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 zero order reaction

Answer»

the TIME TAKEN for HALF the reaction to complete is INVERSELY proportional to its RATE constant
the time taken for half change is directly proportional to its initial concentration
the time taken for completion of the reaction is independent of initial concentration
there is no effect on the rate of reaction if concentration of reactants is doubled

Answer :A::B::D
Discussion
2.

For a zero order chemical reaction, 2NH_(3)(g) to N_(2)(g) + 3H_(2)(g) rate of reaction=0.1 "atm"//"sec".Initially only NH_(3) is present and its pressure =3atm . Claculate total pressure at t=10 sec.

Answer»


ANSWER :5
Discussion
3.

For a weak monobasic acid, if pK_a = 4, then at a concentration of 0.01 M of the acid solution, the van't Hoff factor is

Answer»

1.01
1.02
`1.10`
`1.20`

ANSWER :C
Discussion
4.

For a which reaction does the equilibrium constant depend on the units of concentration?

Answer»

`COCl_(2) HARR CO(G)+Cl_(2)(g)`
`Zn(s)+Cu^(2+)(AQ.) hArr Cu(s)+Zn^(2+)(aq.)`
`C_(2)H_(5)OH(l)+CH_(3)COOH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)(O)(l)`
`NO(g) hArr 1/2 N_(2)(g)+1/2O_(2)(g)`

Answer :A
Discussion
5.

For a yellow coloured solution obtained in column-1, select the only correct option.

Answer»

1, C, P
2,a, P
3, b,S
1,d,S

Solution :`Co^(2+)+"excess"CN^(-)UNDERSET(air)tounderset("yellow coloured solution")([Co(CN)_6]^(3-))`
`Co^(2+)+"excess"CN^(-)underset(air)tounderset("yellow coloured solution")([Co(CN)_6]^(3-))`
`Co^(3+), 3d^6 ,S.F.Limplies d^2sp^3`& OCTAHEDRAL complex
n=0 `implies` Diamagnetic
Discussion
6.

For a weak monobasic acid, if pK_(a)=4. then at a concentration of 0.01 M of the acid solution, the van't Hoff factor is

Answer»

1.01
1.02
`1.10`
`1.20`

Solution :`pK_(a)=4" means "K_(a)" for "HA=10^(-4)`
For weak acid, `HA hArr H^(+)+A^(-)`
`K_(a)=C ALPHA^(2)"(Ostwald's dilutio law)"`
`therefore""alpha=sqrt((K_(a))/(C))=sqrt((10^(-4))/(0.01))=10^(-1)=0.10`
`{:(,HA,hArr,H^(+),+,A^(-)),("Initial","1 mole",,,,),("Moles after ",1-alpha,,alpha,,alpha","):}`
`"dissoc.Total"=1+alpha`
`i=1+alpha=1+0.10=1.10`
Discussion
7.

For a weak acid HA with dissociation constant 10^(-9), pOH of its 0.1 M solutions is

Answer»

9
3
11
10

Solution :Dissociation constant of `HA = 10^(-9)`
`HA hArr H^(+) + A^(-)`
`[H^(+)] = sqrt((K_(a))/(C)) = sqrt((10^(-9))/(0.1)) , [H^(+)] = 10^(-4)`
`:. PH = 4`
`because pH + POH = 14`
`pOH = 14- pH = 14-4 , pOH = 10` .
Discussion
8.

For a weak acid, the incorrect statement is

Answer»

Its dissociation constant is low
Its `pK_(a)` is very low
it is PARTIALLY dissociated
Solution of its sodium salt is EXPECTED to form IONIC

Solution :For a weak acid value of PKA will be very HIGH but in case of strong acid value of pKa will be very low.
Discussion
9.

For a weak base, the concentration of OH^(-) ion at concentration 'c' would be(disociation constant = K_(b))

Answer»

`SQRT((K_(B))/(C ))`
`(K_(b))/(sqrt(c ))`
`sqrt(K_(b)XX c)`
`sqrt(K_(omega)//K_(b)c)`

Answer :C
Discussion
10.

For a weak acid HA, Ostwald's dilution law is represented by the equation

Answer»

`K_(a) = (alpha C)/(1-alpha^(2))`
`K_(a) = (alpha^(2) c)/(1-alpha)`
`K_(a) = (K_(a)c)/(1-c)`
`K_(a) = (alpha^(2)c)/(1-alpha^(2))`

Solution :MATHEMATICAL form of Ostwald' DILUTION law.
Discussion
11.

