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.

Density of a gas is found to be 5.46 g//dm^(3) at 27^(@)C and at 2 bar pressure. What will be its density at STP ?

Answer»

Solution :`d=(MP)/(RT)`. For the same gas at different temperatures and pressure, `(d_(1))/(d_(2))=(P_(1))/(T_(1))XX(T_(2))/(P_(2))`.
Here,`"" d_(1)=5.46 g dm^(-3),T_(1)=27^(@)C=300 K,P_(1)=2" bar"`.
At STP,`"" d_(2)=? "" T_(2)=0^(@)C=273 K, P_(2)=1" bar"`
`:.(5.46 g dm^(-3))/(d_(2))=(2" bar" )/(300 K)xx(273 K)/(1" bar" )`
or `"" d_(2)=3 g dm^(-3)`
Discussion
2.

Density of a gas is found to be 5.46 g//dm_(3) at 27 ^(@)C at 2 bar pressure. What will be its density at STP?

Answer»

<P>

Solution :The molecular mass of a gas is given by `M=(dRT)/P`
Since, molecular mass does not change with temperature or pressure, in two different SETS of conditions, we have
`(d_1 RT_1)/P_1 =(d_2 RT_2)/P_2`
(given condition) (at S.T.P.)
or `(5.46 xx R xx 300)/2 = (d_2 xxRxx 273)/1`
or `d_2 =(5.46xx300)/(2xx273)=3 g/dm^3`
Discussion
3.

Density of a gas is __________

Answer»

DIRECTLY proportional to PRESSURE
indirectly proportional to pressure
directly proportional to volume
both B and c

Answer :A
Discussion
4.

Density of a gas is found to be 5.46 g dm^-1 at 27^@C and 2 bar pressure. What will be density at STP?

Answer»

Solution :d = PM/RT
For the same gas (i.e,, same M) at different temperatures and PRESSURES, we can WRITE `d_1/d_2 = P_1/T_1 XX T_2/P_2`
`d_1 = 5.46g dm^-3,P_1` = 2 bar,`T_1` = 300K
At STP,`P_2` = 1 bar,`T_2` = 273
therefore `5.46/d_2 = 2xx273/300xx1`
` therefore d_2 = (300xx1xx5.46)/(2xx273) = 3g dm^-1`
Discussion
5.

Density of a crystal remains unchanged as a result of

Answer»

IONIC DEFECT
Sohttky defect
Frenkel defect
CRYSTAL defect

Solution :Density o f crystal does not change in Frenkel defect as missing IONS are present in the interstitial sites.
Discussion
6.

Density of a gas at one atm pressure is 1.43xx10^(-2)g c c^(-1). Calculate the RMS velocity of the gas.

Answer»

Solution :DENSITY`=d= 0.0143g c c^(-1)`
Pressure`=P=1atm= 76cmHg= hd g `
` = 76 XX 13.6 xx 980.6`
` = 1.0133 xx 10^6g cm^(-1) s^(-2)`
RMS velocity (C ) `= sqrt(( 3 P)/( d))`
Substituting the VALUES, RMS velocity of the gas = ` sqrt((3 xx 1.0133 xx 10^6 )/(0.143 ) ) = 4.16 xx 10^4cms^(-1)`
` =461ms^(-1)`
Discussion
7.

The density of 3M Na_2S_2O_3 is 1.25 g/cc. Calculate the mole fraction of Na_2S_2O_3 and molality of thiosulphate.

Answer»


ANSWER :(a) 0.065 and (B) 7.73
Discussion
8.

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:

Answer»

`1.14 mol KG^(-1)`
`3.28 mol kg^(-1)`
`2.28 mol K^(-1)`
`0.44 mol kg^(-1)`

Solution :Molality (m) `=M/(1000d - M M_(1)) xx 100`
M = Molarity, MI = Molecular mass, d = density
`therefore m=(2.05)/((1000 xx 1.02)-(2.05 xx 60)) xx 1000`
`=2.28 mol kg^(-1)`
Discussion
9.

Density of 84% pure sulpuric acid in aqueous solution is 1.75 g cc^(-1) . Find the molarity of the acid solution.

