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 solution of dibasic acid the molarity (M) and normality (N) are related as

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N=M/2
2 M = N
M =N
M GT N

Solution :For dibasic ACID there are 2 H, GROUPS Hence basicity=2 and normality = 2 M.
2.

For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here x_L and x_M represent mole fraction of L and M, respectively, in the solution. The correct statement(s) applicable to this system is(are)

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The POINT Z represents VAPOUR pressure of pure liquid M and Raoult's LAW is obeyed from `x_L` = 0 to `x_L` = 1
Attractive intermolecular interactions between L–L in pure liquid L and M-M in pure liquid M are stronger than those between L–M when mixed in solution
The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when `x)L rarr 0`
The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when `x_L rarr 1`

Answer :B::D
3.

For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown is the following figure. Here x_(L) and x_(M) represent mole fractions of L and M, respectively, in the solution. The correct statement (s) applicable to this system is (are)

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The point Z represents vapour pressure of pure LIQUID M and Raoult's law is obeyed from `x_(L) = 0` to `x_(L) = 1`
Attractive intermolecular interaction between `L-L` in pure liquid L and `M-M` in pure liquid M are stronger than those between L-M when mixed in solution
The point Z represents vapour of pure liquid M and Raoult's law is obeyed when `x_(L) rarr 0`
The point Z represents vapour of pure liquid L and Raoult's law is obeyed when `x_(L) rarr 1`

ANSWER :B::C
4.

For a second order reaction, t_(75%)=x t_(50%) Find the value of x.

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ANSWER :`3`
5.

For a second order reaction, t_(75%) = x t _(50%) find the value of x

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SOLUTION :`(t_(3//4))/(t_(1//2)) = 2^(n-1) + 1 = 2^(2 -1) + 1 = 3`
6.

For a second order reaction, 2" A"toProducts, a plot of log t_(1//2) vs log a (where a is the initial concentration) will give an intercept equal to which of the following ?

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`1//k`
`LOG(1//2" k")`
`log (1//k)`
`log k`

SOLUTION :For a second ORDER reaction, `2A to ` PRODUCTS,
`k= (1)/(t)[(1)/([A])-(1)/([A]_(0))]`
When `t = t_(1//2), [A] = ([A]_(0))/(2)`
Hence, `t_(1//2) = (1)/(k)[(2)/([A]_(0))-(1)/([A]_(0))] = (1)/(k[A]_(0))`
`therefore log t_(1//2) = - log k - log [A]_(0) =- log k - log a`
Plot of log `t_(1//2)` vs log a will be linear with intercept = log k.
7.

For a second order reaction rate at a particular time is x . If the initial concentration is tripled , the rate will become :

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3X
`9X^(2)`
9x
27x

Solution :( C ) Rate = `k[A]^(2) = X `
If conc . Is tripled i.e A = [3A]
Rate = `k[3A]^(2) = k.9[A]=x `
`:. (x)/(x)=9` i.e BECOMES nine times (9x)
8.

For a second order reaction, 2Ararr Products, a plot of logt_(1//2) vs log a (where a is initial concentration) will give an intercept equal to which one of the following?

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`(1)/(K)`
`log((1)/(2K))`
`log((1)/(k))`
`LOGK`

ANSWER :D
9.

For a sample of perfect gas when its pressure is changed isothermally from p_(i) to p_(f), the entropy change is given by

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`DeltaS=RTln((p_(i))/(p_(F)))`
`DeltaS=nRln((p_(f))/(p_(i)))`
`DeltaS=nRln((p_(i))/(p_(f)))`
`DeltaS=nRTln((p_(f))/(p_(i)))`

SOLUTION :`DeltaS_("sys")=nRln.(P_(i))/(P_(f))+nCpln.(T_(f))/(T_(i))`
In isothermal PROCESS `T_(i)=T_(f)`
`DeltaS_("sys")=nRln.(P_(i))/(P_(f))`
10.

For a sample of an ideal gas at given temperature (T), speed distribution curve is given as follows. Then the speeds corresponding to point A, B and C are respectively known as:

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Most probable, AVERAGE and root mean SQUARE
Average, root mean square and most probable
Root mean square, average and most probable
Most probable, root mean square and average

SOLUTION :THEORY BASED
11.

