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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. |
The number of state functions the following properties are Temperature, Pressure, Volume, Heat capacity. Density, `pH` of a solution, `EMF` of a cell. Entropy, Free energy. Enthalpy, Surface tension, Viscosity, Boiling point. |
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Answer» Correct Answer - ` to 8` State Functions : Pressure, Volume, Temperature, Density, `pH`, `EMF`, Surface tension, Viscosity, Boiling point. |
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| 2. |
Enthalpy of reaction `DeltaH` is expressed asA. `DeltaH=sumH_(P)-sumH_(R )`B. `DeltaH=dH_(P)+dH_(R )`C. `DeltaH=(dH_(P))/(dH_(R ))`D. `DeltaH=(sumH_(P))/(sumH_(R ))` |
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Answer» Correct Answer - A `DeltaH=["Sum of enthalpy of formation of products"]-["Sum of enthalpy of formation of reactants"]` |
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| 3. |
Heat capacity isA. `(dQ)/(dT)`B. `dQxxdT`C. `sumQ.(1)/(dt)`D. none of these |
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Answer» Correct Answer - A Heat capacity `C=(dQ)/(dT)` (see definition) |
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| 4. |
Given that: `2C(s)+O_(2)(g)to2CO_(2)(g)" "(DeltaH=-787kJ)` . . . (i) `H_(2)(g)+1//2O_(2)(g)toH_(2)O(l)" "(DeltaH=-286kJ)` . . . (ii) `C_(2)H_(2)+2(1)/(2)O_(2)(g)to2CO_(2)(g)+H_(2)O(l)" "(DeltaH=-1310kJ)` . . .(iii) The heat of formation of acetylene is:A. `-1802kJ" " mol^(-1)`B. `+1802kJ" "mol^(-1)`C. `+237kJ" "mol^(-1)`D. `-800kJ" "mol^(-1)` |
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Answer» Correct Answer - C Required equation is `2C(s)+H_(2)(g)toC_(2)H_(2)(g)` It can be adding eqs. (i) and (ii) and then subtracting eq. (iii) from it Heat of formation of acetylene `=(-787)+(-286)-(-1310)` `=+237kJ" "mol^(-1)` |
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| 5. |
A gas can expand from `100 mL` to `250 mL` under a constant pressure of `2` atm. The work done by the gas isA. `30.38` jouleB. `25` jouleC. `5 k` jouleD. `16` joule |
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Answer» Correct Answer - A `DeltaV=250mL-100mL=150mL` `=0.15 L` `P_("ext")=2"atm"` `W=-P_("ext")xxDeltaV=-(0.15 L)xx(2 "atm")` `=-0.3 L "atm"=-0.3xx1.01325xx10^(2)J` `=30.3975 J` |
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| 6. |
The free energy change for the following reactions are given below `C_(2)H_(2)(g)+(5)/(2)O_(2)(g)to2CO_(2)(g)+H_(2)O(l)`, `DeltaG^(@)=-1234 kJ` `C(s)+O_(2)(g)toCO_(2)(g)`, `DeltaG^(@)=-394 kJ` `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)`, `DeltaG^(@)=237 kJ` What is the standard free energy change for the reaction `H_(2)(g)+2C(s)toC_(2)H_(2)(g)` ?A. `209 kJ`B. `-2259 kJ`C. `+2259 kJ`D. `209 kJ` |
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Answer» Correct Answer - D `(1) C_(2)H_(2)(g)+(5)/(2)O_(2)(g)to2CO_(2)(g)+H_(2)O(l)` , `DeltaG^(@)=1234 kJ` `(2) C(s)+O_(2)(g)toCO_(2)(g)` , `DeltaG^(@)=-394 kJ` `(3) H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)` , `DeltaG^(@)=-237 kJ` To get : `H_(2)(g)+2C(s)toC_(2)H_(2)`, `DeltaG=?` Solve Eq. No. `(3)+2xx"Eq. No."(2)-"Eq.No."(1)` `DeltaG=(-237kJ)+2xx(-394kJ)-(-1234 kJ)` `=-237kJ-0.788kJ+1234kJ=209kJ` |
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| 7. |
The specific heat at constant volume for a gas is 0.075 cal/g and at constant pressure it is 0.125 cal/g. Calculate, (i) the molecular weight of gas, (ii). Atomicity of gas. |