Answer»


ANSWER :15mol `L^(-1)`
Discussion
10.

Densities of diamond and graphic are 3.5 and 2.5 gm/ml. C_("daimond") hArr C_("graphite") , Delta_(t)H = -1.9 KJ/ mole favourable conditions for formation of diamond are

Answer»

HIGH PRESSURE and LOW TEMPERATURE
low pressure and high temperature
high pressure and high temperature
low pressure and low temperature

Solution :Le-chatelier.s PRINCIPLE
Discussion
11.

Demericuration allows markownifoff's addition of H,OH without rearrangement. The net result is the addition of H_(2)O. Answer the following question. Q. CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH=CH_(2)underset(ii.CH_(3)OH//NaBH_(4)//NaOH)overset(i.Hg(OAc)_(2).THF)toAoverset(HI(conc.))toB, product B is

Answer»

`(CH_(3))_(3)C UNDERSET(OCH_(3))underset(|)(C)H-CH_(3)`
`CH_(3)I`
`CH_(3)CH_(2)I`
`(CH_(3))_(3)C underset(I)underset(|)(C)H-CH_(3)`

ANSWER :B
Discussion
12.

Demericuration allows markownifoff's addition of H,OH without rearrangement. The net result is the addition of H_(2)O. Answer the following question. Q.

Answer»




Both A and B

Answer :B
Discussion
13.

Demericuration allows markownifoff's addition of H,OH without rearrangement. The net result is the addition of H_(2)O. Answer the following question. Q. The product 'C' formation from B occurs through

Answer»

ENOL formation
`S_(N)1` mechanism
Neighboring GROUP participation
`S_(N)Ar` mechanism

Answer :C
Discussion
14.

DeltaU^(@) of combustion of methane is -X kJmol^(-1). The value of DeltaH^(@) is : (i) = DeltaU^(@)(ii)gt DeltaU^(@)(iii) lt DeltaU^(@)(iv) = 0

Answer»

Solution :The BALANCED equation for combustion of methane will be
`CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l)`
Thus, `Deltan_(g)= (n_(p) -n_(r))_(g)= 1 -3= -2`
`DeltaH^(@) LT DeltaU^(@) + Deltan_(g) RT = - X - 2RT`
Thus, `DeltaH^(@)lt DeltaU^(@)` . Hence, (iii) is the correct answer.
Discussion
15.

DeltaU^(@) of combustion of methane is -X kJ"mol"^(-1).The value of DeltaH is

Answer»

` = DELTAU^(@)`
`GT DeltaU^(@)`
`LT DeltaU^(@)`
` = 0`
Discussion
16.

DeltaU^Ө of combustion of methane is -X "kJ mol"^(-1) . The value of DeltaH^Ө is _____

Answer»

SOLUTION :` LT DeltaU^Ө`
Discussion
17.

DeltaU is equal to

Answer»

ISOCHORIC work
isobaric work
adiabatic work
isothermal work

Solution :For adiabatic PROCESS Q = 0
`DeltaU = q + W` or `DeltaU = w`
Discussion
18.

DeltaU= -10.5 Kj and DeltaS^(@)=-44.2 J/Kelvin for the reaction 2x_((g)) + y_((g)) to 2z_((g)) at 298 K temperature. Find the Delta_(f) G^( @) for the reaction. Reaction will be spontaneous or not ? Why ?

Answer»

SOLUTION :`Delta_(F) G^(@)=+0.193` KJ/mol
Discussion
19.

DeltaS is expected to be maximum for the reaction….........

Answer»

`Ca_((s))+1//2O_(2(g))toCaO_((s))`
`C_((s))+O_(2(g))toCO_(2(g))`
`Na_(2(g))+O_(2(g))to2NO_((g))`
`CaCO_(3(s))toCaO_((s))+CO_(2(g))`

Solution :In `CaCO_(3(s))toCaO_((s))+CO_(2(g))`, entropy change is positive.In (a) and (b) entrophy change is NEGATIVE, in © entrophy change is zero.
Discussion
20.