For a reversible reaction : X_((g))+3Y_((g))iff2Z_((g)), DeltaH=-40kJ the standard entropies of X, Y and Z are 60, 40 and 50 JK^(-1) mol^(-1) respectively. The temperature at which the above reaction attains equilibrium is about

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400 K
500 K
273 K
373 K

Solution : `X_((g))+3Y_((g))iff2Z_((g))`
`DeltaS^(@)=2S^(@)(Z)-{S^(@)(X)+3S^(@)(Y)}`
`=2xx50-{60+3xx40}=100-180 "= - 80 J K"^(-1) mol^(-1)`
`DeltaH^(@)=-40 kJ=-40,000 J,"" DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
At equilibrium, `DeltaG^(@)=0 ""therefore DeltaH^(@)=TDeltaS^(@)`
`or, T = (DeltaH^(@))/(DeltaS^(@))=(40000)/(80)=500 K`.
12.

For a reversible reaction : X_((g))+3Y_((g))hArr 2Z_((g)), Delta H=-40 kJ, the standard entropies of X,Y and Z are 60, 40 and 50 JK^(-1)mol^(-1) respectively. The temperature at which the above reaction attains equilibrium is about

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400 K
500 K
273 K
373 K

Solution :`X_((G))+3Y_((g))hArr 2Z_((g))`
`Delta S^(@)=2S^(@)(Z)-{S^(@)(X)+3S^(@)(Y)}`
`=2xx50-{60+3xx40}=100-180`
`=-80J K^(-1)MOL^(-1)`.
Given `Delta H^(@)=-40 kJ =-40,000 J`
`Delta G^(@)=Delta H^(@)-T Delta S^(@)`
At equilibrium, `Delta G^(@)=0`
`therefore Delta H^(@)=T Delta S^(@)`
or,`T=(Delta H^(@))/(Delta S^(@))=(40000)/(80)=500 K`.
13.

For a reversible reaction the rate constant for the forward reaction is 2.38xx10^(-4) and for the backward reaction is 8.15xx10^(-5) The k_c of the reaction is:

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0.342
2.92
0.292
3.42

Answer :B
14.

For a reversible reaction if the concentration of the reactants are doubled, the equilibrium constant will be :

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HALVED
DOUBLED
the same
one fourth

Answer :C
15.

For a reversible reaction C hArr D, heat of reaction at constant volume is -33.0 kJ mol^(-1), calculate: (i) The equilibrium constant at 300 K. (ii) If E_(f) and E_(b) are energy of activation for forward and backward reactions respectively, calculate E_(f) and E_(b) at 300 K. Given that E_(f) : E_(b)=20:31 Assumw pre exponential factor same for forward and backward reaction.

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SOLUTION :N//A
16.

For a reversible reaction following graph is obtained between log_(10)K_(C) and (1)/(T) hence DeltaH for the reaction is:-

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`-19.14` J/mole
`-8.314` J/mol
`-4.606` J/mol
`+4.606` J/mol

Answer :A
17.

For a reversible reaction ALeftrightarrowBwhich one of the following statements is wrong from the given energy profile diagram ?

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Activation energy of forward REACTION is greater thanbackward reaction
the forward reaction is endothermic
the threshold energy is less than that of activation energy
the energy of activation of forward reaction is equal to the SUM of heat of reaction and the energy of activation of backward reaction

Solution :
where,
`E_(a)` = activation energy of forward reaction
` E_(a)` = activation energy of baclward reaction
the above profile DIAGRAM shows that `E_(a) gt E' _(a)`
the potential energy of the product is greater than that of the reactant , so the reaction is endothemic .
` E_(a)= E'_(a) + DELTAE`
` E_(t)= E_(a) or E_(t) gt E'_(a)`
18.

For a reversible reaction A underset(k_(2)=2 xx 10^(2)s^(-1))overset(k_(1)=4xx10^(2)s^(-1))hArr B initial concentration of A is 21 mol L^(-1) . Select the correct statement:

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Equilibrium CONCENTRATION of A is 14 MOL `L^(-1)`
Equilibrium conentration of B is 14 mol `L^(-1)`.
The concentration of A reduce by 50% of equilibrium concentration after 11.55 sec .
The concentration of A reduce to 50% after 23.1 sec.

