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: 2A + B_2 + C to A_2B + Bc ,the rate law expression has been determined experimentally to be R = k [A]^2[C]withk = 3.0 xx 10^(-4) M^(-2)"min"^(-1)(i) Determine the initial rate of the reaction, started with concentrations:[A] = 0.1 M, [B.] =0.35 M and [C] = 0.25 M.(ii) Determine the rate after 0.04 mole per litre of A has reacted.

Answer»

Solution :`7.5 XX 10^(-7) M "MIN"^(-1) , 2.5 xx 10^(-7) M "min"^(-1)`
2.

For the reaction, 2A+Brarr3C, the reaction rate is equal to

Answer»

`1/2 (d[A])/(DT)`
`1/3 (d[C])/(dt)`
`(d[B])/(dt)`
`(-d[A]^(2))/(dt)`

ANSWER :A::B::C::D
3.

For the reaction 2A+Bto Products, it is found thatdoubling the concentration of both reactants increases the rate by a factor of 8. But doubling the concentration of B alone, only doubles the rate. What is the order of the reaction w.r.t to A?

Answer»

`2`
`3`
`0`
`1`

ANSWER :A
4.

For the reaction : 2A+B to A_(2)B, the rate =k[A][B]^(2) with k=2.0xx10^(-6)"mol"^(-2)L^(2)s^(-1). Calculate the initial rate of the reaction when [A]=0.1"mol L"^(-1) and [B]=0.2"mol L"^(-1). Calculate the rate of reaction after [A] is reduced to 0.06 mol L^(-1).

Answer»

Solution :SUBSTITUTING the values in the rate equation, we get
INITIAL rate `=K[A][B]^(2)`
`=(2.0xx10^(-6)"mol"^(-2)L^(2)s^(-1))(0.1 "mol L"^(-1))(0.2" mol L"^(-1))^(2)`
When [A] is reduced from 0.10 mol `L^(-1)` to 0.06 mol `L^(-1)`, 0.04 mol `L^(-1)` of A has reacted.
Amount of B reacted `=(1)/(2)xx0.04 "mol L"^(-1)=0.02"mol L"^(-1)`
Hence, [B] at that TIME `=0.2-0.02=0.18"mol L"^(-1)`
Now,rate `=(2.0xx10^(-6)"mol"^(-2)L^(2)s^(-1))(0.06"mol L"^(-1))(0.18" mol L"^(-1))^(2)`
`=3.89xx10^(-9)" mol L"^(-1)s^(-1)`.
5.

For the reaction 2A+B rarr A_(2)B. The rate = k[A] [B]^(2) with 2.0xx10^(-6)mol^(-2)L^(2)s^(-1). Calculate the initial rate of the reaction, when [A] = 0.1 "mol L"^(-1), [B] = 0.2 "mol L"^(-1). Calculate the rate of reaction after [A] is reduced to 0.06 "mol L"^(-1).

Answer»

Solution :The initial rate of the REACTION is
Rate `= k[A] [B] 2`
`= (2.0xx10^(-6)MOL^(-2)L^(2)s^(-1))(0.1 "mol L"^(-1)) (0.2 "mol L"^(-1))^(2)`
`= 8.0xx10^(-9)mol^(-2)L^(2)s^(-1)`
When [A] is reduced from `0.1 "mol L"^(-1)` to `0.06 mol^(-1)`, the concentration of A reacted `=(0.1-0.06)"mol L"^(-1)=0.04 "mol L"^(-1)`.
`therefore` The concentration of B reaction
`=(1)/(2)xx0.04 "mol L"^(-1)`.
`=0.02 " mol L"^(-1)`.
Then, concentration of B available `[B]=(0.2-0.02)"mol L"^(-1)`.
`= 0.18 " mol L"^(-1)`.
Aften [A] is reduced to `0.06 "mol L"^(-1)`, the rate of the reaction is given by,
Rate `= k[A][B]2`
`=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.06"mol L"^(-1)) (0.18 "mol L"^(-1))^(2)`
`= 3.89"mol L"^(-1)s^(-1)`.
6.

For the reaction 2A+B rarr product, doubling the initial concentrations of both the reactants increases the rate by a factor of 8 and doubling the concentration of B above double the rate. This rate-law for the reaction is

Answer»

`r=k[A][B]^(2)`
`r=k[A][B]`
`r=k[A]^(2)[B]^(2)`
`r=k[A]^(2)[B]`

Solution : Since doubling the concentration of B doubles the rate therefore the order of REACTION w.r.t. [B] is one and since on doubling the conc. of A and B rate INCREASES 8 times, therefore order of reaction w.r.t. A will be 2.
`i.e.""rprop[A]^(2)[B]`
or`r=k[A]^(2)[B]`
7.

