor is a MEASURE of the rate at which collisions occure, irrespective of their energy

Solution :is incorrect because high activation energy implies a slow reaction (b),(c ) and (d) are correct and follow from Arrhenius equation, viz.,
`k=Ae^(-E_(a)//RT)`.
Discussion
86156.

According to Arrheneius equation rate constant k is equal to Ae^(-E_(a)//RT). Which of the following options represents the graph of ln k vs 1/T?

Answer»




ANSWER :A
Discussion
86157.

According to 2nd law of Faraday's electrolysis the correct one is i) (" wt. of "H_2 " liberated")/("wt. of " Cl_2 " librated") = ("eq. wt . of" H_2 )/(" eq. wt . of " Cl_2) ii) (m_(Ag))/(m_(cu)) = (E_(Ag))/(E_(Cn)) iii) (m_(Ag))/(m_(Cu)) = (E_(Cu))/(E_(Ag))iv) (m_(H_2))/(m_(Cu)) = (E_(H_2))/(E_(Cu)) The correct combination is

Answer»

only II, IV
only i
only I, ii, iv
only ii, III
Discussion
86158.

According produces to the adsorption theory of catalysis, the speed of the reaction increases because

Answer»

Adsorption produces heat which increases the speed of the reaction
Adsorption LOWERS the activation energy of the reaction
The CONCENTRATION of REACTANT molecules at the active centres of the CATALYST becomes high due to adsorption
In the process of adsorption, the activation energy of the molecules becomes large

Answer :B
Discussion
86159.

According the Le-chatelier principle, if heat is given to solidliquid system, then

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Quantity of solid will REDUCE
Quantity of liquid will reduce
INCREASE in temperature
Decrease in temperature

Solution :On increasing temperature EQULIBRIUM will SHIFT in FORWARD direction due to decrease in intemolecular forces of solid.
Discussion
86160.

According for the following pK_b of aniline is more than that of methylamine.

Answer»

SOLUTION :Basic strength of aniline smaller the value of `pK_b`, stronger the base. In aniline, the `NH_2` group is directly ATTACHED to the benzene ring. The LONE pair of electron on nitrogen ATOM in ailine gets delocalised over the benzene ring and hence it is less available for protonation makes the, aromatic amines (aniline ) less basic than `NH_3`. So aniline has a lower `pK_b` value than METHYL amine.
Discussion
86161.

According for the following Gabriel phthalimide synthesis is prefered for synthesising primary amines.

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SOLUTION :Gabriel phthalimide synthesis results in the formation of `1^(@)` amine only. `2^(@)" and "3^(@)` amine are not FORMED by this synthesis. Thus pure PRIMARY amine can be OBTAINED. Therefore Gabriel phthalimide synthesis is prefered for synthesizing primary AMINES.
Discussion
86162.

According for the following Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Answer»

Solution :The stability of arenze DIAZONIUM salt is due to the DISPERSAL of the positive charge over the benzene ring. Resonance stabilised diazonium ion is further stabilised by the involvement of the benzene ring.

So C - N in anyl diazonium salt is stronger than that in ALIPHATIC diazonium salt.
Discussion
86163.

Accordingbaeyer- streaintheorywhichcompoundhasminimum anglestrain ?

Answer»

N- butane
CYCLOPENTANE
cyclopropane
cyclohexane

SOLUTION :(B ) cyclopentane
Discussion
86164.

Accomplishthe following conversions : (i) C_(6)H_(5)NO_(2) rightarrow C_(6)H(5)- COOH (ii) Benzene rightarrow m-bromophenol (iii) C_(6)H_(5)COOH rightarrow C_(6)H_(5)NH_(2) (iv) Aniline rightarrow 2,4,6 tribromoaniline (v) Benzylchloride rightarrow 2-phenyl ethanamine.

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SOLUTION :
Discussion
86165.

Accomplish the following conversions: (i) Nitrobenzene to benzoic acid, (ii) Benzene to m-bromophenol, ltBrgt (iii) Benzoic acid to aniline, (iv) Aniline to 2,4,6-tribromofluorobenzene, (v) Benzyl chloride to 2-phenylethanamine. (vi) Chlorobenzene to p-chloroaniline. (vii) Aniline to p-bromoaniline (viii) Benzamide to toluene (ix). Aniline to benzyl alcohol.

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SOLUTION :

.
Discussion
86166.

Accomplish the following conversions : Nitrobenzene to benzoic acid

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SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86167.

Accomplish the following conversions : Chlorobenzene to p-chloraniline

Answer»

SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86168.

Accomplish the following conversions : Aniline to p-bromoaniline

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SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86169.

Accomplish the following conversions : Benzoic acid to aniline

Answer»

SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86170.

Accomplish the following conversions : Benzyl chloride to 2-phenylethanamine

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SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86171.

Accomplish the following conversions : Benzene to m-bromophenol

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SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86172.

Accomplish the following conversions : Benzamide to toluene

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SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86173.

Accomplish the following conversions : Aniline to benzyl alcohol.

Answer»

SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86174.