Answer :B::C::D
19.

For a reversible reaction, A hArr B, which one of the following statements is wrong from given energy profile diagram ?

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Activation energy of forward REACTION is greater than that of backward reaction.
The threshold energy is less than that of activation energy
The forward reaction is endothermic
Activation energy of forward reaction is EQUAL to the SUM of heat of reaction and the activation energy of backward reaction.

Answer :B
20.

For a reversible first-order reaction, A underset(K_(2))overset(K_(1))(hArr)B, K_(f)=10^(-2) s^(-1) and B_(eq.)/A_(eq.)=4, If A_(0)=0.01 ML^(-1) and B_(0)=0, what will be concentration of B after 30 sec?

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SOLUTION :N//A
21.

For a reversible chemical reaction where the forward process is exothermic which of the following statements is correct

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The backward REACTION has HIGHER activation ENERGY than the forward reaction
The backward and the forward PROCESSES have the same activation energy
The backward reaction has LOWER activation energy
No activation energy is required at all since energy is liberated in the process

Answer :A
22.

For a reversible chemical reaction where the forward process is exothermic. Which of the following statements is correct?

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The backward REACTION has HIGHER activation energy than the forward reaction
The backward and the forward processes have the same activation energy
The backward reaction has LOWER activation energy
No activation energy is required at all SINCE energy is liberated in the process.

Solution :The backward reaction has higher activation energy than the forward reaction
23.

For a real gas y at T=60 k, select the only correct option (a/(bR)=270k)

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SOLUTION :`T_C=(8a)/(27Rb)=8/27xx270 =80`
`because T lt T_c IMPLIES` on applying pressure GAS will liquify
24.

For a redox reaction to proceed in a cell, the e.m.f. must be:

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Positive
Negative
Fixed
Zero

Answer :A
25.

For a redox reaction to proceed spontaneously in a given direction , the emf should

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be ZERO
have + ve sign
have - ve sign
have either +ve or -ve sign

SOLUTION :A cell OPERATES only when `E_("Cell")^(@) gt 0`
26.

For a redox reaction to proceed spontaneously in a given direction, the emf should:

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be zero
have `+ve` sign
have `-ve` sign
have EITHER `+ve or -ve` sign

Answer :B
27.

For a real gas the P -V curve was experimentally plotted and it hed the following appearance. With respect to liquifaction. Choose the correct statement.

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at `T = 500K, P = 40` atm, the STATE will be liquid
at `T = 300K, P = 50` atm, the state will be gas
at `T lt 300K, P gt 20` atm, the state will be gas
at `300 K lt T lt 500 K, P gt 50` atm, the state will be liquid.

Solution :
(a) at `T = 500K, P = 40 atm` corresponds to 'a' SUBSTANCE - ga
(b) at `T= 300 K, P = 50 atm` corresponds to 'b' substance -liquid
(c ) at `T lt 300 K, P gt 20 atm` corresponds to 'c' substance -liquid
(d) at `T lt 500 K, P gt 50 atm` corresponds to 'd' substance -liquid
So, Answer (D)
28.

For a real gas x at its critical temperature the correct option is -

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(II)(iv)(P)
(IV)(iv)(S)
(IV)(i)(S)
(II)(III)(S)

Solution :At crictical POINT, Z=3/8
& liquid & VAPOUR phase become indistinguishable
`:.` Their pysical properties are same
`because Z lt 1` , attractive TENDENCIES dominate.
29.

For a real gas the P-V curve was experimentally plotted and it had the following appearance. With respect to liquefaction, choose the correct statement.

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at `T = 500 K, P = 40` ATM, the STATE will be liquid
at `T = 300 K, P = 50` atm, the state will be gas
at `T lt 300 K, P GT 20` atm, the state will be gas
at `300 K lt T lt 500K, P gt 50` atm, the state will be liquid

Answer :D
30.

For a real gas critical volume (Vc) is '2x' times the actual volume occupied by gas molecules, 'x' value is

Answer»
31.

For a reactions , A+B rarr product , therate law is given by r = k[A]^(1//2)[B]^2. What is the order of the reaction ?

