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251.

The favourable conditions for a spontaneous reaction areA. `TDeltaS gt DeltaH`, `DeltaH=+ve`, `DeltaS=+ve`B. `TDeltaS gt DeltaH`, `DeltaH=+ve`, `DeltaS=-ve`C. `TDeltaS = DeltaH`, `DeltaH=-ve`, `DeltaS=-ve`D. `TDeltaS = DeltaH`, `DeltaH=+ve`, `DeltaS=+ve`

Answer» Correct Answer - A
`DeltaG=DeltaH-TDeltaS`. For a process to be spontaneous, `DeltaG` should be `-ve`. In `(A) DeltaH` is `+ve` and `DeltaS` is `+ve` and `TDeltaS gt DeltaH`
`0 gt DeltaH-TDeltaS`
`0 gt DeltaG`
i.e., process is spontaneous
Discussion
252.

The unit of entropy isA. `Jk^(-1)mol^(-1)`B. `KJ^(-1)mol^(-1)`C. `Kjmol^(-1)`D. `J^(-1)K^(-1)mol^(-1)`

Answer» Correct Answer - A
Entropy change `DeltaS=(q_(rev))/(T)`
`:.` Unit of `S` is `JK^(-1)mol^(-1)`
Discussion
253.

Which of the following neutralisation reasons,the heat of neutralisation will be highest?A. `NH_(4)OH` and `CH_(3)COOHH`B. `NH_(4)OH` and HClC. `NaOH` and `CH_(3)COOH`D. `NaOH` and `HCl`

Answer» Correct Answer - C
Heat of neutralisation between strong acid and a strong base is about `-13.7Kcal`
Discussion
254.

Which of the following has `DeltaS^(@)` greater than zeroA. `CaO(s)+CO_(2)(g)hArrCaCO_(3)(s)`B. `NaCl(aq)hArrNaCl(s)`C. `NaNO_(3)(s)hArrNa^(+)(aq)+NO_(3)^(-)(aq)`D. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`

Answer» Correct Answer - C
For choice `(C )` i.e.,
`NaNO_(3)hArrNa^(+)(aq)+NO_(3)^(-)(aq)`
`DeltaS^(@)` is positive as only in this process entropy of the system increases.
Discussion
255.

When `1`pentyne `(A)` is treated with `4N` alcoholic `KOH` at `175^(@)C`, it is slowly converted into an equilibrium mixture of `1.3%` of `1`pentyne `(A), 95.2% 2`-pentyne `(B)` and `3.5%` of `1,2`-pentandiene `(C )`. The equilibrium was maintained at `175^(@)C`. calculate `DeltaG^(Theta)` for the following equilibria: `B hArr A, DeltaG^(Theta)underset(1) = ?` `B hArr C, DeltaG^(Theta)underset(2) =?` From the calculated value of `DeltaG^(Theta)underset(1)`and `DeltaG^(Theta)underset(2)`, indicate the order of stability of `A,B` and `C`.

Answer» `{:(,"Pentyne"-1,hArr,"Pentyne"-2,+,1","2-"pentadiene"),(,(A),,(B),,(C)),(t_(eq),1.3,,95.2,,3.5):}`
`K_(eq)=([B][C])/([A])=(95.2xx3.5)/(1.3)=256.31`
for `BhArrA`
`K_(1)=([A])/([B])=([C])/(K_(eq))=(3.5)/(256.31)=0.013`
`DeltaG_(1)^(@)=-2.303RT" log "K_(1)`
`=-2.303xx8.314xx448" log "0.013`
`=16178J=16.178kJ`
for `BhArrC`
`K_(2)=([C])/([B])=(K_(eq)[A])/([B]^(2))=(256.31xx1.3)/((95.2)^(2))=0.037`
`DeltaG_(2)^(@)=-2.303RT" log "K_(2)`
`=-2.303xx8.314xx448" log "0.037`
`=12282J=12.282kJ`
Stability will lie in the order
`B gt C gt A`
Discussion
256.