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Answer» (i) `C_(P)-C_(V)=(R)/(M)` where, M=molecular of gas `0.125-0.075=(1.987)/(M)` `M=39.74~~40` (ii) `(C_(P))/(C_(V))=gamma` `therefore=(0.125)/(0.075)=1.66` `therefore` 1.66 value of `gamma` shows that the gas is monoatomic. |
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| 8. |
Assertion:Absolute values of intenal energy of substances cannot be determined. Reason:It is impossible to determine exact values of constituent energies of the substances.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. |
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Answer» Correct Answer - A it is that absolute values of internal energy of substances cannot be determined. It is also ture that to determine exact values opf constituent energies of the substance is impossible. |
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| 9. |
`F_(2)C=CF-CF = CF_(2) rarr F_(2)underset(FC=)underset(|)(C)-underset(CF)underset(|)(CF_(2))` For this reaction (ring closure), `DeltaH =- 49 kJ mol^(-1), DeltaS =- 40.2 J K^(-1) mol^(-1).Up` to what temperature is the forward reaction spontaneous?A. `1492^(@)C`B. `1219^(@)C`C. `946^(@)C`D. `1080^(@)C` |
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Answer» Correct Answer - C For spontaneous process, `DeltaG=DeltaH-TDeltaS` (should e negative) `thereforeT gt (DeltaH)/(DeltaS)` `T gt (-49xx1000)/(-40.2)` `T gt 1219K`, i.e., `946^(@)C`. |
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| 10. |
For the reation `F_(2)(g)+2HCL(g)rarr2HF(g)+C1_(2)(g)` `DeltaH^(@)` at `298K` is `-84.4Lcal` , `DeltaH^(@)f(HF)=-64.2Kcal//mol` `DeltaH^(@)f` for the `HCL(g)` per gram isA. `-0.603Kcal`B. `-0603cal`C. `0.0603Kcal`D. `6.03Kcal` |
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Answer» Correct Answer - B `F_(2)(g)+2HCL(g)rarr2HF(g)+C1_(2)(g)` `DeltaH=-84.4=-128-2DeltaH_(f)(HCL1)` `DeltaH_(f)(HC1)=-22Kcal//mol` `DeltaH_(f)(HC1)=-0.603Kcal//g` |
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| 11. |
The values of `DeltaH` and `DeltaS` for the reacrtion, `C_("graphite")+CO_(2)(g)rarr2CO(g)` are `170KJ` and `170JJK^(-)` respectively. This reaction will be spontaneous atA. `510K`B. `710K`C. `910K`D. `1110K` |
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Answer» Correct Answer - A `DeltaH=170KJ=170xx10^(3)J` `DeltaS=170JK^(-1),T=`? `DeltaG=DeltaH-TDeltaS` For spontaneous reaction, `DeltaGlt0` `implies 0lt170xx10^(3)-Txx170,Tgt1000` `:. T=1110K` |
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| 12. |
The values of `DeltaH` and `DeltaS` for the reaction `C("graphite")+CO_(2)(g)to2CO(g)` are `170 kJ` and `170 JK^(-1)` respectively. The reaction will be spontaneous atA. `990 K`B. `1140 K`C. `1100 K`D. `1110 K` |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS` `DeltaH=170xx1000=170000J` `:. DeltaG=DeltaH-TDeltaS` For equilibrium `DeltaG=0` `:. 0=170000J-Txx170 J` or `T=(170000)/(170)=1000K` `T` should be less than `1000 K` for spontaneity. |
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| 13. |
How much heat will be required to make `2kg` of calcium carbide `(CaC_(2))` according to the following reaction? `CaO(s) +3C(s) rarr CaC_(2)(s) +CO(g)` The heats of formations of `caO(s), CaC_(2)(s)`, and `CO(g)` are `-151.0,-14.0`,and `-26.0 kcal`, respectively. |
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Answer» `DeltaH=sumDeltaH_(f("products"))^(@)-sumDeltaH_(f("reactants"))^(@)` `=[DeltaH_(f(CaC_(2)))^(@)+DeltaH_(f(CO))^(@)][DeltaH_(f(CaO))^(@)+3DeltaH_(f(c))^(@)]` `=[-14.2-26.4]-[-151.4+3xx0]` `=40.6+151.6=111.0kcal` For formation of 64g of `CaC_(2)` 111.0 kcal of heat is required so, heat required for making 2000 g of `CaC_(2)=(111.0)/(64)xx2000=3468.75kcal` |