Accomplish the following conversions : Aniline to 2, 4, 6-tribromofluorobenzene

Answer»

SOLUTION :The CONVERSIONS are CARRIED out as under :
Discussion
86175.

Accepting the definition that an acid a proton donor, the acid in the following reaction NH_(3)+H_(2)O rarr NH_(4)^(+) + OH^(-) is

Answer»

`NH_(3)`
`H^(+)`
`NH_(4)^(+)`
`H_(2)O`

SOLUTION :In this REACTION `H_(2)O` acts as a ACID.
Discussion
86176.

Acc. to V.B. Theory, the following complexes will have same geometry.

Answer»

`[NI(CN)_(4)]^(2-)`
`[Fe(CN)_(6)]^(-3)`
`[FeF_(6)]^(3-)`
`[CR(NH_(3))_(6)]^(3+)`

Solution :`[Ni(CN)_(4)]^(2-)`- SQUARE planar`[FeF_(6)]^(3-)`- octahedral
`[Fe(CN)_(6)]^(4-)`- octahedral`[Cr(NH_(3))_(6)]^(3+)`- octahedral
Discussion
86177.

(A)Catalyst increases the rate of reaction. (R ) Catalyst functions by lowering the energy of activation.

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Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A)
Both (A)and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is FALSE
Both (A) and (R ) are false

ANSWER :A
Discussion
86178.

A:Carbon is used in blast furnace for reuction of Fe_2O_3B:This process is called smelting.

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If both ASSERTION & Reason are true and the reason is the correct explanation of the assertion, then mark (1).
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
If Assertion is true STATEMENT but Reason is false, then mark
 If both Assertion and Reason are false statements, then mark (4)

Answer :B
Discussion
86179.

A(C_(7)H_(7)Cl)underset((ii)"sodalime"//Delta)overset((i)KMnO_(4))(rarr) chlorobenzene. In this sequence , 'A' can be-

Answer»




NONE of these

ANSWER :B::C
Discussion
86180.

A(C_(6)H_(12))overset(HCl)to underset((C_(6)H_(13)C))(B+C) B overset(alc KOH)toD (isomer of A) D overset("ozonolysis")toE (it gives negative test with Fehling solution but responds to iodoform test). A overset("ozolysis")toF+G (Both give positive Tollen's but do not give iodoform test) F+G overset("conc. NaOH")to HCOONa +a primary alcohol Identify to A to G

Answer»


ANSWER :A::C
Discussion
86181.

(A)(C_(4)H_(8)O)overset(H_(3)O^(o+))rarr(B)overset(CrO_(3))underset("acetic acid")rarr(C )overset(CH_(2)N_(2))underset(Delta)rarr(D) The compound 'D' is

Answer»




ANSWER :A
Discussion
86182.

(A)C_(4)H_(10)Ounderset(Delta)overset("Conc."H_(2)SO_(4))rarrCH_(3)-overset(CH_(3))overset(|)(C)=CH_(2)overset("dil."H_(2)SO_(4))rarr(B)C_(4)H_(10)O What is relationship between A & B ?

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A and B may be POSITION ISOMER
A and B may be CHAIN isomers
A and B may be stereoisomers
All of the above

Answer :A
Discussion
86183.

A(C_(3)H_(9)N) reacts with benzenesulphonyl chloride to give a solid insoluble in alkali. Give the structure of A.

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Solution :Since `A(C_(3)H_(9)N)` reacts with benzenesulphonyl chloride to give a SOLID insoluble in alkali, (A) must be a `2^(@)` amine: i.e., `CH_(3)-CH-CH_(2)CH_(3)` (ethylmethylamine).
Discussion
86184.

Ac^(227) has a half- life of 22.0 years with respect to one leading . To Th^(227) and the other to Fr^(227). The percentage yields of these two daughter nuclides are 2.0 and 98.0, respectively . What are the decay constants (lambda) for each of the separate paths ?

Answer»

Solution :For the radioactive decay of `A` into `B` and `C` by two PARALLEL paths
So, `(-d[A])/(dt)=lambdaN....(i)`
`(+d[B])/(dt)=lambda_(1)N.....(ii)`
`(+d[C])/(dt)=lambda_(2)N....(III)`
where `lambda,lambda_(1) , `and `lambda_(2)` are radioactive decay constants, respectively, and `N` is the number of atoms of `A` at any GIVEN time.
THUS, `(+d[A])/(dt)=(d[B])/(dt)+(d[C])/(dt)`
`:. lambdaN=lambda_(1)N+lambda_(2)N`
`:. lambda=lambda_(1)+lambda_(2)N`
From eqs. `(ii)` and `(iii)`, we get
`(d[B])/(d[C])=(lambda_(1))/(lambda_(2))`
On integration ,we get
`([B])/([C])=(lambda_(1))/(lambda_(2))`
For decay of `Ac^(227)` into `TH^(227)` and `Fr^(223)` , on the basis of given data
`(lambda_(1))/(lambda_(2))=(2.0)/(98.0) ....(iv)`
and `lambda=(0.693)/(t_(1//2))=(0.693)/(22)=0.0315year^(-1)` ltbr. So,` 0.0315=lambda_(1)+lambda_(2) ...(v)`
On solving Eqs. `(iv)` and `(v)`, we get
`lambda_(1)=6.3xx10^(-4)year^(-1)`
and `lambda_(2)=0.03087year^(-1)`
Discussion
86185.

