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 a hypothetical reactionA + B toproducts The rate law is , R = R[A]^(@) [B] , The order of reaction is

Answer»

1
2
1.5
Zero

ANSWER :a
2.

For a hypothetical reaction, A+3BrarrP""DeltaH=-ex Kj/ "mole" of A&Mrarr2Q+R"" DeltaH=+xx kJ/ "mole" of M These reactions are carried simultaneously in a reactor such that temperature is not changing If rate of disppearance of B isy sec^(-1) then rate of formation (in Msec^(-1)) of Q is :

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`(2)/(3)y`
`(3)/(2)y`
`(4)/(3)y`
`(3)/(4)y`

ANSWER :C
3.

For a hypothetic reaction ArarrB, the activation energies for forward and backward reactions are 19 kJ/mole and 9 kJ/mole respectively. The heat of reaction is

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28 KJ
19 kJ
10 kJ
9 kJ

Solution :`DeltaH=E_(a)` for FORWARD REACTION `-E_(a)` for backward reaction = 19-9=10 kJ.
4.

For a hypotherical elementary reaction where k_1/k_2=1/2 Initailly only 2 moles of A are present. The total number of moles A,B and C at the end of 50% reaction are :

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2
3
5
None

Solution :When 1 mole of A is PRESENT then 2 mole of B and Cwill forms after completion of the reaction but when 50% is COMPLETE then the TOTAL mole =1+2x1=3
5.

For a hypotherical elementary reaction where k_1/k_2=1/2 Initailly only 2 moles of A are present. Number of moles of B are:

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2
1
0.666
0.333

Solution :MOLE of B `=2/(1+2)=0.666`
6.

For a hypotherical elementary reaction where k_1/k_2=1/2 Initailly only 2 moles of A are present. The sum of mole of (B) are ( C) is :

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2
3
1
4

Solution :NA
7.

For a hcp lattice, the edge length is equal to

Answer»

`2sqrt(2)r `
`2R`
`(sqrt3)/(4) r`
`4/(sqrt3) r`

ANSWER :B
8.

For a H_2 like gas (at 298 K) having T_B=108 k, select the only correct option-

Answer»

(III)(iii)(P)<BR>(IV)(iii)(Q)
(III)(ii)(Q)
(III)(i)(S)

Solution :for `H_2` the LIKE gas, `a=0` `T_B=108 k=a/(bR)`
`implies P(V_m-b)=RT implies T_C=8/24(a/(bR))=8/27xx108=32 k`
`implies Z=1+(Pb)/(RT)`
`because Z gt 1 implies` Repulsive tendencies will dominate.
9.

For a gives compound There are 3 benzenoid isomer of 'X' P, Q and R for which following observation are made : (a) P is a monosubstituted benzene derivative which can give observation are made : (b) Q give position iodoform test. (c) R gives silver mirror with Tollen's reagent. Numberof positive Q :

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1
2
3
4

Solution :
10.

For a gives compound There are 3 benzenoid isomer of 'X' P, Q and R for which following observation are made : (a) P is a monosubstituted benzene derivative which can give observation are made : (b) Q give position iodoform test. (c) R gives silver mirror with Tollen's reagent. Number of possible P :

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1
2
3
4

Solution :
11.

For a given solution pH=6.9 at 60^(@)C where K_W=10^(-12). The solution is :-

Answer»

ACIDIC
ALKALINE
NEUTRAL
Unpredictable

Answer :B
12.

For a given sample of ideal gas

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<P>`V PROP ( T )/( p )`
`V prop PT`
`V prop ( p )/( T ) `
`V prop ( T )/(p)`

ANSWER :D
13.

For a given redox change, E_(RP_2)^@ +E_(OP_1)^@ is equal to ……… where 1 is oxidised and 2 is reduced :

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Coulomb
Faraday
Ampere
Cell potential

Answer :D
14.

For a given reaction rate = K (A)^1 (B)^(2//3), the unit of rate constant K can be given as

Answer»


ANSWER :A
15.

For a given reaction .of first order, it takes 20 minute for the concentration to drop from 1.0 M litre^-1 to 0.6 M litre^-1time required for the concentration to drop from 0.6 M litre^-1to 0.36 M litre^-1will be:

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More than 20 minute
Less than 20 minute
Equal to 20 minute
Infinity

Answer :C
16.

