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.

Consider three flasks in diagram below. Assuming that connecting tube has negligible volume and all three falsks are at same temperature. If only 1 and 2 stopcocks are opened then select correct option(s).

Answer»

`P_(HCl)=(1)/(2)` atm
`P_(NH_(3))=0` atm
`P_(NH_(3))+P_(HCl)=(1)/(2)` atm
`P_(He)=5` atm

Solution :`n_(HCl)=(PV)/(RT)=(2XX1)/(RT)`
`n_(NH_(3))=(PV)/(RT)=(1xx1)/(RT)`
`NH_(3)+HClrarrNH_(4)Cl(s)`
moles `(1)/(RT) (2)/(RT)`
`0 (1)/(RT)`
`n_(HCl)` remain after REACTION =`(1)/(RT)`
For `HCl(g):-P_("NEW")(V+V)=nRT`
`P_("new")xx2=(1)/(RT)xxRT`
`(P_(HCl))_("final")=P_("new")=(1)/(2)atm=P_("total")`
`P_(NH_(3))=0(LR)`
Discussion
2.

Consider this reaction, 2H_(2)(g)+2NO(g)to N_(2)(g)+2H_(2)O(g) The rate law for this for reaction is rate =k[H_(2)][NO]^(2) . Under what conditions could these Steps represent the mechanism? Step1: 2NOtoN_(2)O_(2) Step2: N_(2)O_(2)+H_(2)to N_(2)O+H_(2)O Step3:N_(2)O+H_(2)to N_(2)+H_(2)O

Answer»

These STEPS cannot be the mechanism under any circumstances.
These steps COULD be the mechanism if Step 1 is the SLOW step.
These step could be the mechanism if Setp2 is the slow step.
These Stepscould be the mechanism if Step 3 is the slow step.

Answer :C
Discussion
3.

Consider thiol anion (RS^(Theta)) and alkoxy anion (RO^(Theta)) . Which of the following statement is correct ?

Answer»

`(RS^(Theta))`is less BASIC but more nucleophilic than`(RO^(Theta))`
`(RS^(Theta))`is more basic and more nucleophilic than `(RO^(Theta))`
`(RS^(Theta))`is more basic but less nucleophilic than `(RO^(Theta))`
`(RS^(Theta))`is less basic and less nucleophilic than `(RO^(Theta))`

ANSWER :A
Discussion
4.

Consider thiol anion (RS^(Θ))and alkoxy anion (RS^(Θ)), which of the following statement is correct ?

Answer»

`RS^(Θ)`is less basic but more NUCLEOPHILIC than `RO^(Θ)`
`RS^(Θ)` is more basic but more nucleophilic than `RO^(Θ)`
`RS^(Θ)`​ is more basic but less nucleophilic than `RO^(Θ)`
`RS^(Θ)` is less basic but less nucleophilic than `RO^(Θ)`

Solution :`A.S RARR R-OH lt R-SH`
`B.S rArr R-O^(Θ)gt R-S^(Θ)`
NUCLEOPHILICITY `rArr R-O^(Θ)lt R-S^(Θ) ""` (size)
Discussion
5.

Consider thiol anion (RS^(-)) and alkoxy anion (RO^(-)). Which of the following statements is correct?

Answer»

`RS^(-)` is less BASIC but more nucleophilic than `RO^(-)`
`RS^(-)` is more basic and more nucleophilic than `RO^(-)`
`RS^(-)` is more basic and less nucleophilic than `RO^(-)`
`RS^(-)` is less basic and less nucleophilic than `RO^(-)`

Solution :On moving down a group, the basicity an nucleophilic are inversely RELATED, i.e., NUCLEOPHILICITY increase while basicity decrease, i.e., `RS^(-)` is more nucleophilic but less basic than `RO^(-)`.This OPPOSITE behaviour is because of the fact that basicity and nucleophilic depends upon DIFFERENT is indirectly related to the strength of the H-element-bond. while nucleophilicity in indirectly related to the electronegativity of the atom to which proton is attached.
Discussion
6.