For the reaction : 2A+B+CtoA_(2)B+C, the rate law has been determined to be Rate =k[A][B]^(2)" with "k=2.0xx10^(-6)" mol"^(-2)L^(2)s^(1) For this reaction determine the initial rate of the reaction with [A]=0.1" mol L"^(-1),[B]=0.2" mol L"^(-1),[C]=0.8" mol L"^(-1). Determine the rate after 0.04" mol L"^(-1). Determine the rate after 0.04 "mol L"^(-1) of A has reacted.

Answer»


Solution :Initial rate `=(2.0xx10^(-6)" MOL"^(-2)L^(2)s^(-1))xx0.1" mol L"^(-1)xx(0.2" mol L"^(-1))^(2)=8xx10^(-9)" mol L"^(-1)s^(-1)`
After `0.04" mol L"^(-1)" of A has REACTED. "[A]=0.1-0.04=0.06" mol L"^(-1)`
`[B]=0.2-0.02=0.18" mol L"^(-1)`
Then rate `=(2XX10^(-6))xx(0.06)xx(0.18)^(2)=3.9xx10^(-9)" mol L"^(-1)s^(-1).`
8.

For the reaction, 2A+Bto2C,/_\G^0=2kJ. Mol^(-1) at 500K. Calculate the value of equilibrium constant for the reaction, A+1/2BtoC, at the same temperature.

Answer»

SOLUTION :`/_\G^0=/_\G^0(PCl_3)+/_\G^0(Cl_2)-/_\G^0(PCl_5)`
`=(-68.42+0+77.6)kcal=+9.18kcal=+3.84xx10^4J`
`:.-2.303RT" "logK=3.84xx10^4`
or `logK=-6.73" ":.K=1.86xx10^(-7)`
9.

For the reaction 2A +B rarr 3C +D Which of the following does not express reaction rate ?

Answer»

`-(d[B])/(DT)`
`(d[D])/(dt)`
`-(d[A])/(2dt)`
`-(d[C])/(3dt)`

ANSWER :D
10.

For the reaction 2A+B+Cto2D. The observed rate law is Rate =K[A][B]^(2). Correct statements are :

Answer»

increase of CONC. Of C does not EFFECT RATE
Doubling the conc. Of A, doubles the rate
Triplign the con of B, increases rate by 9 times
Doubling the conc of C, doubles the rate

Solution :rate only depends onconc. Of A and B and not on C.
11.

For the reaction 2A+B +C rarr A_(2)B+C the rate law has been found to be Rate = k[A] [B]^(2)withk = 2.0xx10^(-6) mol^(-2) L^(2)s^(-1) . The initial rate of reaction with [A] =0.1 mol L^(-1) [B] = 0.2 mol L^(-1) and [C] = 0.8 mol L^(-1) s^(-1)

Answer»

`6.4xx10^(-8) "MOL L"^(-1)s^(-1)`
`4xx10^(-3) "mol L"^(-1)s^(-1)`
`8.0xx10^(-9) "mol L"^(-1) s^(-1)`
`4.0 xx10^(-7) "mol L"^(-1)s^(-1)`

Solution :(C ) Rate `= k[A] [B]^(2)`
`= (2.0xx10^(-6))(0.1) (0.2)^(2)`
`=8xx10^(-9)`
12.

For the reaction, 2A + B to C + D, the order of reaction is

Answer»

ONE with respect to [B]
TWO with respect to [A]
Three
Can't be predicted

Solution :ORDER of a reaction is an EXPERIMENTALLY determined value and can not be predicted from the reaction.
13.

For the reaction : 2" NH"_(3)(g)overset("Pt")to"N"_(2)(g)+3" H"_(2)(g), Rate = k (i) Write the order andmolecularity of this reaction. (ii) Write the units of k.

Answer»

SOLUTION :(i) Rate = k shows that the rate of reaction is independent of the concentrations of the reactants. Hence, it is a reaction of zero order. As two molecules of the REACTANT are persent in the reaction, molecularity is two.
(ii) UNITS of k will be same as that of the rate of reaction viz. `"mol L"^(-1)s^(-1)`.
14.