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| 14. |
Calculate the entropy change for the following reaction, `{:(,CaCO_(3)(s),to,CaO(s),+,CO_(2)(g)),(S^(@),92.9,,39.7,,213.6JK^(-1)" "mol^(-1)):}` |
| Answer» Correct Answer - `+160.4JK^(-1)` | |
| 15. |
The entropy change for the reaction given below is `2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l)` is ... At `300K` . Standard entropies of `H_(2)(g),O_(2)(g)` and `H_(2)O(l)` are `126.6,201.20` and `68.0JK^(-1)mol^(-1)` respectively.A. `-318.4JK^(-1)mol^(-1)`B. `318.4xxJK^(-1)mol^(-1)`C. `31.84xxJK^(-1)mol^(-1)`D. None of these |
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Answer» Correct Answer - A `DeltaS_("Reaction")=SigmaS_("product")-SigmaS_("reactant")` `=2xxS_(H_(2)O)-[2xxS_(H_(2))+So_(2)]` `=2xx68-[2xx126.6+201.20]=-318.4JK^(-1)mol^(-1)` |
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| 16. |
Change in entropy for reaction `2H_(2)O_(2)(l)rarr2H_(2)O(l)+O_(2)(g)` if heat of formation of `H_(2)O_(2)(l)` and `H_(2)O(l)` are `-188` and `-286KJ//mol` respectively isA. `-196KJ//mol`B. `+196KJ//mol`C. `+948KJ//mol`D. `-948KJ//mol` |
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Answer» Correct Answer - D `2H_(2)O_(2)(l)rarr2H_(2)O(l)+O_(2)(g)" " DeltaH=`? `DeltaH=[(2xxDeltah_(f)` of `H_(2)O(l)+(DeltaH_(f)` of `O_(2))` `[(2xxDeltaH_(f)` of `H_(2)O_(2)(1))]` `=[(2xx-286)+(0)-(2xx-188)]` `=[-572+376]=-196KJ//mol` |
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| 17. |
If the enthaply change for the transition of liquid water to steam is 30 KJ `"mol"^(-1)` at `27^(@)` C . The entropy change for the process would beA. `0.1 J mol^(-1) K^(-1)`B. `100 J mol^(-1) K^(-1)`C. `10 J mol^(-1) K^(-1)`D. `1.0 J mol^(-1) K^(-1)` |
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Answer» Correct Answer - B `"Liquid water" to"Steam"` `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `=30xx10^(3)-TDeltaS` `:. DeltaS=(30xx10^(3))/(300)=100 J mol^(-1)K^(-1)` |
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| 18. |
heat of formation of ` H_(2)O is -188 kJ // mol and H_(2)O_(2) is -286 kJ // mol` . The enthaply change for the reaction ` 2H_(2)O_(2) to 2H_(2)O+O_(2)` isA. 196 kJB. `-196 kJ`C. 984 kJD. `-984 kj` |
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Answer» Correct Answer - a `H_(2)+1/2 O_(2)to H_(2)O,DeltaH=-188kJ mol^(-1)` `H_(2) + O_(2) to H_(2)O_(2) , DeltaH =286 kJ mol ^(-1)` multiply Eqs.(i) and (ii) by 2 `2H_(2)+O_(2) to 2H_(2)O, DeltaH=-376 kJ mol^(-1)` `2H_(2)+2O_(2) to 2H_(2)O_(2) , DeltaH =- 572 kJ mol^(-1)` Eq .(iii) - Eq.(iv) `2H_(2)O_(2) to 2H_(2)O + O_(2), DeltaH_(r) =+196 kJ` |
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| 19. |
The enthaply at `298K` of the reaction `H_(2)O_(2)(l)rarr+(1)/(2)O_(2)(g)` is `-23.5Kcal mol^(-1)` and the enthaply of formation of `H_(2)O_(2)(l)` is `-44.8Kcal mol^(-1)` . The enthaply of formatiom of `H_(2)O(l)` isA. `-68.3Kcalmol^(-1)`B. `68.3Kcalmol^(-1)`C. `-91.8Kcalmol^(-1)`D. `91.8Kcalmol^(-1)` |
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Answer» Correct Answer - B `DeltaH_(reaction)=DeltaH_(f)^(@)(H_(2)O_(2))-DeltaH_(f)^(@)(H_(2)O_(2))` `-23.5=x-(-44.8)` or `x=-23.5-44.8=-68.3` |
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| 20. |
Given that `C+2SrarrCS_(2)DeltaH^(@)f=+117.0KJmol^(-1)` (1) `C+O_(2)rarrCO_(2)DeltaH^(@)f=-393.0KJmol^(-1)` (2) `S+O_(2)rarrSO_(2)DeltaH^(@)f=+297.0KJmol^(-1)` (3) The heat of combustion of `CS_(2)+3O_(2)rarrCO_(2)+2SO_(2)` isA. `-807KJmol^(-1)`B. `-1104KJmol^(-1)`C. `807KJmol^(-1)`D. `1104KJmol^(-1)` |