Consider the two complexation equilibria in aqueous solution, between the cobalt (II) ion Co^(2+) (aq) and ethylenediamine (en) on the one hand and ammonia NH_(3) on the other. [CO(H_(2)O)_(6)]^(2+)+6NH_(3)hArr[Co(NH_(3))_(6)]^(2+)+6H_(2)O...(1) [Co(H_(2)O_(6))]^(2+)+3enhArr[Co(en)_(3)]^(2+)+6H_(2)O..(2) Electronicaly, the ammonia and en ligands are very similar, since both bond through N and since the liwis base strengths of their nitrogen atoms are similar. This means that DeltaH^(@) must be very similar for the two reactions, since six Co-N bonds are formed in each case. Interestingly however, the equilibrium constant is 100,000 times larger for the second reaction than it is for the first. This is the so called chelate effect: "the enhanced affinity of chelating ligands for a metal ion compared to similar non-chelating (monodentate) ligands for the same metal". The chelate effect is entropy-driven. Q. Which of the following can be classified as a chelating ligand?

Answer»

`edta^(-4)`
`N_(3)^(-)`
`PPh_(3)`
all of these

Answer :B
Discussion
7.

Consider the value of all four quantum number for last electron and spin multiplicity (2s+1) for givenn two element 'X' and 'Y' in their ground state: {:(,n,,l,m,s,|2s+1|),(X:,2,0,0,+1//2,1),(Y:,2,1,-1,-1//2,4):} The according to given information the correct statement is:

Answer»

The BOND angle `(H-Y-H)` of possible hydride of element Y is less than `190"28'`
The possible halide of 'X' has two vacant p-orbitals on its central atom.
Magnetic MOMENT of Y is greater than X
X and Y element exhibits only single oxidation are

Solution :According to given information `X implies Be and Y implies N`
(a0 possible hydride of N is `NH_(3)(H-N-H` bond angle is less than `109"28.`)
(b) In `BeX_(2)`, Be has two vacent p-orbirtals
(c) Magneti cmoment of N>Be. (number of unpaired `e^(-)` in N>Be)
(d) N can exhibits variable oxidation state.
Discussion
8.

Consider the two geometrical isomers of but-2-eneThe isomer with higher melting point and the reason for the same is respetively

Answer»

CIS, symmetric bonding
Trans, close and symmetric packing
Cis, close packing
Trans, NON- symmetric ARRANGEMENT of similar GROUPS AROUND C=C

Answer :B
Discussion
9.

Consider the sulfonation reactions of phenol shown above.For which sulfonation product is the free energy of activation lower?

Answer»

Solution : The free energy of activation is LOWER for ORTHO substitution. We know this because at the lower TEMPERATURE, where the reaction is under kinetic CONTROL. it is formed faster.
Discussion
10.

Consider the sulfonation reactions of phenol shown above. Which sulfonation product is more stable, ortho or para?

Answer»

Solution :The para-sulfonated phenol is more STABLE. We KNOW this because at the higher temperature, where the reaction is under equilibrium CONTROL, it is the MAJOR PRODUCT
Discussion
11.

Consider the successive ionisation energy for an element 'A' . IE_(1) , IE_(2) , IE_(3) , IE_(4), IE_(5) are 100 eV , 150 eV , 181eV, 2000 eV , 2200 eV . Select correct statement (s) for element 'A' :a)Element 'A' may be metalb)Element 'A' may from trivalent cationc)Oxideof element 'A' may be amphotericd)Element 'A' may be non-metal

Answer»

ELEMENT 'A' MAY be metal
Element 'a' may FORM trivalent cation
Oxide of element 'A' may be amphoteric
Element 'A' may be non-metal

Answer :A::B::C::D
Discussion
12.

Consider the structure of the complex compound, [Cu_(4)OCI_(6)(Ph_(3)P)_(4)], as given below : src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ ##GRB_ORG_CHM_V01_QB_C05_E01_068_Q01.png## Which of the following is the correct IUPAC name of the complex ?

Answer»

`hexa-MU-chloride-mu_(4)-"oxidotetrakis (TRIPHENYLPHOSPHINE) tetracopper (II)"`
`mu_(4)-"oxidohexa"-mu_(2)-"chloridotetra (triphenylphosphine)tetracopper" (II)`
`hexa-mu-"chlorido"-mu-"oxidotetrakis (triphenylphosphine)tetracopper"(II)`
All of these

Solution :N//A
Discussion
13.

Consider the structure of the complex compound, [Cu_(4)OCI_(6)(Ph_(3)P)_(4)], as given below : ##GRB_ORG_CHM_V01_QB_C05_E01_068_Q01## Which of the following is the correct IUPAC name of the complex ?