For a given reaction of first order, it takes 20 min, for the concentration to drop from 1 ML^(-1) to 0.6 ML^(-1). The time required for the momentum to drop from 0.6ML^(-1) to 0.36ML^(-1) will be

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` gt 20` MIN
` LT 20` min
` = 20` min
`OO`

ANSWER :C
17.

For a given reaction of first order, it takes 20 min for the concentration to drop from 1.0M to 0.6. The time required for the concentration to drop from 0.6M to 0.36M will be

Answer»

more than 20 minutes
less than 20 minutes
equal to 20 minute
infinity

Answer :C
18.

For a given reaction half life period was found to be directly proportional to the initial concentration of the reactant. The order is

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zero
1
2
3

Solution :For a reaction `t_(1//2) PROP (a)^(1-N)` where n is ORDER .
If `t_(1//2) prop a ` then `1 - n = 1` or n = 0
19.

For a given reaction half life period was found to be directly proportional to the initial concentration of the reactant. The order is:

Answer»

Zero
1
2
3

Answer :A
20.

For a given reaction, DeltaH=35.5 kJ mol^(-1) and DeltaS=83.6 JK^(-1) mol^(-1). The reaction is spontaneous at : (Assume that DeltaH and DeltaS do not very with temperature)

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`T gt 425 K`
All TEMPERATURES
`T gt 298 K`
`T LT 425 K`

SOLUTION :`DeltaG=DeltaH-TDeltaS`
for equilibrium `DeltaG=0`
`DeltaH-TDeltaS`
`T_(eq).=(DeltaH)/(DELTAS)=(35.5xx1000)/(83.6)=425 K`
Since the reaction is endothermic it will be spontaneous at `T gt 425 K`.
21.

For a given reaction, Ararr"Product, rate is" 1xx10^(-4)"M s^(-1) when [A] = 0.01 M and rate is 1.41xx10^(-4)"M s"^(-1) when[A] = 0.02 M. Hence, rate law is :

Answer»

`-(d[A])/(DT)=K[A]^(2)`
`-(d[A])/(dt)=k[A]`
`-(d[A])/(dt)=(k)/(4)[A]`
`-(d[A])/(dt)=k[A]^(1//2)`

ANSWER :D
22.

For a given reaction 3 A + B to C + D the rate of reaction can be represented by

Answer»

`-(1)/(3) (d[A])/(DT) = (-d[B])/(dt) = (+d[C])/(dt) = (+d[D])/(dt)`
`-(1)/(3) (d[A])/(dt) = (d[C])/(dt) = K[A]^(m) [ B]^(n)`
`+(1)/(3) (d[A])/(dt) = (-d[C])/(dt) = K[A]^(n) [ B]^(m)`
NONE of these

Solution :`-(1)/(3) (d[A])/(dt) = - (d[B])/(dt) = (+d[C])/(dt) = (+d(D))/(dt)` .
23.

For a given reaction ArarrB , then time required for 75% disappearance ofAis twice that required for 50% disappearance of A . The order of the reaction with respect to A is -

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0
1
2
3

Answer :B
24.

For a given one mole of ideal gas kept at 6.5 atm ina container of capacity 2.463 litre. The Avogadro proportionality constant for the hypothesis is (see figure)

Answer»

0.406
2.46
22.4
None of the above

Answer :A
25.

For a given period, the most non metallic p-block element belongs to group:

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1. 13
2. 17
3. 15
4. 14

Answer :B
26.

For a given mass of gas, if pressure is reduced to half and temperature is increased. two times, then the volume would become

Answer»

`(V)/(4)`
`2V^(2)`
6V
4V

Answer :D
27.

For a given mass of a gas , if pressure is reduced to half and its temperature is doubled, then volume V will become :

Answer»

4V
`2V^(2)`
`V//4`
8V

Solution :`p_(1) = p , p_(2) = p //2, T_(1) = T, T_(2) = 2T`,
`V _(1) = V, V_(2) = ? `
`( p_(1) V_(1))/( T_(1)) = ( p_(2)V_(2))/(T_(2))`
or `V_(2) = ( p _(1) V_(1) T_(2))/(p_(2) T_(1))`
`= ( p XX V xx 2T)/( p //2 xx T ) = 4V`
28.