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SOLUTION :ORDER of the REACTION `=1/2+2=2(1)/2" or "0.5`
32.

For a real gas, deviations from ideal gas behaviour are maximum at :

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`-10^@C` and 5.0 ATM
`0^@C` and 2.0 atm
`0^@ C` and 1.0 atm
`100^@ C` and 2.0 atm

Answer :A
33.

For a reactionAtoB with activation energy E_(a) and rate constant k=Ae^(-Ea//RT). The rate of the reaction (Rate =k[A]) increases by increasing the temperature because

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ACTIVATION ENERGY decreases with increase in TEMPERATURE
The FACTOR -Ea/RT increases
less number of collision take place
the value of [A] increases

Solution :If temperature increases, then `((-Ea)/(RT))` increases
34.

For a reaction , X+Y rarr Product , quadrupling [x] . Increases the rate by afactor of 8 . Quadrupling both [x] and [y] , increases the rate by a factor of 16. Find the order of the reaction with respect to x and y . What is the overall order of the reaction.

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Solution :`{:(,X+yrarr,"Product(Z)"),(,x+yrarr,z),("condition"-1implies,4x+y,8),("condition"-2implies,4x+4yrarr,16):}`
`z=k[x]^m[y]^n""(1)`
`8Z=k[4x]^m[y]^n""(2)`
`16z=k[4x]^m[4y]^n""...(3)`
Dividing Eq (2) by Eq (1) we get
`(8z)/z=(k[4x]^m[y]^n)/(k[x]^m[y]^n)`
`8=4^mrArr2^3=(2^2)^mimplies2^3=2^(2m),2m=3,m=3//2`
1.5 ORDER with respect to x .
Dividing Eq (3) by Eq (1) we get ,
`(16z)/z = (k[4x]^m[4y]^n)/(k[x]^m[y]^n)`
`16=4^m.4^n`
`16=4^2.4^n`
`16/16=4^n`
`1 = 4n `
`:. n = 0 ` [Zero order with respect to y ]
OVERALL order of the reaction , `k[x]^m[y]^n`
`k[x]^(1.5)[y]^0`
Order `=(1.5+0)=1.5`
35.

For a reaction(1)/(2)A to 2B rate of disappearance of A is related to rate of apperarance of B by the expression

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`(-d[A])/(DT)=(1)/(2)(d[B])/(dt)`
`(-d[A])/(dt)=4(d[B])/(dt)`
`(-d[A])/(dt)=(1)/(4)(d[B])/(dt)`
`(-d[A])/(dt)=(d[B])/(dt)`

Answer :C
36.

For a reaction X + Y to Z, rate prop [X]. What is (i) molecularity and (ii) order of reaction ?

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(i) 2, (II) 1
(i) 2, (ii) 2
(i) 1, (ii) 1
(i) 1, (ii) 2

Answer :A
37.

For a reaction, X+Y rarr Product , quadrupling [x], increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with suspect to x and y. what is the overall order of the reaction?

Answer»

Solution :`""x+y RARR "Product (z)"`
`"" x+y rarr z`
Condition - 1 `rArr 4x+y rarr 8`
Condition - 2 `rArrr 4x+4y rarr 16`
`""z=k[x]^(m)[y]^(n)"………………….(1)"`
`""8Z=k[4x]^(m)[y]^(n)"......................(2)"`
`""16z=k[4x]^(m)[4y]^(n)".....................(3)"`
Dividing EQ (2) by Eq (1) we get,
`(8z)/z=(k[4x]^(m)[y]^(n))/(k[x]^(m)[y]^(n))`
`8=4^(m) rArr 2^(3)=(2^(2))^(m) rArr 2^(3)=2^(2m), 2m=3, m=3//2`
1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,
`""(16z)/z=(k[4x]^(m)[4y]^(n))/(k[x]^(m)[y]^(n))`
`""16=4^(m).4^(n)`
`""16=4^(2).4^(n)`
`""16/16=4^(n)`
`""1=4n`
`""THEREFORE n=0` [Zero order with respect to y]
Overall order of the reaction,
`""k[x]^(m)[y]^(n)`
`""k[x]^(1.5]y^(0)`
Order `=(1.5+0)=1.5`
38.

