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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. |
Which of the following statement is true. The entropy of the universeA. Increases and tends towars maximum valueB. Decreases and tends to be zeroC. Remains constantD. Decreases and increases with a periodic rate |
| Answer» Correct Answer - A | |
| 2. |
For the homogenous reaction `xA+yB iff 1Y + mZ DeltaH^(@) = -30kJ mol^(-1) and DeltaS=-100JK mol^(-1)` At what temperature the reaction is at equilibriumA. `50^(@)C`B. `250^(@)C`C. 100 KD. `27^(@)C` |
| Answer» Correct Answer - D | |
| 3. |
The enthalpy change for the reaction `C_(2)H_(2)(g)+(5)/(2)O_(2)rarr2CO_(2)(g)+H_(2)O(g)` is known as enthalpy ofA. FormationB. FusionC. VaporizationD. Combustion |
| Answer» Correct Answer - D | |
| 4. |
The free energy change for the following reactions are given below, `C_(2)H_(2)(g)+(5)/(2)O_(2)(g)rarr2CO_(2)(g)+H_(2)O(l),DeltaG^(@)=-1234 kJ` `C(s)+O_(2)(g)rarrCO_(2)(g)DeltaG^(@)=-394 kJ` `H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l)DeltaG^(@)=-237 kJ` What is the standard free energy change for the reaction `H_(2)(g)+2C(s)rarrC_(2)H_(2)(g)`A. `-209 kJ`B. `-2259 kJ`C. `+2259 kJ`D. 209 kJ |
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Answer» Correct Answer - D By `2xx(ii)-(i)+(iii)` `H_(2(g))+2C_((s))rarrC_(2)H_(2(g)), DeltaG^(@)=209 kJ`. |
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| 5. |
When 100 mL of 1.0 M HCl was mixed 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase `5.7^(@)C` was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant `(-57.0 kJmol^(-1))`, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt.2) 100 mL of 2.0 M acetic acid `(K_(a)=2.0xx10^(-5))` was mixed with 100 mL of 1.0 M NaOH (under indentical conditions to Expt. 1) where a temperature rise of `5.6^(@)C` was measured (Consider heat capacity of all solutions as `4.2 Jg^(-1)K^(-1)` and density of al solutions as 1.0 gm`L^(-1)`) The pH of the solution after Expt. 2 isA. `2.8`B. `4.7`C. `5.0`D. `7.0` |
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Answer» Correct Answer - B Final solution contain 0.1 mole of `CH_(3)COOH` and `CH_(3)COONa` each Hence it is buffer solution `pH=pK_(a)+log.([CH_(3)COO^(-)])/([CH_(3)COOH])=5-log2+log.(0.1)/(0.1)=4.7` |
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| 6. |
When 100 mL of 1.0 M HCl was mixed 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase `5.7^(@)C` was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant `(-57.0 kJmol^(-1))`, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt.2) 100 mL of 2.0 M acetic acid `(K_(a)=2.0xx10^(-5))` was mixed with 100 mL of 1.0 M NaOH (under indentical conditions to Expt. 1) where a temperature rise of `5.6^(@)C` was measured (Consider heat capacity of all solutions as `4.2 Jg^(-1)K^(-1)` and density of al solutions as 1.0 gm`L^(-1)`) Enthalpy of dissociation (in `KJmol^(-1)`) of acetic acid obtained from the Expt.2 isA. `1.0`B. `10.0`C. `24.5`D. `51.4` |
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Answer» Correct Answer - A Let the heat capacity of insulated beaker be C Mass of aqueous content in expt. `1=(100+100)xx1=200g` implies Total heat capacity = `(C+200xx4.2)J//K` Moles of acid, base neutralised in expt. `1=0.1xx1=0.1` implies Heat released in expt. `1=0.1xx57=5.7 KJ` `implies5.7xx1000=(C+200xx4.2)xxDeltaT` `5.7xx1000=(C+200xx4.2)xx5.7` `implies (C+200xx4.2)=1000` In second experiment, `n_(CH_(3)COOH)=0.2, n_(NaOH)=0.1` Total mass of aqueous content = 200 g implies Total heat capacity = (C+200`xx`4.2)=1000 implies Heat released = `1000xx5.6=5600 J` Overall, only 0.1 mol of `CH_(3)COOH` undergo neutralization `implies DeltaH_("neutralization")" of "CH_(3)COOH=(-5600)/(0.1)` `=-56000 J//mol=-56 KJ//mol` `impliesDeltaH_("ionization")" of "CH_(3)COOH=57-56=1KJ//mol` |
