InterviewSolution
Saved Bookmarks
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.
| 401. |
Enthalpy of neutralisation of the reaction between `CH_(3)COOH(aq)` and `NaOH(aq)` is `-13.2kcal Eq^(-1)` and that of the reaction between `H_(2)SO_(4)(aq)` and `KOH (aq)` is `-13.7 kcal Eq^(-1)`. The enthalpy of dissociation of `CH_(3)COOH(aq)` isA. `-0.5` kcal `eq^(-1)`B. `+0.5` kcal `eq^(-1)`C. `-26.9kcal` `eq^(-1)`D. `+13.45kcal` `eq^(-1)` |
|
Answer» Correct Answer - B Dissociation enthalpy of `CH_(3)COOH=13.7-13.2=0.5kcal` `eq^(-1)`. Thus, 0.5 kcal `eq^(-1)` heat will be used to dissociate `CH_(3)COOH` completely. |
|
| 402. |
The enthalpy change `DeltaH` for the neutralisation fo `1M HCI` by caustic potash in dilute solution at `298 K` isA. 68kJB. 65 kJC. 57.3 kJD. 50 kJ |
|
Answer» Correct Answer - C Since, both HCl and KOH are strong, 57.3 kJ heat will be released. |
|
| 403. |
Standard enthalpy of vaporisation`DeltaV_(vap).H^(Theta)` for water at `100^(@)C`is`40.66kJmol^(-1)`.Teh internal energy of Vaporization of water at` 100^(@)C("in kJ mol"^(-1))`isA. `+4376`B. `+40.66`C. `+37.56`D. `-43.76` |
|
Answer» Correct Answer - D `H_(2)O(l)rarrH_(2)O(v) :. Deltan=1` `DeltaH^(@)=DeltaU^(@)+DeltanRT` `40.66xx10^(3)=DeltaU^(@)+1xx8.314xx373` `:. DeltaU^(2)=37559J=37.56KJ` |
|
| 404. |
Standard enthalpy of vaporisation`DeltaV_(vap).H^(Theta)` for water at `100^(@)C`is`40.66kJmol^(-1)`.The internal energy of Vaporization of water at` 100^(@)C("in kJ mol"^(-1))`isA. `+37.56`B. `-43.76`C. `+43.76`D. `40.66` |
|
Answer» Correct Answer - A `DeltaH_(vap)^(@)=40.66 kJ mol^(-1)` `T=273+100=373 K`, `DeltaU=?` `DeltaH=DeltaU+Deltan_(g) RT `or `DeltaU=DeltaH-Deltan_(g)RT` `Deltan_(1)(g)=1-0=1` `:. DeltaU=40.66 kJ mol^(-1)-1xx8.314xx10^(-8)JK^(-1)mol^(-1)xx373 K` `=(40.66-3.101) kJ mol^(-1)` `=37.559 kJ mol^(-1)` |
|
| 405. |
Assertion: Where a mole of electron is transported across a potential difference of `1V` from positive electrode to negative electrode, then work done is `-96500J` . Reason: Work done due to transportation of electrons depends on the product of potential difference between the two points and amount to charge carried.A. If both the 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 Assertion and reason are both correct and reason is the correct explanation of assertion. |
|
| 406. |
Assertion: For any gas `DeltaW=-PDeltaV,P` represents the external pressure. Reason: Throughout the process, external pressure must be remain constant.A. If both the 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 Assertion and reason are both correct and reason is the correct explanation of assertion. |
|
| 407. |
One litre-atmosphere is approximately equal toA. 101.3 JB. 8.314 JC. 931 JD. 19.2 J |
|
Answer» Correct Answer - A 0.0821 litre-atm `=8.314J` (values of gas constant) `therefore1"litre"-atm=101.3J` |
|
| 408. |
Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2Ag(s), E_(cell)^(@)=0.46 V`. |
|