A(C_(10) H_(14))+ Cl_(2), Delta toC_(10) H_(13)Cl( one isomaer) A +Cl_(2)//FeCl_(3) to C_(10)H_(13)Cl (two isomers) Possible structure of A is :

Answer»




ANSWER :D
Discussion
86186.

Ac^(227) has a half-life of 22 years . The decays follows two parallel paths What are the decay constant (lambda) for Th and Fr respectively ?

Answer»

0.03087,0.00063
0.00063,0.03087
0.02,0.98
None of these

Answer :B
Discussion
86187.

Ac^(227) has a half - life of 22 years . The decay follows two parallel paths . What are the decay constants (lambda) for Th and Fr respectively ?

Answer»

0.03087 , 0.00063
0.00063 , 0.03087
0.02 , 0.98
none of these

Solution :`Y_(th) = (lambda_(1))/(lambda) xx 100 = 2 ……. (1) , Y_(Fr) = (lambda_2)/(lambda) xx 100 = 98 ……. (2)`
`lambda _(2) = (98 lambda_(1))/(2), (lambda_(1))/(lambda_(2)) = (2)/(98) ….. (1) & lambda = (lambda_(1) + lambda_(2)) = (0.693)/(22) ….. (2) , (1) & (2)`
`lambda_(1) + (98 lambda_1)/(2) = (0.693)/(22) , (2 lambda_(1) + 98 lambda_(1))/(2) = (0.693)/(22) , (100 lambda_(1))/(2) = (0.693)/(22) IMPLIES lambda_(1)= (0.693)/(100) = 0.00063` YEARS
but `(lambda_(1))/(lambda_(2)) = (2)/(98) , lambda = (98 lambda_(1))/(2) = 49 lambda_1 = 49 xx 0 .00063 ` = 0.03087 years
Discussion
86188.

Abtacids are those

Answer»

whichrise the PH of stomach
which RISE the pH of herat
which DECREASES the pH of stomach
which rise the pH of liver

Answer :A
Discussion
86189.

Absolute zero is the temperature at which

Answer»

WATER FREEZES completely
all SUBSTANCES exist as solids
all GASES become liquids
molecular motion ceases

Answer :D
Discussion
86190.

Absorption and adsorption if simultaneously occurs, it is called ..............

Answer»

OCCLUSION
SORPTION
desorption
dissolution

Solution : sorption
Discussion
86191.

Absolute zero is defined as the temperature

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At which all MOLECULAR MOTION ceases
At which LIQUID HELIUM boils
At which etherboils
All of the above

Answer :A
Discussion
86192.

Absolute entropies of solids, liquids and gases can be determined by-

Answer»

Measuring HEAT capacity of substances at various temperatures
Subtracting standard entropy of reactants from products.
Measuring VIBRATIONAL motion of molecules.
Using FORMULA `DELTAS^(@) =S_(T)^(@) - S_(0)^(@)`.

Answer :A
Discussion
86193.

Absolute alcohol is prepared from rectified spirit by which distillation?

Answer»


ANSWER :C
Discussion
86194.

Absolute alcohol is prepared form rectified spirit by

Answer»

STEAM distillation
Fractional distillation
distillation with CAO & Ca
simple distillation

Answer :C
Discussion
86195.

Absolute alcohol is prepared by :

Answer»

FRACTIONAL distillation
Kolbe's method
azeotropic distillation
vacuum distillation

Answer :C
Discussion
86196.

Absolute alcohol contains :

Answer»

`40%H_2O`
`10%H_2O`
`5%H_2O`
`100%C_2H_5OH`

ANSWER :D
Discussion
86197.

Absolute alcohol cannot be prepared from rectified spirit by simple distillation because

Answer»

the boiling POINTS of WATER and alcohol are very close
water and alcohol FORM a constant boiling azeotropic mixture
`C_(2)H_(5)OH` FORMS H-bonds with water
alcohol molecules are highly hydrated.

Solution :Water and alcohol form a constant boiling azeotropic mixture.
Discussion
86198.

Absolute alcohol cannot be prepared by fractional distillation of rectified spirit since:

Answer»

It FORMS AZEOTROPIC mixture
It is USED as POWER alcohol
It is used in wines
None of the above

Answer :A
Discussion
86199.

Absolute alcohol cannot be obtained by simple fractionsal distillation because :

Answer»

pure `C_(2)H_(5)OH` is unstable
`C_(2)H_(5)OH` forms hydrogen bonding with water
BOILING point `C_(2)H_(5)OH` is very close to that of water
constant boiling azeotropic MIXTURE is formed with water

Answer :D
Discussion
86200.

Absolute alcohol can be prepared from rectified spirit by____diistillation.

Answer»

SOLUTION :AZEOTROPIC
Discussion