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.

DeltaH_(f)^(o) ("cyclohexane")=-300u DeltaH_(f)^(o) ("cyclohecene")=-100u DeltaH_(f)^(o)("benzene")=200u Where u is an arbitrary unit of enthalpy. If -X("in u") is the resonance energy of benzene, then find (X)/(100)

Answer»


SOLUTION :
Discussion
2.

DeltaH_(f)^@ (298K) of methanol is given by the chemical equation,

Answer»

`CH_(4)(g)+(1)/(2)O_(2)(g) to CH_(3)OH(g)`
`C("graphite")+(1)/(2)O_(2)(g)+2H_(2)(g)toCH_(3)OH(l)`
`C("graphite")+(1)/(2)O_(2)(g)+2H_(2)(g) to CH_(3)OH(l)`
`CO(g)+2H_(2)(g) to CH_(3)OH (l)`

Solution :Heat of formation of DEFINED as the heat exchanged when one mole of a substance is FORMED from its constituent elements of a substance is formed from its constituent element at STP. In case of allotrpes,most stable ALLOTROPE is taken into account. Among graphite and diamond, graphite is stable. hence, heat of formation of `CH_(3)OH` is represented by equation.
`C("graphite")+(1)/(2)O_(2)(g)+2H_(2)(g) to CH_(3)OH(l)`
Discussion
3.

DeltaH_(f)^(@) (298 K) of methanol is given by the chemical equation

Answer»

`CH_(4)(g)+1//2O_(2)(g)rarrCH_(3)OH(g)`
`C("graphite")+1//2O_(2)(g)+2H_(2)(g)rarrCH_(3)OH(l)`
`C("diamond")+1//2O_(2)(g)+2H_(2)(g)rarrCH_(3)OH(l)`
`CO(g)+2H_(2)(g)rarrCH_(3)OH(l)`

ANSWER :B
Discussion
4.

(DeltaH-DeltaU) for the formation of carbon monoxide (CO) from its elements at 298 K is - (R=8.314JK^(-1)mol^(-1))

Answer»

`1238.78Jmol^(-1)`
`-2477.57Jmol^(-1)`
`2477.57Jmol^(-1)`
`-1238.78Jmol^(-1)`

ANSWER :A
Discussion
5.

(DeltaH - DeltaU) for the formation of carbon monoxide (CO) from its elements at 298 K is (R = 8.314 JK^(-1) mol^(-1))

Answer»

`-1238.78 J MOL^(-1)`
`1238.78 mol^(-1)`
`-2477.57 J mol^(-1)`
`2477.57 J mol^(-1)`

Solution :`C(s)+1//2O_(2)(g)rarrCO(g)`
`{:(DeltaH=DeltaU+Deltan_(g)RT),((DeltaH-DeltaU)=Deltan_(g)RT):}}DELTAN=1-(1)/(2)implies+(1)/(2)`
`=+(1)/(2)xx8.314xx298=+1238.78 J mol^(-1)`.
Discussion
6.

DeltaH and DeltaS for a reaction are +30.558 kJ mol^(-1) and 0.066 kJ mol^(-1) at 1 atm pressure . The temperature at which free energy is equal to zero and the nature of the reaction below this temperature are :

Answer»

483 K , SPONTANEOUS
443 K , non-spontaneous
443 K , spontaneous
463 K , non-spontaneous

ANSWER :D
Discussion
7.

DeltaG^(@)(HI,g)-=+1.7kJ. What is the equilibrium constnt at 25^(@)C"for"2 HI_((g))hArrH_(2(g))+I_(2(g))

Answer»

<P>`24.0`
`3.9`
`2.0`
`0.5`

SOLUTION :`DELTAG^(@)=-2.303xx8.314xxx10^(-3)xx298logK_(p)`
`1.7=-2.303xx8.314xx10^(-3)xx298xxlog K_(p)`
`K_(p)=0.5`
Discussion
8.

DeltaG^o for the reaction X + Y hArr Z IS -4.606 kcal. The equilibrium constant for the reaction at 227^Ocis:

Answer»

100
10
2
0.01

Answer :A
Discussion
9.

DeltaG^(@) versus T plot the Ellingham 's diagram slopes downward for the reaction

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`C+(1)/(2)O_(2)TOCO`
`CO+(1)/(2)O_(2)toCO_(2)`
`2Ag+(1)/(2)O_(2)toAg_(2)O`
`MG+(1)/(2)O_(2)toMgO`

Answer :A
Discussion
10.