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| 7. |
In a constant volume calorimeter, 3.5g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0K to 298.45 K due to the combustion of the gas in kJ `mol^(-1)` is |
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Answer» Correct Answer - 9 `n=(3.5)/(28)` `DeltaT=T_(2)-T_(1)=298.45-298=0.45` `C_(P)=C_(V)+R=2500+8.314=2508.314 JK^(-1)` `Q_(P)=C_(P)DeltaT=1128.74 J` `DeltaH=(Q_(P))/(n)=(1128.74)/(3.5//28)` implies 9030 J `mol^(-1)`=9.030 kJ `mol^(-1)` = 9 kJ `mol^(-1)`. |
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| 8. |
The amount of heat evolved when `500cm^(3)` of 0.1M HCl is mixed with `200cm^(3)` of 0.2 M NaOH isA. 2.292 kJB. 1.292 kJC. 0.292 kJD. 3.392 kJ |
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Answer» Correct Answer - A Mili-equivalent of HCl = `500xx0.1` =50 milli-moles = 0.05 moles Milli-equivalent of NaOH= `200xx0.2` = 40 milli-moles = 0.04 moles When 1 mole of `H^(+)` ions and 1 mole of `OH^(-)` ions are neutralized, 1 mole of water is formed and 57.1 kJ of energy is released. 0.04 moles of `H^(+)` ions react with 0.04 moles of `OH^(-)` ions to form 0.04 moles of water molecules. Heat evolved = `0.04xx57.1=2.29 kJ`. |
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| 9. |
In which of the following reactions, the heat liberated is known as 'heat of combustion'.A. `H_((aq))^(+)+OH_((aq))^(-)rarrH_(2)O_((l))+"heat"`B. `C("graphite")+(1)/(2)O_(2(g))rarrCO_((g))+"heat"`C. `CH_(4(g))+2O_(2(g))rarrCO_(2(g))+2H_(2)O_((l))+"heat"`D. `H_(2)SO_(4(aq))+H_(2)O_((l))rarrH_(2)SO_(4(aq))+"heat"` |
| Answer» Correct Answer - C | |
| 10. |
When 0.5g of sulphur is burnt to `SO_(2), 4, 6 kJ of heat is liberated. What is the enthalpy of formation of sulphur dioxideA. `+ 147.2 kJ`B. `- 147.2 kJ`C. `- 294.4 kJ`D. `+ 294.4 kJ` |
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Answer» Correct Answer - C `S+O_(2)rarrSO_(2), DeltaH_(f)=-4.6 kJ` `because` 0.5 gm of sulphur on burning produce 1 gm of `SO_(2)` `therefore` 32 gm of sulphur on burning produce 64 gm of `SO_(2)` `therefore DeltaH=(-4.6 kJ)xx64=-294.4 kJ` |
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| 11. |
When `50cm^(3)` of `0.2N H_(2)SO_(4)` is mixed with `50cm^(3)` of `1 N KOH`, the heat liberated isA. 11.46 kJB. 57.3 kJC. 573 kJD. 573 J |
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Answer» Correct Answer - D For complete neutralization of strong acid and strong base energy released is 57.32 KJ/mol No. of mole of `H_(2)SO_(4)=(0.2xx50)/(1000)=10^(2)` No. of mole of KOH = `(1)/(1000)xx50=5xx10^(-2)` `"So = "57.32xx10^(-2)=0.5732 kJ = 573.2" Joule"`. |
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| 12. |
When the aqueous solution of 0.5 mole `HNO_(3)` is mixed with the 0.3 mole of `OH^(-)` solution, then what will be the liberated heat (Enthalpy of neutralization 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 0.3 mole `OH^(-)`, neutralize 0.3 mole of `HNO_(3)` Evolved heat, `=57.1xx0.3=17.13 kJ`. |
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| 13. |
The entropy of a perfectly crystalline solid at absolute zero isA. PositiveB. NegativeC. ZeroD. Not definite |
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Answer» Correct Answer - C According to III law of thermodynamics. |
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| 14. |
