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23901.

The reactions with low activation energy are always

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ADIABATIC
SLOW
Non-spontaneous
Fast

SOLUTION :Less is the ACTIVATION energy is the reaction or greater is the activation energy SLOWER is the reaction .
23902.

The reactions that occurs at the cathode of a common dry cell is

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`2ZnO_(2)+Mn^(2+)+2e^(-)rarrMnZn_(2)O_(4)`
`MnrarrMn^(2+)+2e^(-)`
`2MnO_(2)+Zn^(2+)+2e^(-)rarrZnMn_(2)O_(4)`
`ZnrarrZSn^(2+)+2e^(-)`

Answer :C
23903.

The reactions taking place in one step is called ............... reactions.

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SOLUTION :ELEMENTARY
23904.

The reactions of Cl_(2) gas with cold dilute and hot concentrated NaOH in water give sodium salt of two (different) oxoacids of chlorine P and Q, respectively. The Cl_(2) gas reacts with SO_(2) gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. P and Q, respectively, are the sodium salts of

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hypochlorus and chloric acids
hypochlorus and chlorus acids
chloric and PERCHLORIC acids
chloric and hypochlorus acid

Solution :`{:(Cl_(2)+2NaOH overset("Cold")rarr NACL + UNDERSET("Sod. hypochlorite (P)")(NaClO)),(3Cl_(2)+6NaOh overset("Hot")rarr 5 NaCl + underset("Sod. CHLORATE (Q)")(NaClO_(3))+3H(2)O):}`
Thus, NaClO (P) and `NaClO_(3)` (Q) are the SODIUM salts of hypochlorus and chloric acid respectively and hence option (a) is coreect.
23905.

The reactions taking place al anode and cathode are

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OXIDATION and REDUCTION
Reduction and Oxidation
Oxidation and HYDROLYSIS
Reduction and Hydrolysis

23906.

The reactions of sugars are generally carried out in presence of acidic or neutral medium but not in a alkaline medium because in alkaline medium sugars undergo

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DECOMPOSITION
Racemization
REARRANGEMENT
INVERSION

SOLUTION :Rearrangement
23907.

The reactions of Cl_(2) gas with cold dilute and hot concentrated NaOH in water give sodium salt of two (different) oxoacids of chlorine P and Q, respectively. The Cl_(2) gas reacts with SO_(2) gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. R, S and T, respectively are

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<P>`SO_(2)Cl_(2), PCl_(5) and H_(3)PO_(4)`
`SO_(2)Cl_(2), PCl_(3) and H_(3)PO_(3)`
`SOCl_(2), PCl_(5) and H_(3)PO_(2)`
`SOCl_(2), PCl_(5) and H_(3)PO_(4)`

Solution :`{:(Cl_(2)+SO_(2)overset("charcoal")rarr UNDERSET("Sulphuryl CHLORIDE (R)")(SO_(2)Cl_(2))),(10 SO_(2)Cl_(2)+P_(4) rarr 10 SO_(2)+4PCl_(5)(S)),(underset((S))(PCl_(5))+4H_(2)O rarr 5 HCl + underset("Orthophosphoric ACID (T)")(H_(3)PO_(4))):}`
Thus, R, S, and T are respectively `SO_(2)Cl_(2), PCl_(5) and H_(3)PO_(4)` and hence OPTION (a) is correct.
23908.

The reactions of heavy water are slow. The reason is

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HEARY WATER is associated
Heavy water is dissociated
Heavy water is of HIGHER mass
Heavy water is of LOWER mass

Answer :C
23909.

The reactions of (a) oxygen and (b) carbon monoxide with heme (the prosthetic group of haemoglobin) give

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only oxygen-heme complex.
only carbon monoxide-heme complex.
both oxygen-heme and carbon monoxide-heme COMPLEXES but oxygen-heme complex is more STABLE.
both oxygen-heme and carbon monoxide-heme complexes but carbon monoxide-heme complex is more stable.

Solution :`HB + CO to underset(" Carboxyhaemoglobin ( stable )")(HBCO)`
`Hb + O_2 iffunderset(" Oxyhaemoglobin ( UNSTABLE)")(HbO_2)`.
23910.

The reaction(s) leading to the formation of 1,3,5-trimethylbenzene is (are)

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SOLUTION :
23911.

