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 the reaction C(s)+ CO_2(g) hArr 2CO(g) the partial pressure of CO_2 and CO are 4 and 8 atm respectively K_p For the reaction is :

Answer»

16 atm
2 atm
5 atm
4 atm

Answer :A
2.

For the reaction C_((s))+CO_(2(g)) hArr 2CO_((g)) the partial pressures fo CO_2" and CO are 2.0 and 4.0 atm respectively at equilibrium. Kp for the reaction is

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`0.5`
`8.0`
`4.0`
`32`

ANSWER :B
3.

For the reaction CrO_(4)^(2-)+?rarrCr_(2)O_(7)^(-2)the missing ion is

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`OH^(-)`
`H^(+)`
`OH^(+)`
`O^(2-)`

ANSWER :B
4.

For the reaction: CO_((g))+Cl_(2(g))iffCOCl_(2(g)),K_(p)//K_(c) equal to

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`1//RT`
`RT`
`sqrt(RT)`
`1.0`

SOLUTION :`Deltan_(g)=1-2=-1`
`K_(p)=K_(c)(RT)^(Deltan)=K_(c)(RT)^(-1)=K_(c)//RT`
or, `K_(p)//K_(c)=1//RT`
The given reaction is endothermic. So on increasing TEMPERATURE, it will SHIFT in the forward direction.
5.

For the reaction CO_((g))+H_(2)OhArrCO_(2(g))+H_(2(g)) at a given temperature, the equlibrium amount of CO_(2(g))can be increased by

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Adding a suitable catalyst
Adding an inert GAS
DECREASING the volume of the container
Increasing the amount CO(g)

Solution :Accoeding to Le-chatelier's principle.
6.

For the reaction, CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g), at a given temperature, the equilibrium amount of CO_(2)(g) can be increased by

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ADDING a suitable catalyst
adding an inert gas
decreasing the VOLUME of the container
increasing the amount of CO(g)

Answer :D
7.

For the reaction,CO_((g))+Cl_(2(g))hArrCOCl_(2(g))the K_(p)//K_(c) is equal to

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`sqrt(RT)`
RT
`1//RT`
`1.0`

Solution :`CO_((g))+Cl_(2(g))hArrCOCl_(2(g))`
`Deltan=1-2=-1`
`K_(p)=K_(C)[RT]^(Deltan), therefore(K_(p))/(K_(c))=[RT]^(-1)=(1)/(RT)`
8.

For the reaction , CO(g)+Cl(g)hArrCOCl_2(g) then K_p//K_c is equal to :

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RT
`SQRT(RT)`
`1/(RT)`
`1.0`

ANSWER :C
9.

For the reaction CO_((g))+2H_(2(g))hArrCH_(3)OH_((g)), true condition is

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`K_(p)=K_(C)`
`K_(p)gtK_(c)`
`K_(p)ltK_(c)`
`K_(c)=0` but ` K_(p)ne0`

Solution :When `n_(r)gt n_(p)then K_(p)ltK_(c) where n_(r)=` no. of MOLES of reactant `n_(p)=` no. of moles of PRODUCT.
10.

For the reaction :CO(g)+Cl_(2)(g)hArr COCl_(2)(g)the K_(p)//K_(c )is equal to :

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<P>`sqrt(RT)`
RT
1/RT
`1.0`

SOLUTION :`K_(p)=K_(C )(RT)^(Delta n_(G))`
`Delta n_(g)=1-(1+1)=-1`
`(K_(p))/(K_(c ))=(RT)^(-1)=(1)/(RT)`
11.

For the reaction CO_((g))+(1)/(2)O_(2(g))rarrCO_(2(g)),DeltaH and DeltaS " are "-283 kJ and -87 JK^(-1).Respectively. It was intended to carry out this reaction at 1000, 1500, 3000 and 3500 K. At which of these temperature would this reaction be thermodynamically spontaneous

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1500 and 3500 K
3000 and 3500 K
1000, 1500 and 3000 K
1500, 3000 and 3500 K

Solution :`DELTAG=DeltaH-TDeltaS`
For SPONTANEOUS reaction `DeltaG` should be negative
When T=1000 K, `DeltaG`=(-283 -1000 `xx` 0.087) kJ =-370 kJ
When `T=1500 K, DeltaG=(-283-1500xx0.087) kJ =-413.5 kJ`
`T=3000 K, DeltaG=(-283-3000xx0.087) kJ=-544kJ`
When `T=3500 K, DeltaG=(-283-3500xx0.087) kJ=+21.5kJ`
Hence at 3500 K, reaction will be non-spontaneous.
12.

