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.

For the galvanic cell : Pt(s) abs(Fe^(2+), Fe^(2+) ) abs(Fe^(2+))Fecell reaction will be :

Answer»

`3FE^(2+) + FE(s)rarr 2Fe^(3+)`
`3Fe^(2+) rarr Fe (s) + 2 Fe^(3+)`
`Fe(s) + 2Fe^(3+) rarr 3 Fe^(2+)`
None of these

ANSWER :B
2.

For the galvanic cell: Fe(s) | Fe^(3+) (0.02M) || Cu^(2+) (0.1 M)| Cu(s) The E.M.F of cell at 298 K is 0.404 V. The value of DeltaG^(@), (in kJ) for the cell reaction per mole of Fe(S) reacted at 298 K is [(2.303 xx R xx 298)/(96500) = 0.06, 1F = 96500 C] (log 2=0.3)

Answer»


ANSWER :A
3.

For the Galvanic cell, Ag|AgCl(s) |KCl (0.2 M)|| KBr (0.001 M) |AgBr(s) |Ag Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell at 25^(@) C. Given K_(sp)(AgCl) = 2.8 xx 10^(-10), K_(sp)(AgBr) = 3.3 xx 10^(-13)

Answer»

Solution :`AgCl `K_(SP)(AgCl) `therefore [Ag^(+)]_(LHS) = (K_(sp)[AgCl])/[Cl^(-)] = (2.8 xx 10^(-10))/0.2 = 1.4 xx 10^(-9)` M
`AGBR `K_(sp)(AgBr) = [Ag^(+)][Br^(-)]`
`[Ag^(+)]_(LHS) = (K_(sp)(AgBr))/[Br^(-)] = (3.3 xx 10^(-13))/(0.001) = 3.3 xx 10^(-10)` M
`{:("Cell", "Cell Reaction", E^(@)),(LHS,Ag to Ag^(+)+e^(-),E_(Ag//Ag^(+))^(@) =xv),(RHS, Ag^(+) + e^(-) to Ag, E_(Ag^(+)//Ag)=-E_(Ag//Ag^(+))^(@) =-xV):}`
`K= ([Ag^(+)]_(LHS))/([Ag^(+)]_(RHS))`
`E_("cell") = E_("cell"^(@)) -(0.0591)/N log K = 0-(0.0591)/1 log (1.4 xx 10^(-9))/(3.3 xx 10^(-10)) = -0.0371` V
To make `E_("cell")` POSITIVE, LHS cell should be cathode (+ve half cell).
4.

For the galvanic cellAg | AgCl (s), KCl(0.2 M)||KBr(0.001 M), AgBr(s)|Ag,calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at 25^@C.K_(sp)(AgCl) = 1.8 xx 10^(-10), K_(sp)(AgBr) = 3.3 xx10^(-13)

Answer»

SOLUTION :Calculate `[AG^+]` from `K_(sp)`values for both the HALF cells and then calculate
`E_("cell") ` for `Ag|Ag^(+) (c_1) | Ag^(+) (c_2) | Ag `
0.037V
5.

Compare the stability of + 2 oxidation state of the elements of the first transition series.

Answer»

`Mn gt Fe gt CR gt CO`
`Fe gt Mn gt Co gt Cr`
`Co gt Mn gt Fe gt Cr`
`Cr gt Mn gt Co gt Fe`

ANSWER :A
6.

For the four successive transition elements [Cr, Mn, Fe and CO) the stablility of +2 oxidation states will be there in which of the following order ?

Answer»

`Mn GT FE gt Cr gt CO`
`Fe gt Mn gt Co gt Cr`
`Co gt Mn gt Fe gt Cr`
`Cr gt Mn gt Co gt Fe`

ANSWER :A
7.

For the four successive transition elements (Cr, Mn, Fe and Co) ,the stability of +2 oxidation state will be there in which of the following order ?