For the reaction : 2 NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g). What is the % of NH_(3) converted if the mixture diffuses twice as fast as that SO_(2) under similar conditions.

Answer»

3.125
6.25
12.5
None of these

Solution :`{:(,2NH_(3) (g),rarr,N_(2) (g),+,3H_(2) (g),),("Initially",1,,,,,),("After",1 - 2x,,x,,3x,):}`
decomposition
`(r_("mix"))/(r_(SO_(2))) = 2 = SQRT((M_(SO_(2)))/(M_("Mix")))`
`M_("mix") = (64)/(4) = 16 = M_("avg")`
`M_(avg) = (1 xx 17)/(1 + 2x) = 16`
`17 = 16 + 32 x`
1 = 32 x
`x = (1)/(32)`
`% NH_(3) = (2x)/(1) xx 100`
`= (1)/(16) xx 100 = 6.25 %`
15.

For the reaction : 2 NH_(3)(g) rarr N_(2)(g) + 3 H_(2)(g)

Answer»

`q_(p) = (q_(V))^(2)`
`q_(p) = q_(v) - 2 RT`
`q_(p) = q_(v) + 2 RT`
`q_(p) = 2 q_(v) - R`.

Solution :`q_(p) = q_(v) + Delta n_(g)RT = q_(v) + 2RT`
16.

For the reaction 2 N_(2) O_(5(g)) to 4 NO_(2 (g)) to O_(2(g)) , if concentration of NO_(2) in 100 seconds is increased by 5.2 xx 10^(-3) m . Then the rate of reaction will be

Answer»

`1.3 xx 10^(-5) ms^(-1)`
`5 xx 10^(-4) ms^(-1)`
`7.6 xx 10^(-4) ms^(-1)`
`2 xx 10^(-3) ms^(-1)`

Solution :`2N_(2)O_(5(G)) to 4NO_(2 (g)) + O_(2(g))`
Rate of REACTION with RESPECT to `NO_(2)`
`=(1)/(4) (d[NO_(2)])/(DT) = (1)/(4) xx (5.2 xx 10^(-3))/(100) = 1.3 xx 10^(-5) ms^(-1)`.
17.

For the reaction 2" A"+" B"toA_(2)B," rate "=k[A][B]^(2) with k=2.0xx10^(-6)mol^(-2)L^(2)s^(-1). Calculate the initial rate of the reaction when [A]=0.1molL^(-1)" and "[B]=0.2molL^(-1). Calculate the rate of reaction after [A] is reduced to 0.06molL^(-1).

Answer»

Solution :Initial rate `=K[A][B]^(2)=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.1molL^(-1))(0.2molL^(-1))^(2)=8xx10^(-9)MOLL^(-1)s^(-1)` When [A] is reduced from `0.10molL^(-1)" to "0.06molL^(-1)`, i.e., `0.04molL^(-1)` of A has reacted, B reacted
`=(1)/(2)xx0.04molL^(-1)=0.02molL^(-1).`
Hence, new `[B]=0.2-0.02=0.18molL^(-1).`
Now,rate `=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.06molL^(-1))(0.18molL^(-1))^(2)=3.89xx10^(-9)molL^(-1)s^(-1).`
18.

For the reaction 2 HI hArr H_(2)+ I_(2) the equilibrium constant K at 440^(@)C is 0.022 . The equilibrium constant for I_(2) + H_(2) hArr 2 HI is

Answer»

`45.45`
`0.050`
`0.022`
NONE of these

Answer :A
19.

For the reaction 2 A (s) + B^(2+) (aq) to 2 A^(+) (aq) + B(s) , Nernst equation for the EMF of the cell is

Answer»

`E = E^(@) - (RT)/(2F) "ln" ([A^(+)])/([B^(2+)])`
`E = E^(@) - (RT)/(2F) "ln" ([A^(+)])/([B^(2+)]^(2))`
`E = E^(@) - (RT)/(F) "ln" ([A^(+)])/(SQRT([B^(2+)]))`
`E = E^(@) +(RT)/(F) "ln" ([A^(+)])/(sqrt([B^(2+)]))`

Solution :`E = E^(@) - (RT)/(2F) "ln" ([A^(+)]^(2+))/([B^(2+)]) = E^(@) - (RT)/(F) "ln" ([A^(+)])/(sqrt([B^(2+)]))`
20.

