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24351.

The reaction A to B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B ?

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1 hour
0.5 hour
0.25 hour
2 hours

Solution :(a) : The FREACTION of A REACTED in each case is same `(0.2//0.8=1//4),(0.9-0.675)//0.90=0.225//0.90=1//4)`. HENCE, time TAKEN is same.
24352.

The reaction, A + 2B to C +D obeys rate equation, Rate =k[A]^(x)[B]^(y) What would be the order of the reaction?

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SOLUTION :SELF EXPLANATORY
24353.

The reaction : 3O_(2)hArr2O3, DeltaH= + 69 , 000 calories is aided by :

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high TEMPERATURE and low pressure
high temperature and high pressure
low temperature and high pressure
low temperature and low pressure

Solution :ACCORDING to LE Chatalier PRINCIPLE formation of ozone is FAVOURED by high temperature (endothermic reaction ) and high pressure.
24354.

The reaction A overset(k)(to) Product, is zero order while the reaction Boverset(k)(to) Product, is first order reaction. For what initial concentration of A are the half lives of the two reactions equal ?

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`(log_(e)4)M`
`2M`
`log 2M`
`ln 2M`

Solution :For ZERO order reaction, `x=kt`
`:.(a)/(2)xxkxxt_(1//2)`, i.e. `t_(1//2)=(a)/(2K)`……..`(i)`
For first order reaction, `t_(1//2)=(log_(e )2)/(K)` ……….`(ii)`
From `(i)` and `(ii)` `(a)/(2k)=(log_(e)2)/(k)`
`a=log_(e )4M`
24355.

The reaction : A rarr Bfollows first order kinetics . The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 hour . What is the time taken for conversion of 0.9 mol of A to produce 0.675 molof B ?

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1 HOUR
0.5 hour
0.25 hour
2 hours

Solution :(A) The FRACTION of A reacted in each case is
same `i.e (0.6)/(0.8) = (3)/(4) ` and `(0.675)/(0.90) = (3)/(4)`
`:.` Time taken is same .
24356.

The reaction, 3ClO^(-) (aq) rarr ClO_(3)^(-) (aq)+2Cl^(-) (aq), is an example of

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OXIDATION reaction
Reduction reaction
Disproportionation reaction
Decomposition reaction

Solution :In the given reaction, the oxidtion state of the chlorine atom change from +1 (in `ClO^(-)`) to +5 (in `ClO_(3)^(-)`) and -1 (in `Cl^(-)`). This means that chlorine is getting oxidized as WELL as reduced. Such REACTIONS are CALLED disproportionation reactions
24357.

The reaction, A + B to P can never be

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SECOND ORDER reaction
First order reaction
Unimolecular reaction
zero order reaction

Answer :D
24358.

The reaction A+BhArrC+D+ heat has reached equlibrium. The reaction may be made to proceed forward by

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ADDING more C
Adding more D
Decreasing the temperature
INCREASING the temperature

Solution :EXOTHERMIC reaction is faboured by low temperature to proceed in forward direction.
24359.

The reaction A+BtoC has zero order. What is the rate equation ?

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Solution :Rate `=((DX)/(DT))=k[A]^(0)[B]^(0)=k` (rate CONST.)
24360.

The reaction, 3Br_(2)+6CO_(3)^(2-)+3H_(2)Oto5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-)

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decolourization of reddish brown colour of `Br_(2)`
addition of `H_(2)SO_(4)` will increase INTENSITY of reddish brown colour of `Br_(2)`
bromine is REDUCED
it is disproportionation reaction or autoredox change

Solution :`3Br_(2)+6HO^(-)to5Br^(-)+BrO_(3)^(-)+3H_(2)O`,`5Br^(-)+BrO_(3)+6H^(+)to3Br_(2)+3H_(2)O`
Addition of `H_(2)SO_(4)` shift reaction in reverse direction.
So, bromine is GETTING oxidised as well as reduced.
24361.

The reaction3CIO^(-) ""_((aq))to CIO_((aq))^(-) +2CI^(-) ""_((aq))isanexampleof

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OXIDATION REACTION
reduction reaction
DISPROPORTIONATION reaction
decomposition reaction.

SOLUTION :disproportionation reaction
24362.

The reaction, 2SO_(2(g))+O_(2(g))leftrightarrowSO_(3(g)) is carried out in a 1 dm^(3) vessel and 2 dm^(3) vessel separately. The ratio of the reaction velocities will be

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`1:8`
`1:4`
`4:1`
`8:1`

ANSWER :D
24363.