The reactions in which there is no change in volume is

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`K_(C)=K_(p)`
`K_(C)gtK_(p)`
`K_(C)ltK_(p)`
`K_(C)=0`

Solution :At CONSTANT volume equlibrium is not DISTURBED.
23912.

The reaction(s) leading to be formation of 1,3-trimethylbenzene is (are)

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SOLUTION :
23913.

The reactions is an example of …………………. .

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WURTZ REACTION
CYCLIC reaction
WILLIAMSON reaction
KOLBE reactions

Solution :Williamson reaction
23914.

The reaction,RCOOR'+R''OH (excess) overset(H^(+)or OH^(-))to RCOOR''+R'OH is called

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ESTERIFICATION
Trans-esterification
Saponification
Hydrolysis

Solution :When an ESTER CONVERTS into ANOTHER ester, it is CALLED esterification.
23915.

The reaction of nitroprusside anion with sul- phide ion gives purple colouration due to the formation of

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The tetranionic complex of iron (II) coordinating to one `NOS^(-)` ion
The dianionic complex of iron (II) coordinating to one `NOS^(-)` ion
The trianionic complex of iron (III) coordinating to one `NOS^(-)` ion
The tetranionic complex of iron (III) coordinating to one `NCS^(-)` ion

Solution :`Na_(2)S + UNDERSET(" Sod. Nitroprusside")(Na_(2)[Fe(CN)_(5)NO]) RARR underset(" Violet colour")(Na_(4)[Fe(CN)_(5)NOS])`
`[Fe^(+2)(CN)_(5)^(-5)NOS]^(-4) rArr` Tetra ANIONIC complex of iron
(II) co-ordinating to one `NOS^(-)` ion.
23916.

The reaction 2NO+O_(2)to2NO_(2),2CO+O_(2)to2CO_(2) look to be identical, yet the first is faster than the second. The reason is that

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The FIRST reaction has lower enthalpy CHANGE than the second
The first reaction has lower internal energy change than the second
The first reaction has lower activation energy than the second
The first reaction has HIGHER activation energy than the second

Answer :C
23917.

The reactionNO+(1//2)O_2toNO_2 exhibits:

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SMALL negative temperature coefficient
Decrease in VALUE of K with, temperature
Decrease in value of rate^ivith temperature
none

Answer :B
23918.

The reaction,CH_2=CH_2+H_2underset(250-300^@C)overset"Ni"to CH_3-CH_3 is called

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WURTZ reaction
Kolbe reaction
Sabatier and Senderens reaction
Carbylamine reaction

Answer :C
23919.

The reaction.Zn(s)+CuSO_(4)(aq)rarrZnSO_(4)(aq)+Cu(s)is an exampl for a

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SPONTANEOUS process
Non-spontaneous process
Isobaric process
Reversible process

Solution :This is an example of spontaneous reaction because it occurs of its own NAD HENCE `DeltaG` of the reaction must be negative.
23920.

The reaction Zn^(2+) + 2e^(-) to Zn has a standard potential of -0.76 V. This means

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Zn cannot replace HYDROGEN from acids
Zn is a reducing agent
Zn is an oxidising agent
`Zn^(2+)` is a reducing agent

SOLUTION :Since `E_(Zn^(2+)//Zn)^@` is negative, so Zn has a GREATER TENDENCY to be oxidised than hydrogen. Hence it can act as reducing agent.
23921.

The reaction Zn^(2+)+2e^(-)toZn has a standard potential of -0.76V. This means

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Zn can't replace hydrogen from acids
Zn is a REDUCING agent
Zn is an OXIDISING agent
`Zn^(2+)` is a reducing agent

Solution :SINCE `E_(Zn^(++)//Zn)^(o)` is negative, so Zn has greater tendency to be oxidized. Than hydrogen. Hence it can ACT as reducing agent.
23922.