Which one is not a state function ?A. Internal energy `(E)`B. VolumeC. Heat `(q)`D. Enthalpy

Answer» Correct Answer - C
State function are those functions whose values depend only upon the intinal and final state of the system and not on the path.
Discussion
257.

For conversion `C` (graphite)`rarrC` (diamond) the `DeltaS` isA. ZeroB. PositiveC. NegativeD. Unknown

Answer» Correct Answer - A
Conversion of graphite into diamond is an endothermic reaction. So, heat of diamond is higher than that of graphite. But `DeltaS` would be negative for the conversion of graphite into diamond.
Discussion
258.

In an isochoric process the increase in internal energy isA. Equal to the heat absorbedB. Equal to the heat evolvedC. Equal to the work doneD. Equal to the sum of the heat absorbed and work done

Answer» Correct Answer - A
For isochoric process `DeltaV=0` so `qv=DeltaE` , i.e., heat given to a system under constant volume is used up in increasing `DeltaE` .
Discussion
259.

Which of the following describes the criterion of spontaneity?A. `DeltaS_((Total))gt0`B. `DeltaG_((T,P))gt0`C. `DeltaH_((T,P))gt0`D. None of above

Answer» Correct Answer - A
Positive entropy change.
Discussion
260.

Which of the following process has negative value of `DeltaS` ?A. Dissolution of sugar in waterB. Stretching of rubber bandC. Decomposition of lime stoneD. Evaporation of water.

Answer» Correct Answer - B
In rubber the long flexible macromolecules coli up in a random way. When stretched, it gets uncoiled. The uncoiled arrangement has more specific geometry,
Discussion
261.

Choose the reaction with negative `DeltaS` valueA. `2NaHCO_(3)(s)toNa_(2)CO_(3)(s)+CO_(2)(g)+H_(2)O(g)`B. `Cl_(2)(g)to2Cl(g)`C. `2SO_(2)(g)+O_(2)(g)to2SO_(3)(g)`D. `2KCIO_(3)(s)to2KCI(s)+3O_(2)(g)`

Answer» Correct Answer - C
`DeltaS` has a negative value of `Deltan_(g)` is `-ve`
Discussion
262.

A scientist needs a refrigeration machine to maintain temperature of `-13^(@)C` for certain chemical process. How much work must be performed on the system during each cycle of its operation if `3000 J` of heat is to be withdrawn from the `-13^(@)C` reservoir and discharged to the room at `+27^(@)C`? Assume that the machine operates at `100 %` of its theoretical efficiency.A. `65000 J`B. `3000 J`C. `154 J`D. `133 J`

Answer» Correct Answer - B
`eta=(W)/(q_(2))`. Since `eta=1 :. W= q_(2)=3000J`.
Discussion
263.

The bond dissociation energies of`X_(2),Y_(2)andXY` are in the ratio of`1:0.5:1.DeltaH` for the formation of `XY`is `-200kJmol^(-1)`.The bond dissociation energy of `X_(2)`will beA. `200KJmol^(-1)`B. `100KJmol^(-1)`C. `800KJmol^(-1)`D. `400 KJ mol^(-1)`

Answer» Correct Answer - A
Let `B.E.` of `x_(2),y_(2)` & `xy` are `xKJmol^(-1) , 0.5xKJmol^(-1)` and `xKJmol^(-1)` respectivaly
`(1)/(2)x_(2)+(1)/(2)y_(2)rarrxy,Delta=-200KJmol^(-1)`
`DeltaH=-200=Sigma(B.E)_("Reactant")=Sigma(B.E.)_("Product")`
`=(1)/(2)B.E_(x-x)+(1)/(2)B.E_(y-y)-B.E_(x-y)`
`=[(1)/(2)xx(x)+(1)/(2)xx(0.5x)]-[1xx(x)]`
`B.E.` of `x_(2)=x=800KJmol^(-1)` .
Discussion
264.