For a weak acid HA of concentration C(mol l^-1) and degree of dissociation (alpha), Ostwald's dilution law is represented by the equation

Answer»

`K_a=(C^2alpha)/(1-alpha)`
`K_a=(alpha^2C)/(1-alpha)`
`K_a= CALPHA`
`K_a= (Calpha^2)/(1-alpha^2)`

ANSWER :B
Discussion
12.

For a uni - univalent electrolyte, write the Debye - Huckel Onsagar equation.

Answer»

SOLUTION :`wedge_(m)=wedge_(m)^(@)-(A+Bwedge_(m)^(@))SQRT(C)`
Discussion
13.

For a uncleophillic substitution reaction the rate was found in the order RI gt RBr gt RCl gt RF then the reaction could be

Answer»

`S_N1` only
`S_N2` only
either `S_N1 " or " S_N2`
NEITHER `S_N1 " nor " S_n2`

Solution :In both `SN^(1) & SN^2` rate ORDER for DIFFERENT halides is `RI GT RBR gt RCl gt RF`
Discussion
14.

For a van der Waals' gas, a = 4 atm-l^(2)//"mol"^(2) and b = 0.02l//"mol". Select the correct possible graph (s). [Given: R = 0.08 L-atm//"mol'-L]

Answer»




ANSWER :A::D
Discussion
15.

For a third order reaction the ratio of t_(1//3) "and"t_(2//3) is T [t_(1//x) represents the time in which 1/xfraction of reactant get reacted]. The value of '32 T' is :

Answer»


ANSWER :5
Discussion
16.

For A to B, [A] changed from 4.4 xx 10^(-2)M to 3.2 xx 10^(-2) Min 25 min. Now (-Delta[A])/(Delta t) will be

Answer»

`4.8 XX 10^(-4) "min"^(-1)`
`4.8 xx 10^(-4)M.s^(-1)`
`9.6 xx 10^(-3) M.s^(-1)`
`2.4 xx 10^(-4) M."min"^(-1)`

Answer :A
Discussion
17.

For a thermodynamically reversible reaction in a galvanic cell at temperature T, which one of the following is false?

Answer»

`-Delta G = w_(max)`
`Delta G^@ = -RT lnK_c`
`Delta G= - FE_(cell)`
`Delta H = T Delta S `

Solution :Gibbs free energy change is equal to the maximum possible useful work that can be obtained from a process for a reversible change, hence, `- Delta G = W_(max_`
Electrical work done `= nFE_(cell)`
` therefore- Delta G = nFE_(cell)`
Gibbs energy for a reaction in which all reactants and products are in standard state, `Delta G^@`is RELATED to the EQUILIBRIUM constant as,
` Delta G^@- 0RT ln K_c`
when `Delta G^@`is ZERO, process does not occur.
` Delta G = Delta H -T Delta S = 0 , therefore Delta H = T Delta S`
Discussion
18.

For a system at equilibrium which of the following are correct

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`((d"In"K_p)/(dP))_r=(-d)/(dP)((DeltaG)/(RT))_r`
`log K_p =1/(2.303R)(DeltaS-(DeltaH)/T)`
on increasing the TEMPERATURE of an endothermicreaction, the equilibrium shift in forward direction because Kincreases
on increasing the temperature of an ENDOTHERMIC reaction the equilibrium shifts in forward direction because Q decreases

Solution :`DeltaG=RT"In" K_p`
`therefore ((d"In"K_p)/(dP))_T=-d/(dP)((DeltaG)/(RT))_T`
Hence (A) is the CORRECT option.
Also, `DeltaG=DeltaH-TDeltaS=-2.303 RT log K`
`therefore log K=1/(2.303)(DeltaS-(DeltaH)/T)`
Hence (B) is correct option.
In endothermic reaction changing temp. K changes and not Q .
Hence ( C ) is the correct option.
Discussion
19.

For a system in equilibrium, /_\G = 0 under conditions of constant ......

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temperature and pressure
temperature and volume
pressure and volume
energy and volume

Solution :For a SYSTEM in equilibrium, AG = 0, when all the reactants and products are in the STANDARD state (at CONSTANT temperature and pressure) and`K_(C)=1`
Discussion
20.

For a system in equilibrium, DeltaG = 0 under conditions of constant :

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TEMPERATURE and PRESSURE
temperature and volume
Energy and volume
Pressure and volume

Answer :A
Discussion
21.