The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of `10` litres to `20` litres at `25^(@)C` isA. `2.303xx298xx0.082log2`B. `298xx10^(7)xx8.31xx2.303 log2`C. `2.303xx298xx0.082 log0.5`D. `8.31xx10^(7)xx298-2.303 log 0.5` |
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Answer» Correct Answer - B `W=2.303 nRTlog.(V_(2))/(V_(1))` `=2.303xx1xx8.314xx10^(7)xx298log.(20)/(10)` `=298xx10^(7)xx8.314xx2.303 log 2`. |
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| 15. |
When 5 litres of a gas mixture of methane and propane is perfectly combused at `0^(@)C` and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ `[DeltaH_("comb.")(CH_(4))=890 " kJ mol"^(-1), DeltaH_("comb")(C_(3)H_(8))=2220 " kJ mol"^(-1)]` isA. 38B. 317C. 477D. 32 |
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Answer» Correct Answer - B `underset(X)(CH_(4))+underset(2X)(2O_(2))rarrCO_(2)+2H_(2)O` `underset((5-x))(C_(3)H_(8))+underset(5(5-x))(5O_(2))rarr3CO_(2)+4H_(2)O` `2x+5(5-x)=16impliesx=3L` `therefore` Heat released = `(3)/(22.4)xx890+(2)/(22.4)xx2220=317`. |
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| 16. |
The occurrence of a reaction is impossible ifA. `DeltaH" is "+ve, DeltaS " is also + ve but "DeltaHlt TDeltaS`B. `DeltaH" is - ve , "DeltaS " is also - ve but "DeltaH gt TDeltaS`C. `DeltaH" is - ve , "DeltaS " is "+ve`D. `DeltaH" is + ve , "DeltaS " is "- ve` |
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Answer» Correct Answer - D `+ ve DeltaH and - ve DeltaS` both oppose the reaction. |
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| 17. |
For the reaction`H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l),DeltaH=-285.8 kJ mol^(-1)` `DeltaS=-0.163 kJ mol^(-1) K^(-1)`. What is the value of free energy change at `27^(@)C` for the reactionA. `-236.9 kJ mol^(-1)`B. `-281.4 kJ mol^(-1)`C. `-334.7 kJ mol^(-1)`D. `+334.7 kJ mol^(-1)` |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS, T=27+273=300 K` `DeltaG=(-285.8)-(300)(-0.163)=-236.9 kJ mol^(-1)` |
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| 18. |
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 assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - A It is fact that absolute values of internal energy of substances can not be determined. It is also true that to determine exact values of constituent energies of the substance is impossible. |
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| 19. |
The reaction. `Zn(s)+CuSO_(4)(aq)rarrZnSO_(4)(aq)+Cu(s)` is an exampl for aA. Spontaneous processB. Non-spontaneous processC. Isobaric processD. Reversible process |
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Answer» Correct Answer - A This is an example of spontaneous reaction because it occurs of its own nad hence `DeltaG` of the reaction must be negative. |
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| 20. |
Given : `2Fe+(3)/(2)O_(2)rarrFe_(2)O_(3), DeltaH=-193.4 kJ`, `Mg+(1)/(2)O_(2)rarrMgO, DeltaH=-140.2 kJ` What is the `DeltaH` of the reaction `Mg+Fe_(2)O_(3)rarr3MgO+2Fe`A. `-1802 kJ`B. `+1802 kJ`C. `-800 kJ`D. `-228 kJ` |
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Answer» Correct Answer - D `2Fe+(3)/(2)O_(2)rarrFe_(2)O_(3), DeltaH=-193.4 kJ " ...(i)"` `Mg+(1)/(2)O_(2)rarrMgO, DeltaH=-140.2 kJ " ....(ii)"` On multiplying eq. (ii) by 3 `3Mg+(3)/(2)O_(2)rarr3MgO, DeltaH=-420.6 kJ " ....(ii)"` Resulting equation can be obtained by subtracting eq. (i) from (iii) `3Mg+(3)/(2)O_(2)rarr3MgO, DeltaH=-420.6 kJ` `(2Fe+(3)/(2)O_(2)rarrFe_(2)O_(3), DeltaH=-193.4 kJ)/("Subtraction : ")` `3Mg+Fe_(2)O_(3)rarr2Fe+3MgO, DeltaH=-227.2 kJ` |