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Answer» Correct Answer - C The required combustion equation is obtained as `2xxeqn.(3)2S+2O_(2)rarr2SO_(2)2xx(-297.0)KJ` `+eqn.(2)+C+O_(2)rarrCO_(2)+(-393.0)KJ` `-eqn.(1)-C-2Srarr-CS_(2)-(+117.0)KJ` or we get `2S+3O_(2)-2Srarr2SO_(2)+CO_(2)-CS_(2)` or `CS_(2)+SO_(2)rarrCO_(2)+2SO_(2)(-1104KJmol^(-1))` |
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| 21. |
Which of the following equations does not correctly represent the first law of thermodynamcis?A. isothermal process : `q=-w`B. cyclic process : `q=-w`C. isocharic process : `DeltaE=q`D. adiabatic process : `DeltaE=-w` |
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Answer» Correct Answer - D According to `1st` law of thermodynamics, `DeltaE=q+w` For isothermal process, `DeltaE=0`. Hence `q=-w` For cyclic process, `DeltaE=0`. Hence `q=-w` For isobaric process, `DeltaV=0` Hence `DeltaE=q(w=PDeltaV=0)` For adiabatic process, `q=0`. Hence, `DeltaE=w` For expansion in vacuum, `w=0` Hence `DeltaE=q` `:. (D)` is incorrect |
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| 22. |
Identify the correct statement for change of Gibbs energy for a system `(DeltaG_(system))` at constant temperature and pressure:A. if `DeltaG_(system)=0` , the system is still moving in a particular directionB. if `DeltaG_(system)=-ve` , the process is not spontaneousC. if `DeltaG_(system)=+ve` , the process is spontaneousD. if `DeltaG_(system)=0` , the sytem has attained equilibrium |
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Answer» Correct Answer - C `DeltaG_(system)=-ve` , the system is sponteneous, `DeltaG_(system)=0` , the system has attained equilibrim `DeltaG=+ve` , the system is non-spontaneous. |
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| 23. |
Consider the reaction: `N_(2) + 3H_(2) hArr 2NH_(3)` carried out at constant pressure and temperature. If `DeltaH` and `DeltaU` are change in enthalpy and change in internal energy respectively, then:A. `Delta=0`B. `Delta=DeltaU`C. `DeltaH lt DeltaU`D. `DeltaH gt DeltaU` |
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Answer» Correct Answer - C `Deltan_(g)` for this reaction is `-ve` (`Deltan_(g)=-4` if `NH_(3)` is considered as liquid and `Deltan_(g)=-2` if `NH_(3)` is considered as gas) Now `DeltaH=DeltaU+Deltan_(g)RT` `:. DeltaH lt DeltaU` for this reaction. |
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| 24. |
Consider the reaction: `N_(2) + 3H_(2) hArr 2NH_(3)` carried out at constant pressure and temperature. If `DeltaH` and `DeltaU` are change in enthalpy and change in internal energy respectively, then:A. `DeltaH=0`B. `DeltaH=DeltaU`C. `DeltaH lt DeltaU`D. `DeltaH gt DeltaU` |
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Answer» Correct Answer - C `DeltaH=DeltaU+DeltanRT` `:. Deltan=2-4=-2` `:. DeltaHlt DeltaU` |
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| 25. |
Assume each reaction is carried out in an open container. For which reaction will `DeltaH=DeltaU` ?A. `PCl_(3)(g)rarrPCl_(3)(g)+Cl_(2)(g)`B. `2CO(g)+O_(2)(g)rarr2CO_(2)(g)`C. `H_(2)(g)+Br_(2)(g)rarrHBr(g)`D. `C(s)+2H_(2)O(g)rarr2H_(2)(g)+CO_(2)(g)` |
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Answer» Correct Answer - C `Deltan=0` and thus from `DeltaH=DeltaU+DeltanRT` `DeltaH=DeltaU` |
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| 26. |
The `H-H` bond energy is `430KJmol^(-1)` and `Cl-Cl` bond energy is `240KJmol^(-1)` . `DeltaH` for `HCl` is `-90KJ` . The `H-Cl` bond energy is aboutA. `425KJmol^(-1)`B. `213KJmol^(-1)`C. `306KJmol^(-1)`D. `180KJmol^(-1)` |