The reaction Xto Products follows first order kinetics. In 40 minutes, the concentration of X changes from 0.1 M to 0.025M. The rate of reaction when concentration of X is 0.01 M is:

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`1.73 xx 10^(-4) M MIN^(-1)`
`3.47 xx 10^(-5) M min^(-1)`
`3.47 xx 10^(-4) M min^(-1)`
`1.73 xx 10^(-5) M min^(-1)`

SOLUTION :c) K = `(2.303)/t log a/(a-x)`
`=(2.303)/(40 "min") log (0.1M)/(0.025M)`
`(2.303)/(40"min") xx log4 = (2.303 xx 0.6021)/(40"min")`
`=0.03467min^(-1)`
RATE = k[x]
=`(0.03467 min^(-1)) xx (0.01M)`
`3.47 xx 10^(-4) M min^(-1)`
23923.

The reaction Zn^(2+) + 2e^(–)rarrZn(s) has a standard potential of – 0.76 V. This means:

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Zn can't REPLACE hydrogen from ACIDS
Zn is a reducing agent
Zn is an oxidising agent
`Zn^(2+)` IONS are a reducing agent

Answer :B
23924.

The reaction : XrarrProduct follows first order kinetics . In 40 minutes the concentration of X changes from 0.1 to 0.025 M. Then the rate of reaction when concentrartion of X is 0.01 M is :

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`1.73xx10^(-4) "M min"^(-1)`
`3.47xx10^(-5) "M min"^(-1)`
`3.47xx10^(-4)"M min"^(-1)`
`1.73xx10^(-5) "M min"^(-1)`

SOLUTION :( C ) For a first order reaction ,
`k = (k_(2))/(k_(1))= (2.303)/(40) "log" (0.1)/(0.025) `
`= (2.303)/(40) "log" 4 = (2.303)/(40) xx0.6020`
`= 0.0347 "min"^(-1)`
Now at 0.01 M concentration
Rate = k `[A] = 0.347 xx0.01`
`3.47 xx10^(-4) "MOL L"^(-1) "min"^(-1)`
23925.

The reaction, Xe(excess) + F_2 to XeF_2is conducted at

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573 K, 16-70 BAR
273 K, 10 bar
673 K, 1 bar
873 K, 7 bar

Answer :C
23926.

The reaction , X+2Y+Z rarr N occurs by the following mechanism: (P) X + Y hArr M (rapid equilibrium) (Q) M+Z rarr P (slow) (R) O+Y+P rarr N (very fast) What is the rate law for this reaction?

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RATE =`k`[Z]
Rate =`k [X][Y]^(2)[Z]`
Rate =`k` [N]
Rate =`k`[X][Y][Z]

Answer :D
23927.

The reaction, Xe(excess) + F2 to XeF_2 is conducted at

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573 K, 16-70 BAR 
273 K, 10 bar
673 K, 1 bar
873 K, 7 bar

Answer :C
23928.

The reaction X to Product, follows first order kinetics . In 40 minutes the concentration of X changes from 0.1 to 0.025 M . The rate of reaction , when concentration of X is 0.01 M is

Answer»

`1.73 XX 10^(-4) M "min"^(-1)`
`3.47 xx 10^(-5) M "min"^(-1)`
`3.47 xx 10^(-4) M "min"^(-1)`
`1.73 xx 10^(-5) M "min"^(-1)`

Solution :Since 0.1 M of X changes to `0.025` M in 40 minutes , `t_(1//2)` of REACTION = 40/2 = 20 minutes
Rate of reaction of
X = k[X] = `(0.693)/(t_(1//2)) xx [X] = (0.693)/(20) xx 0.01 = 3.47 xx 10^(-4) M "min"^(-1)`.
23929.

The reaction, X+2Y+Z rarr N occurs by the following mechanism

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RATE = k[Z]
Rate `=k[X][Y]^(2)[Z]`
Rate = [N]
Rate = k [X] [Y] [Z]

ANSWER :D
23930.

The reaction : X rarrProduct follows first order kinetics . In 40 minutes the concentration of X changes from 0.1 to 0.025 M. Then the rate of reaction when concentration of X is 0.01 M is

Answer»

`1.73xx10^(-4) "M min"^(-1)`
`3.47xx10^(-5) "M min"^(-1)`
`3.47 xx 10^(-4) "M min"^(-1)`
`1.73 xx10^(-5) " M min"^(-1)`

Solution :For a first ORDER reaction
`k = (2.303)/(t) "log" ([A]_(0))/([A])`
`=(3.303)/(40) "log"(0.1)/(0.025)`
or`=(2.303)/(40) "log"4 `
`=(2.303)/(40)xx0.6020`
`=0.0347 min^(-1)`
Now at 0.01 M CONCENTRATION
Rate = k[A] =`0.0347 xx0.01`
`3.47xx10^(-4) "mol L"^(-1) "min"^(-1)`
23931.