DeltaG^(@) vs T plot in the Ellingham's diagram slopes downwards for the reactions

Answer»

`Mg+(1)/(2)O_(2)rarrMgO`
`2Ag+(1)/(2)O_(2)rarrAg_(2)O`
`C+(1)/(2)O_(2)rarrCO`
`CO+(1)/(2)O_(2)rarrCO_(2)`

Solution :
Reaction (i) : `M_((s))+(1)/(2)O_(2(g))rarrMO_((s))" (where M represents METAL)"`
Reaction (ii) :`CO_((g))+(1)/(2)O_(2(g))rarrCO_(2(g))`
Reaction (iii) :`C_((s))+(1)/(2)O_(2(g))rarrCO_((g))`
`DeltaG^(@)` vs T PLOT in the Ellingham's diagram SLOPES DOWNWARDS for the reaction (i) and reaction (ii). For reaction (iii), `DeltaG^(@)` vs T plot in Ellingham's diagram slopes upwards.
Discussion
11.

DeltaG in Ag_(2)Orarr2Ag+1//2O_(2)at a certain temperature is -10kJ/mole. Pick the correct statement

Answer»

`Ag_(2)O` decomposes to AG and `O_(2)`
Ag and `O_(2)` COMBINES to form `Ag_(2)O`
Reaction is in EQUILIBRIUM
Reaction does not take place

Solution :`DeltaG-ve` means the PROCESS is spontaneous.
Discussion
12.

DeltaG, in process of melting of ice at -15^(@)C,is

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`DeltaG=-ve`
`DeltaG=+ve`
`DeltaG=0`
all of these

Solution :becouse melting of ice at `-15^(@)C` is a NON - SPONTANEOUS process.
Discussion
13.

DeltaG^(@) for the reaction X+Y Z is - 4.606 kcal. The value of equilibrium constant of the reaction at 227^(@)C is (R=2.0 cal. Mol^(-1)k^(-1))

Answer»

100
10
2
`0.01`

SOLUTION :`DELTAG^(@)=-2.303 RT LOG K`
`-4.606=-2.303xx0.002xx500log K`
logK=2, K=100`
Discussion
14.

DeltaG^(@) for the reaction X+Y hArr Z " is " -4.606 kcal. The equilibrium constant for the reaction at 227^(@)C is

Answer»

100
10
2
0.01

Solution :`DELTA G^(@)= -2.303 RT " LOG K"`
`-4.606xx10^(3)= -2.303xx2xx50 log K_(e )`
`THEREFORE K_(e )=100`
Discussion
15.

DeltaG for the reaction (4)/(3)Al+O_(2)to(2)/(3)Al_(2)O_(3) is -772kJ mol^(-1) of O_(2). Calculate the minimum EMF in volts requird to carry out and electrolysis of Al_(2)O_(3)

Answer»

Solution :`AltoAl^(3+)+3E^(-)`
`(4)/(3)MOL" of "Al=(4)/(3)xx3mol" "E^(-)=4mol" "e^(-)`
n=4
`DeltaG=-nFE`
`-772xx1000J=-4xx96500xxE""thereforeE=(772xx1000)/(4xx96500)=2.0V`
Discussion
16.

DeltaG = DeltaH-TdeltaS and DeltaG = DeltaH+ T[(d(DeltaG))/(dT)]" then " ((dE_("cell"))/(dT)) is

Answer»

<P>`(DeltaS)/(nF)`
`(nE)/(DeltaS)`
`-nFE_("cell")`
`+nFE_("cell")`

Solution :`(d(DELTAG))/(DT) = -DeltaS , DeltaS = nFE_("cell") [ (d(DeltaG))/(dT)]_(p) = -nF (dE_("cell"))/(dT), -nF (dE_("cell"))/(dT)=DeltaS , ((dE_("cell"))/(dT)) = (DeltaS)/(nF)`
Discussion
17.

Delta_(f)U^(@) of formation of CH_(4)(g) at certain temperature is - 393 kJ mol^(-1). The value of Delta_(f)H^(@)

Answer»

ZERO
`lt Delta_(f)U^(@)`
`gt Delta_(f)U^(@)`
equal to `Delta_(f)U^(@)`

Solution :`Delta_(f) H^(@) = Delta_(f) U^(@) + DELTA n_(G) RT`
For the reaction,
`C(s) + 2H_(2)(g) rarr CH_(4)(g)`
:. `Delta n_(g) = 1 - 2 = - 1`
:. `Delta_(f) H^(@) = Delta_(f) U^(@) - 1 RT`
or `Delta_(f) U^(@) = Delta_(f) H^(@) + RT`
or `Delta_(f) H^(@) lt Delta_(f) U^(@)`
Discussion
18.