DeltaS is expected to be maximum for the reaction _____

Answer»

`Ca_((s)) + 1//2O_(2(G)) to CaO_((s))`
`C_((s)) + O_(2(g)) to CO_(2(g))`
`N_(2(g)) + O_(2(g)) to 2NO_((g))`
`CaCO_(3(s)) to CaO_((s)) +CO_(2(g))`

Solution :In `CaCO_(3(s)) toCaO_((s)) +CO_(2(g))` , entropy change is positive . In (a) and (B) entropy change is negative , in (C ) entropy change is zero.
Discussion
21.

Delta_(r) G^(@)= - RT " In K. For the same reaction at the same temperature using " K_(p) and K_(p) " the values "Delta _(r)G^(@) "are found to be different . Why ?

Answer»

Solution :The numerical value of equilibrium constant depends upon the standard state chosen for expressing concentrations or pressures. In CASE of pressures, the standard state chosen is 1 bar wheras in case of concentrations, standard state chosen is `1" MOL "L^(-1). " HENCE " , Delta_(r) G^(@)` VALUES are DIFFERENT.
Discussion
22.

DeltaH_(f)^(@) of water is -285.5 KJ mol^(-1). If enthalpy of neutraliztion of monoacidic strong base is -57.3KJ mol^(-1) then DeltaH_(f)^(@) of OH^(-) ion will be :

Answer»

`-285.5KJmol^(-1)`
`285.5KJmol^(-1)`
`114.5KJmol^(-1)`
`-114.5KJmol^(-1)`

ANSWER :a
Discussion
23.

DeltaH_("neutralisation") is always

Answer»

positive
negative
zero
positive or negative

Answer :B
Discussion
24.

DeltaH for the reaction at 298 K CO(g)+1//2O_(2)(g) is 282.85 KJ mol^(-1). Calculate DeltaU of the reaction.

Answer»

SOLUTION :`DeltaU=283.85KJmol^(-1)`
Discussion
25.

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

ANSWER :A
Discussion
26.

DeltaHand Delta Sfor the vaporisation of water at 1 atm pressure are 40.63kJ mol^(-1) and 108 . 8 JK^(-1) mol^(-1) respectively. Calculatethe temperatureat which the freeenergy for this transformation will be zero. Predict, giving reasons, the sign of free energy change (Delta G) above this temperature.

Answer»


ANSWER :373.4 K , NEGATIVE
Discussion
27.

DeltaH and DeltaS for certain reaction are -10KJ and - 44J/K, DeltaG^@ is

Answer»

`+3112 J`, NON -SPONTANEOUS 
`-3112 J`, spontaneous 
`+31.12 J`, non - spontaneous
`-31.12 J`, spontaneous 

ANSWER :A
Discussion
28.

deltaG^(o) vs T plot in the Ellingham's diagram slopes downaward for the reaction.

Answer»

Mg + `1/2 O_2 rArr` MgO
2Ag + `1/2 O_2 rArr Ag_(2)O`
`C+1/2O_(2)toCO`
`CO+1/2O_(2)toCO_(2)`

Solution :For the reaction C + `1/2 O_2` `rarr` CO , the `DELTAS` increase. THEREFORE, T`deltaS` increases with TEMPERATURE. Hence `DELTAG(deltaH - TdeltaS)` decreases. It means the slope of the curve for formation of CO decreases. However, for all other oxides, it increases.
Discussion
29.

DeltaG is net energy available to do useful work and is thus a measure of "free energy " . Show mathematically that DeltaG is a meausre of free energy. Find the unit ofDeltaG. Ifa reaction has positive enthalpy change and positive entropy change, under what condition will condition will the reaction be spontaneous ?

Answer»

Solution :(i) Referto art
(ii)As - `DeltaG = w_("useful"), ` THEREFORE, `DeltaG` has the same units those of work viz. JOULE.
(III) `DeltaG = DeltaH- T DeltaS`. If `DeltaH = +ve` and` DeltaS = +ve` , then `DeltaG` will be negative, i.e., process will be spontaneous only when `T DeltaS GT DeltaH`in magnitude , which will be so when temperature T is high.
Discussion
30.

DeltaG is net energy available to do useful work and is thus a measure of "free energy". Show mathematically that DeltaG is a measure of free energy. Find the unit of DeltaG. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous ?