Answer» Let us apply Nernst equation at equilibrium `E_(cell)^(@)=(0.0591)/(n)log" "K_(C)` `0.47=(0.0591)/(2)logK_(c)` `K_(c)="antilog"[(0.47xx2)/(0.0591)]` `=8.5xx10^(15)`. |
|
| 409. |
The incorrect expression among the following isA. `ln K =( DeltaH^(@)-TDeltaS)/(RT)`B. `K=E^(-Delta^(@)//RT)`C. in isothermal process `W_(rev)=-nRT ln" (V_(f))/(V_(c ))`D. `(DeltaG"system")/(DeltaS"total")=-T` |
|
Answer» Correct Answer - A `DeltaG^(@)=-RT ln K` `DeltaG^(@)=DeltaH^(@)-TDeltaS` or `DeltaH^(@)-TDeltaS^(@)=-RT ln K` `:. lnK=-((DeltaH^(@)-TDeltaS^(@))/(RT))` |
|
| 410. |
An ideal gas is allowed to expand both reversibly and irreversibly in an an islolated system. If `T_(i)` is the initial temperature and `T_(f)` is the final temperature, which of the following statements is correct ?A. (a) `T_(f)` and `T_(i)` are both reversible and irreversible processesB. `(T_(f))_(irrev)gt(T_(i))_(rev)`C. `T_(f)gtT_(i)` for irreversible process but `T_(f)` and `T_(i)` for irreversible processD. `(T_(f))_(rev)gt(T_(i))_(irrev)` |
|
Answer» Correct Answer - D Work done in reversible process is mamximum. Thus `T_(2)lt lt T_(1)` in reversible process or `T_(i)lt lt T_(f)` |
|
| 411. |
The incorrect expression among the following isA. `(DeltaG_("system"))/(DeltaG_("total"))=-T`B. in isothermal process, `W_("reversible")=-nRT"in"(V_(f))/(V_(i))`C. In `K=(DeltaH-TDeltaS^(@))/(RT)`D. `K=e^(-DeltaG^(@)//RT)` |
|
Answer» Correct Answer - C According to Gibbs Helmholtz equation, `DeltaG=DeltaH-TDeltaS` (a) For a system, total entropy change `=DeltaS_("total")` `Delta_("total")=0` `:. DeltaG_("system")=-TDeltaS_("total")` `:. (DeltaG_("system"))/(DeltaS_("total"))=-T` Thus, `(a)` is correct. (b) For isothermal reversible process, `DeltaE=0` By first law of thermodynamics, `DeltaE=q+W` `:. W_("reversible")=-q=-int_(v_(i))^(v_(f))pdV` `implies W_("reversible")=-RT` In `(V_(f))/(V_(i))` Thus, `(b)` is correct (c) `DeltaG^(@)+DeltaH^(@)-TDeltaS^(@)` Also, `DeltaG^(@)=-RT` In `K` In `K=((DeltaH^(@)-TDeltaS^(@)))/(RT)` In `K=(DeltaG^(@))/(RT)` [from Eq. (i)] (d) The standard free energy `(DeltaG^(@))` is related to equilibrium constant `K` as `DeltaG^(@)=-RT` In `K` `:.` In `K=-(DeltaG^(@))/(RT)` `k=-e^(-DeltaG^(@)//Rt)` Thus, `(d)` also correct. |
|
| 412. |
Ionisation energy of `A1=5137KJmol^(-1)(DeltaH)` hydration of `Al^(3+)=-4665KJmol^(-1).(DeltaH)_(hydration)` for `Cl^(-)=-381KJmol^(-1)` . Which of the following statement is correct?A. `AlCl_(3)` would remauin covalent in aqueous solutionB. Only at infinite dilution `AlCl_(3)` undergoes ionisationC. In aqueous solution `AlCl_(3)` becomes ionicD. Nonee of these |
|
Answer» Correct Answer - C If `AlCl_(3)` is present in ionic state in aqueous solution, therefore it has `Al^(3+)` and `3Cl^(-)` inos Standard heat of hydration of `Al^(3+)` and `3Cl^(-)` inos `=-4665+3xx(-381)Kjmol^(-1)` `=-5808KJ//mol` Required energy of ionisation of `Al=5137KJmol^(-1)` `:.` Hydration energy overcomes ionisation energy `:. AlCl_(3)` would be ionic in aqueous solution |