Delta_(f)U^(ө) for the formation of CH_(4)(g) at certain temperature is -393 kJ mol^(-1). The value of Delta_(f)H^(ө) is

Answer»

zero
`LT Delta_(f)U^(ө)`
`gtDelta_(f) U^(ө)`
EQUAL to `Delta_(f)U^(ө)`

Solution :`CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l)`
`Deltan_(g)=(n_(p)-n_(r))=1-3=-2`
`Delta_(f)H^(ө)=Delta_(f)U^(ө)+Deltan_(g)RT`
`Deltan_(g)=2Delta_(f)H^(ө) lt Delta_(f) U^(ө)`
Discussion
19.

Delta_(f)G^(@) at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcal mol^(-1) and +103 kcal mol^(-1) respectively. Vapour pressure of liquid 'S' at 500 K is approximately equal to (R=2 cal k^(-1) mol^(-1))

Answer»

<P>`10` ATM
`0.1` atm
`1` atm
`100` atm

Solution :`DeltaG_("solid" to "liquid")=2.303 RT LOG.(p_(2))/(p_(1)),p_(1)=1 atm,p_(2)=?`
Discussion
20.

Delta_(f)U^(-) of formation of CH_(4) (g) at certain temperature is -393 kJ mol ^(-1) . The value of Delta_(f)H^(-)

Answer»

ZERO
`ltDelta_(f)U^(-)`
`gtDelta_(f)U^(-)`
EQUAL to `Delta_(f)U^(-)`

SOLUTION :`Delta_(f)H^(-) = Delta_(f)U+ pDeltaV. So Delta_(f)H^(-)ltDelta_(f)U^(-)`
Discussion
21.

Delta_(f) H^(theta) of CO_(2)(g), CO(g), N_(2)O(g) and NO_(2)(g) are respectively -393, -110.81 and 34 ("in "kJ mol^(-1)) at 27^(@)C. Delta H (in kJ) for the reaction : 2NO_(2)(g) + 3CO(g) rarr N_(2)O(g) + 3CO_(3)(g) is

Answer»

836
1460
`-836`
`-1460`

Solution :`DELTA H =Delta H_(F^(@)) (N_(2)O) + 3 Delta H_(f^(@)) (CO_(2)) - (2 Delta H_(f^(@)) (NO_(2)) + 3 Delta H_(f^(@)) (CO_(2))`
`= 81 + 3 (-393) - (2 xx 34 + 3 (-110))`
`= - 836 kJ MOL^(-1)`
Discussion
22.

Delta_f H of HCl is __________.

Answer»

`-185`KCAL
185 kcal
92.5 kcal
`-92.5`kcal

Answer :C
Discussion
23.

DeltaE^(@) of combustion of isobutylene is - X kJ mol^(-1). The value of DeltaH^(@) is

Answer»

`=DeltaE^(@)`
`GT DeltaE^(@)`
`=0`
`lt DeltaE^(@)`

SOLUTION :`(CH_(3))_(2)C=CH_(2)(G)+6O_(2)(g)rarr4CO_(2)(g)+4H_(2)O(l)`
`Deltan_(g)=4-6=-2` (i.e., NEGATIVE)
we know that `DeltaH=DeltaE+Deltan_(g)RT`
`=DeltaE-(Deltan_(g))RT (because Deltan_(g)=-ve)thereforeDeltaH lt DeltaE`.
Discussion
24.

(Delta U - Delta H) for the formation of carbon monoxide (CO) from its elements of 298 K is (R = 8.314 JK^(-1) mol^(-1))

Answer»

`1238.78 J mol^(-1)`
`- 2477.57 J mol^(-1)`
`2477.57 J mol^(-1)`
`- 1238.78 J mol^(-1)`

SOLUTION :`C(s) + (1)/(2) O_(2)(G) rarr CO(g)`
`Delta H = Delta U + Delta n_(g) RT`
`(Delta H - Delta U) = (1)/(2) xx 8.314 xx 298`
`= 1238.75 J mol^(-1)`.
Discussion
25.

(Delta U - Delta H) for the formation of NH_(3) from N_(2) and H_(2) is :

Answer»

`- 2 RT`
2 RT
RT
`(1)/(2) RT`.

ANSWER :C
Discussion
26.

Delta S_("surr") for an exothermic reaction is

Answer»

always positive
always negative
zero
may be positive or negative

SOLUTION :In EXOTHERMIC processes, heat released by the reaction increases the entropy of the SURROUNDING. The overall entropy change is CERTAINLY positive when the entropy of the system is positive. In some exothermic reactions, entropy of the system may decrease. If reaction is highly exothermic and increase in entropy of the surroundings is very high, the total entropy change will be positive and the reaction will be spontaneous.
Discussion
27.