Answer»

Solution :Gibbs free energy is that thermodynamic quantity of a system, the decrease in whose value during a process is equal to the maximum POSSIBLE useful work that can be obtained from the system.
Mathematically, this results may be derived as follows :
The relationship between heat absorbed by a system Q, the change in its internal energy, `DeltaU` and the work done by the system is given by the equation of the first law of thermodynamics, therefore,
`q= Delta U + W_("expansion") + W_("non-expansion") ...(i)`
Under constant pressure condition, the expansion work is given by `pDeltaV`.
`therefore Delta U + p Delta V = W_("non-expansion")`
`= Delta H+ W_("non-expansion") ""...(ii)`
For a reversible change taking place at constant temperature,
`DeltaS= (q_("rev") )/( T) "or" q_("rev") = T Delta S""...(iii)`
Substituting the value of q from Eq. (iii) in Eq. (ii), we get
`T Delta S =Delta H + W_("non-expansion")`
OR `Delta H - T Delta S =- W_("non-expansion") ""...(iv)`
Substituting this value in equation (iv), we get
`DeltaG= - W_("non-expansion") ""...(v)`
THUS, free energy change can be taken as a measure of work other than the work of expansion.
For most changes, the work of expansion can not be converted to other useful work, whereas the non-expansion work is convertible to useful work.
Rearranging equation (v), it may be WRITTEN as
`-Delta G= W_("non-expansion")= W_("useful")`
As `- Delta G = W_("useful")` therefore, `Delta G` has the same units as those of work i.e., joule
`Delta G = Delta H - T Delta S`
If `Delta H = ` positive and `Delta S=` positive, then `DeltaG` will be negative i.e., process will be spontaneous only when `T DeltaS gt Delta H` in magnitude, which will be so when temperature is high.
Discussion
31.

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

Answer»

zero
`lt Delta_(F)U^(@)`
`gtDelta_(f)U^(@)`
equal to `Delta_(f)U^(@)`

Solution :`Delta_(f)H^(@) = Delta_(f)U^(@) + Delta_(g)RT`
For the reaction,
`C(s) + 2H_(2)(g) to CH_(4)(g)`
`therefore Deltan_(g) = 1 -2 = -1`
`therefore Delta_(f)H^(@) = Delta_(f)U^(@) - 1RT`
or `Delta_(f)U^(@) = Delta_(f)H^(@) + RT`
or `Delta_(f)H^(@) lt Delta_(f)U^(@)`.
Discussion
32.

Delta_(f)U^(c-) of formationof CH_(4)(g) at certaintemperature is-393kJ mol^(-1) . Thevalue of Delta_(f)H^(c-) is

Answer»

zero
` lt Delta_(f)U^(C-)`
`GT Delta_(f)U^(c-)`
equalto `Delta_(f)U^(c-)`

Solution :`CH_(4)(g) + 2O_(2)(g) RARR CO_(2)(g) + 2H_(2)O(l)`
`Deltan_(g) = ( n_(p)-n_(r))_(g)=1-3= -2`
`Delta_(f)H^(@) =Delta_(f)U^(@) + Deltan_(g) RT`
As`Deltan_(g) = - 2 , Delta_(f)H^(@)lt Delta_(f) U^(@)` .
Discussion
33.

DeltaG^(@)for a reactionis 46.06kcal//mole , K_(p) for the reaction at 300K is

Answer»

<P>`10^(-8)`
` 10^(22.22)`
`10^(-33.33)`
NONE of these

Solution :`DeltaG^(@) = 46.06 kcal mol^(-1)`
`=46.06 xx 1000xx4.184J mol^(-1)`
`DeltaG^(@) = - RT lnK_(p)=- 2.303 RT logK_(p)`
` 46.06 xx 1000 xx 4.184 = - 2.303 xx 8.314 xx 300 log K_(p)`
or `log K_(p) = - 33.55 `or`K_(p) =10^(-33.55)`
Discussion
34.

DeltaE is always positive when

Answer»

system absorbsheat and work id done on it.
system emits heatand work is done on it.
system emits HEAT and nowork is done on it
system ABSORBS heat and work is done by it

Solution :Heat ABSORBED increases the INTERNAL energy and also work done on a system increases the internal energy.
Discussion
35.