The reaction

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WURTZ REACTION
Witting reaction
Ullmann reaction
WILLIAMSON reaction.

SOLUTION :NA
24364.

The reaction, 2SO_(2(g))+O_(2(g))hArr2SO_(3(g)) is carried out in a 1 dm^(3) vessel and 2dm^(3) vessel separtely. The ratio of the reaction velocities will be

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`1:8`
`1:4`
`4:1`
`8:1`

Solution :`2SO_(2(g))+O_(2(g))hArr2SO_(3(g))`
For `1dm^(3)R=k[SO_(2)]^(2)[O_(2)]`
`R=K[(1)/(1)]^(2)[(1)/(1)]=1`
For `2dm^(3)R=K[(1)/(2)]^(2)[(1)/(2)]=1/8`
So, the RATIO is `8:1`
24365.

The reaction, 2SO_2 + O_2 + 2H_2O rarr 2H_2SO_4 is an example of:

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SYNTHESIS of `H_2SO_4`
ANALYSIS of `H_2SO_4`
DISPLACEMENT reaction
Double decomposition

Answer :A
24366.

The reaction 2ROH+2 Na to RONa+H_(2) suggests that alcohols are

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Acidic
Basic
Amphoteric
Neutral in character

Answer :A
24367.

The reaction 2O_(3)(g)rarr3O_(2)(g) is believed to occur according to the mechanism given below. Step -I : O_(3)(g)underset(k_(-1))overset(k_(1))hArrO_(2)(g)+O(g) (fast) Step- II : O_(3)(g)+O(g)overset(k_(2))rarr2O_(2)(g) (slow) Write the rate law for the reaction . Determine the order of the reaction with respect toO_(3)(g) and O_(2)(g).

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Solution :The slow step of reaction is its rate-determiningstep, becauserate of this step determinesthe overall rate of the reaction. So, the rate of the slow step will be the rate of the reaction. Thus,
rate `=k_(2)[O_(3)][O]""...[1]`
The rate equation contains concentration term of reaction INTERMEDIATE O, which does not appear in the overall balanced equation of the reaction. But the rate law of a reaction shouldnot appear in theoverall balanced equation. So, we REQUIRE to eliminate [O] from equation (1). This can be done by making use of the equilibrium CONDITION of step - 1.For this equilibrium, we can write equilibrium constant EXPRESSION as
`k_(1) = ([O_(2)][O])/([O_(3)])`
This gives`[O]=k_(1)[O_(3)]//[O_(2)]`
Substituting the expression for [O] into equation (1)gives
rate `k_(2)K_(1) = ([O_(3)]^(2))/([O_(2)])=k[O_(3)]^(2)[O_(2)]^(-1)[k=k_(2)K_(1)` = constant]
This equation represents the rate law for the GIVEN reaction . It appears from the equation that the reaction has an order 2 with respect to`O_(3)and -1` with respect to`O_(2)`.
24368.

The reaction 2R-Z+2Na overset("dry ether") to R-R+2NaX is an example of

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Williamson'ssynthesis
Wurtz synthesis
Hoffman's synthsis
Cannizaros's reaction

SOLUTION :TREATEMENT of ethernal solution of alkyl halide with sodium metal gives alkane containing TWICE the NUMBER of carbon atoms present in the alkyl grup of PARENT alkyl halide.
24369.

The reaction , 2O_(3) to 3O_(2) , is assigned the following mechanism. (i) O_(3) hArrO_(2)+O (ii) O_(3)+O overset("slow")(to)2O_(2) The ratelaw of the reaction will be

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`R PROP[O_(3)]^(2)[O_(2)]`
`r prop[O_(3)]^(2)[O_(2)]^(-1)`
`r prop [O_(3)]`
`r prop [O_(3)][O_(2)]^(-2)`

ANSWER :D
24370.

Thereaction2NO_((g)) + C1_(2(g))to 2NOC1_((g))is firstorderin C1_(2) andsecondorderin NOWritethe ratelawfor thereaction

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Solution :RATELAW : RATE = `k[NO]^(2) [C1_(2)]`
24371.

The reaction 2NO_(2)(g) to 4NO_(2)(g)+O_(2)(g) follows first order kinetics. The pressure of a vessel containing only N_(2)O_(5) was found to increase from 50mm Hg to 87.5mm Hg in 30min. The pressure exerted by the gases after 60min will be (assume temperature remains constant):

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`106.2mm` HG
`150MM` Hg
`125MM` Hg
`116.5mm` Hg

Answer :A
24372.

