InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
At 1 atm pressure, `DeltaS=75JK^(-1)mol^(-1),DeltaH=30kJ" "mol^(-1)`, the temperature of the reaction at equilibrium isA. 400 KB. 330 KC. 200 KD. 110 K |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS` At equilibrium, `DeltaG=0` `T=(DeltaH)/(DeltaS)=(30xx10^(3))/(75)=400K` |
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| 2. |
The free energy for a reaction having `DeltaH=31400cal,DeltaS=32" cal "K^(-1)mol^(-1)` at `1000^(@)C` isA. `-9336cal`B. `-7386cal`C. `-1936cal`D. `+9336cal` |
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Answer» Correct Answer - A From Gibbs-Helmholtz equation, we know `DeltaG=DeltaH-T*DeltaS=31400-1273xx32` `=31400-40736=-9336cal` |
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| 3. |
From an isolated system, `DeltaU=0`, what will be `DeltaS`? |
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Answer» Correct Answer - B For an isolated system, `DeltaU=0` and for a spontaneous process, total entropy change must be positive. `DeltaS gt 0`. |
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| 4. |
Given, `2Fe+(3)/(2)O_(2)toFe_(2)O_(3),DeltaH=-193.4kJ,Mg+(1)/(2)O_(2)to MgO,DeltaH=-140.2kJ` What is the `DeltaH` for the following reaction? `3Mg+Fe_(2)O_(3) to 3MgO+2Fe`A. `-227.2kJ`B. `-272.3kJ`C. `227.2kJ`D. `272.3kJ` |
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Answer» Correct Answer - A `2Fe+(3)/(2)O_(2)toFe_(2)O_(3),DeltaH=-193.4kJ` . . (i) `Mg+(1)/(2)O_(2)toMgO,DeltaH=-140.2kJ` . .. (ii) On multiplying Eq. (iI) by (3) and subtracting eq. (i) we get the final equation, `3Mg+Fe_(2)O_(3)to2Fe+3MgO,` `therefore DeltaH=(-420.6)+(193.4)=-227.2kJ` |
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| 5. |
Heat of formation of `H_(2)O` is -188kJ/mol and `H_(2)O_(2)` is -286 kJ/mol. The enthalpy change for the reaction, `2H_(2)O_(2) to 2H_(2)O+O_(2)`A. 196 kJB. `-196kJ`C. `984kJ`D. `-984kJ` |
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Answer» Correct Answer - A `H_(2)+(1)/(2)O_(2)toH_(2)O,DeltaH=-188" kJ "mol^(-1)` . . (i) `H_(2)+O_(2) to H_(2)O_(2) to H_(2)O_(2), DeltaH=-286" kJ "mol^(-1)` . . . (ii) Multiply (i) and (ii) by 2 and subtracting, we get `2H_(2)O_(2)to 2H_(2)O +O_(2),DeltaH_(r)=+196kJ`. |
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| 6. |
Assuming enthalpy of combustion of hydrogen at 273 K is -286 kJ and enthalpy of fusion of ice at the same temperature to be +6.0 kJ, calculate enthalpy change during formation of 100 g of ice.A. `+1622 kJ`B. `-1622kJ`C. `+292kJ`D. `-292kJ` |
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Answer» Correct Answer - B (I) `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),Delta_(c)H=-286kJ` (II) `H_(2)O(s)tounderset(18g)(H_(2)O)(l),Delta_(fus)H=+6kJ` On subtracting (II) from (I), we get `H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(s)` `DeltaH=-286-6=-292kJ` `because`On formation of 18 g of ice, change in enthalpy=-292 kJ `therefore`On formation of 100 g of ice, change in enthalpy `=-(292xx100)/(18)=-1622kJ` |
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| 7. |
The (i)_____of a syste can be changed by transfer of heat from the surrounding to the system of cive-verse without expenditure of (ii)_____. Identify (i) and (ii) in order to complete the above statement.A. (i)-enthalpy, (ii)-workB. (i)-internal energy, (ii) heatC. (i)-enthalpy, (ii)- heatD. (i)-internal energy, (ii)-work |
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Answer» Correct Answer - D The internal energy of a system can be changed by transfer of heat from the surroundings to the system or vice-versa wihtout expenditure of work. |
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| 8. |
For the reaction `AtoB,DeltaH=+23kJ//mol and BtoC,DeltaH=-18kJ//mol`, the decreasing order of enthalpy of A,B,C follows the orderA. A,B,CB. B,C,AC. C,B,AD. C,A,B |
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Answer» Correct Answer - B `A to B,DeltaH=+24" kJ "mol^(-1)` `H_(B)-H_(A)=+24`. . .. (i) `B to C, DeltaH=18" kJ "mol^(-1)` `H_(C)-H_(B)=-18` . . (ii) `H_(B)-H_(C)=+18` . . (iii) `H_(C)-H_(A)=6` (from (i) and (ii)) `H_(B) gt H_(C) gt H_(A)`. |
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| 9. |