For a reaction, X + Y rarr Z there is no entropy charge. Enthaply change for the reaction is 100J mol^(-1). Delta G is

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`- 100 J MOL^(-1)`
`100 J mol^(-1)`
0
Infinite

Solution :`Delta G= Delta H- T Delta S or Delta G = 100 - T xx 0`
`= 100 J mol^(-1)`
39.

For a reaction X rarr Y the rate of the reaction has been found to be third order with respect to X . What happens when the concentration of X is doubled :

Answer»

rate becomes DOUBLE
rate becomes three times
rate becomes six times
rate becomes EIGHT times

Solution :(D) Rate =k `[X]^(3)`
when X becomes double rate becoems
`2^(3)` = 8 times
40.

For a reaction to occur spontaneously

Answer»

`DELTA S` MUST be NEGATIVE
`(Delta H - T Delta S)` must be negative
`(Delta H + T Delta S)` must be negative
`Delta H` must be negative

Answer :B
41.

For a reaction, the threshold energy is equal to

Answer»

Activation energy + initial POTENTIAL energy of reactants
Activation energy - norinal initial potential energy of reactants
Activation energy
Normal energy of reactants.

Solution :Activation energy is the energy needed by REACTANT MOLECULES to GAIN threshold energy level.
42.

For a.......... reaction the unit of the reactionrate is atm s^(-1)

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SOLUTION :GAS PHASE
43.

For a reaction ,the value of slope of a plot in Kto (1)/(T)=……….

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`-(E_(a))/(2.303)`
`-(E_(a))/(R)`
`-E_(a)`
`-(E_(a))/(2.303R)`

ANSWER :B
44.

For a reaction to be spontaneous at all temperatures

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`Delta G` and `Delta H` should be NEGATIVE
`Delta G` and `Delta H` should be positive
`Delta G = Delta S =0`
`Delta H lt Delta G`

Solution :From Gibb's Helmholtz equation,
`Delta G=Delta H-T Delta S`
For a REACTION to be spontaneus, `Delta G` must be negative.If `Delta G` and `Delta H` are negative, then reaction will be SPONTANEOUS at all temperatures.
45.

For a reaction, the rate constant is 2.34 s^(-1). The half-life period for the reaction is

Answer»

0.30 s
0.60 s
3.3 s
Date is insufficient

Solution :`t_(1//2) = (0.693)/(K) = (0.693)/(2.34) = 0.296s`
46.

For a reaction ,the rate of reaction was found to increase about .1.8 times when the temperature was increased by 10^@C . The increase in rate is due to :

Answer»

Increase in number of active molecules
Increase in ACTIVATION energy of reactants
Decrease in activation energy of reactants
Increase in the number of COLLISIONS between REACTING molecules

Answer :A
47.

For a reaction, the rate of reaction was found to increase about 1.8 times when the temperature was increased by 10^(@)C. The increase in rate is not due to

Answer»

INCREASE in NUMBER of active MOLECULES
increase in activation energy of reactants
DECREASE in activation energy of reactants
increase in the number of collisions between reacting molecules.

Solution : The increase in temperature would lead to decrease in activation energy, increase in the number of collisions thereby INCREASING the number of active molecules.
48.

For a reaction the free energy change, DeltaG=-RT ln K_(p)+RTlnQ_(p) where K_(P) = equilibrium constant, Q_(P)= reaction quotient. For the reaction to be in equilibrium state

Answer»

`(Q_(p))/(K_(p))gt1`
`(Q_(p))/(K_(p))LT1`
`(Q_(p))/(K_(p))=1`
`Q_(p)K_(p)=1`

ANSWER :C
49.

For a reaction the graph of the rate of the reaction against molar concentration of the reactant is as shown : What is the order of the reaction ?

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SOLUTION :ZERO ORDER.
50.

For a reaction the graph drawn between half life period and reciprocal of initial concentration(1/a) gives a straight line with positive slope. The magnitude of the slope is 500 and iniital concentration of the reactant is 2M. If the initial rate of the reaction is represented as x xx10^(-3) units, then the value of x is _____________

Answer»


SOLUTION :`m=tan theta=1/K=500,r=K.[A]^(2)=(1/500)XX(2)^(2)=8XX10^(-3)`