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Answer» Correct Answer - B `(1)/(2)H_(2)+(1)/(2)Cl_(2)rarrHCl,DeltaH=-90KJ` `:. DeltaH=(1)/(2)e_(H-H)+(1)/(2)e_(Cl-Cl)` or `-90=(1)/(2)xx430+(1)/(2)xx240-e_(H-Cl)` `:. e_(H-Cl)=425KJmol^(-1)` |
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| 27. |
During isothermal expansion of an ideal gas, its:A. internal energy increasesB. enthalpy increasesC. enthalpy reduces to zeroD. enthalpy remains unchanged |
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Answer» Correct Answer - D During isothermal expansion of an ideal gas `DeltaT=0`, `DeltaU=0` `H=U+PV` or `DeltaH=DeltaU+Delta(PV)` `=DeltaU+nRDeltaT=0+0` |
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| 28. |
When a student mixed `50mL` of `1M HCI` and `150mL` of `1M NaOH` in a coffee cup calorimeter, the temperature of the resultant solution increases from `21^(@)C` to `27.5^(@)C`. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the total volume of solution is `100mL`, its density `1gm-mL^(-1)` and that its specific heat is `4.18 J g^(-1)`. calculate: a. The heat change during mixing. b. The enthalpy change for the reaction `HCI(aq) +NaOH(aq) rarr NaCI(aq) +H_(2)O(aq)` |
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Answer» (a). Number of moles of HCl and NaOH added `=(MV)/(1000)=(1xx50)/(1000)=0.05` Mass of mixture `=Vxxd=100xx1=100g` Heat evolved, `q=msDeltaT=100xx4.18xx(27.5-21.0)` `=100xx4.18xx6.5J=2717J=2.717kJ` (b). The involved reaction is: `HCl(aq.)+NaOH(aq.)toNaCl(aq.)+H_(2)O` `DeltaH=`Heat evolved per mol `=(-2.717)/(0.05)=-54.34kJ` |
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| 29. |
1 mol of `H_(2) SO_(4)` in mixed with 2 mol of `NaOH`. The heat evolved will beA. `57.3 kJ`B. `2xx57.3 kJ`C. `57.3//2kJ`D. cannot be predicted. |
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Answer» Correct Answer - B `2NaOH+H_(2)SO_(4)tona_(2)SO_(4)+2H_(2)O` Heat produced when `1` mol of `NaOH` neutralise `0.5 "mol of" H_(2)SO_(4)=57.3 kJ` `:. "Heat produced when " 2 mol "of" NaOH` neutralise `1` mol of `H_(2)SO_(4)=2xx57.3 kJ` |
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| 30. |
calcualate the amount of heat evolved when `500 cm^(3)` of 0.1M HCl is mixed with `200 cm^(3)` of 0.2 M NaOH.A. 57.3 kJB. 2.865 kJC. 2.292 kJD. 0.573 kJ |
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Answer» Correct Answer - c `HCl+NaOHto NaCl+H_(2)O` at t=0, number of moles = `(500xx0.1)/1000 = (200xx0.2)/1000 ` =0.05 =0.04 during neutralisation of 1 mole of NaOH by 1 mole of HCl, heat ecolved = 57. 3 kJ to neutralised 0.04 moles of NaOH by 0.04 molw of NaOH, heat evolved `57.3 xx 0.04` = 2.292 kJ |
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| 31. |
The amount of heat evolved when `500 cm^(3) 0.1 M HCl` is mixed with `200 cm^(3)` of `0.2 M NaOH` isA. `2.292 kJ`B. `1.292 kJ`C. `0.292 kJ`D. `5.292 kJ` |
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Answer» Correct Answer - A `500cm^(3)` of `0.1 M HCl=0.05 mol`. `200 cm^(3)` of `0.2 MNaOH=0.04 mol` `0.04 mol` of `NaOH` neutralize `0.04 mol` of `HCl` and `0.04 mol` of `H_(2)O` is formed. Heat evolved `=57.3xx0.04=2.292 kJ` |
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| 32. |
The following data is known about the melting of KCl: `DeltaH=7.25kJ" "mol^(1) and DeltaS=+0.007JK^(-1)" "mol^(-1)` Calcualt eits melting point. |
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Answer» Correct Answer - 1035.7 K At melting point `DeltaG=0` |
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| 33. |