A system containing some quantity of water in a thermos flask or in an insulated beaker. This would not allow exchange of heat between the system and surrouding through its boundary. This process is called asA. isothermal processB. adiabatic processC. isochoric processD. isobaric process |
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Answer» Correct Answer - B The sytem would not allow exchange of heat between the system and surroundings through its boundary. Hence, this types of system is called adiabatic system. |
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| 10. |
The quantity of heat measured for a reaction in a bomb calorimeter is equal toA. `DeltaG`B. `DeltaH` is negativeC. `pDeltaV`D. `DeltaE` |
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Answer» Correct Answer - D Heat of combustion is usually measured by placing a known mass of a compound in a steel container at contant volume called bomb calorimeter. `q_("reaction")("heat of reaction")=C_(V)Deltat=DeltaE` |
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| 11. |
What would be the heat released when an aqueous solution containing 0.5 mole of `HNO_(3)` is mixed with 0.3 mle of `OH^(-)`? (Enthalpy of neutralisation is -57.1 kJ)A. 28.5 kJB. 17.1 kJC. 45.7 kJD. 1.7 kJ |
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Answer» Correct Answer - B 0.5 mole of `HNO_(3)-=0.5` mole of `H^(+)` and 0.3 mole of `OH^(-)` `0.5H^(+)(aq)+0.3OH^(-)(aq)to0.5H_(2)O(l)` `DeltaH=0.3xx57.1=17.13kJ` |
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| 12. |
An ideal gas expands from `10^(-3) m^(3)` to `10^(-2) m^(3)` at 300 K against a constant pressure of `10^(5) Nm^(-2)`. The workdone isA. `-900J`B. `-900J`C. `270J`D. `900kJ` |
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Answer» Correct Answer - A Work done due to change in volume against constant pressure, `W=-p(V_(2)-V_(1))` `=-1xx10^(5)Nm^(-2)(1xx10^(-2)-1xx10^(-3))m^(3)` `=-900Nm=-900J" "(1Nm=1J)` |
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| 13. |
Identify an extensive property amongst the following:A. ViscosityB. Heat capacityC. DensityD. Surface tension |
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Answer» Correct Answer - B Heat capacity is an extensive property. It is a property that changes when the size of the system changes whereas viscosity, density and surface tension are intensive properties. It is also termed as bulk property that does not depend on the system size. |
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| 14. |
Which of the following is the intensive quantity?A. Enthalpy and temperatureB. Volume and temperatureC. Enthalpy and volumeD. Temperature and refractive index |
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Answer» Correct Answer - D Temperature and refractive index are intensitve quantities. |
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| 15. |
Which of the following is an intensive property?A. TemperatureB. ViscosityC. Surface tensionD. All of these |
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Answer» Correct Answer - D The properties of the system whose value is independent of the amount of substance present in the system are caled intensitve properties, e.g. viscosity, surface tension, temperature, pressure, etc. |
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| 16. |
Heat capacity `(C_(p))` is an extensive property but specific heat (C) is an intensive property. What will be the relation between `C_(p) and C` for 1 mole of water?A. `+4.18JK^(-1)`B. `-4.18JK^(-1)`C. `-75.3JK^(-1)`D. `+75.3JK^(-1)` |
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Answer» Correct Answer - D For water, Molar heat capacity, `C_(p)=18xx`specific heat C Specific heat, `C=4.18Jg^(-1)K^(-1)` (for water) heat capacity, `C_(p)=18xx4.18JK^(-1)` `=75.24JK^(-1)=75*3JK^(-1)` |
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| 17. |
`C(s)+O_(2)(g)toCO_(2)(g),DeltaH=-94` kcal `2CO(g)+O_(2)to2CO_(2),DeltaH=-135.2` kcal The heat of formation of `CO(g)` isA. `-26.4` kcalB. 41.2 kcalC. 26.4 kcalD. 229.2 kcal |