For a system in equilbrium DeltaG=0 under conditions of constant

Answer»

TEMPERATUREE and pressure
TEMPERATUE and VOLUME
Energy and volume
Pressure and volume

Answer :A
Discussion
22.

For a substance undergoes sequential first order reaction calculate ratio calcualate ratio of number of atoms of B to C when steady state is obtained. Aoverset(K_(1))toBoverset(K_(2))toCoverset(K_(3))toD where K_(1)=(In2)/(10), K_(2)=10In2, K_(3)=10^(4)In 2

Answer»

10
`10^(3)`
`10^(-3)`
`10^(4)`

ANSWER :B
Discussion
23.

For a system at equilibrium

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`((dellnK_(p))/(delP))=-(del)/(delP)((DeltaG)/(RT))_(T)`
`DeltaG=0`
`G_((P))=G_((R))`
The free energy of the ststem is minimum

Solution :`DeltaG=-RTlnK_(p)implies-(DeltaG)/(RT)=lnK_(p)`
`implies((dellnK_(p))/(delP))_(T)=-(del)/(delP)((DeltaG)/(RT))_(T)`
HENCE statement (a) is true
At equlibrium `G_(p)=G_(R)`
i.e., free energy of products and reactants are equal.
`impliesDeltaG=G_(p)-G_(R)=0implies` Statement (B) and(c) are CORRECT. At equlibrium, the free energy of system is minimum. Statement (d) is correct.
Discussion
24.

For a system A+2B hArr 2C the equilibrium concentration are[A]=0.06, [B] = 0.12 and [C]=0.216 The K_c for the reaction is :

Answer»

54
415
`4XX10^(-5)`
125

Answer :A
Discussion
25.

For a standard cell, Cu(s)|Cu^(2+)(0.01M)||Ag^(+)(0.02M)|Ag(s) E_(Cu^(2+)|Cu)^0=+0.34V,E_(Ag^+//Ag)^0=+0.80V Write the reactions taking place at the electrodes.

Answer»

Solution :ANODE REACTION: `CU(s) Cathode reaction: `AG^+(aq)+etoAg(s)`
`:.` CELL reaction: `Cu(s)+2Ag^+(aq)toCu^(2+)(aq)+2Ag(s)`
Discussion
26.

For a stable molecule the value of bond order must be

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zero
positive
negative
there is no relationship between STABILITY and bond ORDER.

Solution :For stable MOLECULE the value of bond order must be positive as bond order.
= (no. Bonding electrons) -(no. Of antibonding electrons)/(2)
The electrons in the bonding MOLECULAR orbital leads to lowering in energy and hence stability whereas in antibonding molecular orbital leads to destability.
(i) (-ve) bond order `Rightarrow` greater number of electrons in antibonding molecular orbital than bonding molecular orbital.Therefore molecule is less stable compared to the separated atoms.
0 bond order `rightarrow` number of bonding molecular orbital = number of antibonding molecular orbital.Therefore there is no gain in energyy in forming the molecule.
(+ve) bond order `rightarrow` number of electrons in bonding molecular orbital is more than those in antibonding molecular orbital.Therefore ther is a gain in enrgy while the bonding TAKES place.
Discussion
27.

For a springly soluble salt A_PB_q the relationship between its solubility product (L_s)and its solubility (S) is :

Answer»

`L_S = S^(p+q) p^p q^q`
`L_S = S^(p+q) p^q q^p`
`L_S = S^(PQ) p^p q^q`
`L_S =s^(pq)pq^(q+p)`

ANSWER :A
Discussion
28.

For a spontaneous reaction theDeltaG) equilibrium constnat (K)and E_("Cell")^(@) eill be respectively

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`-ve GT 1, -ve`
`-ve, lt 1, -ve`
`+ve, gt 1, -ve`
`-ve, 1, +ve`

Solution :For spontaneous reaction `DelgaG` should be negative. EQUILIBRIUM constant should be more than one
`(DeltaG =-2.303 RT log K_(e), If K _(C)=1` then `DeltaG=0, If K_(c)lt 1`
then `DeltaG =+ ve).` Again `Delta G=-nFE_(cell)^(@)`
`E _(cell)^(@)` must be + ve have `Delta G-vc.`
Discussion
29.

For a spontaneous reaction the DeltaG, equilibrium constant (K) and E_(cell)^(o) will be respectively

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`-ve,GT1,+ve`
`+ve,gt1,-ve`
`-ve,LT1,-ve`
`-ve,gt1,-ve`

ANSWER :A
Discussion
30.

