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.

In a reversible reaction, the catalyst

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increases the activation energy of the backward REACTION
increases the activation energy of the FORWARD reaction
DECREASES the activation energy of both, forward and backward reaction
decreases the activation energy of forward reaction.

Solution :Decreases the activation energy of both,
forward and backward reaction.
2.

In a reversible reaction if there is no change in total number of molecules , the reaction will be favoured by

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HIGH PRESSURE
LOW pressure
high temperature
higher CONCENTRATION of a reactant

Answer :D
3.

In a reversible reaction a catalyst will affect

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the RATE of forward reaction and REVERSE reaction
neither the forward reaction nor the rate of reverse reaction
the rate of forward reaction
the rate of reverse reaction.

Solution :In a reversible reaction a catalyst does not ALTER the position of EQUILIBRIUM since it effects the rate of forward reaction and reverse reaction to the same EXTENT.
4.

In a reversible reaction, a catalyst :

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INCREASES the rate of the forward REACTION only
Increases the rate of the forward reaction to a greater extent than that of the backward reaction
Increases the rate of the forward reaction and decreases that of the backward reaction extent.
Increases the rate of the forward and backward reaction equally

Answer :D
5.

In a reversible process, DeltaS_(sys)+DeltaS_("surr") is

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`gt0`
`lt0`
`ge0`
`=0`

Solution :Entropy of the system and SURROUNDINGS remains constant in a reeversible EQILIBRIUM process.
`DeltaS_("TOTAL")=DeltaS_("SYS")+DeltaS_("surroundings")=0`
6.

In a reversible chemical reaction having two reactants in equilibrium, if the concentrations of reactants are doubled, then the equilibrium constant will :

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also be doubled
be halved
become ONE - FOURTH
remains the same

ANSWER :D
7.

In a reversible isothermal process the change in internal energy is:

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Zero
Positive
Negative
None

Answer :A
8.

In a reversible adiabatic change Delta Sis

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equal to `NR LN V_2//V_1`
0
equal to `C_V DT`
infinity

Solution : For adiabatic changes q = 0, therefore `DELTA S = 0` as`Delta S`=q/T.
9.

In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is doubled, then the equilibrium constant will

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also be DOUBLED
be halved
remains the same
becomes one-fourth

Solution :EQUILIBRIUM CONSTANT depends only on TEMPERATURE.
10.

In a relation, catalyst changes………..

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PHYSICALLY
qualitatively
chemically
quantitatively

ANSWER :A::B
11.

In a redox reaction Na_(2)S_(2)O_(3) changes to Na_(2)S_(4)O_(6) what is the equivalent weight of Na_(2)S_(2)O_(3)?

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same as its molecularweight
`1//2xx`(its MOLECULAR weight)
`1//2xx`(its molecular weight)
`1//3xx`(its molecular weight )

Solution :In redox REACTION
12.

In a redox reaction, how many moles of electrons are accepted by 1 mol of MnO_(4)^(-) in acidic medium ?

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3
5
43953
6

Solution :`MnO_(4)^(-) + 8H^(+) + 5E^(-) to Mn^(2+) + 4H_(2)O`
13.

In a reaction,the concentration of reactant is increased two times and three times then the increase in rate of reaction were four times and nine times respectively,order of reaction is

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ZERO
1
2
3

Answer :C
14.

In a reaction vessel, the reactants have attained a potential energy of 80K. cal/mole and unable to give even traces of products. The threshold energy of the reaction may be

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60K.cal
70 K.cal
40 K.cal
100K.cal

Answer :D
15.

In a reaction to B, the rate of reaction increases two times on increasing the concentration of the reactant four times, then order of reaction is

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0
2
`1//2`
4

Solution :( c ) : (i) `R="k a"^(ALPHA)` (II) `2r=k(4" a")^(alpha)" Dividing (ii) by (i)," 4^(alpha)=2" or "2^(2alpha)=2" or "2alpha=1" or "alpha=1//2.`
16.

In a reaction vessel, 0.184 g of NaOH is required to be added for completing the reaction. How many millilitres of 0.150 M NaOH solution should be added for this requirement?

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Solution :1000 ML of 0.150 M NaOH CONTAIN NaOH = 0.150 MOL `=0.150xx40=6g`
i.e., For 6 g in 0.150 M M NaOH solution, VOLUME of solution `-=" 1000 mL"`
`therefore"For 0.184 g in 0.150 M NaOH solution, volume "-=(1000)/(6)xx0.184=30.7mL`
17.