Answer»

`Cr gt Mn gt Co gt Fe`
`Mn gtFe gt Cr gt Co`
`Fe gt Mn gt Co gt Cr`
`Co gt Mn gt Fegt Cr`

Solution :`Mn^(2+)(d^(5))` GETS highest stabilization due to HALF filled CONFIGURATION .In case of`Fe^(2+)(d^(6))` , additionof ONE extra electron in the subshelldestabilizes it. Addition of two electronsincase of `Co^(2+)(d^(7))` destabilizes itmore . `Cr^(2+)(d^(4))` has one vacant subshell. `Fe^(2+)` gets more stabilizationas comparedto `Cr^(2+)` through exchange energy . So, the correct order is`Mn gt Fe gt Cr gtCo`.
8.

For the formation of SO_3 in the following reaction, it is given that 2SO_2 + O_2 to 2SO_3 "" E_a= Activation energy SO_2 + 1//2O_2 to SO_3 ""E'_a = Activation energy

Answer»

`E_a GT E_a^1`
`E_a LT E_a^1`
`E_a^1 =E_a^(1//2)`
`E_a = E_a^1`

ANSWER :D
9.

For the formation of NH_3 from N_2 and H_2,(triangleE-triangleH) is :

Answer»

RT
-2RT
2RT
RT/2

Answer :C
10.

For the formation of covalent bond, the difference in the value of electronegativities should be

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EQUAL to or LESS than `1.7`
More than `1.7`
`1.7` or more
NONE

ANSWER :A
11.

For the formation of 3.65 g of hydrogen chloride gas, what volumes of hydrogen gas and chlorine gas are required at N.T.P. conditions?

Answer»

1.12 lit, 1.12 lit
1.12 lit, 2.24 lit
3.65 lit, 1.83 lit.
1 lit., 1 lit.

Solution :`{:(H_(2)(G),+,Cl_(2)(g),rarr,2HCl(g)),("1 MOLE",,"1 mole",,"1 mole"),("22.4 L",,"22.4 L",,2xx36.5g):}`
`therefore" For FORMATION of 3.65 g HCl, "H_(2) or Cl_(2)" required"`
`=(22.4)/(2xx36.5)xx3.65L=1.12L`
12.

For the formation of 3.65 g of hydrogen chloride gas, what volumes of hydrogen and chlorine gas are required at N.T.P conditions ?

Answer»

1.12 L, 1.12 L
1.12 L, 2.24 L
3.65 L, 1.83 L
1 L, 1 L.

Solution :`underset(22.4 L at NTP)(H_2(g) ) + underset(22.4 at NTP)(Cl_2(g) ) to underset(2 XX 36.5 g)(2HCL(g) )`
To obtained `2 xx 36.5 g ` of HCL `H_2` REQUIRED at NTP = 22.4 L
To obtained 3.65 g of HCl `H_2` required at NTP
` = (22.4 xx 3.65)/(2 xx 36.5) = 1.12 L`
`Cl_2 ` required= 1.12L (same)
13.

For the formation of 3.65 g of hydrogen chloride gas, what volume of hydrogen and chlorine gas are required to N.T.P. conditions?

Answer»

1.12L, 1.12L
1.12L, 2.24L
3.65 L, 1.83 L
1L, 1L

Solution :`underset("22.4 L at NTP")(H_(2)(g))+underset("22.4 L at NTP")(Cl_(2)(g))to underset(2xx36.5g)(2HCL(g))`
To prepare `2xx36.5 g` of HCL, `H_(2)` and `Cl_(2)` REQUIRED at N.T.P. are 22.4 L each
`:.` To prepare 3.65 g of HCl, `H_(2)` and `Cl_(2)`
required at N.T.P. are `= (22.4)/(2xx36.5)xx3.65`
`=1.12L`
Thus, 1.12 L of `H_(2)` and 1.12 L of `CL_(2)` are required
14.

For the following type of revrsible reaction N_(2) O_(4(g))hArr 2NO_(2(g))

Answer»

EQUILIBRIUM is possible only in a closed SYSTEM at a given TEMPERATURE
Both the opposing processes occur at the same rate and there is dynamic but stable CONDITION
It can be represented as

(X) REPRESENTS point of dynamic equilibrium

Answer :A::B::C
15.

For the foramation of NH_(3) using Haber's process, identify the correct statement.