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Answer» Correct Answer - A `2CO(g)+O_(2)(g) to 2CO_(2)(g),DeltaH=-135.2kcal` (I) `CO_(2)(g)toCO(g)+(1)/(2)O_(2)(g),DeltaH=(135*2)/(2)` kcal (II) `C(s)+O_(2)(g)toCO_(2)(g),DeltaH=-94kcal` Required equation, `C(s)+(1)/(2)O_(2)(g) to CO(g),` On adding eqs. (I) and (II) `C(s)+(1)/(2)O_(2)(g) to CO(g),DeltaH=-26.4kcal` |
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| 18. |
Which of the following statement is incorrect about the equation? `C_(2)H_(5)OH(l)+3O_(2)(g) to 2CO_(2)(g)+3H_(2)O(l),Delta_(r)H^(-)=-1367" kJ "mol^(-1)`A. the reaction is endothermic in nature.B. The numerical value of `Delta_(r)H^(ө)` refers to the number of moles of substances specified by an equation. Standard enthalpy change `Delta_(r)H^(ө)` will have units as kJ `mol^(-1)`C. The coefficients in a balanced thermochemical equations refer to the number of moles (never molecules) of reactants and products involved in the reactionD. The equation describes the combustion of liquid ethanol at constant temperature and pressure. |
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Answer» Correct Answer - A `C_(2)H_(5)OH(l)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(l)` `Delta_(r)H=-1367" kJ "mol^(-1)` The above equation describes the combustion of liquid ethanol at constant temperature and pressure. The negative sign of enthalpy change indicates that this is an exothermic reaction. |
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| 19. |
If `DeltaE` is the heat of reaction for `C_(2)H_(5)OH(l)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(l)` at constant volume, the `DeltaH` (heat of reaction at constant pressure), at constant temperature isA. `DeltaH=DeltaE+RT`B. `DeltaH=DeltaE-RT`C. `DeltaH=DeltaE-2RT`D. `DeltaH=DeltaE+2RT` |
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Answer» Correct Answer - B We know that, `DeltaH=DeltaE+DeltanRT` where, `Deltan`=number of moles of gaeous products-number of moles of gaseous reactants=2-3=-1 so, `DeltaH=DeltaE-RT` |
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| 20. |
The bond energy is the energy required toA. dissociate oneemole of the substanceB. dissociate bond in 1 kg of the substanceC. break onemole of similar bondsD. break bonds in one mole of substance |
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Answer» Correct Answer - C The energy is the energy required to break one moel of similar bonds. |
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| 21. |
For the equilibrium, `A(g)hArrB(g),DeltaH` is `-40kJ//mol`, if the ratio of the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reactions is 2/3, thenA. `E_(f)=80kJ//mol,E_(b)=120kJ//mol`B. `E_(f)=60kJ//mol,E_(b)=100kJ//mol`C. `E_(f)=30kJ//mol,E_(b)=70kJ//mol`D. `E_(f)=70kJ//mol,E_(b)=30kJ//mol` |
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Answer» Correct Answer - A Given, `A(g) hArr B(g)" "DeltaH=-40kJ//mol` So, `(E_(f))/(E_(b))=(2)/(3)` We know that, `E_(f)-E_(b)=DeltaH` `E_(f)-E_(b)=-40` `E_(b)xx(2)/(3)-E_(b)=-40` Hence, `E_(b)=120kJ//mol,E_(f)=80kJ//mol` |
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| 22. |
In evaporation of water, `DeltaH and DeltaS` areA. `+,+`B. `+,-`C. `-,-`D. `-,+` |
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Answer» Correct Answer - A Evaporation of water is an endothermic reaction. Hence, `DeltaH` value will be positive and when water changes to vapour, randomness increase so, `DeltaS` value will also positive. |
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| 23. |
For the given reaction, `CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l)` Select the correct option for `Delta_(r)H`.A. `[H_(m)(CO_(2),g)+2H_(m)(O_(2),g)]-[2H_(m)(H_(2)O,l)+H_(m)(CH_(4),g)]`B. `[2H_(m)(O_(2),g)+H_(m)(CH_(4),g)]-[H_(m)(CO_(2),g)+2H_(m)(H_(2)O,l)]`C. `[H_(m)(CO_(2),g)+2H_(m)(H_(2)O,l)]-[H_(m)(CH_(4),g)+2H_(m)(O_(2),g)]`D. `[H_(m)(CO_(2),g)+H_(m)(H_(2)O,l)]-[H_(m)(CH_(4),g)+2H_(m)(O_(2),g)]` |
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Answer» Correct Answer - C `CH_(4)(g)+2O_(2)(g) to CO_(2)(g)+2H_(2)O(l)` `Delta_(r)H=underset(i)(sum)sigma_(i)H_("products")-underset(i)(sum)sigma_(i)H_("reactants")` `=[H_(m)(CO_(2),g)+2H_(m)(H_(2)O,l)]-[H_(m)(CH_(4),g)+2H_(m)(O_(2),g)]` where, `H_(m)` is the molar enthalpy. |