Assuming that water vapour is an ideal gas, the internal energy change `(Delta U)` when 1 mol of water is vapourised at 1 bar pressure and `100^(@)C`, (Given : Molar enthalpy of vapourisation of water at 1 abr and 373 K = 41 kJ `mol^(-1)` and `R = 8.3 J mol^(-1)K^(-1)` will be) :-A. ` 41.00kJ" "mol^(-1)`B. `4.100kJ" "mol^(-1)`C. `3.7904kJ" " mol^(-1)`D. `37.904kJ" "mol^(-1)` |
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Answer» Correct Answer - D `DeltaU=DeltaH-DeltanRT` `=41000-1xx3.314xx373` `=37898.878J" "mol^(-1)` `=37.9 kJ" "mol^(-1)` |
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| 34. |
water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from 12 V supply is passed for 300 s through a resistance in thermal contact with it , it found that 0.798 g of water is vaporised . Calculate the molar internal energy change at boiling point (375 .15 K).A. ` 37 .5 kJ mol ^(-1)`B. `3.75 kJ mol ^(-1)`C. `42.6 kJ mol ^(-1)`D. ` 4 .26 kJ mol ^(-1)` |
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Answer» Correct Answer - a `DeltaH`=work done =`ixxVxxt=0.5 A xx12 V xx300 s` `1800 J =+1.8 kJ` molar enthaply of vaporisation, `DeltaH_(m)=(DeltaH)/(moles of H_(2)O)=(DeltaH)/n_(H_(2)^(O))=(1.8kJ)/(0.798/18)=40.6 kJ mol^(-1)` `DeltaH_(m)=DeltaE_(m)+P DeltaV` `DeltaH_(m)=DeltaE_(m)+Deltan_(g)RT` `DeltaH_(m)=DeltaE_(m)+RT` molar internal energy change , `DeltaE_(m)=DeltaH_(m)-RT` `=40.6-8.314xx10^(-3)xx373.15=37.5 kJ mol^(-1)` |
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| 35. |
latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10 .0 kcal/ mol . What will be the change in internal energy (`DeltaE`) of 3 moles of liquid at same temperature ?A. 30 kcalB. `-54 kcal`C. 27.0 kcalD. 50 kcal |
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Answer» Correct Answer - c `DeltaH=DeltaE+Deltan_(g) RT, 30 = 30 Delta + 3xx2xx500xx10^(-3)` `DeltaE=27 kcal` |
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| 36. |
Latent heat of vaporisation of a liquid at `500K` and `1` atm pressure is `10.0 kcal//mol`. What will be the change in internal energy `(DeltaE)` of `3` mol of liquid at same temperature?A. `13.0 kcal`B. `-13.0 kcal`C. `27.0 kcal`D. `-27.0 kcal` |
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Answer» Correct Answer - C Vaporisation of `3` moles of water is `3H_(2)O(l)to3H_(2)O(g)` Here `Deltan=3-0=3` Heat change for `3` moles of water to vapours `=3xx10=30 kJ` `:. DeltaE=DeltaH-DeltanRT` `=30-(3)(0.002)(500)=27 kcal` |
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| 37. |
`(Delta H - Delta U)` for the formation of carbon monoxide `(CO)` from its elements at `298K` is `(R = 8.314 K^(-1)mol^(-1))`A. `2477.57mol^(-1)`B. `-1238.78Jmol^(-1)`C. `1238.78Jmol^(-1)`D. `-2477.57Jmol^(-1)` |
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Answer» Correct Answer - A `C+1//2O_(2)rarrCO` `:. DeltaH-DeltaU=DeltanRT` `=-(1)/(2)xx8.314xx298` `=- 1238.78J"mol"^(-1)` |
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| 38. |
The heat of combustion of liquid ethanol is -327.0 kcal calculate the heat of formation of ethanol. Given that the heats of formation of `CO_(2)(g) and H_(2)O(l)` are -94.0 kcal and -68.4 kcal respectively. |
| Answer» Correct Answer - `-66.2` kcal | |
| 39. |
Heat supplied to a Carnot engine is `453.6 kcal`. How much useful work can be done by the engine that works between `10^(@)C` and `100^(@)C`? |
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Answer» `T_(2)=100+273=373K,T_(1)=10+273=283K`, `q_(2)=453.6xx4.184=1897.86kJ` We know that, `w=q_(2)*(T_(2)-T_(1))/(T_(2))` `=1897.86xx((373-283))/(373)` `=(1897.86xx90)/(373)=457.92kJ` |
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| 40. |