Answer»

`hexa-MU-chloride-mu_(4)-"oxidotetrakis (TRIPHENYLPHOSPHINE) tetracopper (II)"`
`mu_(4)-"oxidohexa"-mu_(2)-"chloridotetra (triphenylphosphine)tetracopper" (II)`
`hexa-mu-"chlorido"-mu-"oxidotetrakis (triphenylphosphine)tetracopper"(II)`
All of these

Solution :N//A
Discussion
14.

Consider the statements: I) Maltose is also known as malt sugar II) Sucrose is also known as cane sugar III) Lactose is also known as grape sugar IV) Starch is also known as Amylum The correct statement is / are

Answer»

I,II and IV
I, II and III
II, III and IV
I and II

Answer :A
Discussion
15.

Consider the statementltbrlt I. Bond length in N_(2)^(+) is 0.02Å greater thanin N_(2). II. Bond length of NO^(+) is 0.09Å less than in NO. III O_(2)^(2_(-)) has shorter bond length than O_(2). Which of the following statements are true?

Answer»

I and II
II and III
I , II and III
I and III

Answer :A
Discussion
16.

Consider the standard electrode potential values (M2+/M) of the elements of the first transition series. {:(Ti,,V,,Cr,,Mn,,Fe,,CO,,Ni,,Cu,,Zn),(-1.63,,-1.18,,-0.90,,-1.18,,-0.44,,-0.28,,-0.25,,+0.34,,-0.76):} Explain : (i) E^(@) value for copper is positive. (ii) E^(@) value of Mn is more negative as expected from the trend. (iii) Cr^(2+) is a stronger reducing agent than Fe^(2+).

Answer»

Solution :(i) HIGH ionisation enthalpy to convert Cu (s) to `Cu^(2+)` (aq) is not balanced by its hydration enthalpy.
(ii) `Mn^(2+)` ion has stable `d^(5)` (half-filled) configuration.
(iii) `d^(4)` to `d^(3)` occurs in the case of `Cr^(2+)` to `Cr^(3+)`, it is more stable. `d^(6)` to `d^(5)` occurs in case of `Fe^(2+)` to `Fe^(3+)`. Therefore `Cr^(2+)` is a STRONGER reducing agent than `Fe^(2+)`.
Discussion
17.

Consider the reversible reaction, HCN(aq.) ⇌H^+ + CN^-(aq.)At equilibrium the addition of CN^-(aq.) would:

Answer»

Reduce HCN(AQ.) concentration
Decrease the `H^+`(aq.) ion concentration
Increase the EQUILIBRIUM constant
Decrease the equilibrium constant

Answer :B
Discussion
18.

Consider the reactions The products A, B, C respectively are

Answer»




SOLUTION :
Discussion
19.

Consider the reactions (i)"" S("rhombic") + 3//2O_(2)(g) rarr SO_(3)(g),DeltaH_(1) (ii) " "S("monoclinic") + 3//2O_(2)(g) rarr SO_(3)(g),DeltaH_(2) (iii) " "S("rhombic") + O_(3)(g) rarr SO_(3)(g),DeltaH_(3) (iv) " "S("monoclinic") + O_(3)(g) rarr SO_(3)(g),DeltaH_(4)

Answer»

`DeltaH_(1) LT DeltaH_(2) lt DeltaH_(4)` (MAGNITUDE only)
`DeltaH_(1) lt DeltaH_(3) lt DeltaH_(4)` (magnitude only)
`DeltaH_(1) lt DeltaH_(2) = DeltaH_(3) lt DeltaH_(4)` (magnitude only )
`DeltaH_(1) + DeltaH_(4) = DeltaH_(2) + DeltaH_(3)`

Solution :N//A
Discussion
20.

Consider the reactions identify X, A, C and D

Answer»

X= n-propanol, A = PROPANALDEHYDE, C=sodium acetate, D=ketoxime
X=isopropanol, A=acetone, C=sodium propanoate, D=semicarbazone
X=n-propanol, A = PROPENE, C=sodium propanoate, D=semicarbazone
X=iso - propanol, A=Acetone, C=sodium acetate, D=2,4 - dinitrophenylhydrazone

SOLUTION :Generally ketone do not give response to Tollen's reagent
Discussion
21.