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| 24. |
Which one of the following is correct?A. `-DeltaG=DeltaH-TDeltaS`B. `DeltaH=DeltaG-TDeltaS`C. `DeltaS=(1)/(T)[DeltaG-DeltaH]`D. `DeltaS=(1)/(T)[DeltaH-DeltaG]` |
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Answer» Correct Answer - D Gibbs energy change, `DeltaG=DeltaH-TDeltaS` `TDeltaS=DeltaH-DeltaG` `DeltaS=(1)/(T)(DeltaH-DeltaG)` |
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| 25. |
The standard free energy change `(DeltaG^(@))` is related to equilibrium constant (K) asA. `DeltaG^(@)=-2.303` RT log KB. `DeltaG^(@)=`RT log KC. `DeltaG^(@)=`2.303 RT log KD. `DeltaG^(@)=`-2.303 RT ln K |
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Answer» Correct Answer - A Standard free energy change `(DeltaG^(@))` `=-2.303 RT " log K"`. |
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| 26. |
The equilibrium constant for a reaction is 10. what will be the value of `DeltaG^(@)`? Given, `R=8.314JK^(-1)mol^(-1),T=300K`.A. `-574.414Jmol^(-1)`B. `-5744.14Jmol^(-1)`C. `-57.4414Jmol^(-1)`D. `57441.4Jmol^(-1)` |
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Answer» Correct Answer - B `DeltaG^(2)=-2.303RTlogK_(c)` `DeltaG^(@)=-2.303xx8.314JK^(-1)mol^(-1)xx300Kxxlog10` `=-5744.14J" "mol^(-1)`. |
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| 27. |
If `K lt 1,0,` what will be the value of `DeltaG^(@)`?A. ZeroB. 1C. PositiveD. Negative |
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Answer» Correct Answer - C If `K lt 1` then `DeltaG^(@) gt 0`. Hence the value of `DeltaG^(@)` is positive if `K lt 1`. |
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| 28. |
Find the equilibrium constant `K_(p)` in `log K_(p)` if the standard free energy change of a reaction `DeltaG^(@)=-115kJ` at 298K isA. 2.303B. 13.83C. 2.016D. 20.16 |
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Answer» Correct Answer - D `DeltaG^(@)=-115xx10^(3)J,T=298K` and `R=8.314JK^(-1)mol^(-1)` `-DeltaG^(@)=2.303RTlog_(10)K_(p)` `-(-115xx10^(3))=2.303xx8.314xx298xxlog_(10)K_(p)` `log_(10)K_(p)=(115000)/(2.303xx8.314xx298)` `log_(10)K_(p)=20.6` |
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| 29. |
Find the value of `DeltaG^(@)` if the `DeltaH^(@)=-29.8` kcal and `DeltaS^(@)=-0.100` kcal `K^(-1)` is given at 298K. |
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Answer» Correct Answer - A `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)=-29.8-298xx(-0.1)` `=-29.8+29.8=0` |
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| 30. |
What is `Deltan_(g)` for the combusion of 1 mole of benzene, when both reactants and products are gases at 298K |
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Answer» Correct Answer - C `C_(6)H_(6)(g)+(15)/(2)O_(2)(g) to 6CO_(2)(g)+3H_(2)O(g)` `Deltan=6+3-1-(15)/(2)=+0.5` |
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| 31. |
At the same conditions of pressure, volume and temperature, work done is maximum for which gas if all gases have equal masses?A. `NH_(3)`B. `N_(2)`C. `Cl_(2)`D. `H_(2)S` |
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Answer» Correct Answer - A When p,V and T are same and mass is also same, work done depends only upon molecular mass. `W prop (1)/(M)` (where M=molecular mass) Among the given gases, `NH_(3)` has lowest molecular mass, so work done is maximum for it. |
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| 32. |
In______process, work is done at the expense of internal energy.A. IsothermalB. isochoricC. adiabaticD. isobaric |
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Answer» Correct Answer - C We know, that `DeltaE=q+W` ltBrgt For adiabatic process, q=0 `therefore DeltaE=W` i.e., work is done at the expense of only internal energy and Q= for adiabatic process. |
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| 33. |
The heat of formation of water is 260 kJ. How much `H_(2)O` is decomposed by 130 kJ of heat?A. 0.25 molB. 1 molC. 0.5 molD. 2 mol |
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Answer» Correct Answer - C The reaction for the formation of water is as `H_(2)+(1)/(2)O_(2) to H_(2)O,DeltaH=260kJ` . . (i) On reversing Eq. (i), we get `underset(1mol)(H_(2)O) to H_(2)+(1)/(2)O_(2),DeltaH=-260kJ` . . (ii) `because` By 260 kJ heat, water decomposed=1 mol `therefore `130 kJ heat will decompose water`=(1xx130)/(260)=0.5`mol |