Which of the following is not a state function?A. Internal energyB. EnthalpyC. WorkD. Entropy

Answer» Correct Answer - C
Work is not a state function as during a process its value depends on the path followed. The value of enthalpy, internal energy and entropy depends on the state functions.
Discussion
265.

An athelete is given `100g` of glucose `(C_(6)H_(12)O_(6))` of energy equivalent to `1560 kJ`. He utilises `50%` of this gained enegry in the event. In order to avoid storage of enegry in the body, calculate the weight of water he would need to perspire. The enthalpy of evaporation of water is `441kJ//mol`.

Answer» Unused energy`=(1560)/(2)=780kJ`
Mass of water needed for perspiration `=18xx(780)/(44)=318.96g`
Discussion
266.

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 parameters isA. `DeltaG=0`, `DeltaS=+ve`B. `DeltaG=0`, `DeltaS=-ve`C. `DeltaG=+ve`, `DeltaS=+ve`D. `DeltaG=-ve`, `DeltaS=+ve`

Answer» Correct Answer - A
As the product is in equilibrium at `373 K`,
`DeltaG=0`
A liquid changes into gas, `DeltaS +ve`
Discussion
267.

The heat released when `NH_(4)OH` and `HCl` neutralise isA. `13.7 kcal`B. ` gt 13.7 kcal`C. ` lt 13.7 kcal`D. None of the above

Answer» Correct Answer - C
Factual questions.
Discussion
268.

Which of the following pairs of a chermical reaction is centain to result in a spontaneous reaction ?A. Exothermic and decreasesing disorderB. Exothermic and increasesing disorderC. Exothermic and increasesing disorderD. Exothermic and decreasesing disorder

Answer» Correct Answer - B
If reaction is exothermic, therefore `DeltaH` is `-ve` and on increasing disorder, `DeltaS` is `+ve` , thus, at these conditions, `DeltaG` is negative according to the following equation.
`DeltaG=DeltaH-TDeltaS`
`Delta=-ve` , and for spontaneous reation `DeltaG` must `be-ve` .
Discussion
269.

the equillibrium constant `K_(P)` for the reaction , `N_(2)(g)+ 3H_(2)(g)Leftrightarrow2NH_(3)(g)` `1.6 xx 10^(-4) (atm)^(-2) at 400^(@)C` if heat of the reaction in this temoerture range is -25.14 kcal ?A. ` 1.231 xx 10^(-4)(atm)^(-2)`B. `1.876 xx 10^(-7)(atm)^(-2)`C. `1.462 xx 10^(-5)(atm)^(-2)`D. `3.462 xx 10^(-5)(atm)^(-2)`

Answer» Correct Answer - c
Equilibrium constants at different temperature and heat of the reaction are related by the equation
`2.303 log(K_(P_(2))/(K_(p_(1))))= (DeltaH^(@))/R[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log K_(P_(2))=-4.835`
`K_(P_(2))=1.462xx10^(5)(atm)^(-2)`
Discussion
270.

Assuming that `50%` of the heat is useful, how many `kg` of water at `15^(@)C` can be heated to `95^(@)C` by burning 200 litre of methane at `STP` ? The heat of combustion of methane is `211kcal//mol`. Specific heat of water is `1.0 kcal kg^(-1)K^(-1)`.

Answer» Correct Answer - 11.76 kg
Discussion
271.

The system in which there is no exchange of matter, work, or energy from the surroundings isA. closedB. isolatedC. adiabaticD. isothermal

Answer» Correct Answer - B
An isolated system does not exchange matter or energy (or work) with the surroundings.
Discussion
272.