Answer»

Both yield of the reaction as well as rate of ammonia formation will be more at HIGH temperature.
Yield of the reaction will be more in a LARGE container at lower temperature as compared to SMALL container at high temperature.
While yield of the reaction will increase with increase in temperature rate of formation will decrease with increase in temperature.
For COMMERCIAL production of ammonia, high PRESSURE and high temperature are maintained

Answer :D
16.

For the following three reaction (i), (ii) and (iii), equilibrium constants are given (i) CO_((g))+H_(2)O_((g))iffCO_(2(g))+H_(2(g)),K_(1) (ii) CH_(4(g))+H_(2)O_((g))iffCO_((g))+3H_(2(g)),K_(2) (iii) CH_(4(g))+2H_(2)O_((g))iffCO_(2(g))+4H_(2(g)),K_(3) Which of the following relations is correct?

Answer»

`K_(3)K_(2)^(3)=K_(1)^(2)`
`K_(1)sqrt(K_(2))=K_(3)`
`K_(2)K_(3)=K_(1)`
`K_(3)=K_(1)K_(2)`

Solution :`CO_((g))+H_(2)O_((g))iffCO_(2(g))+H_(2(g))`
`K_(1)=([CO_(2)][H_(2)])/([CO][H_(2)O])""...(i)`
`CH_(4(g))+H_(2)O_((g))iffCO_((g))+3H_(2(g))`
`K_(2)=([CO][H_(2)]^(3))/([CH_(4)][H_(2)O])""...(ii)`
`CH_(4(g))+2H_(2)O_((g))iffCO_(2(g))+4H_(2(g))`
`K_(3)=([CO_(2)][H_(2)]^(4))/([CH_(4)][H_(2)O]^(2))""...(iii)`
From EQUATIONS (i), (ii) and (iii) , `K_(3)=K_(1)xxK_(2)`
17.

For the following three reaction a, b and c, equilibrium constants are given : (i) CO(g) * H_(2)O_(g) hArr CO_(2)(g) * H_(2)(g), K_(1) (ii) CH_(4)(g) * H_(2)O(g) hArr CO(g) * 3H_(2)(g), K_(2) (iii) CH_(4)(g)+2H_(2)O(g) hArr CO_(2)(g)+4H_(2)(g), K_(3)

Answer»

`K_(1)sqrt(K_(2))=K_(3)`
`K_(2)K_(3)=K_(1)`
`K_(3)=K_(1)K_(2)`
`K_(3).K_(2)^(3)=K_(1)^(2)`

Solution :Reaction (C) can be obtained by adding reactions (a) and (B) therefore `K_(3)=K_(1)K_(2)`
Hence (c ) is the CORRECT answer.
18.

For the following reactions: (A). CH_(3)CH_(2)CH_(2)Br+KOHoverset((alc))toCH_(3)CH=CH_(2)+KBr+H_(2)O (B). Which of the following statements is correct?

Answer»

(A) is ELIMINATION, (B) and (C) are substitution reactions.
(A) is substitution, (B) and (C) are addition reactions.
(A) and (B) are elimination reaction. (C) is addition reaction.
(A) is elimination, (B) is substitution and (C) is addition reaction.

Solution :It is the CORRECT ANSWER.
19.

For the followingreactions. .A overset( 600K)to " Product"A underset( " catalyst") overset( 400K) to " Product"It was found that the E_ais decreased by 40kJ / mol in the presence of catalyst. If the rateremains uncharged the activations energyfor catalyzed reactions is [Assume per exponnetial factor is same:]

Answer»

105 kJ/ mol
120 kJ / mol
75 kJ/mol
135 kJ/mol

Answer :B
20.

For the following reactions The correct decreasing order of enthalpy of formation of carbocation is

Answer»

`DeltaH_1 GT DeltaH_2 gt DeltaH_3 gt DeltaH_4`
`DeltaH_4 gt DeltaH_1 gt DeltaH_2 gt DeltaH_3`
`DeltaH_3 gt DeltaH_2 gt DeltaH_1 gt DeltaH_4`
`DeltaH_2 gt DeltaH_1gt DeltaH_4 gt DeltaH_3`

ANSWER :B
21.