The reaction which suggest that glucose has one carbonyl group is I) Glucose + HCN II) Glucose + NH_2OH III) Glucose + HI IV) Glucose + conc. HNO_3

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All
I, II are CORRECT
I only
I, II, III are correct

Answer :B
23932.

The reaction which shows as autoreduction is ________.

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`Cu_((aq))^(2+) + Zn_((s)) to Cu_((s)) + Zn_((aq))^(2+)`
`Cu_(2)O + C to 2Cu + CO`
`Cu_(2)O + 1/2 CU + 1/2SO_(2)`
`Fe_(3)O_(4) + 4CO to 3Fe + 4CO_(2)`

Answer :C
23933.

the reaction which proceeds in the forwards direaction is

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`Fe_(2)O_(3) + 6HCL to 2FeCl_(3) + 3H_(2)O`
`NH_(3)+ H_(2)O + NaCl to NH_(4)Cl + NaOH`
`SnCl_(4)+ Hg_(2)Cl_(2) to SnCl_(2) + 2HgCl_(2)`
`2"Cul"+l_(2)+4K^(+) to 2Cu^(+)+4Kl`

Solution :in (d) , ONDINE ACTS as oxidising agent , so this reaction proceeds in forward direaction.
23934.

The reaction which proceeds in the forward direction is

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`Fe_(2)O_(3)+6HCl=2FeCl_(3)+3H_(2)O`
`NH_(3)+H^(2)O+NaCl=NH_(4)Cl+NaOH`
`SnCl_(4)+Hg_(2)Cl_(2)=SnCl_(2)=SnCl_(2)+2HgCl_(2)`
`2CuI+I_(2)+4K^(+)=2Cu^(2+)+4KI`

Solution :Due to absence of hydrolystion of `FeCl_(3)` BACKWARD reaction will not take place.
23935.

The reaction which proceed in the forward direction is :

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`Fe_2O_3 + 6HCl = 2FeCl_3 + 3H_2O`
`SnCL_4 + Hg_2Cl_2 = SnCl_2 + 2HgCl_2`
`NH_3 + H_2O + NaCl = NH_4Cl + NaOH`
`2CuI + I_2 +4K^+ = 2CU^(2+) 3KI`

Answer :A
23936.

The reaction which is not nucleophilic substitution is

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`C_2H_5Br + C_2H_5SNa toC_2 H_5SC_2H_5 + NABR`
`C_2H_5Br + 2[H] toC_2H_6 + HBR`
`CH_3Br + AGCN to CH_3NC + AGBR`
`CH_3Br + NaSH to CH_3SH + NaBr`

Solution :It is a REDUCTION reaction
23937.

The reaction which involves dichlorocarbene as an electrophile is,

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Reimer - Tiemann REACTION
KOLBE's reaction
Friedel - Crafts' ACYLATION
Fittig's reaction

Solution :
(ii) Generation of electrophile
`HO^(-) + CHCl_(3)hArrH_(2)O+C Cl_(3)^(-) rightarrow:C Cl_(2)+Cl^(-)`
(ii) Electrophilic substitution
23938.

The reaction which involves dichlorocarbene as an electrophile is

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Reimer-Tiemann REACTION
KOLBE's reaction
Friedel-Crafts' acylation
Fittig's reaction

Answer :C
23939.

The reaction , water gas ( CO + H_(2) ) + H_(2) 673K, 300 atmosphere in presence of the catalyst Cr_(2)O_(3)/ ZnO is used for the manufacture of

Answer»

HCHO
HCOOH
`CH_(3)OH`
`CH_(3)COOH`

SOLUTION :`underset("water gas")ubrace(CO+H_(2))+ H_(2)underset(672 , 200 "atm")overset(Cr_(2)O_(3)//ZnO)to CH_(3)OH`
23940.

The reaction which distintegrates neutron is or neutron is emitted (which completes first)

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`._(96)Am^(240) + ._(2)He^(4) rarr ._(97)Bk^(244) + ._(+1)e^(0)`
`._(15)P^(30) rarr ._(14)Si^(30) + ._(1)e^(0)`
`._(6)C^(12) + ._(1)H^(1) rarr ._(7)N^(13)`
`._(13)Al^(27) + ._(2)He^(4) rarr ._(15)P^(30)`

Solution :Equate MASS no. and atomic no
23941.