For an ideal gas, consider only `P-V` work in going from an initial state `X` to the final state `Z`. The final state `Z` can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [Take `DeltaS` as change in entropy and `w` as work done] A. `DeltaS_(xtoz)=DeltaS_(xtoy)+Delta_(ytoz)`B. `w_(xtoz)=w_(xtoy)+w_(ytoz)`C. `w_(xtoytoz)=w_(xtoy)`D. `DeltaS_(xtoytoz)=DeltaS(xtoy)`

Answer» Correct Answer - A::C
Entropy is state function, while work is the path dependent function
`DeltaS_(xtoz)=DeltaS_(xtoy)+DeltaS_(ytoz)`
`w_(xtoytoz)=w_(xtoy)`.
Discussion
273.

Among the following, the intensive property is (properties are) :A. molar conductivityB. electromotive forceC. resistanceD. heat capacity

Answer» Correct Answer - A::B
Resistance and heat capacity are mass dependent properties, hence they are extensive properties.
Discussion
274.

Which of the following is correct option for the free expansion of an ideal gas under adiabatic condition ?A. `q=0`, `DeltaT lt 0`, `w ne 0`B. `q=0`, `DeltaT ne 0`, `w = 0`C. `qne0`, `DeltaT = 0`, `w = 0`D. `q=0`, `DeltaT = 0`, `w = 0`

Answer» Correct Answer - D
For adiabatic process, `q=0`
For free expansion `P_(ext)=0 :. W=0`
For adiabatic expansion of ideal gas, `DeltaT=0`
Discussion
275.

`AtoB` , `DeltaH=4 kcal mol^(-1)`, `DeltaS=10 cal mol^(-1)K^(-1)`. Reaction is spontaneous when temperature isA. `400 K`B. `300 K`C. `500 K`D. none of these

Answer» Correct Answer - C
`DeltaG=DeltaH-TDeltaS`. `DeltaG` will be negatively charged only when `T=500 K`.
Discussion
276.

For a reaction `DeltaH=+29kJ" "mol^(-1),DeltaS=-35KJ^(-1)" "mol^(-1)` at what temperature, the reaction will be spontaneous?A. `828.7^(@)C`B. `828.7K`C. Spontaneous at all temperatureD. non possible

Answer» Correct Answer - D
`DeltaG=DeltaH-TDeltaS`
When `DeltaH=+ve,DeltaS=-ve` then `DeltaG` will be positive and the reaction is non-spontaneous.
Discussion
277.

The `DeltaH` value for reaction `C+(1)/(2)O_(2)toCO` and `CO+(1)/(2)O_(2)toCO_(2)` are `100` and `200 kJ` respectively. The heat of reaction for `C+O_(2)toCO_(2)` will beA. `50 kJ`B. `100 kJ`C. `150 kJ`D. `300 kJ`

Answer» Correct Answer - D
Add eqn. `(i)` and `(ii)`.
Discussion
278.

internal energy and pressure of a gas of unit valume are related asA. `P =2/3E`B. `P=E/2`C. `P= 3/2E`D. P=2E

Answer» Correct Answer - a
internal energy `E=3/2RT=3/2P RightarrowP2/3 E`
Discussion
279.

The work done in erge for the reversible expansion of 1 mole of an ideal gas from a volume of 10 litres to 20 litres at `25^(@)C` isA. `-2.303xx298xx0.082log2`B. `-298xx10^(7)xx8.31xx2.3031 log2`C. `2.303xx298xx0.082log0.5`D. `-8.31xx10^(7)xx298xx2.303log0.5`

Answer» Correct Answer - D
`W=-2.303" nRTlog"(V_(2))/(V_(1))`
`=-2.303xx1xxx8.314xx107xx298" log"(20)/(10)`
`=-298xx107xx8.314xx2.303log2`
Discussion
280.

Which of the follwing has lowest fusion temperature ?A. NaphthaleneB. DiamondC. `NaCl`D. `Mn`

Answer» Correct Answer - C
The fusion temperature of naphthalene is minimum, because it non-polar covalent compound and has less fusion temperature
Discussion
281.

Calculate the work done during the process, when one mole of gas is allowed to expand freely into vacuum.A. `0`B. `+ve`C. `-ve`D. any of there

Answer» Correct Answer - A
`w=P_(ext)(V_(2)-V_(1))`
` :. P_(ext)=0`
`:. w=0`
Discussion
282.