In a reaction, triangleH and triangleS both are more than zero. In which of the following cases, the reaction would not be spontaneous:

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`triangleHgtTtriangleS`
`triangleS=triangleH/T`
`triangleH=TtriangleS`
All

Answer :D
18.

In a reaction the threshold energy is equal to

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AVERAGE ENERGY of the reactants
activation energy
activation energy `+` average energy of the reactants
activation energy `-` average energy of the reactants

Answer :B
19.

In a reaction , the time required to complete half of the reaction was found to increase 16 times when the initial concentration of the reactant was reduced to 1//4^(th). What is the order of the reaction ?

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Solution :`3`
`(t_(2))/(t_(1))=((a_(1))/(a_(2)))^(n-1)`
`implies16=((1)/((1)/(4)))^(n-1)`
`n=3`
THIRD order REACTION.
20.

In a reaction, threshold energy is equal to

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activation energy
normal energy of the REACTANTS
activation energy + energy or reactants
activation energy - energy of reactants

ANSWER :C
21.

In a reaction the rate of reaction is proporational to its active mass, this statement is known as

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Law of mass action
Le-chatelier principle
FARADAY law of electrolysis
LOW of constant proportion

ANSWER :A
22.

In a reaction, the rate expression is Rate =k[A][B]^(2//3), the order of reaction is

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1
2
`5//3`
zero

Solution :Order of REACTION is SUM of power raised on concentration terms in order to write RATE expression.
Order of reaction `= 1+2/3=5/3`
23.

In a reaction, the threshold energy is equal to: .

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ACTIVATION ENERGY + NORMAL energy of reactants
Activation energy- normal energy of reactants
Activation energy
Normal energy of reactants

Answer :A
24.

In a reaction, the rate expression is, rate = k[A][B]^(2//3)[C]^0,the order of reaction is:

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1
2
`5//3`
Zero

Answer :C
25.

In a reaction the initial concentration of the reactants increase four fold and the rate becomes eight times its initial value . The order of the reaction is

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`2.0`
`3.5`
`2.5`
`1.5`

Solution :Rate = `KC^(n)`
Or r = `KC^(n) ""… (i)`
`8R = K(4C)^(n) "" … (ii)`
Dividing EQ. (ii) by (i) , we GET
`2^(3) = 2^(2N)`
or 2 n =3
n = 1.5.
26.

In a reaction the following equation holds good as per chemical kinetics is viewed. logk=4-4000/(2.303RT). The frequency factor is 10^(x). What is x.

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SOLUTION :`logK=logA-(E_(a))/(2.303RT),logA=4,A=10^(4)`
27.

In a reaction, the concentration of reactant is increased two times and three times than the increases in rate of reaction were four times and nine times respectively, order of reaction is

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ZERO
1
2
3

Solution :R = `k (A)^(2) , R' = k(2A)^(2), THEREFORE (R')/(R) = 4`
` R = k (A)^(2) , R' = k (3A)^(2) , (R')/(R) = 9`.
28.

In a reaction, RCHO is reduced to RCHO_(3) using amalgamated zinc and concentrated HCl and warming the solution. The reaction is known as

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Meerwein-Ponndorf VERLEY formate
Clemmensen reduction
Wolff-Kishner reduction
Schiff's reaction

Answer :B
29.

In a reaction of the typeA(s) + 2B(g) iff C(g) + D(g)the equilibrium concentrations of A, B, C and D are 1, 2, 5 and 6 mol/litre respectively. Argon is then introduced at equilibrium at constant volume. Calculate the concentrations of A, B, C and D at the new equilibrium position.

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SOLUTION :Eqb. concs, do not CHANGE
30.

In a reaction K_(2)MnO_(4) is converted into KMnO_(4)The change in the oxidation number of Mn is :

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zero
`+1`
`-1`
`+7 `

Solution :`K_(2)MnO_(4) (+6) to KMnO_(4) (+7)`
Changes in OXIDATION number= +1
31.

In a reaction N_(2)+3H_(2)to2NH_(3), the rate of appearance of ammonia is 2.5xx10^(-4)" M sec"^(-1) the rate of disappearance of N_(2) will be

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`7.5XX10^(-4)`
`1.25xx10^(-4)`
`5xx10^(-4)`
`2.5xx10^(-4)`

ANSWER :B
32.

In a reaction if the concentration of reactant A is tripled,the rate of reaction becomes twenty seven times.What is the order of the reaction?

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Solution :Reaction ORDER=3
For any reaction `RtoP` and order =n so rate
`r_(1)k[R]^(n)`
If the concentration becomes 3 times i.e. 3R then rate ,`r_(2)k[]3R^(n)`
So ,`(r_(2))/(r_(1))=(k[3R]^(n))/(k[R]^(n))=(27)/(1)`
`THEREFORE (""^(27)r_(1))/(r_(1))=27`
`therefore 27=3^(n)`
`therefore 3=3^(n)`
So, n=3
33.