The reaction used for the preparation of phenol is

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All of the above

Solution :All the given METHODS are USED to PREPARE phenol.
23942.

The reaction that takes place in the cathode half cell in a Galvanic cell is _______.

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oxidation
reduction
redox
hydrolysis

Answer :B
23943.

The reaction underset"Ethylene"(H_2C=CH_2)+underset"water"(H_2O)underset(300^@//60 atm)overset(H_3PO_4)tounderset"Ethyl alcohol"(C_2H_5OH) is called

Answer»

Hydration
Sublimation
Dehydration
Substitution

Solution :Alkenes react with water in the presence of ACID and form alcohols. This REACTION is called as hydration .
`underset"Ethylene"(H-undersetundersetH|C=undersetundersetH|C-H+H_2O) underset(300^@C//"60 atm")OVERSET(H_3PO_4)to underset"Ethyl alcohol"(H-undersetunderset(H)(|)oversetoverset(H)(|)C-undersetunderset(H)(|)oversetoverset(H)(|)C-OH)`
23944.

The reaction that take place is aryl halide is ……

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NUCLEOPHILIC substitution
ELECTROPHILIC addition
Electrophilic substitution
NUCLEOPHILI addition

Answer :C
23945.

The reaction taking place in the cell Pg|H_(2)(g)|HCl(1.0M)|AgCl|Ag is 1 atm

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`AGCL+(1//2)H_(2)toAg+H^(+)+Cl^(-)`
`Ag+H^(+)+Cl^(-)toAgCl+(1//2)H_(2)`
`2Ag^(+)+H_(2)to2Ag+2H^(+)`
`2Ag+2H^(+)to2Ag^(+)+H_(2)`.

SOLUTION :LHS ELECTRODE (oxidation half REACTION)
`(1)/(2)H_(2)TOH^(+)+e^(-)`
RHS electroe (reduction half reaction)
`AgCl+e^(-)toAg+Cl^(-)`
Adding the two half ractions, we get
`AgCl+e^(-)toAg+H^(+)+Cl^(-)`
23946.

The reaction taking place at cathode is

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HYDROLYSIS
neutrolisation
OXIDATION
reduction

Solution :Cathode is negatively charge, where positive ION gains the electron which is reduction REACTION.
23947.

The reaction taking place at anode when an aqueous solution of CuSO_4 is electrolysed using inert Pt electrode:

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`2SO_4^(2-) rarrS_2O_3^(2-) +2E`
`CU^(2+) +2e rarrCu`
`2H_2O rarrO_2+4H^+ +4e`
`2H^+ +2e rarrH_2`

ANSWER :C
23948.

The reaction takes place in two steps as (i) NO_2Cl(g)overset(K_1)rarrNO_2(g)+Cl(g) (ii) NO_2Cl(g)+Cl(g)overset(K_1)rarrNO_2(g)+Cl_2(g)

Answer»

`NO_2Cl(g)`
`NO_2(g)`
`Cl_2(g)`
`Cl(g)`

Solution :CI (g) is the REACTION intermediate involved in the formation of `NO_2(g)` and `Cl_2(g)`. Reaction INTERMEDIATES are those which are formed within the reaction and has no PRESENCE in the products. CI (g) is involved in both the reactions.
`NO_2Cl(g)OVERSET(k_1)toNO_2(g)+Cl(g)`
`NO_2Cl(g)+Cloverset(k_2)toNO_2(g)+Cl(g)`
23949.

The reaction take place when ethers are exposed to oxygen is……………………… .

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SOLUTION :AUTO OXIDATION
23950.

The reaction ,Sucroseoverset(H^(+))rarr Glucose + Fructose, take place at certain temperature while the volume of solution is maintained at 1 litre. At time zerothe initial rotation of the mixture is 34^(@). After 30 minutes the total rotation of solutions is 19^(@) and after a very long time, the total roation is -11^(@) C Find the time when solution was potically inaactive.

Answer»

135 MIN
`103.7 min`
`38.7 min`
45 min

Answer :B