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| 21. |
Standard entropy of `X_(2),Y_(2)` and `XY_(3)` are 60, 40 and 50 `JK^(-1)mol^(-1)`, respectively. For the reaction, `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30kJ` to be at equilibrium, the temperature will beA. 500 KB. 750 KC. 1000 KD. 1250 K |
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Answer» Correct Answer - B `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3)` `DeltaS=50-((60)/(2)+(3)/(2)xx40)=50-(30+60)=-40J//k, mol` at equilibrium `DeltaG=0` `DeltaH=TDeltaS, T=(DeltaH)/(DeltaS)=(-30xx10^(3))/(-40)=750 K` |
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| 22. |
For a reversible reaction : `X_((g))+3Y_((g))iff2Z_((g)), DeltaH=-40kJ` the standard entropies of X, Y and Z are 60, 40 and 50 `JK^(-1) mol^(-1)` respectively. The temperature at which the above reaction attains equilibrium is aboutA. 400 KB. 500 KC. 273 KD. 373 K |
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Answer» Correct Answer - B `X_((g))+3Y_((g))iff2Z_((g))` `DeltaS^(@)=2S^(@)(Z)-{S^(@)(X)+3S^(@)(Y)}` `=2xx50-{60+3xx40}=100-180 " = - 80 J K"^(-1) mol^(-1)` `DeltaH^(@)=-40 kJ=-40,000 J," " DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` At equilibrium, `DeltaG^(@)=0 " "therefore DeltaH^(@)=TDeltaS^(@)` `or, T = (DeltaH^(@))/(DeltaS^(@))=(40000)/(80)=500 K`. |
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| 23. |
The standard entropies of `CO_(2)(g),C(s)` and `O_(2)(g)` are 213.5, 5.690 and `205JK^(-1)` respectively. The standard entropy of formation of `CO_(2)(g)` isA. `1.86 JK^(-1)`B. `1.96 JK^(-1)`C. `2.81 JK^(-1)`D. `2.86 JK^(-1)` |
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Answer» Correct Answer - C Formation of `CO_(2)` is, `C_((s))+O_(2(g))rarrCO_(2(g))` `DeltaS^(@)=213.5-5.690-205=2.81 JK^(-1)` |
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| 24. |
A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If `DeltaH` is the enthalpy change and `DeltaE` is the change in internal energy, thenA. `DeltaH gt DeltaE`B. `DeltaH lt DeltaE`C. `DeltaH = DeltaE`D. The relationship depends on the capacity of the vessel |
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Answer» Correct Answer - B Formation of `CO_(2)` from CO is an exotheric reaction, heat is evolved from the system, i.e., energy is lowered. Thus, exothermic reactions occur spontaneously on account of decrease in enthalpy of system. Thus, `DeltaE gt DeltaH`. |
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| 25. |
The enthalpy of hydrogenation of cyclohexene is `-119.5 kJ mol^(-1)`. If resonance energy of benzene is `-150.4kJ mol^(-1)`. Its enthalpy of hydrogention would beA. `-269.9 kJ mol^(-1)`B. `-358.5 kJ mol^(-1)`C. `-508.9 kJ mol^(-1)`D. `-208.1 kJ mol^(-1)` |
| Answer» Correct Answer - D | |
| 26. |
A piston filled with 0.04 mol of an ideal gas expands eversibly from 50.0 mL to 375 mL at a constant temperature of `37.0^(@)C`. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be `(R=8.314 J//mol K) (ln 7.5 = 2.01)`A. q=+208 J, w=-208 JB. q=-208 J, w=-208 JC. q=-208 J, w=+208 JD. q=+208 J, w=+208 J |
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Answer» Correct Answer - A The process is isothermal expansion Hence, q =-w `Deltau=0` q=+208 J w=-208 J (expansion work). |
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| 27. |
An ideal gas expands in volume from `1xx10^(-3)m^(3) " to "1xx10^(-2)m^(3)` at 300 K against a constant pressure of `1xx10^(5)Nm^(-2)`. The work done isA. 270 kJB. `-900 kJ`C. `-900 J`D. 900 kJ |