For the reaction CO (g) + (1)/(2) O_(2) (g) rarr CO_(2) (g) Delta H and Delta S are -283 kJ and -87 J K^(-1), respectively. It was intended to carry out this reaction at 1000,1500,3000, and 3500 K. At which of these temperatures would this reaction be thermodynamically spontaneous?

Answer»

1500 and 3500K
3000 and 3500 K
1000, 1500 and 3000 K
1500, 3000 and 3500 K

Solution :`:.DeltaG-DeltaH-TDeltaS`
For a spontaneous reaction `DeltaG` should be negative
`DeltaH=-234kJ,DeltaS=-87JK^(-1)`
Hence reaction will be spontaneous when `DeltaHgtT.DeltaS`.
THEREFORE at 1000, 1500 and 3000 K the reaction would be spontaneous.
13.

For the reaction :CO(g)+2H_(2)(g)hArr CH_(3)OH(g)

Answer»

<P>`K_(p)=K_(c )`
`K_(p)GT K_(c )`
`K_(p)LT K_(c )`
`K_(c ) = 0` but`K_(p) ne 0`

Solution :`K_(p)=K_(c )(RT)^(Delta n), Delta n =1-(1+2)=-2`
`K_(p)=K_(c )(RT)^(-2)`
`therefore K_(p)lt K_(c )`
14.

For the reaction CO_((g))+1/2O_(2(g))hArrCO_(2(g)),(K_(p))/(K_(c)) is equivalent to

Answer»

1
RT
`(1)/(sqrt(RT))`
`(RT)^(1//2)`

Solution :For `CO+1/2O_(2)hArrCO_(2)`
`K_(p)=K_(C)(RT)^(1-1(1)/(2))=K_(c)(RT)^(-1/2),(K_(p))/(K_(c))=sqrt((1)/(RT))`
15.

For the reaction CO(g)+1/2 O_(2) (g) rarr CO_(2)(g) using data given in table find out incorrect statement(s) among the following. {:(,DeltaH_(f)^(@)"(kJ/mole)",S^(@)"(J/Kmole)"),(CO(g),-110,+197),(O_(2)(g),0,+205),(CO_(2)(g),-395,+213):} Assume vibration modes of motion do not contribute to heat capacity at low temperature.

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<P>`DeltaH^(@) gt DeltaU^(@)` for the reaction at 298 K.
In standard state condition, the reaction `CO(g)+1/2 O_(2) (g) rarr CO_(2)(g)` attain equilibrium at very high temperature.
At low temperature `(D(DELTA H)^(@))/(dT) = - ve`
In a `CO, O_(2)` fuel cell electrical ENERGY obtained by cell `gt |Delta H_("COMBUSTION")^(@) [CO (g)]|`

Solution :`DeltaH^(@)-DeltaU^(@)=Deltan_(g)RT`
since `Deltan_(g)=-ve`
`rArr DeltaH^(@)-DeltaU^(@) le 0`
`DeltaH^(@)=-ve`
`DeltaS^(@)=-ve"",""C_(p)=3/2R+R"",""=5/2 R"",""Delta_(r)C_(p)=-ve`
16.

For the reaction CO(g)+(1//2)O_(2)(g)=CO_(2)(g),K_(p)//k_(e ) is

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<P>RT
`(RT)^(-1)`
`(RT)^(-1//2)`
`(RT)^(1//2)`

Solution :`K_(p)=K_(E )(RT)^(DELTAN)`,
`Delta n=1-(1+(1)/(2))=1-(3)/(2)=-(1)/(2) THEREFORE (K_(p))/(K_(e ))=(RT)^(-1//2)`
17.

For the reaction :CO(g)+1//2O_(2)(g)hArr CO_(2)(g)the partial pressures are :p(O_(2))=0.24atm, p(CO)=0.4 atm and p(CO)=0.04 atm.The equilibrium constant K_(p)is :

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<P>`0.3`
`3.0`
`9.0`
`0.03`

SOLUTION :`K_(p)=((p CO_(2)))/(p(CO)xx [p(O_(2))]^(1//2))`
`=(0.24)/(0.4xx(0.04)^(1//2))=3.0`
18.