Consider the reactions: (i) (CH_(3))_(2)CH-CH_(2)Broverset(C_(2)H_(5)OH)to(CH_(3))_(2)CH-CH_(2)OC_(2)Hd_(5)+Br^(-) (ii) (CH_(3))_(2)CH-CH_(2)Broverset(C_(2)H_(5)O^(-))to(CH_(3))_(2)CH-CH-(2)OCH_(2)H_(5)+Br^(-) The mechanism of reactions (i) and (ii) are respectively-

Answer»

`S_(N)1andS_(N)2`
`S_(N)1andS_(N)1`
`S_(N)2andS_(N)2`
`S_(N)2andS_(N)1`

Solution :Carbocation INTERMEDIATE is formed in case of `S_(N)1` reaction and the corbocations attains stability via rearrangements have taken place. Therefore, it MAY be ASSUMED that they follow `S_(N)2` mechanism.
Discussion
22.

Consider the reactionsC(s)+2H_(2)(g)rarrCH_(4)(g),DeltaH=-x kcalC(g)+4H(g)rarrCH_(4)(g), DeltaH=-x_(1) kcalCH_(4)(g)rarrCH_(3)(g)+H(g),DeltaH=+y kcal The bond energy of C - H bond is

Answer»

y KCAL `mol^(-1)`
`x_(1)` kcal `mol^(-1)`
`X//4` kcal `mol^(-1)`
`x_(1)//4` kcal`mol^(-1)`

SOLUTION :The BOND energy of C-H bond is y kcal `mol^(-1)`.
Discussion
23.

Consider the reactions : (1) (CH_(3))_(2)CH-CH_(2)Br overset(C_(2)H_(5)OH)rarr (CH_(3))_(2)CH-CH_(2)OC_(2)H_(5)+HBr(2)(CH_(3))_(2)CH-CH_(2)Br overset(C_(2)H_(5)O^(-))rarr (CH_(3))_(2)CH-CH_(2)OC_(2)H_(5)+Br^(-)The mechanisms of reactions (i) and (ii) are respectively :

Answer»

`S_(N)1` and `S_(N)2`
`S_(N)1` and `S_(N)1`
`S_(N)2` and `S_(N)2`
`S_(N)2` and `S_(N)1`

Solution :A strong NUCLEOPHILE favours the `SN^(2)` reaction and a weak nucleophile favours the `S_(N)1` reaction.
First reaction is `S_(N)1` reaction because `C_(2)H_(5)OH` used as SOLVENT which US a weak nucleophile.
Second reaction is `S_(N)2` reaction because `C_(2)H_(5)O^(-)` is strong nucleophile.
Discussion
24.

Consider the reaction,M^(n+) (aq)+ne rarrM(s) The standard reduction potential value of the metals M_1,M_2 and M_3 are -0.34 V,-3.05 V and -1.66V respectively.The order of their reducing power will be :

Answer»

`M_1 GT M_2 gt M_3`
`M_3 gt M_2 gt M_1`
`M_1 gt M_3 gt M_2`
`M_2 gt M_3 gt M_1`

ANSWER :D
Discussion
25.

Consider the reaction,M^(n+) (aq)+nerarrM^@(s) The standard reduction potential value of the metals M_1,M_2 and M_3 are -0.34 V,-3.05 V and -1.66V respectively.The order of their reducing power will be :

Answer»

`M_1 GT M_2 gt M_3`
`M_3 gt M_2 gt M_1`
`M_1 gt M_3 gt M_2`
`M_2 gt M_3 gt M_1`

ANSWER :D
Discussion
26.

Consider the reaction where K_(p)=0.497at 500K. PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g)) If the thre gases are mixed in a rigid containeer so that the partial pressure of each gas is initially 1 atm, which is true

Answer»

<P>More `PCl_(5)` will be produced
More `PCl_(3)` will be proudced
Equlibrium will be ESTABLISHED when `50%` REACTION is complete
None of the above

Solution :`Q_(p)=(pPCl_(3)xxpCl_(2))/(pPCl_(5))=(1atmxx1atm)/(1atm)=1atm`
Since, `Q_(p)gtK_(p),` the equlibrium shifts towards left.
Discussion
27.

Consider the reaction where K_(p)=0.497at 500 K.PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially 1 atm, which is true ?