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Answer» Correct Answer - C `W=-PDeltaV=-1xx10^(5)(1xx10^(-2)-1xx10^(-3))` `=-1xx10^(5)xx9xx10^(-3)=-900 J` |
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| 28. |
A heat engine absorbs heat `Q_(1)` at temperature `T_(1)` and heat `Q_(2)` at temperature `T_(2)`. Work done by the engine is `(Q_(1)+Q_(2))`. This dataA. Violates `I^(st)` law of thermodynamicsB. Violates `I^(st)` law of thermodynamics if `Q_(1)` is -veC. Violates `I^(st)` law of thermodynamics if `Q_(2)` is -veD. Does not violate `I^(st)` law of thermodynamics |
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Answer» Correct Answer - D It does not violates the first law of thermodynamics but violates the II law of thermodynamics. |
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| 29. |
Fermentation is a reaction calledA. EndothermicB. ExothermicC. IsotemperatureD. Reversible |
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Answer» Correct Answer - B Fermentation is exothermic reaction. |
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| 30. |
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below : `(1)/(2)Cl_(2)(g)overset((1)/(2)Delta_(diss)H^(Theta))rarrCl(g)overset(Delta_(eg)H^(Theta))rarrCl^(-)(g)overset(Delta_(hyd)H^(Theta))rarrCl^(-)(aq)` The energy involved in the conversion of `(1)/(2)Cl_(2)(g) " to " Cl^(-)(aq)` (Using the data, `Delta_(diss)H_(Cl_(2))^(Theta)=240 kJ mol^(-1)`, `Delta_(eg)H_(Cl)^(Theta)=-349 kJ mol^(-1)`, `Delta_(hyd)H_(Cl)^(Theta)=-381 kJ mol^(-1))` will beA. `-610 kJ mol^(-1)`B. `-850 kJ mol^(-1)`C. `+120 kJ mol^(-1)`D. `+152 kJ mol^(-1)` |
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Answer» Correct Answer - A `DeltaH=(240)/(2)-349-381` `=120-349-381=-610kJ//mol`. |
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| 31. |
Match the entries listed in Column I with appropriate entries listed in Column II. `{:(,"Column I",,"Column II"),((A),"Isothermal process",(p),((delU)/(delV))_(T)=0),((B),-nFoverset(Theta)E,(q),W=-DeltaU),((C),"Adiabatic reaction",(r),DeltaU=0),((D),"van der waals gas",(s),DeltaG^(Theta)),((E),"Ideal gas",(t),((delT)/(delP))_(H)ne0):}` |
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Answer» Correct Answer - A `rarr` t; B `rarr` r; C `rarr` p; D `rarr` p (A) For ideal gas, `((delU)/(delV))_(T)=0` (B) In adiabatic process, q = 0. Therefore, from the first law of thermodynamic, `DeltaU=q-w and q=0` `therefore DeltaU=-w` (C ) In isothermal process, `DeltaU=0`, because internal energy is the function of temperature (D) `DeltaG^(Theta)=-nFE^(Theta)` (E) `((delT)/(delP))_(H)ne0` for real gases. |
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| 32. |
From the following bond energies H-H bond energy : 431.37 kJ `mol^(-1)` C=C bond energy : 606.10 kJ `mol^(-1)` C-C bond energy : 336.49 kJ `mol^(-1)` C-H bond energy : 410.50 kJ `mol^(-1)` Enthalpy for the reaction. `underset(H)underset(|)overset(H)overset(|)(C)=underset(H)underset(|)overset(H)overset(|)(C)+H-HrarrH-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-H` will beA. 1523.6 kJ `mol^(-1)`B. `-243.6 kJ mol^(-1)`C. `-120.0 kJ mol^(-1)`D. 553.0 kJ `mol^(-1)` |
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Answer» Correct Answer - C `[(4xx410.5)xx606.1+431.3)]-[(6xx410.5)+336.49)]` `=-120.0 kJ mol^(-1)`. |
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| 33. |
Energy required to dissociate 4 gm of gaseous hydrogen into free gaseous atoms is 208 kcal at `25^(@)C`. The bond energy of H-H bond will beA. 104 kcalB. 10.4 kcalC. 1040 kcalD. 104 cal |
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Answer» Correct Answer - A 4g `H_(2)=2` moles. Bond energy for 1 mole of `H_(2)=208//2=104` kcal. |