For a spontaneous reaction, the DeltaG, the equilibrium constant (K) and E_("cell")^(@) will be respectively ………. .

Answer»

`-ve , GT 1 , +ve`
`+ve , gt 1 , -ve`
`-ve , LT 1 , -ve`
`-ve , gt 1 , -ve`

ANSWER :A
Discussion
31.

For a spontaneous reaction the DeltaG , equilibrium constant (K) and E_("cell")^(0) will be respectively .

Answer»

`-ve, GT 1, +ve`
`-ve, gt 1, -ve`
`-ve, LT 1, -ve`
`-ve, gt 1, -ve`
Discussion
32.

For a spontaneous reaction DeltaG^(@), Equilibrium constant (K) and E_(cell) will be respectively.

Answer»

`-ve GT 1 gt +ve`
`+ve gt 1 gt -ve`
`-ve LT 1 lt -ve`
`-ve gt 1 gt -ve`.

SOLUTION :(a) is the CORRECT answer.
Discussion
33.

For a spontaneous reaction, DeltaG, equilibrium constant (K) and E_(cell)^(@) will be respectively:

Answer»

`-ve, LT 1, - ve`
`-ve, GT 1, -ve`
`-ve, gt 1, +ve`
`+ve, gt 1, -ve`

ANSWER :C
Discussion
34.

For a spontaneous reaction, Delta G should be :

Answer»

POSITIVE
negative
equal to zero
may be positive or negative

Answer :B
Discussion
35.

For a spontaneous process, the correct statement(s) is

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`(Delta G_(system))_T,PGT 0`
`(Delta S_(system)) + (Delta S_(surr)) gt 0`
`(Delta S_(system))_T ,P LT 0`
`(Delta U_(system))_T, V gt 0`

Solution :For a spontaneous process,
(i) `Delta G_(system) lt 0`
(ii) `Delta G_("TOTAL") = Delta S_("univ") = Delta S_(sys) + Delta S_(surr) gt 0`
This is a king of the mathematical expression of the 2nd law of thermodynamics .
Discussion
36.

For a spontaneous reaction, Delta G, equilibrium constant (K) and E_(cell)^(@) will be respectively.

Answer»

`- ve, GT 1, + ve`
`+ ve, gt 1, - ve`
`- ve, lt 1, - ve`
`- ve, gt 1, - ve`

SOLUTION :For a spontaneous REACTION :
`DELTA H = - ve, K gt 1 and E_("cell")^(@) = + ve`
Discussion
37.

For a spontaneous process the correct statement is -

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ENTROPY of the SYSTEM ALWAYS increases
Free energy of the system always increases
Total entropy change is always negative
Total entropy change is always positive

Solution :DEFINITION BASED.
Discussion
38.

For a spontaneous process :-

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`DeltaG=0`
`DeltaltG`
`DeltaGlt0`
Any of the above

Answer :B
Discussion
39.

Fora spontaneous endothermic reaction :

Answer»

`DELTA G GT 0`
`Delta G = 0`
`Delta H lt 0`
`Delta S gt (Delta H)/(T)`

Answer :D
Discussion
40.

For a spontaneous change, free energy change DeltaG is

Answer»

POSITIVE
Negative
Zero
Can be positive or negative

Solution :For SPONTANEOUS CHANGE `DeltaG=-ve`.
Discussion
41.

For a sponaneous reaction, the free energy change must be negative, Delta G=Delta H-T Delta S, Delta H is the enthalpy change during the reaction. T is the absolute temperature, and Delta S is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide M+O_(2) to MO Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence Delta S is negative. Thus, if the temperature is raised then T Delta S becomes more negative,Since, TDelta S is substracted in the equation, then Delta G becomes less negative. Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals. Free energy change of Hg and Mg for the convertion to oxides the slpe of Delta G vsT has been changed above the boiling points of the given metal because :

Answer»

above the BOILING point of the METAL entropy is increased
above the boiling point of the metal the entropy is decreased
above the boiling point of the metal the entropy change is equal to zero
All of these

Answer :A
Discussion
42.

For a sponaneous reaction, the free energy change must be negative, Delta G=Delta H-T Delta S, Delta H is the enthalpy change during the reaction. T is the absolute temperature, and Delta S is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide M+O_(2) to MO Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence Delta S is negative. Thus, if the temperature is raised then T Delta S becomes more negative,Since, TDelta S is substracted in the equation, then Delta G becomes less negative. Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals. As per the Ellingham diagram of oxides which of the following conclusion is true ?