For the following reactions A. CH_3CH_2CH_2Br + KOH rarr CH_3CH = CH_2 + KBr + H_2O Which of the following statements is correct?

Answer»

A is elimination, B and C are substitution reactions 
A is substitution, B and C are ADDITION reactions 
A and B are elimination reactions and C is addition reaction 
A is elimination, B is substitution and C is addition reaction 

ANSWER :D
22.

For the following reaction occuring in dilute aqueous solution at 298K. [Ni(H_(2)O)_(6)]^(2+) +2NH_(3) hArr [Ni(NH_(3))_(2)(H_(2)O)_(4)]^(2+)+2H_(2)O ""…….(i) In K_(c)=11.60 and DeltaH^(@)=-33.5 KJ"mol"^(-1) [Ni(H_(2)O)_(6)]^(2+) +en hArr[NI(en)(H_(2)O)_(4)]^(2+) + 2H_(2)O "".......(ii) InK_(c) =17.78and DeltaH^(@)=-37.2 KJ "mol"^(-1) Note : en is ethylenediamine (a neutral bidentate ligand) (R=8.314510 JK-1 "mol"^(-1)=0.0820584 L atm K_(1)"mol"^(-1)) Calculate the value of DeltaG^(@), DeltaS^(@) and DeltaS^(@) . at 298 K for reaction(iii) occuring in a dilute equeous solution: [Ni(NH_(3))_(2) (H_(2)O)_(4)]^(2+)+ en hArr [Ni(en) (H_(2)O)_(4)]^(2+) + 2NH_(3) ""........(iii)

Answer»


SOLUTION :N//A
23.

For the following reaction, the moles ofrequired to produce 990 gm H_2O are __________. 2C_(57)H_(110)O_(6) + 163O_(2)(g) to 114CO_(2)(g) + 110 H_(2)O(l)

Answer»


Solution :`2C_(57)H_(110)O_(6)(s)+1630_(2)(G)rarr114CO_(2)(g)+110H_(2)O(l)`
MOLE of` C_(57)H_(110)O_(6)=(2)/(110)xx "moles of "H_(2)O`
24.

For the following reaction, the mass of water produced from 445 g of C_(57)H_(110)O_(6) is : 2C_(57)H_(110)O_(6)(s) + 163O_(2)(g) to 114CO_(2)(g) + 110 H_(2)O(l)

Answer»

`495 G`
`490 g`
`890 g`
`445 g`

ANSWER :A
25.

For the following reaction : N_(2)O_(5)(g)+O_(2)(g)overset(40%)rarr2NO_(2)(g)+O_(3)(g) NO_(2)(g)+O_(2)(g)overset(50%)rarrNO(g)+O_(3)(g) If initially 20 moles of N_(2)O_(5) and 30 moles of O_(2) are taken then calculate sum of moles of O_(2) and O_(3) after the reaction.

Answer»

16
21
27
30

Solution :`N_(2)O_(5)(g)+O_(2)(g)overset(40%)rarr2NO_(2)(g)+O_(3)(g)`
`NO_(2)(g)+O_(2)(g)overset(50%)rarrNO(g)+O_(3)(g)`
For `1^(st)` reaction: `N_(2)O_(5)(LR) `
`n_(N_(2)O_(5))/(1)xx0.4=n_(O_(3))=n_(O_(3))` FORMED `=n_(NO_(2))` formed/2
mole 16 22
`n_(NO_(2))+0.5=n_(O_(3))` formed `=16xx0.5=8` mole
`n_(O_(2))` used `=n_(NO_(2))` used `=16xx0.5=8` mole
`n_(O_(3))` total form =8+8=16
`n_(O_(2))+n_(O_(3))` after reaction =14+16=30
26.

For the following reaction R - CN overset(H_(3)O^(+)) to R-COOH

Answer»

there is protonation of electronegative NITROGEN
an amide is FORMED as an intermediate
nitrogen atom is expelled as ammonia
all are CORRECT

Answer :D
27.