For the reaction, (1)/(2)A_(2)+(1)/(2)B_(2) to AB,DeltaH=-50 kcal If the bond energies of A_(2),B_(2) and AB are respectively x,(x)/(2) and x kcal, the value of x is

Answer»

50
100
200
400

Answer :C
21.

For the reaction2 A + B to A_(2) B, the rate law given is

Answer»

`K[2A][B]`
`k[A]^(2) [B]`
`k[A] [B]^(3)`
`k [A]^(2) [B]`

Solution :Rate of reaction is directly PROPORTIONAL to the ACTIVE mass of reactant
22.

For the reaction 2 A + B to 3 C + D which of the following does not express the reaction rate

Answer»

`(d [D])/(DT)`
`-(d[A])/(2 dt)`
`-(d[C])/(3dt )`
`- (d[B])/(dt)`

ANSWER :c
23.

For the reactin uNO_((g))+Cl_(2(g))hArr2NOCl_((g)) which is true

Answer»

<P>`K_(p)=K_(c)xxRT`
`K_(p)=K_(c)(RT)^(2)`
`K_(p)=(K_(c))/(RT)`
`K_(p)=(K_(c))/((RT)^(2))`

Solution :We know that
`K_(p)=K_(c)(RT)^(Delta_(NG))`
Given `Delta_(ng)=2-3=-1`
Puttinvlaue of `Delta_(ng)` we GET `K_(p)=K_(c)(RT)^(-1)`
or ` K_(p)(K_(c))/(RT).`
24.

Forthe reactin C_((s))+CO_(2(g))hArr2CO_((g)), The partial pressure of CO_(2) and CO are 2.0 and 4.0 atm respectively at equlibrium. The K_(p) or the reaction is

Answer»

`0.5`
`4.0`
`8.0`
`32.0`

SOLUTION :`K_(p)=([P_(co)]^(2))/([P_(co_(2))])=(4xx4)/(2)=8.`
25.

For the reaction N_2+3H_2=2NH_3, triangleH=?

Answer»

`triangleE=2RT`
`triangleE-2RT`
`triangleE+RT`
`triangleE+2RT`

ANSWER :B
26.

For the reaaction, CO(g) +H_2O(g)

Answer»

Increasing the pressure
Adding an inert gas at CONSTANT pressure
Increasing the VOLUME of the container
Increasing the AMOUNT of CO(g)

ANSWER :D
27.

For the rate of the reaction 2H_(2)O_(2)to2H_(2)O+O_(2),r=K[H_(2)O_(2)] it is

Answer»

Zero order reaction
FIRST order reaction
SECOND order reaction
THIRD order reaction

ANSWER :B
28.

For the purpose of systematic qualitative analysis, cations are classified into various groups on the basisof their behaviour against some reagents. The group reagents used for the classification of most common cations are hydrochloric acid, hydrogen sulphide, ammonium hydroxide, and ammonium carbonate. Classification id based on wheather a certain reacts with these reagents by the formation of precipitates or not. To avoid the precipitation of hydroxides of Ni^(2+),Co^(2+),Mn^(2+) along with those of the third group cations, the solutions should be :

Answer»

Heated with few drops of concentrated `HNO_3`
Boiled with excess of AMMONIUM chloride
Concentrated to small volume
None of these

Solution :Function of strong electrolyte `NH_4Cl` is to suppress the ionisation of `NH_4OH` so that the concentration of `OH^-`ions in the solution is decreased but it is SUFFICIENT to precipitate the third GROUP basic radicals because the solubility prouduct of group III hydrooxides is lower than IV, V and VI group hydroxides. The `Cr(OH)_3darr` is slightly soluble in excess of precipitant , upon boiling the solution, `Cr(OH)_3` is precipitated.
29.

For the raction A+3Bto2C+D, which one of the following is not correct ?

Answer»

Rate of DISAPPEARANCE of A = Rate of FORMATION of D
Rate of formation of C = `2/3xx` Rate of disappearance of B
Rate of formation of D = `1/3xx` Rate of disappearance of B
Rate of disappearance of A = `2XX` Rate of formation of C

ANSWER :D
30.

For the purpose of systematic qualitative analysis, cations are classified into various groups on the basisof their behaviour against some reagents. The group reagents used for the classification of most common cations are hydrochloric acid, hydrogen sulphide, ammonium hydroxide, and ammonium carbonate. Classification id based on wheather a certain reacts with these reagents by the formation of precipitates or not. An aqueous solution contains Hg^(2+),Hg_2^(2+),Pb^(2+) and Cd^(2+).The addition of 2 M HCl will precipitate :

Answer»

`HgCl_2` only
`PbCl_2` only
`PbCl_2 and Hg_2Cl_2`
`PbCl_2 and CdCl_2`

Solution :2M HCl is group REAGENTS for group `I^(st)` cations. `PbCl_2 and Hg_2Cl_2` will GET PRECIPITATE, as their solubility products `(K_(sp))` are less than that of other RADICALS.
31.