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| 34. |
The enthalpy and entropy change for the reaction `Br_(2)(l)+Cl_(2)(g)rarr2BrCl(g)` are 30 kJ `mol^(-1)` and `105 JK^(-1) mol^(-1)` respectively. The temperature at which the reaction will be in equilibrium isA. 450 KB. 300 KC. 285.7 KD. 273 K |
| Answer» Correct Answer - C | |
| 35. |
Assertion : Molar entropy of vaporization of water is different from ethanol. Reason : Water is more polar than ethanol.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - B The molar entropy of vaporization of water differ from ethanol due to hydrogen bonding according to VSEPR theory water molecule and show some polarity which is higher than that of ethanol so both assertion and reason are correct but reason is not explanation assertion. |
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| 36. |
Molar heat of vaporization of a liquid is `6kJ mol^(-1)`. If the entropy change is `16 J mol^(-1) K^(-1)`, the boiling point of the liquid isA. `375^(@)C`B. 375 KC. 273 KD. `102^(@)C` |
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Answer» Correct Answer - B `DeltaS=16 J mol^(-1) K^(-1)` `T_(b.p.)=(DeltaH_("vapour"))/(DeltaS_("vapour"))=(6xx1000)/(16)=375 K`. |
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| 37. |
If molar heat of vaporization is 9698 cals `mol^(-1)` then entropy of vaporization of water at `100^(@)C` will beA. 20.0 cals `mol^(-1) k^(-1)`B. 24.0 cals `mol^(-1) k^(-1)`C. 26.0 cals `mol^(-1) k^(-1)`D. 28.0 cals `mol^(-1) k^(-1)` |
| Answer» Correct Answer - C | |
| 38. |
The entropy change, in the conversion of one mole of liquid water at 373 K to vapour at the same temperature is (Latent heat of vaporization of water, `DeltaH_("vap")=2.257 kJ//g`)A. `105.9 JK^(-1)`B. `107.9 JK^(-1)`C. `108.9 JK^(-1)`D. `109.9 JK^(-1)` |
| Answer» Correct Answer - C | |
| 39. |
If enthalpies of formation of `C_(2)H_(4)(g), CO_(2)(g)` and `H_(2)O(l)` at `25^(@)C` and 1 atm pressure be 52, - 394 and - 286 kJ `mol^(-1)` respectively, the enthalpy of combustion of `C_(2)H_(4)(g)` will beA. `+1412 kJ mol^(-1)`B. `-1412 kJ mol^(-1)`C. `+141.2 kJ mol^(-1)`D. `141.2 kJ mol^(-1)` |
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Answer» Correct Answer - B `C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O` `DeltaH_("reaction")=[2xxDeltaH_(f)^(@)(CO_(2))+2xxDeltaH_(f)^(@)(H_(2)O)]-DeltaH_(f)^(@)(C_(2)H_(4))+3xxDeltaH_(f)^(@)(O_(2))]` `=[2(-394)+2(-286)]-[52+0]=-1412 kJ`. |
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| 40. |
If the bond dissociation energies of XY, `X_(2)` and `Y_(2)` (all diatomic molecules) are in the ratio of `1 : 1 : 0.5` and `Delta_(f) H` for the formation of XY is `- 200 kJ " mole"^(-1)`. The bond dissociation energy of `X_(2)` will beA. 100 kJ `mol^(-1)`B. 800 kJ `mol^(-1)`C. 300 kJ `mol^(-1)`D. 400 kJ `mol^(-1)` |
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Answer» Correct Answer - B `XYrarrX_((g))+Y_((g)), DeltaH=+"a kJ/mole"^(-1)" .........(i)"` `X_(2)rarr2X, DeltaH=+" a kJ/mole .....(ii)"` `Y_(2)rarr2Y, DeltaH=+0.5" a kJ/mole .......(iii)"` `(1)/(2)xx(ii)+(1)/(2)xx(iii)-(i)`, gives `(1)/(2)X_(2)+(1)/(2)Y_(2)rarrXY,` `DeltaH=(+(a)/(2)+(0.5)/(2)a-a)"kJ/mole"` `+(a)/(2)+(0.5a)/(2)-a=-200` a = 800 |
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| 41. |
The `H_(2)O(g)` molecule dissociates as (i) `H_(2)O(g)rarrH(g)+OH(g),DeltaH=490 kJ` (ii) `OH(g)rarrH(g)+O(g),DeltaH=424 kJ` The average bond energy (in kJ) for water isA. 490B. 424C. 914D. `914//2` |
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Answer» Correct Answer - D eq. (i) + eq. (ii) find the required result and divide by 2. |