The reaction 2NO_(2)Cl to 2NO_(2)+Cl_(2) , has experimental data that Rate =k[NO_(2)Cl]. Propose the mechanism.

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Solution :`{:(NO_(2)Cloverset("slow")(to)NO_(2)+Cl),(UL(NO_(2)Cl+Cloverset("FAST")(to)NO_(2)+Cl_(2))),(2NO_(2)Cl to2NO_(2)+Cl_(2)):}`
24373.

The reaction 2NO + 2H_(2) to N_(2) +2H_(2)O follows the mechanism: I) NO + NO_(2) ltimplies N_(2)O_(2) (fast) II) N_(2)O_(2) + H_(2) to N_(2)O + H_(2)O (slow) III) N_(2)O + H_(2) to N_(2) + H_(2)O (fast) the rate constant for the slow step (II) is 1.2 xx 10^(-4) mol^(-1)Lmin^(-1) while the equilibrium constant for step (I) is 1.4 xx 10^(-2) what is the rate of reaction when the concentration of NO and H_(2) each is 0.5 mol^(-1)L?

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`2.1 xx 10^(-7) mol L^(-1)min^(-1)`
`3.2 xx 10^(-6)mol L^(-1)min^(-1)`
`3.5 xx 10^(-4)mol L^(-1)min^(-1)`
None of the above.

Solution :`k_("obs") = k xx k_(c) = 1.2 xx 10^(-4)`
`=1.2 xx 10^(-4)mol^(-1)L min^(-1) xx (1.4 xx 10^(-2))`
`=1.68 xx 10^(-6) mol^(-1)L min^(-1)`
RATE =`k_("obs")[NO]^(2)[H_(2)]`
`=(1.68 xx 10^(-6) mol^(-1)L min^(-1)) xx (0.5 molL^(-1))^(3)`
24374.

The reaction, 2NO+br_(2)to2NOBris supposed to have the meachanism Step -I NO+Br_(2)hArrNOBr_(2) Step-II NOBr_(2)+NOoverset("slow")(to)2NOBr Predict the rate expression for the reaction

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Solution :The RATE expression is derived by step-II of the reaction mechanism which is a slow step.thus
Rate `=K[NOBr_(2)][NO]`………`(i)`
HOWEVER, `NOBr_(2)` is an intermediate and thus `[NOBr_(2)]` should be eliminated from Eq. `(i)`
Also for, step `(i)` Equilibrium constant `K_(c )` is
`K_(c )=([NOBr_(2)])/([NO][Br_(2)])`...........`(ii)`
`:.[NOBr_(2)]=K_(c )[NO][Br_(2)]`
Then from Eqs. `(i)` and `(ii)`
Rate `=k.K_(c )[NO]^(2)[Br]^(2)`
Rate `=K.[NO]^(2)[Br_(2)]`
24375.

The reaction 2NO+Br_2to2NOBr follows the mechanism : (1)NO+Br_2hoverset("Fast")hArrNOBr_2 (2)NOBr_2+Nooverset("Slow")to2NOBr Which of the following is/are true regarding this :

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The order of the REACTION with respect to NO is two :The molecularity of the steps (1) and (2) are two each
The molecularity of the OVERALL reaction is three.
The overall order of the reaction is three

Solution :NA
24376.

The reaction 2NO + Br_2rarr 2NOBr, obey.s the following mechanism:

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`R=[NO]^2[Br_2]`
`r=K[NO][Br_2]`
`r=K[NO][Br_2]^2`
`r=K[NOBr_2]`

ANSWER :A
24377.

The reaction : 2N_(2)O_(5)(g) to 2NO_(2)(g) + (O_(2))_(g) was studied and the following data were collected. Determine: (i) the order, (ii) the rate law and (iii) rate constant for the reaction.

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Solution :Let the order of the REACTION = n
`Rate propto [N_(2)O_(5)]^(n)`
`(34 xx 10^(-5)) propto (1.13 xx 0^(-2))^(n)`
`(25 xx 10^(-5)) propto (0.84 xx 10^(-2))^(n)`
DIVIDING eqn. (i) by eqn. (ii), we have
`(34 xx 10^(-5))/(25 xx 10^(-5)) = (1.13 xx 10^(-2))^(n)/(0.84 xx 10^(-2))^(n), (1.36)^(1) = (1.35)^(n)` or n=1
i) The order of the reaction = 1,
ii) Rate law, The rate of reaction depends UPON one concentration team of `N_(2)O_(5)` because of the reaction is 1.
ii) Rate law. The rate of reaction depends upon one concentration term of `N_(2)O_(5)` because order of the reaction is 1.
Reaction Rate = `k[N_(2)O_(5)]`
III) Rate constant, `k = "Reaction rate"/([N_(2)O_(5)])= (18 xx 10^(-5) mol L^(-1)min^(-1))/(0.62 xx 10^(-2)mol L^(-1))= 0.0.29 min^(-1)`
24378.