Delta U^( Theta ) of combustion of methane is -X kj mol""^(-1). The value of Delta H^( Theta ) is….

Answer»

`= DELTA U^( Theta )`
`gt Delta U^( Theta )`
`lt Delta U^( Theta )`
`=0`

SOLUTION :`CH_(4(g)) + 2O_(2(g)) to CO_(2(g)) + 2H_(2)O_((l))`
`therefore Deltan_((g)) = -2`
`therefore Delta H^( Theta ) lt DeltaU^( Theta )`
Discussion
36.

DeltaS_("says") "for" 4Fe_((s)) + 3O_(2) rarr 2Fe_(2) O_(3(s)) " is" -550 J//k//mol at 298K. If enthalpy change for same process is -1600kJ//mol, DS_("total") (in J/mol/K) is

Answer»

`[(1600)/(298) XX 10^(3)] + 550 GT 0`
`550- [(1600)/(298)[ lt 0`
`[(1600)/(298) xx 10^(3)] -550 gt 0`
`[(1600 + 550)/(298)] gt 0`

Solution :`Delta S_("total") = Delta S_("sys") + Delta S_("sourrounding") gt 0`
`= Delta S+ ((-Delta H)/(T)) = -550 + ((-(-1600))/(298) xx 10^(3))`
`= (1600)/(298) xx 10^(3) - 550 gt 0`
Discussion
37.

DeltaS_("surr") " for " H_(2) + 1//2O_(2) rarr H_(2)O, DeltaH -280 kJ at 400K is

Answer»

700 J/g/K
700 KJ/mol/J
700 J/mol/K
`0.7J//mol//K`

ANSWER :C
Discussion
38.

DeltaS for vapousization of 900 g water (in KJ//K) is [DeltaH_(vap) = 40 KJ//mol]

Answer»

`(900 XX 40)`
`(50 xx 40)/(373)`
`(900 xx 40)/(373)`
`(18 xx 40)/(373)`

ANSWER :B
Discussion
39.

Delta S for 4Fe _((s)) + 3 O _(2(g)) to 2 Fe _(2) O _(3 (s)) is -550 J/mole/K.The process is found to be spontaneous even at 298K because [Delta H =- 1660kJ]

Answer»

`Delta S _("TOTAL ") =- 2000 J `
`Delta S _("total") =+ 1650J`
`Delta S _("total ") =+ 4980J`
`Delta S _("total") =-4980J`

Answer :C
Discussion
40.

Standard entropies ofH_(2)(g),O_(2)(g) " and "H_(2)O(l) are respectively 126.6,201.2 and 68JK^(-1)mol^(-1) Determine DeltaS" for "2H_(2)(g)+O_(2)(g) to 2H_(2)O(l) at 25^(@)C.

Answer»

`-574`
`+574`
`+1148`
`-1148`

ANSWER :A
Discussion
41.

Delta molecular orbitals are formed by _________ a

Answer»

`d_(xy) and d_(yz)`
`d_(xy) and d_(X^(2)-y^(2))`
`d_(x^(2)-y^(2)) and d_(z^2)`
`d_(xy) and d_(XZ)`

Answer :B
Discussion
42.

Delta HI^(@) of diamond is

Answer»

0 KJ `m o l^(-1)`
`1.90 kj m o l^(-1)`
38.1 kj `m o l ^(-1)`
20 kj `m o l ^(-1)`

Solution :`DELTA HF^(@)` of diamond is 1.90 kJ ` m o l^(-1)`
Discussion
43.

Delta H_(f) or BaCO _(3 (s)), CO _(2(g)), HCl _((aq)) and H _(2) O _((l))respectively are -1216, -393, -176 and - 286 KJ//mol. Then Delta H _(f) of BaCl _(2 (aq)) from BCo _(3 (s)) + 2HCl _((aq)) to BaCl _(2 (aq)) + CO _(2 (g)) + H _(2) O_((l))’ Delta H =- QKJ

Answer»

`(889-Q)KJ`
`(Q+889)KJ`
`-(Q+889)KJ`
`(Q-889)KJ`

ANSWER :C
Discussion
44.