For a given mass of a gas, if pressure increases :

Answer»

VOLUME and TEMPERATURE remain constant
Volume decreases provided temperature remains constant.
Temperature INCREASES provided volume remains constant
Temperature decreases provided volume remains constant.

Answer :C
29.

For a given mass of a gas at constant temperature. If the volume V becomes four times, the pressure p will become :

Answer»

4p
`p//4`
2p
`4p //T`

SOLUTION :PV = constant
If volume becomes 4 times, pressure will BECOME `p //4` so that pV is constant
30.

For a given halogen atom, the reactivity is maximum for :

Answer»

METHYL HALIDE
PRIMARY halide
secondary halide
tertiary halide

Answer :D
31.

For a given exothermic reaction , K_(p) and k'_(p) are the equilibrium constants at temperatures T_(1) and T_(2) respectively. Assuming that heat of reaction is constant in temperaturerange between T_(1) and T_(2) , it is readily observed that

Answer»

`K_(p)=K'_(p)`
`K_(p)=(1)/(K'_(p))`
`K_(p)gtK'_(p)`
`K_(p)LTK'_(p)`

Solution :Assuming `T_(2)gtT_(1).`
32.

For a given complex [Co(NH_3)_5 NO_2 ] Cl_2. Write its IUPAC name and Linkage isomer.

Answer»

SOLUTION :pentaamminenitrito -N- COBALT(III) chloride.
`[CO(NH_3)_5 (ONO)]Cl_2`
33.

For a given cell reaction , Cr +3H_2O +OCI^- rarrCr^(3+)+3CI^- 6OH^-, the species undergoing reduction is :

Answer»

`CR`
`Cr^(6+)`
`OCI^-`
`CI^-`

Answer :C
34.

For a general substance A the phase diagram is represented as shown. Identify the option (s) which is/are correct.

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Triple point of the SUBSTANCE is 200 K and 3.8 MM of Hg.
Standard boiling point of the substances should be slightly less than 400 K
Above 500 K, gas-liquid transition cannot occur WITHOUT CHANGE in temperature
Melting point of the substance will increase with increase in pressure.

Answer :A::B::C
35.

For a given alkyl group, the boiling point are in the order:

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`RI lt RBT lt RCI`
`RI lt Rd lt RBR`
`RBr lt RT lt RQ`
`RG lt RBr lt RI`

ANSWER :D
36.

For a given alcohol the order of reactivity with halogen acid is

Answer»

HI `GT` HCL `gt` HBR
HCl`gt`HBr`gt`HI
HCl`gt`HI`gt`HBr
HI`gt`HBr`gt`HCl

Answer :D
37.

For a general reaction AtoB. plot of concentrating of A vs time is given in fig. Answer the following questions on the basis of this graph. a) What is the order of the reaction? b) What is the slope of the curve? c) What are the units of rate constant?

Answer»

SOLUTION :a) Reaction is of zero order.
b) Slope of curve = `-k`
C) UNITS of RATE constant = mol `L^(-1)s^(-1)`
38.

For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-)] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals What is the molar solubility of Cu(OH)_(2), in 1.0 M NH_(3) if the deep blue complex ion [Cu(NH_(3))_(4)]^(2+) is formed. The K_(sp), of Cu(OH)_(2), is 1.6xx 10^(-19) and K_(3), of[Cu(NH_(3))_(4) is 1.1 xx 10^(13)

Answer»

`7.1xx10^(-4)`M
`7.1xx10^(-4)`M
`7.6xx10^(-3)`M
`5.6xx10^(-4)`M

Answer :B
39.

For a general nth order reaction A to P with initial concentration of the reactant 'a' and rate constant 'k', derive expression for time for 75% completion of the reaction in terms of a, n and k.