For the following reaction: Initial concentration: overset(10"mol"//L)(2A)+overset(2"mol"//L)(B)to"product" t_(1//2) of the overall reaction is the time when

Answer»

HALF of A CHANGES to product
half of B changes to product
half of each of A and B changes to product
6 MOLES of A and B changes to product

Answer :B
28.

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. H_(2)O_(2(aq))+3I_((aq))^(-)+2H^(+)rarr 2H_(2)O_((l))+I_(3)^(-)" Rate" = K[H_(2)O_(2)][I^(-)]

Answer»

`"2, L MOL"^(-1)s^(-1)`
`1, s^(-1)`
`(3)/(2), L^(1//2)mol^(-1//2)s^(-1)`
None of these

ANSWER :A
29.

For the following reaction in gaseous phase CO(g)+1/2O_(2) rarr CO_(2) K_(P)/K_(c) is

Answer»

<P>`(RT)^(1//2)`
`(RT)^(-1//2)`
(RT)
`(RT)^(-1)`

Solution :`K_(p)=K_(c)[RT]^(Deltan_(G))`
`Deltan_(g)=1-1.5=-0.5`
`K_(p)=K_(c)[RT]^(1//2)THEREFORE(K_(p))/(K_(c))=[RT]^(-1//2)`
30.

For the following rate law determine the unit of rate constant. Rate k[A]^((1)/(2))[B]^(2)=[R]^((5)/(2))

Answer»

Solution :The total order of reaction `n = 1/2 + 2 = 5/2 = 2.5 `
Rate `K[A]^(1/2) [B]^(2)= [R]^(5/2)`
`therefore k=(Rate)/([5]^((5)/(2)))`
`therefore` UNIT of k=`("unit of rate")/(("unit of concentration")^((5)/(2)))`
`=((mol L^(-1))^(1)s^(-1))/((mol L^(-1))^((5)/(2)))`
`(mol L^(-1))^(-(3)/(2))s^(-1)`
`=(mol)^(-(3)/(2))(L^(-1))^(-(3)/(2))S^(-1)`
`=(mol)^(+(3)/(2))(L)^(-(3)/(2))S^(-1)`
If the order of reaction =`(5)/(2)` then unit of rate CONSTANT k is `L^((+3)/(2)) mol^((-3)/(2))s^(-1)`
31.

For the following question, enter the correct numerical value, (in decimal-notation, truncated/rounded-off to the second decimal place, e.g., 6.25, 7.00, - 0.33, 30.27, - 127.30) using the mouse and the onscreen virtual numeric keypad in the place designated to enter the answer. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl_(2).6H_(2)O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of NiCl_(2).6H_(2)O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is ........... (Atomic weights in g mol^(-1): H = a, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

Answer»


Solution :`{:(UNDERSET((1584)/(132)"mol = 12 mol")((NH_(4))_(2)SO_(4))+Ca(OH)_(3)overset(100%)rarr underset("12 mol")underset("(Gypsum)")(CaSO_(4).2H_(2)O)+2NH_(3)),(underset((952)/(238)"mol = 4 mol")(NiCl_(2).6H_(2)O)+6NH_(3)overset(100%)rarr underset("4 mol")([NI(NH_(3))_(6)]Cl_(2))+6H_(2)O):}`
`{:("Mass of CaSO"_(4).2H_(2)O" (gypsum)"=12xx172g=2064g,"(Molecular mass of "CaSO_(4).2H_(2)=172")"),("Mass of "[Ni(NH_(3))_(6)]Cl_(2)=4xx232g=928g,"(Molecular mass of "[Ni(NH_(3))_(6)]Cl_(2)=232")"),("combined weight (in G) "=2064+928=2992,):}`
32.

For the following Newman projection

Answer»




SOLUTION :
33.

For the following homogeneous gas reaction 4NH_(3)+5O_(2)hArr 4NO+6H_(2)O, the equilibrium constant k_(c) has the dimension of

Answer»

CONC `""^*+(10)`
Conc `""^(+1)`
Conc`""^(-1)`
It is dimensionless

Solution :K has the units of `(conc.)^(DELTAN), where Deltan=10-+1`
34.