For the production of equalamounts of hydrogen fromthe following reactions, which metal, An or Al,is less expensive if Zn costsabouthalf as much as Al an a mass basisand by how much ? {:(An + 2HCl to ZnCl_(2)+H_(2) ),(2Al+6HCl to2AlCl_(3)+3H_(2)):}

Answer»

Solution :1 mole of `H_(2)` is produced by 1 mole, i.e., 65 g of Zn and
1 mole of `H_(2)`is produced by `(2)/(3)` mole, i.e., `(2 xx 27)/( 3)` g of AL = 18 g of Al
Now that Zn costs about half as MUCHAS Al, to purchase Zn and Al to PRODUCE the same AMOUNT of `H_(2)` the costratio of Znand Al will be 65:36
Al is thus lessexpensive by `(65-36)/( 65) xx 100` ,i.e.,`44.61 %`
32.

For the production of y L H_2 at STP at cathode, cost of electricity is x then cost of production of y LO_2 at STP at anode will be

Answer»

X
`x/2`
2x
4x

Answer :C
33.

For the process X(g)toproducts, (order ne 0), rates of disappearances of X at t = 0,t = 50sec & t = 30 sec respectively are p, q&r mol/lit/sec then

Answer»

`pltqltr`
`rltpltq`
`qgtrgtp`
`pgtrgtq`

ANSWER :D
34.

For the process, melting of ice at 260 K theDelta H is -

Answer»

NEGATIVE
POSITIVE
Zero
Cannot be predicted

Answer :B
35.

For the process, melting of ice at 260 K the DeltaH is -

Answer»

Negative
Positive
Zero
Cannot be predicted

Answer :B
36.

For the process H_(2)O(l)rarrH_(2)O(g)at T=100^(@)C and 1 atmosphere pressure, the correct choice is

Answer»

`DeltaS_("system")gt0and DeltaS_("surroundin gs")gt0`
`DeltaS_("system")gt0and DeltaS_("surroundin gs")lt0`
`DeltaS_("system")lt0and DeltaS_("surroundin gs")gt0`
`DeltaS_("system")lt0and DeltaS_("surroundin gs")lt0`

Solution :At `100^(@)C` and 1 ATMOSPHERE PRESSURE `H_(2)O(l)iffH_(2)O(g)` is at equilibrium. For equilibrium `DeltaS_("total")=0 and DeltaS_("system")+DeltaS_("SURROUNDING")=0`
`therefore DeltaS_("system")gt0 and DeltaS_("surrounding")lt0`.
37.

For the process H_(2)O(l) to H_(2)O(g) at T=100^(@)C and p=1 atm, the correct choice is

Answer»

DeltaS_(sys)gt0"and"Delta_(SUR)gt0`
DeltaS_(sys)gt0"and"DeltaS_(sur)LT0`
DeltaS_(sys)lt0"and"Delta_(sur)gt0`
`DeltaS_(sys)lt0"and"DeltaS_(sur)lt0`

Answer :B
38.

For the process H_(2)O(l) (1 bar, 373 K) rarr H_(2)O (1 bar, 373 K), the correct set of thermodynamic parameters is

Answer»

`DeltaG=0, DeltaS=+ve`
`DeltaG=0, DeltaS=-ve`
`DeltaG=+ve, DeltaS=0`
`DeltaG=-ve, DeltaS=+ve`

Solution :Since, liquid is PASSING in to gaseous phase so ENTROPY will increase and at 373 K the phase TRANSFORMATION REMAINS at equilibrium. So `DeltaG=0`.
39.

For the process H_2O(l) (1 bar , 373 K)hArr H_2O(g) (1 bar , 373 K), the correct set of thermodynamic parameters is :

Answer»

`DELTAG=0`
`DeltaSgt0`
`DeltaHgt0`
`DeltaG= -ve`

Solution :`H_2O`(l bar, 373 K)`to H_2O`(G, 1 bar,373 K)
`DELTAS` gt0
`DeltaH` gt 0
`DeltaG` gt 0
40.