Delta S^(0) will be highest for the reaction

Answer»

`N_(2)(g) + O_(2)(g) rarr 2NO (g)`
`CaCO_(3)(s) rarr CaO(s) + CO_(2)(g)`
`C(s) + O_(2)(g) rarr CO_(2)(g)`
`Ca(s) + (1)/(2) O_(2)(g) rarr CaO (s)`

ANSWER :A
Discussion
28.

Delta is formed by the combination of

Answer»

COLLOIDAL waters of RIVER and SEA.
electrolytic waters of river and sea.
colloidal river water and electrolytic sea water.
electrolytic river water and colloidal sea water.

Solution :colloidal river water and electrolytic sea water.
Discussion
29.

Delta H_(f) (H_(2)O) = X, Heat of neutralisation of CH_(3)COOH and NaOH will be

Answer»

LESS than 2X
Less THA X
X
Between X and 2X

Solution :Acetic ACID is a weak acid.
Discussion
30.

Delta H (vap) for water is 40.7 kJ mol^(-1). The entropy of vaporisation of water is :

Answer»

`- 40.7 kJ mol^(-1) K^(-1)`
`407 J mol^(-1) K^(-1)`
`109 J mol^(-1) K^(-1)`
`722 J mol^(-1)`

Solution :`DELTA S = (Delta_("vap") H)/(T_(b)) = (40.7 xx 10^(3))/(373)`
`= 109 JK^(-1) mol^(-1)`
Discussion
31.

Delta H for transition of carbon in the diamond form to carbon in the graphite form is -453.5 cal. This suggests that :

Answer»

GRAPHITE is CHEMICALLY DIFFERENT from DIAMOND
Graphite is as STABLE as diamond
Graphite is more stable than diamond
Diamond is more stable than graphite

Solution :`C_("Diamond")rarr C_("Graphite") , Delta H =-453.5` Cal
In exothermic reactions products are more stable than reactans, so graphite is more stable than diamond.
Discussion
32.

Delta H for the reaction, SO_(2) (g) + (1)/(2) O_(2)(g) hArr SO_(3)(g) " " Delta H = -98.3kJ If the enthalpy of formation of SO_(3)(g) is -395.4kJ, then enthalpy of formation of SO_(2)(g) is:

Answer»

`-297.1kJ`
`493.7kJ`
`-493.7kJ`
`297.1kJ`

Solution :`Delta H = Delta_(f) H(SO_(3)) - [Delta_(f) H(SO_(2)) + (1)/(2) Delta_(f) H(O_(2))]`
`-98.3 = -395.4 - [Delta_(f) H(SO_(2)) + 0]`
or `Delta_(f)H(SO_(2)) = -395.4 + 98.3 = -297.1 kJ`
Discussion
33.

Delta G^(0) value for the formation of A_(2)O and B_(2)O are - 827 kJ and - 540 kJ. Which one of them is more stable ?

Answer»

SOLUTION :`A_(2) O`
Discussion
34.

Delta G^(@) vs T plot in the Ellingham's diagram slopes downward the reaction

Answer»

`Mg + (1)/(2)O_(2) rarr MgO`
`C + (1)/(2)O_(2) rarr CO`
`2Ag + (1)/(2)O_(2) rarr Ag_(2)O`
`CO + (1)/(2)O_(2) rarr CO_(2)`

ANSWER :B
Discussion
35.

DeltaG^@ vs T plot in the Ellingham's diagram slopes downward for the reaction

Answer»

`Mg+(1)/(2)O_(2)to MgO`
`2AG+(1)/(2)O_(2)to Ag_(2)O`
`C+(1)/(2)O_(2)to CO`
`CO+(1)/(2)O_(2)to CO_(2)`

Solution :
Reaction (i) : `M_((s))+(1)/(2)O_(2(g))rarr MO_((s))`(where M represents metal)
Reaction (ii): `CO_((g))+(1)/(2)O_(2(g))rarr CO_(2(g))`
Reaction (III) : `C_((s))+(1)/(2)O_(2(g))`
`Delta G^(@) vs` T PLOT in the Ellingham's diagram slopes downwards for the reaction (i) and reaction (ii).For reaction (iii), `Delta G^(@)vs` T plot in Ellingham's diagram slopes upwards.
Discussion
36.