|
| 413. |
The standard enthalpies of formation of `SF_(6)(g)` , `S(g)` and `F(g)` are `-1150` , `+280` and `+85KJmol^(-1)` . Thus, average bond energy of `(S-F)` in `SF_(6)` isA. `309.16KJmol^(-1)`B. `1855KJmol^(-1)`C. `11.130xx10^(3)KJmol^(-1)`D. `323.33KJmol^(-1)` |
|
Answer» Correct Answer - B `S(g)+6F(g)rarrSF_(6)(g)` Based on `BE` values `Delta_(f)H^(@)(SF_(6))=-1100=Delta_(f)H^(@)(S,g)` `+6Delta_(f)H^(@)(F,g)-6(BE)_(S-F)` `-1150=280+6xx85-6(BE)_(S-F)` `:. (BE)_(S-F)=(280+510+1150)/(6)=323.33KJmol^(-1)` |
|
| 414. |
Bond energy of `(N-H)` bond is `yKJ mol^(-1)` under standard state. Thus,change in internal energy in the following process is `NH_(3)(g)rarrN(g)+3H(g)`A. `-3yKJmol^(-1)`B. `-yKJmol^(-1)`C. `3yKJmol^(-1)`D. `yKJmol^(-1)` |
|
Answer» Correct Answer - D `H-underset(H)underset(|)(N)-H` (three `N-H`) bonds Thus, `DeltaE=3yKJmol^(-1)` |
|
| 415. |
`DeltaH` for `CaCO_(3)(s)rarrCaO(s)+CO_(2)(g)` is `176 kJ mol^(-1)` at `1240 K`. The `DeltaE` for the change is equal toA. `160KJ`B. `165.6KJ`C. `186.3KJ`D. `180.0KJ` |
|
Answer» Correct Answer - A `Delta H = Delta E + Delta nRT` `:. Delta E = 176 - 1 xx 8.314xx8.314xx1240xx10^(-3)` `= 165.6 kJ` |
|
| 416. |
When no heat energy is allowed to enter or leave the system, it is calledA. Isothermal proecessB. Reversible processC. Adiabatic processD. Irreversible process |
|
Answer» Correct Answer - D In adiabatic process, there is no exchange of heat. i.e., `q=0` . |
|
| 417. |
Which of the following statements is true ?A. `DeltaG` may be lesser or greater or equal to `DeltaH`B. `DeltaG` is always proportional to `DeltaH`C. `DeltaG` is always greater than `DeltaH`D. `DeltaG` is always less than `DeltaH` |
|
Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS`. Depending upon `TDeltaS`, `DeltaG` can be smaller or greater or equal to `DeltaH`. |
|
| 418. |
Assertion: Molar entropy of vaporization of water is different from ethanol. Reason: Water is more polar than ethanol.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 The molar entropy of vaporization of water is different from enthanol due to hydrogen bonding. According to VSERP theory water molecule hastwo lone pairs of electronsand has angular shape and show some polarity which is higher than that of ethanol. So both assertion and reason are correct but reason is not the expanation of assertion. |
|
| 419. |
For the following process, `H_(2)(g)rarr2H(g)`. It absorbs `438kJmol^(-1)` . Thus,A. change in internal energy of the system is `219KJmol^(-1)`B. internal energy of the system is `438KJmol^(-1)`C. change in internal energy if `438KJmol^(-1)`D. internal energy of the system is `219KJmol^(-1)` |
|
Answer» Correct Answer - D Internal energy cannot be measured. However, change in internal energy can be measured `H-H` covalent bond breaks by absorbing `438KJmol^(-1)` energy. Thus, increases in internal energy `(DeltaE)=438KJmol^(-1)` . |
|
| 420. |