Answer»

<P>More `PCl_(5)` will be produced
More `PCl_(3)` will be produced
Equilibrium will be established when 50% REACTION is complete
None of the above

Solution :`PCl_(5)hArr PCl_(3)+Cl_(2)`
`K_(p)=0.497`
`Q=(P_(PCl_(3))xx P_(Cl_(2)))/(P_(PCl_(5)))=(1xx1)/(1)=1`
Since `Q gt K_(p)`, the reaction will PROCEED in backward DIRECTION. Hence, more `PCl_(5)` will be produced.
Discussion
28.

Consider the reaction. The rate constant for two parallel reactions were found to be 10^(-2) dm^(-2)mol^(-1)s^(-1) and 4xx10^(-2)dm^(-3)mol^(-1)s^(-1) . If the corresponding energies of activation of the parallel reaction are 100 and 120 kJ/mol respectively, what is the net energy of activation (E_(a)) of A

Answer»

100 kJ/mol
120 kJ/mol
116 kJ/mol
220 kJ/mol

Answer :C
Discussion
29.

Consider the reaction The correct explanation is

Answer»

The product is formed due to nucleophilic substituion
The product is formed according to Saytzeff's rule
Conjugated double bond product is formed due to higher stability because of resonance stabilization
`(CH_(3))_(3)CO^(-)` is a BETTER LEAVING group

Solution :`(CH_(3))_(3)CO^(-)` is a better BASE than a nucleophile.
Hence elimination occurs. The product formed is resonance stabilised.
Discussion
30.

Consider the reaction sequence B and C respectively are

Answer»

Anisole, isobutylene
Anisole , t-butyl benzene
Phenetole , isobutylene
Phenetole , t-butyl benzene

Solution :GENERALLY, TERTIARY lakyl HALIDE undergoes elimination with a BASE on heating
Discussion
31.

Consider the reaction sequence below:

Answer»




SOLUTION :N//A
Discussion
32.

Consider the reaction: RCHO+NH_(2)NH_(2) to RCH=N-NH_(2) What sort of reaction is it?

Answer»

ELECTROPHILIC addition-elimination REACTION
Free RADICAL addition-elimination reaction
Electrophilic substitution-elimination reaction
NUCLEOPHILIC addition-elimination reaction

Solution :
This reaction is an EXAMPLE of nucleophilic addition-elimination reaction.
Discussion
33.

Consider the reaction : RCHO + NH_(2)NH_(2) rarr RCH = N-NH_(2) What sort of reaction is it ?

Answer»

Electrophilic addition - elimination REACTION.
Free RADICAL addition - elimination reaction
Electrophilic substitution-elimination reaction
NUCLEOPHILIC addition-elimination reaction

ANSWER :D
Discussion
34.

Consider the reaction : Provide a complete mechanism for the formation of the major product.

Answer»

SOLUTION :
Discussion
35.

Consider the reaction , PbO_2 to PbO , DeltaG_(298) lt 0 l SnO_2 to SnO , DeltaG_(298) gt 0 the most probable oxidation state of Pb and Sn will be

Answer»

`PH^(+4) , SN^(+2) `
`Pb^(+4) , Sn^(+4)`
`Pb^(+2), Sn^(+3) `
`Pb^(+2) , Sn^(+4)`

Solution :`PbO_2 to PbO , SnO_2 to SnO`
` UNDERSET(Pb^(+2))(DeltaG_(298) LT 0)"" underset(Sn^(+4))(DeltaG_(298) gt 0`
Discussion
36.

Consider the reaction {:((R)-CH_(3)CHICH_(2)CH_(3)+^(137)T^(-),rarrCH_(3)CHICH_(2)CH_(3),,,),(alpha_(obs)=-15.90^(@),"containing" 2%137T^(-),,,),(,alpha_(obs)=-15.26^(@),,,):} The percentage of racemic from in the product is

Answer»

`10%`
`8%`
`4%`
`12%`

Solution :Note that replacement of `I` with `.^(137)I` does not change he sign or value of the rotation. The optical purity of the recovered iodide is `(15.26//15.90)(100%) = 96%`. HENCE `4%` of the iodide is RACEMIC. This MEANS thet `2%`, (the same percentage that contains isostopic `I)`. has been converted to the `(S)` configuraiton. It is concluded that every REACTION with `.^(137)T^(-)` resulted in an INVERSION at chiral carbon.
Discussion
37.