Calculate the maximum efficiency of an engine operating between `100^(@)C` and `25^(@)C` |
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Answer» Efficiency `=(T_(2)-T_(1))/(T_(2))` `T_(2)=100+273=373K` `T_(1)=25+273=298K` Efficiency `=(373-298)/(373)` `=(75)/(373)=0.20=20%`. |
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| 41. |
Calculate the theoretical maximum efficiency of a heat engine operating between 373 K and 173K. |
| Answer» Correct Answer - 0.536 or 53.6% | |
| 42. |
Calculate the amount of work done by 2 mole of an ideal gas at 298 K in reversible isothermal expansion from 10 litre to 20 litre. |
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Answer» Amount of work done in reversible isothermal expansion, `w=-2.303nRTlog((V_(2))/(V_(1)))` given, `n=2,R=8.314K^(-1)mol^(-1),T=298K,V_(2)=20L and V_(1)=10L`. ltBrgt Substituting the values in above equation, `w=-2.303xx2xx8.314xx298log((20)/(10))` `=-2.303xx2xx8.314xx298xx0.3010=-3434.9J` i.e., work is done by the system. |
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| 43. |
The enthalpy of combustion of `H_(2)` , cyclohexene `(C_(6)H_(10))` and cyclohexane `(C_(6)H_(12))` are `-241` , `-3800` and `-3920KJ` per mol respectively. Heat of hydrogenation of cyclohexene isA. `-121kJ//mol`B. `+121kJ//mol`C. `-242kJ//mol`D. `+242kJ//mol` |
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Answer» Correct Answer - A `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)(DeltaH=-241kJ)` . . .(i) `C_(6)H_(10)+(17)/(2)O_(2)(g)to6CO_(2)(g)+5H_(2)O(l)" "(DeltaH=-3800kJ)` `C_(6)H_(12)+9O_(2)(g)to6CO_(2)+6H_(2)O(l)" "(DeltaH=-3920kJ)` . . .(iii) eq. (i) eq. (ii)-eq.(iii) gives for `DeltaH=-241-3800-(-3920)=-121kJ` for `C_(6)H_(10)+H_(2)toC_(6)H_(12)`. |
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| 44. |
One mole of a gas is heated at constant pressure to raise its temperature by `1^(@)C`. The work done in joules isA. `-4.3`B. `-8.314`C. `-16.62`D. unpredictable |
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Answer» Correct Answer - B `w=-nRTDeltaT` or `w=-PDeltaV` `=-1xx8.314xx1=-P((nRT_(2))/(P)-(nRT_(1))/(P))` `=-8.314J` |
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| 45. |
The standard ethelpy of combustion at `25^(@)C` of hydrogen, cyclohexene `(C_(6)H_(10))`, and cyclohexane `(C_(6)H_(12))` are `-241,-3800`, and `-3920 kJ mol^(-1)` repectively. Calculate the heat of hydrogenation of cyclohexane. |
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Answer» Correct Answer - `-121kJ" " mol^(-1)` `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g)," "(DeltaH=-241kj" "mol^(-1))` . . (i) `C_(6)H_(10)(g)+(17)/(2)O_(2)(g)to6CO_(2)(g)+5H_(2)O(g),(DeltaH=-3800kJ" "mol^(-1))` . . .(ii) `C_(6)H_(12)9g)+9O_(2)(g)to6CO_(2)(g)+6H_(2)O,(DeltaH=-3920kJ" "mol^(-1))` . . (iii) Adding eqs. (i) and (ii) and subtracting eq. (iii), `C_(6)H_(10)(s)+H_(2)(g)toC_(6)H_(12)(s),` `DeltaH=-241-3800-(-3920)=-121kJ" "mol^(-1)` |
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| 46. |
The enthalpy of neutralization of oxitic acid by strong acid is `25.4 kcal mol^(-1)`. The enthalpy of neutralization of strong acid and strong base is `-13.7 kcal equil^(-1)`. The enthalpy of dissociation of `H_(2)C_(2)O_(4)hArr2H^(+)+C_(@)O_(4)^(2-)` isA. `1.0 kcal mol^(-1)`B. `2.0 kcal mol^(-1)`C. `18.55 kcal mol^(-1)`D. `11.7 kcal mol^(-1)` |
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Answer» Correct Answer - B Enthalpy of dissociation of oxalic acid `=` Heat of neutralization of a strong acid with a strong base `-` Heat of neutralization of oxalic acid with a strong base `=-25.4 kcal mol^(-1) -(-2xx13.7 kcal mol^(-1))` `=-2.0 k cal mol^(-1)` |