For the given heat of reaction, (i) `C(s)+O_(2)(g)CO_(2)(g)+97Kcal` (ii) `CO_(2)(g)+C(s)=2CO(g)-39Kcal` the heat of combustion of `CO(g)` is:A. `68Kcal`B. `-68Kcal`C. `+489Kcal`D. None |
|
Answer» Correct Answer - A Subtracting equattion (ii) from equation (i), we get `C(s)+O_(2)(g)+97Kcal` `(CO_(2)(g)+C(s)=2CO(g)-39Kcal)/(or -CO_(2)(g)+O_(2)(g)-2CO(g)+136Kcal)` or, `2CO(g)+O_(2)=CO_(2)(g)+136Kcal` or, `CO(g)+1//2O_(2)(g)=CO_(2)(g)68Kcal` ltgbrgt Required value `=68Kcal` |
|
| 421. |
What is the amount of heat to be supplied to prepare `128g` of `CaCO_(3)` by using `CaCO_(3)` and followed by reduction with carbon, Reactions are `CaCO_(3)(s)=CaO(s)+CO_(2)(g),DeltaH^(@)=42.8Kcal` `CaCO(s)=3C(s)=CaC_(2)+CO(g),DeltaH^(@)=111Kcal`A. `102.6Kcal`B. `221.78Kcal`C. `307.6Kcal`D. `453.46Kcal` |
|
Answer» Correct Answer - A `CaCO_(3)(s)=CaO(s)+O_(2)(g)DeltaH^(@)=42.8Kcal` `(CaO(s)+3C(s)=CaC_(2)(s)+CO(g)DeltaH^(@)=111Kcal)/(CaCO_(3)(s)+3C(s)=CaC_(2)(s)+CO(g)+CO_(2)(g))` `DeltaH=153.8Kcal` Molecular weight of `CaC_(2)=40+24=64` `64g` of `CaC_(@)` requires `307.6Kcal` of heat `128g` of ` CaC_(2)` requires `307.6Kcal` of heat |
|
| 422. |
Which among the following is not a state function?A. Internal energyB. Free energyC. WorkD. Enthalpy |
|
Answer» Correct Answer - C Work is not a state function as it depends upon the path followed. |
|
| 423. |
The free energy change for a reversible reaction at equilibrium isA. `gt 0`B. `lt 0`C. equal to zeroD. unpredictable |
|
Answer» Correct Answer - C Reversible process is carried out under equilibrium conditions. Hence `DeltaG_(T,P)=0` |
|
| 424. |
Which of the following is the intensive property?A. TemperatureB. ViscosityC. DensityD. All of these |
|
Answer» Correct Answer - C Those properties which depend only on the nuture of the substance and are independent of the amount of the substance present in the system e.g., `T` , `P` , density, viscosity, etc. |
|
| 425. |
Which of the following is an intensive property?A. VolumeB. MassC. DensityD. Energy |
|
Answer» Correct Answer - C Density is an itensive property as it does not depend upon the amount of the substance. |
|
| 426. |
Which of the following is the intensive property?A. boiling pointB. molarityC. freezing pointD. all of these |
|
Answer» Correct Answer - A An extensive property of a system is that property which depends upon the amount of the substance or substances present in the system, e.g., mass, volume, energy, entropy, Gibbs potential enthaply etc. whereas intensive properties are those properties of a system, which are characteristic of the substance or substances present, and are idenpendent of its mass. Temperature pressure density, heat capacity, boiling point, etc., are intensive property. |
|
| 427. |
The intensive property among these quantities isA. massB. volumeC. enthalpyD. mass/volume |
|
Answer» Correct Answer - D Mass/volume (density) is the intensive property. |
|
| 428. |
In an adiabatic expansion of an ideal gas -A. `W=-DeltaE`B. `W=DeltaE`C. `DeltaE=0`D. `W=0` |