For the following : I^(-), Cl^(-), Br^(-), the increasing order of nucleophilicity would be :

Answer»

`CL^(-)ltBr^(-)LTI^(-)`
`I^(-)ltCl^(-)ltBr^(-)`
`Br^(-)ltCl^(-)LTF^(-)`
`I^(-)ltBr^(-)ltCl^(-)`

Answer :A
35.

From the following graph, identify order of reaction and mention the unit of its rate constant.

Answer»

SOLUTION :
It is a FIRST ORDER REACTION and its UNIT of rate constant is `s^(-1)`
36.

For the following gaseous fraction H_(2)+I_(2)hArr2HI, the equlibrium constant

Answer»

<P>`K_(p)gtK_(C)`
`K_(p)ltK_(c)`
`K_(p)=K_(c)`
`K_(p)=1//K_(c)`

Solution :`Deltan=0then K_(p)=K_(c)`
37.

For the following gases equilibrium, N_(2)O_(4) (g)hArr2N_(2) (g), K_(p) is found to be equal to K_(c). This is attained when:

Answer»

`0^(@)C`
273 K
1 K
12.19 K

Answer :D
38.

For the following flow diagram : Which of the following option describes the reagents, products and the reaction conditions given in parentheses as small alphabets ?

Answer»


SOLUTION :
39.

For the following equilibrium reaction , N_2O_4(g) hArr 2NO_2(g) NO_2 is 50% of total volume at given temperature . Hence vapour density of the equilibrium mixture is

Answer»

34.5
25
`23.0`
20

Answer :A
40.

For the following equilibrium in a closed rigid vessel A(g)hArrB(g)+C(g) D(g)hArrE(g)+B(g) If some E(g) is introduced into the vessel, then at the new equilibrium.

Answer»

[A] increaes
[C] DECREASES
[A] decreases
[B] increases

Solution :THEORY BASED
41.

For the following electrochemical cell at 298 K Pt(s)|H_(2),(g,1" bar")|H^(+)(aq,1M)||M^(4+)(aq)|M^(2+)(aq)|Pt(s) E_(cell)=0.092V when ([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x) Given E_(M^(4+)//M^(2+))^(@)=0.151V,2.303(RT)/(F)=0.059V the value of x is

Answer»

`-2`
`-1`
`1`
`2`

SOLUTION :The reactions OCCURRING in the cell are
At anode: `H_(2)(g)to2H^(+)(aq)+2E^(-)`
`underline("At cathode:"M^(4+)(aq)+2e^(-)TOM^(2+)(aq))`
Overall reaction: `H_(2)(g)+M^(4+)(aq)toM^(2+)(aq)+2H^(+)(aq)`
`E_(cell)=E_(cell)^(@)-(0.059)/(2)"LOG"([M^(2+)][H^(+)]^(2))/([M^(4+)])`
`E_(cell)^(@)-E_(M^(4+)//M^(2+))^(@)-E_(H^(+)//H_(2))^(@)`
`=0.151-0=0.151V`
`therefore0.092=0.151-(0.059)/(2)log(10^(x)xx1^(2))`
or `0.092=0.151-0.0295log10^(x)`
or `0.0295log10^(x)=0.151-0.092=0.059`
or `log10^(x)=(0.059)/(0.0295)=2`
`therefore10^(x)`=Antilog `2=10^(2)thereforex=2`
42.

For the following elementary step (CH_(3))_(3)CHr_((aq))to(CH_(3))C_((aq))^(+)+Br_((aq))^(-) the molecularity is

Answer»

ZERO
`1`
`2`
fractional

Answer :B
43.

For the following elecrochemical cell at 298K Pt(s)|H_(2)(g,1"bar")|H^(+)(aq,1M)||M^(4+)(aq),M^(2+)(aq)|Pt(s) E_(cell)=0.092V when ([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x) Given: E_(M^(4+)//M^(2+))^(0)=0.151V,2.303(RT)/(F)=0.059V the value of x is

Answer»

`-2`
`-1`
`1`
`2`

Solution :ANODE: `H_(2)-2ehArr2H^(+)`
`underline("Cathode:"M^(4+)+2ehArrM^(2+))`
Net CELL reaction: `H_(2)+M^(4+)hArr2H^(+)+M^(2+)`
`E_(cell)=E_(cell)^(0)-(0.059)/(2)"log"([H^(+)]^(2)[M^(2+)])/([M^(4+)]xxP_(H_(2)))`
`0.092=0.151-(0.059)/(2)log10^(X)`
`0.092=0.151-(0.059)/(2)x`
`(0.059x)/(2)=0.151-0.092`
`0.059x=0.059xx2impliesx=2`.
44.