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| 34. |
In the following process, `H_(2)(g) to 2H(g), " "Delta_(H-H)H^(ө)=435.0" kJ "mol^(-1)` the enthalpy change involved is theA. bond dissociation enthalpy of H-H bondB. bond association enthalpy of H-H bondC. mean bond dissociation enthalpy of H-H bondD. mean bond association enthalpy of H-H bond |
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Answer» Correct Answer - A In the given equation, `H_(2)(g) to 2H(g),Delta_(H-H)H^(ө)=435.0kJ" "mol^(-1)` the enthalpy change involved is the bond dissociation enthalpy of H-H bond. |
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| 35. |
A system consisting of 1 mole of an ideal gas undergoes an isothermal expansion at `25^(@)` from 1.0 bar to a lower pressure while generating 100 J of work. What is the final pressure of the system if the external pressure of the system is constant of 0.1 bar?A. 1.23 barB. 0.712 barC. 0.958 barD. 0.664 bar |
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Answer» Correct Answer - B Given, `p_(ext)=0.1," bar ",R=8.314JK^(-1)mol^(-1),T=298K,` `p_(1)=1" bar",p_(2)=?` `W=-p_(ext)(V_(2)-V_(1))` `=-p_(ext)((nRT)/(p_(2))-(nRT)/(p_(1)))` `=-nRT" "p_(ext)((1)/(p_(2))-(1)/(p_(1)))` or `-100=-1xx8.314xx298xx0.1((1)/(p_(2))-(1)/(1))` `therefore p_(2)=0.712` bar |
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| 36. |
During the adiabatic process,A. pressure is maintained constantB. gas is isothermally expandedC. there is a perssure volume workD. The system changes heat with surrounding |
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Answer» Correct Answer - C dq=0, for adiabatic process, which means perfect heat insulation. |
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| 37. |
Which among the following is a feature of adiabatic expansion?A. `DeltaV lt 0`B. `DeltaU lt 0`C. `Delta U gt0`D. `DeltaT=0` |
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Answer» Correct Answer - D For adiabatic expansion, there is no change in heat, which means no change in temperature also. `dq=0,T=`constant `therefore dT=0` |
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| 38. |
Which of the following statement is correct for the spontaneous adsorption of a gas?A. `DeltaS` is negative and therefore, `DeltaH` should be highly positiveB. `DeltaS` is negative and therefore, `DeltaH` should be highly negativeC. `DeltaS` is positive and therefore, `DeltaH` should be negativeD. `DeltaS` is positive and therefore, `DeltaH` should also be highly positive |
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Answer» Correct Answer - B `DeltaS` (change in entropy) and `DeltaH` (change in enthalpy) are related by the equation `DeltaG=DeltaH-TDeltaS` (here, `DeltaG`=change in Gibbs free energy) for adsorption of a gas, `DeltaS` is negative because randomness decreases. Thus, in order to make `DeltaG` negative (for spontaneous reaction), `DeltaH` must be highly negative. |
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| 39. |
In which case, a spontaneous reaction is possible at any temperatureA. `DeltaH(-ve),DeltaS(+ve)`B. `DeltaH(-ve),DeltaS-(ve)`C. `DeltaH+(ve),DeltaS+(ve)`D. none of these |
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Answer» Correct Answer - A For a spontaneous process, `DeltaG=-ve=DeltaH-TDeltaS` `therefore DeltaH=-ve, DeltaS=+ve` |
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| 40. |
Which of the following does not have zero entropy even at absolute zer0o? `CO,CO_(2),NaCl,NO`A. `CO,CO_(2)`B. `CO,NO`C. `CO_(2),NaCl`D. `NaCl` |
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Answer» Correct Answer - B CO and NO molcules in solid states at 0 K adopt a nearly random arrangement indicating a positive value of entropy. It is due to their dipole ment which result in disorder. |
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| 41. |
Select the correct statement(s).A. `q=+ve` means heat is transferred from the system to the surroundingsB. q=-ve means heat is transferred from the surroundings to the systemC. q=+ve means heat is transferred from the suroundings to the systemD. all of these |