Under the same conditions, how many `mL` of `1M KOH` and `0.5M H_(2)SO_(4)`solutions, respectively, when mixed to form a total volume of `100mL`, produces the highest rise in temperature?A. `67,33`B. `33,67`C. `40,60`D. `50,50`

Answer» Correct Answer - D
`50ml` of `1 M KOH+50 ml` of `0.5 M H_(2)SO_(4)` will involve complete neutralisation and hence produce highest rise in temperature.
Discussion
283.

Under the same conditions, how many `mL` of `1M KOH` and `0.5M H_(2)SO_(4)`solutions, respectively, when mixed to form a total volume of `100mL`, produces the highest rise in temperature?A. `67:33`B. `33:67`C. `40:60`D. `50:50`

Answer» Correct Answer - C
`50 Meq` . Of `KOF` and `50Meq` . Of `H_(2)SO_(4)` will produce maximum heat.
Discussion
284.

The free energy change `DeltaG = 0`, whenA. the system is at equilibriumB. catalyst is addedC. reactants are intially mixed throughlyD. the reactants are completely consumed.

Answer» Correct Answer - A
Factual question.
Discussion
285.

Which of the following is true for a reaction `H_(2)O(l)toH_(2)O(g)` at `100^(@)C`, `1` atm. PressureA. `DeltaH=DeltaE`B. `DeltaE=0`C. `DeltaH=0`D. `DeltaH=TDeltaS`

Answer» Correct Answer - D
At equilibrium `DeltaG=0` or `DeltaH=T DeltaS`
Discussion
286.

`DeltaH` for solid to liquid transitions for protein `A` and `B` are `2.73 kcal//mol` and `3.0 kcal//mol` .The two melting points are `0^(@)C` and `30^(@)C` respectively. The entropy changes `DeltaS_(A)` and `DeltaS_(B)` at two transition temperatures are related asA. `DeltaS_(A)=DeltaS_(B)`B. `DeltaS_(A)ltDeltaS_(B)`C. `DeltaS_(B)ltDeltaS_(A)`D. `DeltaS_(B)(300DeltaS_(A))/(273)`

Answer» Correct Answer - C
`DeltaS_(A)=(2.73)/(273)=0.01` , `DeltaS_(B)=(3.0)/(303)=0.0099`
Hence `DeltaS_(B) lt DeltaS_(A)`
Discussion
287.

Three moles of an ideal gas expanded spotaneously into vacuum. The work done will beA. infinityB. `10 J`C. zeroD. `5 J`

Answer» Correct Answer - C
If an ideal gas expand against vacuum, the irreversible expansion is called free expansion. As `P_(ext)=0`, therefore, for reversible as well as irrversible expansion, work done `=0`
`W_(irre)=-P_(ext)DeltaV=0xxDeltaV=0`
Discussion
288.

find the entropy change when 2 moles of ideal gas at `27 ^(@)C` temperature is expanded reversiby from 2 L to 20 L .A. 92 . 1B. 0C. 4D. 9.2

Answer» Correct Answer - d
`DeltaS=2.303nR log((V_(2))/V_(1))= 2.303xx2xx2xxlog(20/2)=9.2`
Discussion
289.

A gas expands , isthemally and reversiby.the work done by the gas isA. zeroB. maximumC. minimumD. cannot be determined

Answer» Correct Answer - b
work done during isothemal reversible expansion of a gas is maximum
Discussion
290.

which of the following statement is correct ?A. the presence of reacting species in a covered beaker is an example of open systemB. there is an exchange of energy as well as matter between the system and the surrounding in a closed systemC. the presence of reactants in a closed vessel made up of copper is an example of a closed systemD. the presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system

Answer» Correct Answer - c
the presence of reactants in a closed vessel madeup of conducting material.e.g. copper or steel is an example of a closed system.
Discussion
291.

which among the following is intensive quantity ?A. EnthalpyB. TemperatureC. VolumeD. Refractive index

Answer» Correct Answer - B::D
Both temperature and refractive index are intensive properties.
Discussion
292.