Consider the reaction: P_(4)(s)+F_(2)(g)rarrPF_(3)(g) How many gram of F_(2) are needed to produce 11.2 L of PF_(3) at NTP?

Answer»

28.5gm
48gm
57gm
85.5gm

Solution :`P_(4)+6F_(2)rarr4PF_(3)`
`11.2` lit of `PF_(3)rArr0.5` mole of `PF_(3)`
mole of `F_(2)` required `=(6)/(4)xx0.5=(3)/(4)` moles
mass of `F_(2)` required `=(3)/(4)xx38=28-5gm`
Discussion
38.

Consider the reaction of with aqueous HI. Give the mechanism and product of the faster reaction

Answer»

Solution :
Several variations of this mechanism are acceptable, as LONG as they MAKE CHEMICAL sense and LEAD to the same diiodide.
Discussion
39.

Consider the reaction of with aqueous HI. Explain your choice of the faster reaction.

Answer»

SOLUTION :A significant difference between these TWO REACTANTS is the presence of a ring strain in the four-membered ring ETHER (called an oxetane). This ring strain is absent in the six-membered ring ether (called a tetrahydropyran). Relief of this ring strain would make the oxetane CLEAVAGE reaction faster than the tetrahydropyran cleavage reaction.
Discussion
40.

Consider the reaction of chloromethane with OH^(-) in aqueous solution CH_(3)Cl(aq)+OH(aq)underset(k_(r))overset(k_(f))hArrCH_(3)OH(aq)+Cl(aq) At 25°C, the rate constant for the forward reaction is 6x× 10^(-6) M^(-1) s^(-1), and the equilibrium constant K_(c) is 1 x× 10^(16). Calculate the rate constant for the reverse reaction at 25°C.

Answer»

SOLUTION :`k_(b)=(k_(k))/(k_(eq))=(6xx10^(-6))/(1XX10^(16))=6xx10^(-22)`
Discussion
41.

Consider the reaction NO_(3^(-))+3F^(2+)4H^(+) to NO+3Fe^(3+)+2H_(2)O [Fe(H_(2)O)_(6)]^(2+)+NO to [Fe(H_(2)O)_(5)NO]^(2+)+H_2O The brown complex is formed. IUPAC name of the complex is

Answer»

Pentaaquanitrosyliron(II)
Pentaaquanitrosyliron(III)
Pentaaquanitrosylferrate(II)
Pentaaquanitrosylferrate(III)

ANSWER :A
Discussion
42.

Consider the reaction, N_(2)(g)+3H_(2)(g) to 2NH_(3)(g) The equility relationship between (d[NH_(3)])/(dt) and -(d[H_(2)])/(dt) is

Answer»

`+(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)`
`+(d[NH_(3)])/(dt)=-(3)/(2)(d[H_(2)])/(dt)`
`(d[NH_(3)])/(dt)=(d[H_(2)])/(dt)`
`+(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`

SOLUTION :If we write rate of REACTION in terms of concentration of `NH_(3) and H_(2)`, then
Rate of reaction`=(1)/(2)(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`
So, `(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)`
Discussion
43.

Consider the reaction nA+nB toProducts, when concentration of both reactant A and B is doubled the rate increases by eight times. However, when concentration of A is doubled keeping the concentration of B fixed the rate is doubled. The overall order of reaction is :

Answer»

`0`
`1`
`2`
`3`

Solution :Rate `(r )=K[A]^(alpha)[B]^(beta)=K[x]^(alpha)[y]^(beta)`
`r=K(2x)^(alpha)(2Y)^(beta)=2^((alpha+beta)).Kx^(alpha)y^(beta)`…….`(i)`
`2r=K(2x)^(alpha)y^(beta)=2^(alpha)Kx^(alpha)y^(beta)`…………`(II)`
`(iii)//(i) ` gives `2=2^(alpha) :.alpha=1`
`(ii)//(iii)` gives `4=2^(beta) :.beta=2`
OVERALL order `=1+2=3`
Discussion
44.

Consider the reaction : N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g) The equality relationship between (d[NH_(3)])/(dt) and -(d[H_(2)])/(dt) is

Answer»

`+(d[NH_(3)])/(DT)=-(2)/(3) (d[H_(2)])/(dt)`
`+(d[NH_(3)])/(dt)= -(3)/(2) (d[H_(2)])/(dt)`
`(d[NH_(3)])/(dt)= -d(H_(2)])/(dt)`
`(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`

SOLUTION :(A) `-(1)/(2) (d[N_(2)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)=(1)/(2) (d[NH_(3)])/(dt)`
or `(d[NH_(3)])/(dt) =-(2)/(3)(d[H_(2)])/(dt)`
Discussion
45.