For the process : H_(2)O(l) (1 bar, 373 K) rarr H_(2)O(g) (1 bar, 373 K), the correct set of thermodynamic parameter is :

Answer»

`Delta G = 0, Delta S = + ve`
`Delta G = 0, Delta S = - ve`
`Delta G = + ve, Delta S = 0`
`Delta G = - ve, Delta S = + ve`

SOLUTION :This is a PHASE transformation and hence `Delta G = 0`. SINCE `H_(2)` is CHANGING to `H_(2)O (g)` at the same temperature and pressure. `Delta S = +ve`
41.

For the process H_(2)O(l) (1 bar, 373 K) to H_(2)O(g) (1 bar, 373 K), the correct set of thermodynamic parametes is

Answer»

`DeltaG=0,DeltaS=+ve`
`DeltaG=0,DeltaS=-ve`
`DeltaG=+ve,DeltaS=0`
`DeltaG=-ve,DeltaS=+ve`

ANSWER :A
42.

For the process: H_(2)O(1)(1"bar",373K)toH_(2)O(g)(1"bar",373K), the correct set of the thermodynamic parameters is

Answer»

`DeltaG=0,DeltaS=+ve`
`DeltaG=0,DeltaS=-ve`
`DeltaG=+ve,DeltaS=0`
`DeltaG=-ve,DeltaS=+ve`

ANSWER :A
43.

For the process dry ice rarr CO_(2)(g)

Answer»

`DELTAH` is positivewhile `Deltarho` is NEGATIVE
Both `DeltaH` and `Deltarho` are negative
Both `DeltaH` and `Deltarho` are POSITIVE
`DeltaH` is negative while `Deltarho` is positive

Solution :Both `DeltaH` and `DELTAP` are positive.
44.

For the process - CO_(2)(s)toCO_(2)(g)

Answer»

Both `DeltaH and DELTAS` are POSITIVE
`DeltaH` is negative and `DeltaS` is positive
`DeltaH` is positive and `DeltaS` is negative
Both `DELTAHANDDELTAS` are negative

Answer :A
45.

For the process : CO_(2)(s) rarr CO_(2)(g)

Answer»

Both `DELTA H and Delta S` are +ve
`Delta H` is -ve, `Delta S` is +ve
`Delta H` is +ve, `Delta S` is -ve
Both `Delta H` and `Delta S` are -ve

Answer :A
46.

For the process, CO_2(s)rarrCO_2(g):

Answer»

Both `TRIANGLEH` and `TRIANGLES` are +ve
`triangleH` is NEGATIVE and `triangleS` is +ve`
`triangleH` is +ve and `triangleS` is -ve`
Both `triangleH` and `triangleS` are -ve`

ANSWER :A
47.

For the process 2A toproducts, rate of reaction w.r.t A at 10thsecond is 2xx10^(-2)M-s^(-1)then rates of same process at 5th & 15th seconds (order ne 0) respectively are (in M/s)

Answer»

`10^(-4) & 4 XX 10^(-2)`
`2.7 xx 10^(-2) & 1.6 xx 10^(-2)`
`1.6 xx 10^(-2) & 2.7 xx 10^(-2)`
`2 xx 10^(-2) & 2 xx 10^(-2)`

ANSWER :B
48.

For the process,

Answer»

The efficiency of the cyclic process `= 50%`
Net WORK done `=- 6 KJ`
Heat supplied during the process `= +12 kJ`
More work is done (MAGNITUDE WISE) in BC than DA

Answer :A::B::C
49.

For the proces H_(2)O(l)toH_(2)O(g) at T=100^(@)C and 1 atomsphere pressure, the correct choice is :

Answer»

`DeltaS_("system")gt0andDeltaS_("SURROUNDINGS")gt0`
`DeltaS_("system")gt0andDeltaS_("surroundings")LT0`
`DeltaS_("system")lt0andDeltaS_("surroundings")gt0`
`DeltaS_("system")gt0andDeltaS_("surroundings")lt0`

ANSWER :B
50.

For the preparation of sodium phiosulphate by "springs reaction", the reactants used are

Answer»

`Na_(2)S+Na_(2)SO_(3)+Cl_(2)`
`Na_(2)S+SO_(2)`
`Na_(2)SO_(3)+S`
`Na_(2)S+Na_(2)SO_(3)+I_(2)`

SOLUTION :`Na_(2)S+I_(2)+Na_(2)SO_(3)toNa_(2)S_(2)O_(3)+2NaI`