A well stoppered thermo flask containing some ice cubse is an example ofA. Closed systemB. Open systemC. Isolated systemD. None of these

Answer» Correct Answer - C
An isolated system neither shows exchange of heat nor matter with surrooundings.
Discussion
293.

For an ideal gas expanding adiabatically in vacuum,A. `DeltaH=0`B. `DeltaH gt0`C. `DeltaH lt0`D. None of these

Answer» Correct Answer - A
`DeltaH=DeltaU+PDeltaV`. For expansion of an ideal gas, `DeltaU=0`. For expansion against vacum, `PDeltaV=0` Hence, `DeltaH=0+0=0`
Discussion
294.

The enthalpy combustion of a substanceA. is always positiveB. is always negativeC. can be either zero or greater than zeroD. is unpredictable till calculations are done.

Answer» Correct Answer - B
Combustion reaction is akways exothermic.
Discussion
295.

1 mole of naphthalene `(C_(10)H_(8))` was burnt in oxygen gas at `25^(@)C` at constant volume. The heat evolved was found to be 5138.8kJ. Calculate the heat of reaction at constant pressure. `(R=8.3JK^(-1)" "mol^(-1))`

Answer» Correct Answer - 5143.8kJ
Water is present in liquid state at `25^(@)C` and naphthalene in solid state.
Discussion
296.

For a reaction to occur spontaneouslyA. `(DeltaH-TDeltaS))` must be negativeB. `(DeltaH+TDeltaS))` must be negativeC. `DeltaH` must be nagative.D. `DeltaS` must be nagative.

Answer» Correct Answer - A
`DeltaG`, which equal to `(DeltaH-TDeltaS)` must be negative for spontaneity.
Discussion
297.

for the dissociation reaction , `H_(2)(g)Leftrightarrow2H(g),DeltaH= 162 kcal` heat of atomisation of H isA. 81 kcalB. 162 kcalC. 208 kcalD. 218 kcal

Answer» Correct Answer - a
`DeltaH=DeltaH_("product")DeltaH_(("reactant"))`
` 162 =2 xx DeltaH_(H)-DeltaH_(H_(2))`
`DeltaH_(H)=162/2=81 kcal`
`(DeltaH_(H_(H_2))=0)`
Discussion
298.

If `H_(2)(g)=2H(g),DeltaH=104cal` , then heat of atomisation of hydrogen isA. `52Kcal`B. `104cal`C. `208Kcal`D. None of these

Answer» Correct Answer - B
Heat of atomisation of hydrogen is derived for 1 mole of atoms of hydrogen formed.
Discussion
299.

The bond energies of `H--H` , `Br--Br` and `H--Br` are `433,, 192` and `364KJmol^(-1)` respectively. The `DeltaH^(@)` for the reaction `H_(2)(g)+Br_(2)(g)rarr2HBr(g)` isA. `-261kJ`B. `+103kJ`C. `+261kJ`D. `-103kJ`d

Answer» The given reaction is:
`H-H(g)+Br-Br(g)to2H-Br(g)`
`DeltaH^(@)=sum(B.E.)_("Reactants")-sum(B.E.)_("products")`
`=[433+192]-[2xx364]`
`=-103kJ" "mol^(-1)`
Discussion
300.

Assertion:The change in entropy during melting of ice is negligible in comparison to change in entropy during vaporisation. Reason:The volume occupied by solid and liquids is too less in comparison to volume occupied by gas.A. If both ther assertion and reason are true and the reason is the correct explanation of the assertion.B. If both the assertion and reason are true but reason is not the correct explantion of the assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is true.

Answer» Correct Answer - A
`DeltaS=(q_(rev))/(T)` , thus, unit is `JK^(-1)mol^(-1)` .
Discussion