The reaction: 2MnO_(4)^(-) + 3Mn^(2+)+ 2H_(2)O rarr 5MnO_(2) + 4H^(+) is catalysed by…..

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`NiSO_(4)`
`ZnSO_(4)`
`Mn^(2+)`
`FE^(3+)`

ANSWER :B
24379.

The reaction 2H_(2)O_(2)to 2H_(2)O + O_(2) shows that

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it ACTS as reducing agent
it acts as oxidising agent
`H_(2)O_(2)`is decomposed
none of these

Solution :`H_(2)O_(2)` DECOMPOSES to EVOLVE `O_(2)` gas.
24380.

The reaction 2KI+Cl_(2) to 2KCl+I_(2)

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is possible
depends UPON the STATE of products
is not possible
depends upon the temperature

Solution :`Cl_(2)` is more POWERFUL OXIDIZING agent, while `I^(-)` is reducing agent
24381.

The reaction 2H_(2) O_(2) to 2 H_(2)O+ O_(2) is a

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ZERO order reaction
First order reaction
SECOND order reaction
Third order reaction

Solution :As r = k `(H_(2)O_(2))` , it is a reaction of `1^(ST)` order .
24382.

The reaction, 2H_2O_2rarr 2H_2O + O_2, shows that H_2O_2:

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Is decomposed
Acts as OXIDISING agent
Acts as REDUCING agent
all of the above

Answer :A
24383.

The reaction 2ClCH_2 CH_2 OH overset((i) KOH) underset((ii) "Conc." H_2 SO_4) to will produce

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`CH_(2)=CH_(2)OH`
DIOXANE
`HOCH_(2)CH_(2)OH`
`OVERSET(+)(K) BAR(O)CH_(2)CH_(2)bar(O)K^(+)`

SOLUTION :
24384.

The reaction 2CH_(3)-C-underset(O)underset(||)(C)C_(2)H_(5) overset(C_(2)H_(5)ON a)toCH_(3)-underset(O)underset(||)(C)-CH_(2)-underset(O)underset(||)(C)-OC_(2)H_(5)+C_(2)H_(5)OH is called

Answer»

Etard reaction
Perkin's reaction
Claisen condensation
Claisen SCHMIDT reaction

Answer :C
24385.

The reaction 2C_6H_5I+2Na overset(ether, Delta)(to) C_6H_5-C_6H_5+2NaI

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FITTING REACTION
WURTZ-fitting reaction
Ulmann reaction
Wurtz reaction

SOLUTION :Fitting reaction
24386.

The reaction 2C_(6)H_(5)CHOoverset(50% NaOH)rarrC_(6)H_(5)COONA+C_(6)H_(5)CH_(2)OHis an example of

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perkine REACTION
KOLBE' reaction
etard reaction
CANNIZZARO reaction

ANSWER :B
24387.

The reaction 2C + O_2 rarr 2CO_2 is carried out by taking 24 g carbon and 96 gm o_2 which one is limiting reagent :

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C
`O_2`
`CO_2`
NONE

Answer :A
24388.

The reaction 2A_((g))+B_((g))hArr3C_((g))+D_((g)) is begun with the concentrations of A and B both at an initial value of 1.00 M. when equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

Answer»

`"["(0.75)^(3)(0.25)"]"DIV"["(1.00)^(2)(1.00)"]"`
`"["(0.75)^(3)(0.25)"]"div"["(0.50)^(2)(0.75)"]"`
`"["(0.75)^(3)(0.25)"]"div"["(0.50)^(2)(0.25)"]"`
`"["(0.75)^(3)(0.25)"]"div"["(0.75)^(2)(0.25)"]"`

Solution :`2A+BhArr3C+D`
`{:("INITIAL",1,1,0,0),(Eq.,1-0.50,1-0.25,0.75,0.25):}`
`K=((0.75)^(3)(0.25))/((0.50)^(2)(0.75))`
24389.

The reaction, 2ArarrB+Cfollow zero order kinetics. The differential rate equation for the reaction is:

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`(DX)/dt=K[A]^0`
`(dx)/dt=K[A]^2`
`(dx)/dt=K[B][C]`
`(dx)/dt=K[A]`

ANSWER :A
24390.