|
Answer» Correct Answer - A In an adiabatic expansion of ideal gas, work done, `W=-DeltaE` |
|
| 429. |
An adiabatic expansion of an Ideal gas always hasA. decrease in temperatureB. `q=0`C. `W=0`D. `DeltaH=0` |
|
Answer» Correct Answer - B Since during adiabatic expansion of an ideal gas, no heat enters or leaves the system, `q=0`. |
|
| 430. |
Which of the following is the correct option for the free expansion of an ideal gas under adiabatic condition?A. `q=0,DeltaTlt0,wne (=)0`B. `q=0,DeltaTne 0,w=0`C. `qne 0,DeltaT=0,w=0`D. `q=0,DeltaT=0,w=0` |
|
Answer» Correct Answer - D For addiabatic expansion, `q=0` For free expansion, `P_(ext)=0` , Hence, `w=0` `DeltaU=q+w=0+0=0` . As `U` is a function of temperature `(U+-(3)/(2)nRT)` and as `DeltaU=0` `:. DeltaT=0` . |
|
| 431. |
The standard enthalpy of formation of `NH_(3)` is `-46.0KJ mol^(-1)` . If the enthalpy of formation of `H_(2)` from its atoms is `-436KJ mol^(-1)` and that of `N_(2)` is `-712KJ mol^(-1)` , the average bond enthalpy of `N-H` bond in `NH_(3)` isA. `-964KJmol^(-1)`B. `+352KJmol^(-1)`C. `+105KJmol^(-1)`D. `-1102KJmol^(-1)` |
|
Answer» Correct Answer - C `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)rarrNH_(3)(g),Delta_(f)H=-46KJmol^(-1)` Bond enthalpy of `H_(2)=436KJmol^(-1)` [`.^(+)ve` sing is taken because energy is supplied to break the `H-H` bond into its atoms] Similary, bond enthalpy of `N_(2)=+712KJmol^(-1)` `Delta_(f)H=[(1)/(2)BE(N_(2))+(3)/(2)BE(H_(2))]-3BE(N-H)` `-46=[(1)/(2)xx712+(3)/(2)xx436]-3BE(N-H)` `3BE(N-H)=(1010)+46` `3BE(N-H)=1056` `BE(N-H)=1056//3=352KJmol^(-1)` |
|
| 432. |
Which of the following acid will release maximum amount of heat when completely neutralised by strong base `NaOH`?A. 1 M HClB. 1 M `HNO_(3)`C. `1M` `HClO_(4)`D. `1M` `H_(2)SO_(4)` |
|
Answer» Correct Answer - D Ionisation of `H_(2)SO_(4)` gives double amount of `H^(+)` ions as compared to other acids. `H_(2)SO_(4)to2H^(+)+SO_(4)^(2-)` |
|
| 433. |
find out the standard free energy change at `60^(@)C`and at 1 atn if the `N_(2)O_(4)` is 50 % dissociatedA. `-800 .0 kJ mol^(-1)`B. `+ 800.0 kJ mol^(-1)`C. `789.89 JK^(-1) mol^(-1)`D. `+ 789.98 JK^(-1)mol ^(-1)` |
|
Answer» Correct Answer - c `N_(2)O_(4)(g)Leftrightarrow2NO_(2)(g)` ` N_(2)O_(4)` is 50 % dissociated , the mole fraction of the substance is given by `x_(n2O4)= (1-05)/(1 + 0.5), x_(no2)= (2xx0.5)/(1+0.5)` `P_(n2O4)=0.5/1.5xx 1 atm , p_(no2)1/1.5 xx 1atm` the equilibrium constant ` K_(p)` is given by `k_(p) =(P_(no_(2)))^(2)/(P_(N_(2)O_(4)))=1.5/((1.5)^(2)(0.5))=1/0.75=1.33` ` Delta_(a)G^(Theta)=-RT in K_(p)` `=- 8.314 Jk^(-1)mol^(-1)xx333 K xx 2.303xx 0.1239` `=-789.89 JK^(-1) mol^(-1)` |
|
| 434. |
moles of an ideal gas expand isothermallty anad reversibly from pressure of 5 atm to 1 atm at 300 K . Calculate the largest mass than can be lifted through a height of 1 M by this expansion .A. 4092.76 kgB. 8730.9368 kgC. 4492.76 kgD. 8170.2344 kg |
|