Consider the reaction :N_(2)+3H_(2)rarr2NH_(3) carried out at constant temperature and pressure. If DeltaH and DeltaU are the enthalpy and internal energy changes for the reactions, which of the following expression is true

Answer»

<P>`DELTAH = 0`
`DeltaH = DeltaU`
`DeltaH LT DeltaU`
`DeltaH gt DeltaU`

Solution :At constant P or T
`DeltaH=DeltaU+DeltanRTimplies Deltan=n_(p)-n_(R)=2-4=-2`
`therefore DeltaHltDeltaU`.
Discussion
46.

Consider the reaction : N_(2)+3H_(2)to2NH_(3) carried out at constant temperature and pressure. If DeltaHandDeltaU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

Answer»

`DeltaH=DeltaU`
`DeltaHgtDeltaU`
`DeltaHltDeltaU`
`DeltaH=0`

ANSWER :D
Discussion
47.

Consider the reaction: N_2(g) + 3H_2(g)to2NH_3(g) carried out at constant temperature and pressure. If Delta H and Delta Eare the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

Answer»

`Delta H = 0 `
`Delta H = Delta E`
`Delta H lt Delta E`
`Delta H GT Delta E`

Solution :`N_2(g) + 3H_2(g)to 2NH_3(g)`
` Delta N(g) = 2 -4= -2, Delta H = Delta E + Delta n(g) RT = Delta E - 2RT`
` therefore Delta H lt Delta E`
Discussion
48.

Consider the reaction M_((aq))^(n+) + "ne"^(-) to M(s) The standard reduction potential values of the elements M_(1) , M_(2) and M_(3) " are " -0.34 V , -3.05 Vand -1.66 V respectively . The order of their reducing power will be

Answer»

`M_(1) GT M_(2) gt M_(3)`
`M_(3) gt M_(2) gt M_(1)`
`M_(1) gt M_(3) gt M_(2)`
`M_(2) gt M_(3) gt M_(1)`.

Solution :LESS the reduction potential , greater is the REDUCING POWER
Discussion
49.

Consider the reaction: (i). (CH_(3))_(2)CH-CH_(2)Broverset(C_(2)H_(5)OH)to(CH_(3))_(2)CH-CH_(2)-O-C_(2)H_(5)+HBr (ii). (CH_(3))_(2)CH-CH_(2)Broverset(C_(2)H_(5)O^(-))to(CH_(3))_(2)CH-CH_(2)-OC_(2)H_(5)+Br^(-) The mechanisms of reaction (i) and (ii) are respectively.

Answer»

`S_(N^(1)) and S_(N^(2))`
`S_(N^(1)) and S_(N^(1))`
`S_(N^(2)) and S_(N^(2))`
`S_(N^(2)) and S_(N^(1))`

Solution :It is the correct answer. In case the reaction is to PROCEED by `S_(N^(1))` mechanism in branched chain ALKYL halides, rearrangement of the carbocation is expected in the slow STEP. Since no such rearrangement is actually noticed in the products in both the REACTIOIN, this means that they both follows `S_(N^(2))` mechanism.
Discussion
50.

Consider the reaction: (i) (CH_(3))_(2)CH-CH_(2)Br overset(C_(2)H_(5)OH)to(CH_(3))_(2)CH-CH_(2)OC_(2)H_(5)+HBr (ii) (CH_(3))_(2)CH-CH_(2)Br overset(C_(2)H_(5)O^(-))to(CH_(3))_(2)CH-CH_(2)OC_(2)H_(5)+Br^(-) The mechanisms of reactions (i) and (ii) are respectively:

Answer»

`S_(N)1 and S_(N)2`
`S_(N)1 and S_(N)1`
`S_(N)2 and S_(N)2`
`S_(N)2 and S_(N)1`

SOLUTION :Since rearrangements do not occur in the given nucleophilic SUBSTITUTION reactions, THEREFORE, carbocations are not the immediates in these reactions. Thus, both the reactions occur by `S_(N)2` mechanism, i.e., option (C) is correct.
Discussion