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Answer» Correct Answer - C When q is positive, heat is transferred from the surroundings to the system. |
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| 42. |
If a gas, at constant temperature and pressure expands, then itsA. internal energy increases and then decreasesB. internal energy increasesC. internal energy remains the sameD. internal energy decreases |
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Answer» Correct Answer - C We know that internal energy of a gas depends upon its pressure and temperature. Thus, if a gas exapnd at constant temperature and pressure then its internal energy remains same. |
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| 43. |
In which process net work done is zero ?A. CyclicB. IsobaricC. AdiabaticD. Free expansion |
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Answer» Correct Answer - D `W=-pDeltaV` (p=zero iin vacuum) `thereforeW=0` |
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| 44. |
Select the correct formula for `Delta_(sol)H^(-)`.A. `Delta_(sol)H^(ө)=Delta_("lattice")H^(ө)-Delta_(hyd)H^(ө)`B. `Delta_(sol)H^(ө)=Delta_("lattice")H^(ө)+Delta_(hyd)H^(ө)`C. `Delta_("lattice")H^(ө)=Delta_(sol)H^(ө)+Delta_(hyd)H^(ө)`D. `Delta_(hyd)H^(ө)=Delta_(sol)H^(ө)+Delta_("lattice")H^(ө)`. |
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Answer» Correct Answer - B `Delta_(sol)H^(ө)=Delta_("lattice")H^(ө)+Delta_(hyd)H^(ө)`. |
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| 45. |
The change in the internal energy for an isolated system at constant volume isA. `DeltaU ne0`B. `DeltaU=DeltaE+DeltaW`C. `DeltaU=0`D. none of these |
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Answer» Correct Answer - C For an isolated system, there is not transfer of energy as heat work, i.e., W=0 and q=0 `DeltaU=q+W=0 implies DeltaU=0` |
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| 46. |
Which of the given properties depends on the manner in which a change is brought about? (i) `DeltaE` (ii) q (iii) WA. Only (i)B. Both (i) and (ii)C. Both (ii) and (iii)D. Only (iii) |
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Answer» Correct Answer - C `DeltaE` is a state variable, i.e. it is independent of the manner in which change is brough about while q(heat) and W(work done) are pah dependent. |
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| 47. |
Which is correct for an endothermic reaction ?A. `DeltaH` is positiveB. `DeltaH` is negativeC. `DeltaE` is negativeD. `DeltaH` is zero |
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Answer» Correct Answer - A For an endothermic reaction, `DeltaH` is positive because in endothermic reaction, heat is always absorbed. `DeltaH=H_(P)-H_(R)` |
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| 48. |
Which of the following statement is correct regarding melting of ice?A. It is the phase transformationB. It takes places at constant pressure and temperatureC. Both (a) and (b)D. none of these |
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Answer» Correct Answer - C Phase transofmrations also involve in the energy changes. For example, ice requires heat for melting. Normally, this melting takes place at constant pressure and during phase change, temperature remains constant. |
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| 49. |
The gibbs free energy for a reversible reaction at equilibrium isA. positiveB. negativeC. zeroD. can be positive or negative |
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Answer» Correct Answer - C Gibbs free energy is given by `DeltaG=DeltaH-TDeltaS` For reversible reaction at equilibrium, `DeltaG=0` |
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| 50. |
Which of the following is not an endothermic reaction?A. DehydrogenationB. Ethane to etheneC. Combustion of propaneD. Change of chlorine molecule into chlorine atoms |
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Answer» Correct Answer - C Combustion of propane is an exothermic reaction because the enthalpy of combustion (i.e., `DeltaH`) is always negative. |
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