The enthalpy of combustion of cyclohexane cyclohexene and H_2 are respectively -3920, -3800 and -241 KJ mol^(-1). The heat of hydrogenation of cyclohexene in KJ mol^(-1) is

Answer»

`-205 KJ `
`+205KJ`
`=53.5KJ`
`+53.5KJ`

ANSWER :D
Discussion
45.

In which of the following equations Delta H^(0) reaction equal to Delta H_(f)^(0) for the product?

Answer»

`2CO _((g)) + O_(2(g)) to 2 CO _(2(g))`
`N _(2 (g)) + O _(3(g)) to N _(2 ) O _(3 (g))`
`CH _(4 (g)) + 2 Cl _(2 (g)) to CH _(2) Cl _(2 (L)) + 2 HCl _((g))`
`XE _((g)) + 2 F _(2 (g)) to Xe F _(4 (g))`

Answer :D
Discussion
46.

DeltaH of combustion of yellow P and red P are - 11 K.J and -9.78 KJ respectively DeltaH of transition of yellow P to Red P is

Answer»

`-20.78 K.J`
`-1.22K.J`
`+1.22 K.J`
`+20.78 K.J`

ANSWER :B
Discussion
47.

Delta Hfor the reaction H-C-= N(g) + 2H_(2)(g) rarr H - underset(H) underset(|)overset(H ) overset(|)(C) - overset(H) overset(|)(N) - H (g) is - 150 kJ . Calculate the bond energy ofC -= Nbond. [ Given bond energies of C-H= 414 kJ mol^(-1), H-H= 435kJ mol^(-1), C-N = 293kJ mol^(-1), N-H =396 kJ mol^(-1)]

Answer»


SOLUTION :`Delta_(r)H^(@) = SIGMA ` B.E. ( Reactants ) `-Sigma ` B.E. ( Produts)
`- 150 = [ B.E. ( C-H) + B.E.(C-=N) + 2B.E.( H-H) ]-[3 xx B.E. (C-H) + B.E.(C-N) + 2 xx B.E. (N-H)]`
Discussion
48.

Zn_((s)) + Cu_((aq))^(+2) -> Cu_((s)) + Zn_((aq))^(+2), DeltaG = -XKJ, Cu_((s)) + 2Ag_((aq))^(+) rarr 2Ag_((s)) + Cu_((aq))^(+2), DeltaG = -Y KJ, then DeltaG for Zn_((s)) + 2Ag_((aq))^(+) rarr 2Ag_((s)) + Zn_((aq))^(+2) will be (in KJ)

Answer»

`- 82KJ`
`-22KJ`
`+226 KJ`
ZERO

ANSWER :A
Discussion
49.

DeltaH for the formation of XY is -200 kJ mol^(-1).The bond enthalpies of X_2, Y_2, and XY are in the ratio 1:0.5:1. Then determine the bond enthalpies.

Answer»

Solution :Let the BOND enthalpy of `X _(2)` is a, `Y_(2)` is (a/2) and XY is `a. (1)/(2) X _(2) + (1)/(T) Y_(2) to XY: Delta H =- 200 KJ`
Heat of reaction `, Delta H = ` (Enthlpy ofbond DISSOCIATION ) - (Enthalpy of bond forming)
`= ((a)/(2) +(a)/(4)) - (a) =- 200 kJ =- (a)/(4) (or) a =800 kJ`
The bond enthalpy of `X _(2) = 800 kJ mol ^(-1), Y_(2) = 400 kJ mol ^(-1) and XY = 800 kJ mol ^(-1)`
Discussion
50.

In the conversion of lime stone to lime, CaCO_(3(s)) rarr CaO_((s)) + CO_(2(g)), the values of Delta H^(@) and Delta S^(@) are + 179.1 kJ mol^(-1) and 160.2 JK^(-1) mol^(-1) respectively at 298K and 1 bar. Assuming, Delta H^(@) and Delta S^(@) do not change with temperature, temperature above which conversion of lime stone to lime will be spontaneous is

Answer»

`+2734`
`-2734`
`-2984`
`-838`

ANSWER :B
Discussion