For the reaction CO (g) + (1)/(2) O_(2) (g) rarr CO_(2) (g) using data given in table find out incorrect statement (s) among the following {:(,Delta H_(f)^(@) (kJ//"mole"),S^(@) (J//K"mole"),),(CO (g),-110,+197,),(O_(2) (g),0,+205,),(CO_(2) (g),-395,+213,):} Assume vibration modes of motion do not contribute to heat capacity at low temperature.

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`Delta H^(@) GT Delta U^(@)` for the reaction at 298 K
In standard state conditin, the rection `CO (g) + (1)/(2) O_(2) (g) rarr CO_(2) (g)` attain equilibrium at very HIGHT temperature.
At low temperature `(d(Delta H)^(@))/(DT) = - ve`
In a `CO, O_(2)` fuel cell electrical energy obtained by cell `gt |Delta H_("combustion")^(@) [CO (g)]|`

Solution :`DeltaH^(@)-DeltaU^(@)=Deltan_(g)RT`
since `Deltan_(g)=-ve`
`rArr DeltaH^(@)-DeltaU^(@) lt 0`
`DeltaH^(@)=-ve`
`DeltaS^(@)=-ve"",""C_(p)=3/2 R+R" ",""=5/2 R`
`Delta_(r)C_(p)=-ve`
19.

For the reaction CO_2(g) +H_2 hArr CO(g) +H_2O(g) The K_p for the reaction is 0.11 if the reaction was started with 0.45 moles of CO_2 and 0.45 moles of H_2 at 700k ,the concentration of CO_2 when 0.34 mole of CO_2 and 0.34 mole of H_2 are added when the first equilibrium is attained is

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0.52 M
0.17 M
0.34 M
0.26 M

Answer :B
20.

For the reaction : Cl_(2(g))rarr2Cl_((g))

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`DELTA H` is POSITIVE, `Delta S` is positive
`Delta H` is positive, `DeltaS` is NEGATIVE
`DELTAH` is negative, `DeltaS` is negative
`DeltaH` is negative, `DeltaS` is positive

Answer :A::D
21.

For the reaction: Cl_(2)(g) + 2NO(g) to 2NO_(2)F(g) The following mechanism is suggested. i) NO_(2) + F_(2) to NO_(2)+F (slow) ii) NO_(2) + F to NO_(2)F+F (fast) What is the predicted rate law?

Answer»

SOLUTION :ORDER of REACTION =3
22.

For the reaction Cl_(2)(g)+2NO(g) to 2NOCl (g), the rate law is expressed as : rate =k[Cl_(2)][NO]^(2). What is the overall order of this reaction?

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SOLUTION :ORDER = 3.
23.

For the reaction : Cl_(2)(g) to 2Cl_(g)

Answer»

`DELTAH` is POSITIVE, `DELTAS` is positive.
`DeltaH` is POSTIVE, `DeltaS` is negative.
`DeltaH` is negative, `DeltaS` is negative
`DeltaH` is negative, `DeltaS` is positive

ANSWER :A
24.

For the reaction Cl_(2)+2I^(-)toI_(2)+2CI^(-), the initial concentration of I^(-) was 0.20 mol "lit"^(-1) and the concentration after 20 min was 0.18 mol "lit"^(-1). If the rate of formation of I_(2) in mol "lit"^(-1) "min"^(-1) is x xx 10^(-4) then the value of x is

Answer»


SOLUTION :`(r_(I^(-)))/2=(r_(I_(2)))/1,r_(I_(2))=0.001/2=5xx10^(-4),r_(I^(-))=(0.20-0.18)/20=0.02/20=0.001`
25.

For the reaction CH_(4(g))+2O_(2(g))hArrCO_(2(g))+2H_(2)O_((1)), ""DeltaH=-170.8kJmol^(-1) Which of the following statements is not true

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Adoeon of `CH_(4(g))on O_(2(g))` at equlibrium will cause a shift to the right
The reaction is EXOTHERMIC
At equlibrium, the CONCENTRATIONS of `CO_(2(g))and H_(2)O_((1))` are not equal
The equlibrium CONSTANT for the reaction is given by `K_(p)=([CO_(2)])/([CH_(4)][O_(2)])`

Answer :D
26.

For the reaction CH_3COCH_3+I_2 overset(H^+)to products. The rate equation is,(dx)/(dt) =k[CH_3COCH_3][H^+]. The order of the reaction with respect to iodine is

Answer»

ONE
two
half
zero

SOLUTION :`I_2` does not involved in RATE EQUATION Hence order to REACTION with respect to `I_2` is zero
27.