The reaction , 2AB(g) + 2C(g) to A_(2(g)) + 2BC_((g)) proceeds according to the mechanism . I. 2AB hArr A_(2)B_(2)(fast) II. A_(2)B_(2)+C to A_(2)B + BC(slow ) III. A_(2)B+C to A_(2)+BC (fast) what will be the initial rate taking [AB] = 0.2 M and [C] = 0.5 M ?The K_(c)for the step I is10^(2) M^(-1)and rate constant for the step II is3.0 xx 10^(-3) mol^(-1) min^(-1)

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`0.0716 M MIN^(-1)`
`0.0891 M min^(-1)`
`0.006 M min^(-1)`
`0.0257 M min^(-1)`

ANSWER :C
24391.

The reaction, 2A to B+C follows zero order kinetics. The differential rate equation for the reaction is

Answer»

`(dx)/(dt) =k[A]^0`
`(dx)/(dt) =k[A]^2`
`(dx)/(dt) = k[B][C]`
`(dx)/(dt) =k[A]`

Solution :For ZERO order REACTIONS `(dx)/(dt) = k [A]^(0)`
24392.

The reaction , 2 NO(g) + O_(2) (g) to 2NO_(2) (g) , is second order with respect to NO and first order with respect to O_(2) . If the volume of reactants is suddenly reduced to half value , the rate of reaction would be

Answer»

One-fourth of ORIGINAL VALUE
One-eighth of original value
EIGHT times of original value
Four times of original value

Answer :c
24393.

The reaction 2" NO"(g)+O_(2)(g)to2" NO"_(2)(g)" and "2" CO "(g)+O_(2)(g)to2" CO"_(2)(g) look to be similar. Yet the former is faster than the latter at the same temperature. Explain why.

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SOLUTION :The former has LOWER ACTIVATION ENERGY than the latter REACTION. Lower the activation energy, faster is the reaction.
24394.

The reaction ._(1)H^(2) + ._(1)H^(3) rarr ._(2)He^(4) + ._(0)n^(1)+ energy represents

Answer»

Nuclear fission
Nuclear fusion
Artifical disintegration
Transmutation of element

Solution :In hydrogen bomb, the following REACTION occurs,
`._(1)H^(2) + ._(1)H^(3) RARR ._(2)He^(4) + ._(0)^(1) N +` energy
24395.

The reaction 2 Fe Cl_(3) + SnCl_(2) to 2 FeCl_(2) + SnCl_(4) is an example of

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First order REACTION
Second order reaction
Third order reaction
None of these

Solution :R = `K [FeCl_(3)]^(2) [SnCl_(2)]^(1)`. Order = 2 + 1 = 3 .
24396.

The reaction ._(13)Al^(27) + ._(2)He^(4) rarr ._(14)Si^(30) + ._(1)H^(1) is of the type

Answer»

NUCLEAR fusion
Nuclear fission
Chemical reaction
Transmutation

ANSWER :D
24397.

The reaction

Answer»

WURTZ reaction
Kolbe reaction
Rimer-tiemann reaction
Schotten-Baumann reaction

Answer :D
24398.

The reaction : ""_(1)^(2)D+""_(1)^(3)Tto""_(1)^(2)He+""_(0)^(1)n is an example of

Answer»

NUCLEAR FISSION
nuclear fusion
artificial radioactivity
radioactive DISINTEGRATION.

ANSWER :B
24399.

The reaction, (1//2)H_(2)(g) + AgCl(s) to H^(+)(aq) + Cl^(-)(aq) + Ag(s), occurs in the galvanic cell:

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`Ag|AgCl(s) |KCl(soln)|AgNO_(3)(soln)|Ag`
`Pt|H_(2)(g) | HCL(soln)|AgNO_(3)(soln.)|Ag`
`Pt|H_(2)(g)|HCl(soln.)|AgCl(s)|Ag`
`Pt|H_(2)(g) | KCl(soln.)|AgCl(s) | Ag`

Answer :C
24400.

The reaction ,1//2H_(2(g)) + AgCl_((s)) to H_((aq))^(+) + Cl_((aq))^(-) + Ag_((s)) occurs in the galvanic cell :

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`Ag | AgCl_((s)) |KCl_((So LN))|AgNO_(3(so ln))|Ag`
`Pt, H_(2(s))|HCl_(So ln )|AgNO_(3(so ln))|Ag`
`Pt,H_(2(G))||HCl_((So ln))||AgCl_((s))|Ag`
`Pt, H_(2(g))||KCl_((So ln))||AgCl_((s))|Ag`