Delta G^(@)vs T plot in theEllingham's diagram slopes downward for thereaction

Answer»

` Mg +(1)/(2) O_2toMg O `
` 2Ag +(1)/(2) O_2toAg_ 2O `
` C +(1)/(2)O_2 to CO`
` CO+ (1)/(2) O_2 toCO_ 2`

Solution :Forthereaction,
` C + 1//2O_ 2toCO`,the` Delta S `INCREASES. Therefore,as thetemperatureincreases ` T Delta S `increases andhence `Delta G(Delta H -T Delta S) `decreases. Inother WORDS,the SLOPE ofthecurveforformation of CO decreases. However, FORALL otheroxides, itincreases.
Discussion
37.

Delta G^(@) vs T plot in the Ellingham's diagram slopes downward for the reaction

Answer»

`MG+(1)/(2)O_(2) to MgO`
`2AG+(1)/(2)O_(2)to Ag_(2)O`
`C+(1)/(2)O_(2)to CO`
`CO+(1)/(2)O_(2)to CO_(2)`

ANSWER :C
Discussion
38.

Delta G^(@) for the reaction Ag_((aq))^(+)+1/2H_(2(g))rarrH_((aq))^(+)+Ag_((s)), where standardpotential for silver half cell reaction in 0.8 V, will be

Answer»

`-77.2` KJ
`+77.2` kJ
`154.4` kJ
`-38.6` kJ

Answer :A::B
Discussion
39.

Delta G = Delta H -T Delta S was given by

Answer»

Faraday
Kirchhoff
Einstein
Gibbs-Helmholtz

Solution :Gibbs-Helmholtz
Discussion
40.

Delta at the rivers are formed due to ....

Answer»

peptisation
COAGULATION
hydrolysis
precipitation

Solution :coagulation
Discussion
41.

Deionised water is produced by

Answer»

Calgon process
Ion-exchange resin process
Clark's process
Permutit process

Solution :Deionised water is water which has no solids DISSOLVED in it . One way it is to distil the water . The water which comes over is CALLED the distillate and it has no ions in it hence is deionised water . A second way to make deionised water is to pass water through an ion-exchange column . The latter is composed of insoluble CHEMICALS which remove both positive and negative ions from the water leaving the RESULTING water without ions.
Discussion
42.

Dehydrohalogenation will not occur in

Answer»




SOLUTION :Because of the ABSENCE of `BETA`-HYDROGEN.
Discussion
43.

Dehydrohalogenation of 3-bromohexane gives a mixture of cis-2-hexene and trans-2-hexene. How can this mixture be converted to pure (a) cis-2-hexene? (b) trans-2-hexene?

Answer»

Solution :Relatively pure alkenes (geometric ISOMERS) are prepared by stereoselective reduction of ALKYNES. <BR> (a) Hydrogenation of 2-hexyne with Lindlar.s catalyst gives cis-2-hexene.
`CH_3CH=CHCH_2CH_2CH_3overset(Br_2)rarrCH_3underset(Br)underset(|)CH-underset(Br)underset(|)CHCH_2CH_2CH_3overset(2NaNH_2)rarr`

(b) Reduction with Na in liquid `NH_3` gives the trans PRODUCT.
Discussion
44.

Dehydrohalogenation of 1,2-dibromo-butane with alc KOH gives:

Answer»

1-butyne
2-butene
1-butene
1-bromo-1-butene

Answer :A
Discussion
45.

Dehydrohalogenation in monohaloalkanes produces

Answer»

A SINGLE bond
A double bond
A TRIPLE bond
Fragmentation

Solution :`H-underset(H)underset(|)overset(H)overset(|)(C)-underset(Cl)underset(|)overset(H)overset(|)(C)-H underset(-HCl)overset("Dehydrohalogenation")tounderset("Alkene")(H-overset(H)overset(|)(C)=overset(H)overset(|)(C)-H+HCl)`
Discussion
46.

Dehydrohalogenation in haloalkanes produces:

Answer»

A SINGLE BOND
A DOUBLE bond
A TRIPLE bond
Fragmentation

Answer :B
Discussion
47.

Dehydrohalogenation by strong base is slowest in

Answer»




ANSWER :C
Discussion
48.

Dehydrohalogenation by strong base is slowest in :

Answer»




Solution :Compound (C) has all EQUITORIAL substituent hence it ismost STABLE and UNDERGO ELIMINATION with slowest STATE.
Discussion
49.

Dehydrogenation of isopropyl alcohol gives

Answer»

METHANOL
METHANAL
ETHANAL
PROPANONE

ANSWER :D
Discussion
50.

Dehydration of 2-butanol gives

Answer»

2-butene
Butanone
Butyraldehyde
1-butene

Answer :B
Discussion