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| 47. |
The heats fo neutralization of `HCl` with `NH_(4) OH` and `NaOH` with `CH_(3) COOH` are `- 51.4 kJ eq^(-1)` and `- 50.6 kJ eq^(-1)`, respectively. The heat of neutralization of acetic acid with `NH_(4) OH` will beA. `-44.6 kJ eq^(-1)`B. `-50.6 kJ eq^(-1)`C. `-51.4 kJ eq^(-1)`D. `-57.4 kJ eq^(-1)` |
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Answer» Correct Answer - A `(i) HCl+NH_(4)OHtoNH_(4)Cl+H_(2)O`, `DeltaH=-51.4 kJ` `(ii) CH_(3)COOH+NaOHtoCH_(3)COONa+H_(2)O`, `DeltaH=-50.6 kJ` `(iii) CH_(3)COOH+NH_(4)OHtoCH_(3)COONH_(4)+H_(2)O`, `DeltaH=?` We know `(iv) HCl+NaOH to NaCl+H_(2)O`, `DeltaH=-57.4 kJ` To get `(iii)` add `(i)` and `(ii)` and subtract `(iv)`. (Please note that salts formed in case `(i)` and `(ii)` are almost completely ionised. Also suppose that the salt formed in case `(iii)` is also ionised). `DeltaH=(-51.4)+(-50.6)-(-57.4)kJ` `=-51.4-50.6+57.4=-44.6 kJ` |
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| 48. |
Using the bond enthalpy data given below, calculate the enthalpy of formation of acetone (g). Bond enegry `C-H = 413.4 kJ mol^(-1)`, Bond enegry `C-C = 347.0 kJ mol^(-1)`, Bond enegry `C=O = 728.0 kJ mol^(-1)`, Bond enegry `O=O = 495.0 kJ mol^(-1)`, Bond enegry `H-H = 435.8 kJ mol^(-1)`, `Delta_("sub")H^(Theta)C(s) = 718.4 kJ mol^(-1)` |
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Answer» `3C(g)+6H(g)+O(g)toCH_(3)COCH_(3)(g)` In acetone, six C-H bonds, one C=O bond and two C-C Bonds are present. Energy released in the formation of these bonds is `=-6xx413.4-728.0-2xx347.0=3902.4kJ" "mol^(-1)` The equation of the enthalpy of formation of acetone is `3C_(("graphite"))+3H_(2)(g)+(1)/(2)O_(2)(g)toCH_(3)COCH_(3)(g),DeltaH=?` ltbr. This equation can be obtained from the following equations by adding, `3C(g)+6H(g)+O(g)toCH_(3)COCH_(3)(g), " "DeltaH=-3902.4kJ" "mol^(-1)` `3C(s)to3C(g)," "Delta=2155.2kJ" "mol^(-1)` `3H_(2)(g)to6H(g)," "DeltaH=1307.4kJ" "mol^(-1)` and `(1)/(2)O_(2)(g)toO(g)," "DeltaH=247.5kJ" "mol^(-1)` `overline(3C(s)+3H_(2)(g)+(1)/(2)(g)toCH_(3)COCH_(3)(g),DeltaH=-192.3kJ" "mol^(-1))` |
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| 49. |
Standard enthalpies of formation of `O_(3),CO_(3),NH_(3) and HI` are 142.2,-383.2,-46.2 and +25.9kJ `mol^(-1)` respectively. The order of their increasing stabilities will be:A. `O_(3),CO_(2),NH_(3),HI`B. `CO_(2),NH_(3),HI,O_(3)`C. `O_(3),HI,NH_(3),CO_(2)`D. `NH_(2),HI,CO_(2),O_(3)` |
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Answer» Correct Answer - C (i) Exothermic compounds are more stable than endothermic compounds (ii) Greater is the amount of heat evolved in the formation of a compound, more will be its stability. |
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| 50. |
The approxiamte standard enthalpies of formation of methanol and octane are determiend to be -1.5 kJ/mol and -10.9 kJ/mol respectively. The standard enthalpies of combustion of octane is denoted as `DeltaH("octane")` and that fo methanol as `DeltaH` (methanol). the correct statement is:A. `DeltaH` (octane) is more negative than `DeltaH` (methanol)B. `DeltaH`(octane) is less negative than `DeltaH` (methanol)C. `DeltaH` (octane) is equal to `DeltaH` (methanol)D. `DeltaH` (octane)`+DeltaH`(methanol)=0 |
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Answer» Correct Answer - A Heat of combustion of octane will be more negative than methanol because octane has greaterr number of carbon than methanol. |
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