In a reaction, if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction ?

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Solution :`r_(1)=k[A]^(ALPHA)=ka^(alpha),r_(2)=k(3" a")^(alpha),(r_(2))/(r_(1))=3^(alpha)=27("GIVEN")=3^(3):.alpha=3.`
34.

In a reaction, if Delta Ult 0, then heat __________ .

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CHANGE is zero
is absorbed
is evolved
change is abnormal

Answer :C
35.

In a reaction, if half of the double bond is broken and two new bonds are formed, this is a case of

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Elimination
Addition
Displacement
Rearrangement

Solution :e.g. `CH_2=CH_2+Br_2 to undersetunderset(Br)(|)CH_2-undersetunderset(Br)(|)CH_2`
HALF of the DOUBLE bond is broken. Its MEANS `PI` bond is broken while sigma bond is retained also two new C-Br bonds are formed
36.

In a reaction if half of the double bond is broken and two new bonds are formed, this is a case of:

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Elimination
Addition
Displacement
Rearrangement

Answer :A
37.

In a reaction , DeltaH and DeltaS both are more than zeron. In which of the following cases would the reaction be spontaneous ?

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`DeltaHgtTDeltaS`
`TDeltaSgtDeltaH`
`DeltaH=TDeltaS`
NONE of these

Answer :B
38.

In a reaction, Cu_2 S is oxidized by reacting with oxygen to given Cu^(2+) and SO_2. If the equivalents of O_2 used were x, the equivalent of Cu^(2+) and SO_2. Produced w.r.t. Cu_2 S would be:

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X and x
x and 0.25 x
0.25 x and 0.75 x
0.75 x and 0.25 x

Answer :C
39.

In a reaction CH_2=CH_2underset"acid"overset"Hypochlorous"to MoversetRto underset(CH_2-OH)overset(CH_2-OH)| Where M=molecule, R=reagent , M and R are

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`CH_3CH_2Cl` and NAOH
`CH_2Cl-CH_2OH` and AQ. `NaHCO_3`
`CH_3CH_2OH` and HCL
`CH_2=CH_2` and heat

Solution :
40.

In a reaction CH_(2)=CH_(2) underset("acid")overset("Hypochlorous")to M overset(R )to underset(CH_(2)-OH)overset(CH_(2)-OH)| Where M =molecule, R=reagent, M and R are

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`CH_(3)CH_(2)Cl and NaOH`
`CH_(2)Cl-CH_(2)OH and AQ. NaHCO_(3)`
`CH_(3)CH_(2)OH and HCL`
`CH_(2)=CH_(2)` and heat

Solution :
41.

In a reaction, catalyst changes ___________.

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physically
QUALITATIVELY
chemically
QUANTITATIVELY

ANSWER :C::D
42.

In a reaction, catalyst changes ..........

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Physically
qualitatively 
CHEMICALLY 
quantitatively 

Solution :Catalyst does not get CONSUME in the reaction. HENCE, it does not UNDERGO a chemical change.
43.

In a reaction between A and B, the initial rate of reaction was measured for different intial concentrations of A and B as given below : {:(A//"mol L"^(1),,,0.20,,,0.20,,,0.40),(B//"mol L"^(-1),,,0.30,,,0.10,,,0.05),(r_(0)//"mol L"^(-1)s^(-1),,,5.07xx10^(-5),,,5.07xx10^(-5),,,7.16xx10^(-5)):} Whatis the order of reaction with respect to Aand B ?

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SOLUTION :`r_(0)=[A]^(alpha)[B]^(beta)`
`(r_(0))_(1)=5.07xx10^(-5)=(0.20)^(alpha)(0.30)^(beta)""...(i)`
`(r_(0))_(2)=5.07xx10^(-5)=(0.20)^(alpha)(0.10)^(beta)""...(ii)`
`(r_(0))_(3)=7.16xx10^(-5)=(0.40)^(alpha)(0.05)^(beta)""...(iii)`
`((r_(0))_(1))/((r_(0))_(2))=1=((0.30)/(0.10))^(beta)=(3)^(beta):.beta=0`
`((r_(0))_(3))/((r_(0))_(2))=(7.16)/(5.07)=((0.40)/(0.20))^(alpha)((0.05)/(0.10))^(beta)" or "1.412=2^(alpha)((1)/(2))^(beta)=2^(alpha)( :.beta=0)`
`log1.412=alphalog2"or"alpha=0.1523//0.3010=0.5.`
`("or directly "1.412=sqrt(2)=2^(1//2):.2^(alpha)=2^(1//2)" or "alpha=1//2)`
THUS, order w.r.t. A = 0.5, orderw.r.t. B = 0
44.