For the reaction (CH_(3))_(2)CH-CH=CH_(2)+C_(2)H_(5)OH to Product the catayst that can be used is

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DILUTE `H_(2)SO_(4)`
Dilute `NAOH`
Dilute NaCl
None of these

Solution :
28.

For the reaction CH_(2)=CH_(2)+HXtoCH_(3)CH_(2)X the orther of reactivity of HX is ,

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`HIgtHClgtHBr`
`HClgtHBrgtHI`
`HIgtHBrgtHCl`
`HBrgtHClgtHI`

Solution :Reactivity DEPENDS upon the BOND length. HCl bond is SHORTER and stronger, hence it is LESS reactive. HI bond islonger than HCl and HBR, hence HI is more reactiv than HCl and HBr.
29.

For the reaction, C_("graphite")+(1)/(2)O_(2)(g)=CO(g) at 298 K and 1 atm, DeltaH=-26416 cal. If the molar volume of graphite is 0.0053 litre, calculate DeltaH.

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ANSWER :(-26712 CAL)
30.

For the reaction, CaCO_(3)(s) iff CaO(s)+CO_(2)(g)partial pressure of CO_(2) at 1000 K is 0.003 atm. DeltaG^(@)=27.2 kcal. Calculate the value of DelaG

Answer»

12.6 kcal
15.6 kcal
13.4 kcal
14.2 kcal

Solution :`CaCO_(3)(s)iffCaO(s)+CO_(2)(G)`
`K_(p)=P_(CO_(2))=0.003 atm`
`DeltaG=DeltaG^(@)+2.303 RT LOG K_(p)`
`=27.2+2.303xx2xx1000xxlog 0.003`
`=15.6 kcal (R=2 cal mol^(-1)K^(-1))`.
31.

For the reaction CaCO_3(s) hArr CaO(s) +CO_2(g), the pressure of CO_2 depends on

Answer»

the MASS of `CaCO_3(s)`
the mass of `CAO(s)`
the MASSES of the `CaCO_3(s) and CaO(s)`
TEMPERATURE of the system

ANSWER :D
32.

For the reaction, CaCO_3(s)rarrCaO(s)+CO_2(g)

Answer»

`TRIANGLEH=TRIANGLEU`
`triangleHlttriangleU`
`triangleH!=triangleU`
`triangleH=0`

ANSWER :C
33.

For the reaction C_(6)H_(12)(l)+9O_(2)(g) rarr 6H_(2)O(L)+6CO_(2)(g) DeltaH_(298)=936kcal which of the following is true

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`-936.9 =Delta E -(2 xx 10^(-3)xx298 xx 3)KCAL`
`-936.9 =Delta E -(2 xx 10^(-3)xx298 xx 2)kcal`
`-936.9 =Delta E +(2 xx 10^(-3)xx298 xx 2)kcal`
`-936.9 =Delta E -(0.0821xx 10^(-3)xx298 xx 3)kcal`

SOLUTION :Factual question.
34.

For the reaction, CaCO_3 (s) to CaO(s) +CO_2(g) , which is correct representation ?

Answer»

<P>`K_p =(P_(CO))_2`
`K_p=K_c(RT)`
`K_c=(CO_2)//1`
All of these

Answer :D
35.

For the reaction. C_(3)H_(8)(g) + 5O_(2)(g) rarr3CO_(2)(g)+4H_(2)O(l) at constant temperature, DeltaH - DeltaE is

Answer»

`- RT`
`+ RT`
`- 3 RT`
`+ 3 RT`

SOLUTION :`DeltaH-DeltaE=DeltanRT,Deltan=-3`
`"so, "DeltaH-DeltaE=-3RT`
36.

For the reaction : C_(3)H_(8)(g) + 5O_(2) rarr 3CO_(2)(g) + 4H_(2)O(l) a constant temperature, Delta H - Delta U is

Answer»

`+ RT`
`- 3RT`
`+ 3RT`
`- RT`

SOLUTION :`DELTA n_(g) = n_(p) - n_(r) = 3 - 6 = - 3`
`Delta H = Delta U + Delta n_(g) RT`
`Delta H - Delta U = - 3RT`
37.

For the reaction : C_2H_4(g)+5O_2(g)→3CO_2(g)+4H_2O(l)at constanttemperature. triangle H - triangleU is:

Answer»

RT
-3RT
`=3RT`
-RT

Answer :B
38.

For the reaction C_2H_5OH + HX overset(ZnX_2)to C_2H_5X. The order of reactivity is

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

Answer :C
39.