Answer»

Solution :For REACTIONS of 2nd order, `k=(1)/(t)[(1)/(C_(t))-(1)/(C_(0))]`
For reactions of 3RD order, `k=(1)/(2t)[(1)/(C_(t)^(2))-(1)/(C_(0)^(2))]`
For reaction of nth order, `t=(1)/((n-1)k)[(1)/(C_(t)^(n-1))-(1)/(C_(0)^(n-1))]`
When `75%` of the reaction is COMPLETE,
`C_(t)=25%" or "C_(0)=(1)/(4)C_(0)`
REPLACING `C_(0)` by a,
`t_(75%)=(1)/((n-1)k)[(1)/((a//4)^(n-1))-(1)/(a^(n-1))]=(1)/((n-1)k)[(4^(n-1))/(a^(n-1))-(1)/(a^(n-1))]=(1)/((n-1)k)[(2^(2n-2)-1)/(a^(n-1))]`
40.

For a general chemical change 2A+3B rarrProducts, the rates with respect to A is r_1 and that with respect toB is r_2 The rates r_1 and r_2are related as

Answer»

`3r_1=2r_2`
`r_1=r_2`
`2r_1=3r_2`
`r_1^2= 3r_2`

Solution :For the reaction `2A + 3B to` PRODUCTS ,
`-1/2. (d[A])/(dt)=+1/3.(d[B])/(dt)`
GIVEN: `(d[A])/(dt)=r_1, (d[B])/(dt)=r_2`
or `1/2r_1=1/3r_2or 3r_1=2r_2`
41.

For a gaseous reaction the unit of rate of reaction is

Answer»

L ATM s 1
atm MOL 's 1
T-S tup
mol s

SOLUTION :(iv) (C) atm s
42.

For a gaseous reaction the unit of rate of reactions is

Answer»

L ATM `s^(-1)`
atm `"MOL"^(-1) s^(-1)`
atm `s^(-1)`
mol s

Answer :C
43.

For a general gaseous reaction of the type R to P , if the initial concentration of R is doubled, half life of the reaction is also doubled, the order of that reaction is

Answer»

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

ANSWER :A
44.

For a gaseous reaction, the rate expression is k[A][B]. If the volume of the reaction vessel is reduced to 1//4^(th) of the initial volume, the reaction rate, relating to original rate will be... . .. times

Answer»

10
8
`1//10`
16

Answer :D
45.

for a gaseous reaction, the unit of rate of reaction are

Answer»

L atm `s^(-1)`
atm `s^(-1)`
atm `mol^(-1)s^(-1)`
`mols^(-1)`

SOLUTION :UNITS of RATE for GASEOUS reaction are atm `s^(-1)`
46.

For a gaseous reaction the rate =K[A][B]. The volume of the vessel containing the gas is suddenly reduced to 1/4 th of the initial volume. The rate of reaction relative to initial rate would be

Answer»

`1/16`
16
8
same

Answer :B
47.

For a gaseous reaction at 300K, DeltaH - DeltaU = - 4.98 kJ assuming that R=8.3 JK^(-1) mol^(-1), Deltan_((g)) is

Answer»

1
2
`-2`
0

Solution :`DeltaH=DeltaE+DeltanRT`
`DeltanRT=DeltaH-DeltaE`
`Deltanxx8.3xx300=-4.98xx10^(3)`
`Deltan=-2`.
48.

For a gaseous reaction, A(g)+3B(g)rarr3C(g)+3D(g) triangleU is 17 kcal at 27^@C. Assuming R=2cal K^-1 mol^-1, the value of triangleH for the above reaction will be:

Answer»

15.8 kcal
16.4 kcal
18.2 kcal
20.0 kcal

Answer :C
49.

For a gaseous reaction : 2A(g)rarrB(g)+2C(g), the pressure changes from 10 atm to 15 atm in 10 min. Order of the reaction may be :

Answer»

1
`1.2`
`1.5`
`0.75`

ANSWER :D
50.

For a gaseous reaction, 2A(g) + B(g) to C(g) , DeltaH =- 30 "Kcal"//"mole at"300K If the reaction follows the rate law (d[C])/(dt) =4xx10^(-3)[A]^(2) [B] M //min and initially concentration ofB "is" [B] = 10^(-3) M and concentration of A is [A] =1 M, then calculate the rate at which heat will be liberatedper minute initially if reaction occurs in a rigid container of volume 10 litres. [Express answer in 10^(-1) "cal/min"] ["Use": R =2 "cal"// "mol" K ]

Answer»


ANSWER :12