In a reaction between zinc and iodine in which zinc iodide is formed, which is oxidized ?

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ZINC ions
Iodide ions
Zinc atom
Iodine

Answer :C
45.

In a reaction between A and B, the initial rate of reaction was measured for different initial concentration of A and B as given ahead: what is the order of reaction with respect to A and B?

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Solution :The rate law equation may be expressed as:
Rate = `k[A]^(p)[B]^(q)`
Comparing experiments 1 and 2.
`("Rate")_(1) = k[0.2]^(p)[0.3]^(q)= 5.07 XX 10^(-5)`..........(i)
`("Rate")_(2) = k[0.2]^(p)[0.1]^(q) = 5.07 xx 10^(-5)`...........(ii)
Dividing EQN. (i) by eqn. (ii),
`(Rate)_(1)/(Rate)_(2) = (k[0.2]^(p)[0.3]^(q))/(k[0.2]^(p)[0.1]^(q)=(5.07 xx 10^(-5))/(5.07 xx 10^(-5)`
`[3]^(q) = [1]^(0)` or q=0
By comparing experiments 2 and 3.
`("Rate")_(2)= k[0.2]^(p)[0.1]^(q) = 5.07 xx 10^(-5)`...........(i)
`("Rate")_(3) = k[0.4]^(P)[0.05]^(q)= 7.56 xx 10^(-5)`..............(iv)
Dividing eqn. (iv) by eqn. (iii),
`("Rate")_(3)/("Rate")_(2) = (k[0.4]^(p)[0.05]^(q))/(k[0.2]^(p)[0.1]^(q))= (7.56 xx 10^(-5))/(5.07 xx 10^(-5))`
`[2]^(p) = [1.49]=[2]^(3//2)` or p `=3//2`
Order with RESPECT to A = `3//2`
Order with respect to B = 0
46.

In a reaction, between A and B, the initial rate of reaction (r_(0)) was measured for different initial concentrations of A and B as given below : What is the order of reaction with respect to A and B ?

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SOLUTION :`r_(0)= k[A]^(ALPHA)[B]^(beta)`
`(r_(0))_(1)=k XX 5.07 xx 10^(-5)=(0.20)^(alpha) (0.30)^(beta) ""…(i)`
`(r_(0))_(2)= k xx 5.07 xx 10^(-5)= (0.20)^(alpha) (0.10)^(beta)"" …(ii)`
`(r_(0))_(3)=k xx 1.43xx10^(-4)=(0.40)^(alpha) (0.05)^(beta) ""...(iii)`
Dividing (i) by (ii)
`(r_(0))_(1)//(r_(0))_(2)=1=(0.30//0.10)^(beta)=(3)^(beta)`
`therefore beta=0`
Dividing (iii) by (ii)
`(r_(0))_(3)//(r_(0))_(2)=(7.16)/(5.07)=(0.40//0.20)^(alpha) (0.05//0.10)^(beta)`
or `1.412=2^(alpha)((1)/(2))^(beta)=2^(alpha) ""[because beta=0]`
or `LOG 1.412=alpha log 2`
or `alpha=0.1523//0.3010=0.5`
Thus, ORDER w.r.t. A = 0.5, order w.r.t. B = 0.
47.

In a reaction at equillibrium X mole of the reactant A decompose to give 1 mole each ofC and D. if the fraction of A decomposed at equillibrium is independent of initial concentration of A then what will be the value pf X ?

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1
3
2
4

Answer :C
48.

In a reaction at equilibrium x mole of the reactants A decompose to give 1 mole each of C and D. it has been found that the fraction of A decomposed at equilibrium is independent of initial concentration of A. find the value of x.

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ANSWER :2
49.

In a reaction A toproducts, when start is made from 8.0times10^(-2)M of A, half-life is found to be 120minute. For the initial concentration 4.0times10^(-2)M, the half-life of the reaction becomes 240 minutes. The order of the reaction is :

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zero
one
two
0.5

Solution :`((t_((1)/(2)))_(1))/((t_((1)/(2)))_(2))=((a_(2))/(a_(1)))^(n-1)`, `(120)/(140)=((4times10^(-2))/(8times10^(-2)))^(n-1) , n=2`
50.

In a reaction AtoB when the concentration of reaction is made 8 times, the rate got doubled. The order of reaction is

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`1//3`
`1`
`1//2`
`2`

ANSWER :A