For the reaction, C_(2)H_(5)OH + HX overset(ZnCl_(2))to C_(2)H_(5)X+ H_(2)O, where HX is a halogen acid. The order of reactivity of halogen acids for their reaction is :

Answer»

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

ANSWER :D
40.

For the reaction, C_2H_4(g)+3O_2(g)rarr2CO_2(g)+2H_2O(l), triangleU=-1414kJ. Then triangleH at 27^@C is:

Answer»

`-1410 KJ`
`-1420 kJ`
`+1420 kJ`
`+1410 kJ`

ANSWER :B
41.

For the reaction : C_(2)H_(4)(g) + 3 O_(2)(g) rarr 2 CO_(2)(g) + 2 H_(2)O(l) at 298 K, Delta U = - 1415 kJ. If R = 0.0084 kJ K^(-1). Then Delta H is equal to

Answer»

`- 1400 kJ`
`- 1410 kJ`
`- 1420 kJ`
`- 1430 kJ`.

Solution :`DELTA H = Delta U + Delta n_(G) RT`
`Delta n_(g) = 2 -1 -3 = -2`
`Delta H = - 1415 - 2 xx 0.0084 xx 298`
`= - 1420 kJ`
42.

For the reaction, C_2H_5OH+HX overset(ZnX_2)rarrC_2H_5X,the order of reactivity is:

Answer»

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

Answer :B
43.

For the reaction, C_(2) H_(4) (g) + 3O(2) (g) to 2 CO_(2) (g) + 2H_(2) O(I), Delta E = -1415 KJ. then Delta H at 27^(@) C is

Answer»

`+ 140 KJ`
`- 1420 kJ`
`+ 1420 kJ`
`- 1410 kJ`

ANSWER :C
44.

For the reaction: C_(12) H_(22)O_(11) +H_(2)O overset(H^(+))to C_(6)H_(12)O_(6)+C_(6)H_(12)O_(6) write: (i) rate of reaction expression. (ii) rate law equation. (iii) molecularity. (iv) order of reaction.

Answer»

Solution :(i) RATE `=(-d[C_(12)H_(22)O_(11)])/(dt)= -(d[H_(2)O])/(dt)= (d[C_(6)H_(12)O_(6)])/(dt)= (d[C_(6)H_(12)O_(6)])/(dt)`
(ii) Rate `=k[C_(12)H_(22)O_(11)]`
(iii) Two
(iv) The reaction is of first ORDER because `H_(2)O` is taken in such large EXCESS that its concentration does not change with time.
45.

For the reaction, bondenergies are given as under – (i) C—C, 346 kJ/mol (ii) C—H, 413 kJ/mol (iii) H—H, 437 kJ/mol and (iv) C = C, 611 kJ/mol What will be the value of Delta H at 25^(@) C for the above reaction ?

Answer»

`-289 kJ mol^(-1)`
`-124 kJ mol^(-1)`
`+124 K J mol^(-1)`
`+ 289 k J mol^(-1)`

ANSWER :B
46.

for the reactionC+ O_(2) to CO_(2)

Answer»

`DeltaH gt DELTAE`
`DeltaHlt DeltaE`
`DeltaH=DeltaE`
NONE of these

Solution :for the REACTION `Deltan=1-1=0 `
`DeltaH=DeltaE`
47.

For the reaction between CO_(2) and graphite CO_(2)(g)+C(s)to2CO(g) DeltaH=+170.0kJandDeltaS=170JK^(-1). The reaction is spontaneous at -

Answer»

298 K
500 K
900 K
1200 K

Answer :D
48.

For the reaction between CO_2and graphite,CO_2(g) +C(s) to 2CO(g) , DeltaH = 170 kJ " and " Delta S = 179 JK^(-1)At equilibrium, the reaction will be non-spontaneous at

Answer»

300K
500K
900K
1100K

Solution :For EQUILIBRIUM `DELTA G = 0= Delta H- T Delta S`
` T = (Delta H)/(Delta S) = (170 xx 10^3)/(170) = 1000 K`
The reaction would be non-spontaneous at TEMPERATURE less than 1000 K, so as to have `Delta G`positive. 5
49.

For the reaction between A andB,the relationship between the concentration of A and B and rate is given by the reaction is

Answer»

first order with RESPECT to A
second order with respect to A
first order with respect to B
third order overall

Answer :B::C::D
50.

For the reaction below, the product is Q. The compound Q is

Answer»




SOLUTION :