For the following data answer the questions : Reaction : A+B to P The value of rate constant at 300 K is (M^(-2) sec^(-1)):

Answer»

`2.667xx10^(8)`
`2.667xx10^(5)`
`2.667xx10^(6)`
`2.667xx10^(9)`

SOLUTION :`k_(300)=(5xx10^(-4))/((2.5xx10^(-4))^2(3XX10^(-5)))=2.677xx10^8`
45.

For the following disproportionation reaction 5Br_(2)+6PH^(-) to 5Br^(-)+BrO_(3)^(-)+3H_(2)O correct statements are :

Answer»

EQUIVALENT weigt of `Br_(2)` when it is reduced to `Br^(-)` is 80
equivalent weigt of `Br_(2)` when it is oxidized to `Br^(-)` is 80
equivalent weigt`Br_(2)` in the net REACTION is 96
equivalent weigt of `Br_(2)` is 80

Answer :A::B::C::D
46.

For the following data answer the questions : Reaction : A+B to P The order w.r.t A is :

Answer»


Solution :`R=K[A]^m[B]_n`
`5xx10^(-4)=k[2.5xx10^(-4)]^m[3xx10^(-5)]^n`…(i)
`4XX10^(-3)=k[5xx10^(-4)]^m[6xx10^(-5)]^n`…(ii)
`1.6xx10^(-2)=k[1XX10^(-3)]^m[3xx10^(-4)]^n`…(iii)
From it m=2
47.

For the following conversion which can be usedR_(2) C =O to R_(2) CH_(2)

Answer»

Clemmenson's REACTION
Wolff Kishner reaction
`ZN + CONC. HCL`
both 'a' and 'b'

Answer :D
48.

For the following data answer the questions : Reaction : A+B to P The energy of activation for reaction (KJ/mol) is :

Answer»

20.83
13.83
15.23
10.23

Solution :`k=A.e^(-Ea//RT)`
`lnk_2/k_1=E_a/R[1/300-1/400]`
`E_a=13.83`
49.

For the following concentration cell, to be spontancous Pt(H_2)P_1atm. |HCl |Pt(H_2) P_2atm which of the following is correct ?

Answer»

`P_1 = P_2 `
`P_1 lt P_2 `
`P_1 gt P_2 `
can't be predicted.

Solution :PT `(H_(2)) P_(1)` atm `|P(H_2)P_2`atm for SPONTANEOUS reaction. `E_eth` should be POSITIVE so `P_1 gt p_2`
`E_("cell")=0.059/2 log (p_1)/(P_2)`
Also if ` p_1gt p_2 ` OXIDATION of R.H.S electrode.
50.

For the following cell, Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)|Cu(s) When the concentration of Zn^(2+) is 10 times the concentration of Cu^(2+), the expression for DeltaG ( in Jmol^(-1)) is [F is faraday constant: R is gas constant, T is temperature, E^(@)(cell)=1.1V]

Answer»

2.303RT+1.1F
1.1F
2.303RT-2.2F
`-2.2F`

Solution :`DeltaG=DeltaG^(o)+2.303RTlog_(10)Q,Q=([Zn^(2+)])/([Cu^(2+)])`
`=-2F(1.1)+2.303RTlog_(10)10`
`=2.303RT-2.2F`
GIVEN
`E_(Cl_(2)//Cl^(-))^(o)=1.36V,E_(Cr^(3+)//Cr)^(o)=-0.74V`
`E_(Cr_(2)O_(7)^(2-)//Cr_(3+))^(o)=1.33V,E_(MnO_(4)^(-))^(o)//Mn^(2+)=1.51V`