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| 42. |
If the bond energies of H - H, Br - Br and HBr are 433, 192 and 364 kJ `mol^(-1)` respectively, the `DeltaH^(@)` for the reaction, `H_(2)(g)+Br_(2)(g)rarr2HBr(g)` isA. `+ 261` kJB. `- 103` kJC. `- 261` kJD. `+ 103` kJ |
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Answer» Correct Answer - B `H-H+Br-Brrarr2H-Br` `{:(433+192,2xx364),(625,728):}` Energy absorbed = Energy released Not energy released = 728-625=103 kJ i.e., = `DeltaH=-103 kJ`. |
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| 43. |
Given that `C(g)+4H(g)rarrCH_(4)(g),DeltaH=-166 kJ` The bond energy C-H will beA. 208 kJ/moleB. `- 41.6` kJ/moleC. 832 kJ/moleD. None of these |
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Answer» Correct Answer - B `(-166)/(4)=-41.5 kJ//"mole"`. |
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| 44. |
Consider the reactions `C(s)+2H_(2)(g)rarrCH_(4)(g),DeltaH=-x kcal` `C(g)+4H(g)rarrCH_(4)(g), DeltaH=-x_(1) kcal` `CH_(4)(g)rarrCH_(3)(g)+H(g),DeltaH=+y kcal` The bond energy of C - H bond isA. y kcal `mol^(-1)`B. `x_(1)` kcal `mol^(-1)`C. `x//4` kcal `mol^(-1)`D. `x_(1)//4` kcal `mol^(-1)` |
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Answer» Correct Answer - A The bond energy of C-H bond is y kcal `mol^(-1)`. |
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| 45. |
The enthalpy of formation of ammonia is `-46.0 kJ mol^(-1)`. The enthalpy change for the reaction `2NH_(3)(g)rarr2N_(2)(g)+3H_(2)(g)` isA. 46.0 kJ `mol^(-1)`B. `92.0 kJ mol^(-1)`C. `-23.0 kJ mol^(-1)`D. `-92.0 kJ mol^(-1)` |
| Answer» Correct Answer - B | |
| 46. |
The enthalpy change `(DeltaH)` for the reaction, `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) " is "-92.38 kJ` at 298 K. The internal energy change `DeltaU` at 298 K isA. `-92.38 kJ`B. `-87.42 kJ`C. `-97.34 kJ`D. `-89.9 kJ` |
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Answer» Correct Answer - B `DeltaH=DeltaU+Deltan_(g)RT` `Deltan_(g)=-2`. |
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| 47. |
For a given reaction, `DeltaH=35.5 kJ mol^(-1)` and `DeltaS=83.6 JK^(-1) mol^(-1)`. The reaction is spontaneous at : (Assume that `DeltaH` and `DeltaS` do not very with temperature)A. `T gt 425 K`B. All temperaturesC. `T gt 298 K`D. `T lt 425 K` |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS` for equilibrium `DeltaG=0` `DeltaH-TDeltaS` `T_(eq).=(DeltaH)/(DeltaS)=(35.5xx1000)/(83.6)=425 K` Since the reaction is endothermic it will be spontaneous at `T gt 425 K`. |
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| 48. |
An endothermic reaction has a positive internal energy change `DeltaU`. In such a case, what is the minimum value that activation energy can have ?A. `DeltaU`B. `DeltaU=DeltaH+DeltanRT`C. `DeltaU=DeltaH-DeltanRT`D. `DeltaU=E_(a)+RT` |
| Answer» Correct Answer - C | |
| 49. |
A system is changed from state A to state B by one path and from B to A another path. If `E_(1)` and `E_(2)` are the corresponding changes in internal energy, thenA. `E_(1)+E_(2)=-ve`B. `E_(1)+E_(2)=+ve`C. `E_(1)+E_(2)=0`D. None of these |
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Answer» Correct Answer - C `DeltaE=0` for a cyclic process. |
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| 50. |
The correct relation between equilibrium constant (K), standard free energy `(DeltaG^(@))` and temperature (T) isA. `DeltaG^(@)=RT lnK`B. `K=e^(-DeltaG^(@)//2.303 RT)`C. `DeltaG^(@)=-RTlog_(10)K`D. `K=10^(-DeltaG^(@)//2.303RT)` |
| Answer» Correct Answer - D | |