Answer»

Al reduces `Fe_(2)O_(3)`, WHEREAS MgO cannot be reduced by Al at `1500^(@)C`
Fe reduces `Al_(2)O_(3)`, whereas MgO cannot be reduced by Al at `1500^(@)C`
Al reduces `Fe_(2)O_(3)`, whereas MgO cannot be reduced by CA at `1500^(@)C`
Al can reduce both `Fe_(2)O_(3)` and MgO to the corresponding METAL at `1500^(@)C`

Answer :A
Discussion
43.

For a sponaneous reaction, the free energy change must be negative, Delta G=Delta H-T Delta S, Delta H is the enthalpy change during the reaction. T is the absolute temperature, and Delta S is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide M+O_(2) to MO Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence Delta S is negative. Thus, if the temperature is raised then T Delta S becomes more negative,Since, TDelta S is substracted in the equation, then Delta G becomes less negative. Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals. Which of the following elements can be prepared by heating the oxide above 400^(@)C ?

Answer»

Hg
Mg
Fe
Al

Answer :A
Discussion
44.

For a spontaneous cell reaction, the DeltaG should be .................... .

Answer»

SOLUTION :NEGATIVE
Discussion
45.

For a sparingly soluble salt A_(p)B_(q), the relationship of its solubility product (L_(S)) with its solubility (S) is

Answer»

<P>`L_(S) = S^(p+q).p^(p).q^(q)`
`L_(S) = S^(p+q).p^(q).q^(p)`
`L_(S) = S^(pq). p^(p). q^(q)`
`L_(s) = S^(pq). (pq)^(p+q)`

SOLUTION :`{:(A_(p)B_(q)(s), hArr, pA^(+q) +, qB^(-p)),(,,PS,qS):}`
`L_(S) = (pS)^(p).(qS)^(q) = p^(p).q^(q).S^(p+q)`.
Discussion
46.

For a solution of weak triprotic acid H_(3)A(K_(a_(1))gtgtK_(a_(2)),K_(a_(3)):K_(a_(2))=10^(-8),K_(a_(3))=10^(-13),[A^(3-)]=10^(-17)M.Determine pH of solutions. Report your answer as 0 if you find data insufficient.

Answer»

Solution :`{:(,HA^(2-),HARR,H^(+),+,A^(3-),,,K_(a_(3))=10^(-13)),(t=eq,,10^(-8),,,,10^(-17),):}`
Discussion
47.

For a solvent K_(b)=5 kg mol^(-1) using this solvent, the solution records the elevation of boiling point of 0.5 ^(@)C The molality of the solution is

Answer»

0.25
0.1
10
unpredictable.

Solution :`DELTA T_(B) = K_(b) XX m `
`:. m=(Delta T_(b))/K_(b)=0.5/5 = 0.1`
Discussion
48.

For a sparingly soluble salt A_(p)B_(q) the relationship between its solubility product (L_(S)) and its solubility (S) is :

Answer»

<P>`L_(S)=S^(p+Q)p^(q)q^(q)`
`L_(S)=S^(p+q)p^(q)q^(p)`
`L_(S)=S^(PQ)p^(p)q^(p)`
`L_(S)=S^(pq)(pq)^(q+p)`

ANSWER :A
Discussion
49.

For a solution of two liquids A and B it was proved that P _(S)=x_(A)(p_(a)^(@)-p_(a)^(@)+ P ^(@)B. The recsulting solution will be

Answer»

<P>Non-idcal
idcal
semi-ideal
None of these

SOLUTION :`p_(s) =X_(A) (p_(A)^(@)-p_(B)^(@) )+ p_(B)^(@),`
`p _(s) =P_(A)^(@)xx X_(A)- p_(B)^(@)xx X_(A) + p_(B)^(@)`
`p _(s)=p_(A)^(@)xx x_(A)+p_(B)^(@) (1-x_(n)) +p_(a)^(@)`
`THEREFORE p_(s) =p_(A)^(@) xx x_(A) + p_(B)^(@) xx x_(B).`
This is condifion for ideal solution.
Discussion
50.

For a solution of volatileliquids the partial vapour pressure of eachcomponent in solution is directly proportional to

Answer»

MOLARITY
Mole fraction
Molality
NORMALITY

Solution :Raoult's LAW.
Discussion