Answer» Correct Answer - d `W_(exp)=-2.303nRTlogP_(1)//P_(2)` ` 2.303xx 20 xx 8.314 JK^(-1)mol^(-1)xx300 K log 5/1` `80.15 xx 10^(3 J)` work done = mgh `80.15 xx 10^(3) = m xx 9.81 ms^(-2) xx1m` ` m = (80.15xx10^(3) kgm^(2)s^(-2))/(9.81 ms ^(-2)xx1m) = 8.17023xx10^(3)` ` 8170 . 2344 kg ` |
|
| 435. |
One moles of anhydrous `AB` dissolves in water and liberates `21.0 J mol^(-1)` of heat. The valueof `DeltaH^(Theta)` (hydration) of `AB` is `-29.4 J mol^(-1)`. The heat of dissolution of hydrated salt `AB.2H_(2)O(s)` isA. `50.4 J mol^(-1)`B. `8.4 J mol^(-1)`C. `-50.4 J mol^(-1)`D. `-8.4 J mol^(-1)` |
|
Answer» Correct Answer - B `DeltaH_(sol)` (anhyd. Salt) `=[DeltaH_(sol)("hydrated salt") + DeltaH_("hydration")]` |
|
| 436. |
Assertion: Adsorption is exothermic and spontaneous in spite of the fact that adsorption is accompanied with deccrease in entropy. Reason: The factor `TDeltaS` is lesser than `DeltaH`A. If both the 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 - D `DeltaS` is `-ve` because adsorption leads to more orderd arrangement. Thus `DeltaG=DeltaH-TDeltaS` `Delta=-ve` only when `DeltaHgtTDeltaS` . |
|
| 437. |
The lattice enthalpy and hydration enthalpy of four compounds are given below: the pair of compounds which is soluble in water is:A. P and QB. Q and RC. R and SD. P and R |
|
Answer» Correct Answer - D For solubility of ionic compouds, hydration energy m ust be greater than lattice energy. |
|
| 438. |
The standard enthalpies fo formation of `CO_(2) (g), H_(2) O (1)`, and glucose (s) at `25^(@)C` are `- 400 kJ mol^(-1), - 300 kJ mol^(-)`, and `- 1300 kJ mol^(-1)`, respectively. The standard enthalply of combustion per gram of glucose at `25^(@)C` isA. `+2900kJ`B. `-2900kJ`C. `-16.11kJ`D. `+16.11kJ` |
|
Answer» Correct Answer - C `C_(6)H_(12)O_(6)(s)+6O_(2)(g)to6CO_(2)(g)+6H_(2)O(l)` `DeltaH[6Delta_(f)H_(CO_(2))+6Delta_(f)H_(H_(2)O)]-[Delta_(f)H_(C_(6)H_(12)O_(6)+6Delta_(f)H_(O_(2))]` `=[6(-400)+6(-300)]-[-1300+0]` `=-4200+1300=-2900kJ//mol` Enthalpy of combustion per gram `=(-2900)/(180)=-16.11kJ//g` |
|
| 439. |
A piston filled with `0.04` mole of an ideal gas expands reversible from `50.0mL` at a constant temperature of `37.0^(@)C` . As it does so, it absorbs `208J` of heat. The value of `q` and `W` for the process will be `(R=8.314J//molK` , `1n7.5=2.01)`A. `q=+208 J`, `w=+208 J`B. `q=208 J`, `w=-208 J`C. `q=-208 J`, `w=-208 J`D. `q=-208 J`, `w=+208 J` |
|
Answer» Correct Answer - B For isothermal expansion of an ideal gas, `DeltaV=0` Hence, from first law of thermodynamics, `DeltaU=q+w`, we have `q=-w`. A process involves adsoption of heat, i.e., it is endothermic `:. Q=-208 J`. Hence `w=-208 J` |
|
| 440. |
The standard enthalpies fo formation of `CO_(2) (g), H_(2) O (1)`, and glucose (s) at `25^(@)C` are `- 400 kJ mol^(-1), - 300 kJ mol^(-)`, and `- 1300 kJ mol^(-1)`, respectively. The standard enthalply of combustion per gram of glucose at `25^(@)C` isA. `+2900 kJ`B. `-2900 kJ`C. `-16.11 kJ`D. `+16.11 kJ` |
|
Answer» Correct Answer - C `C_(6)H_(12)O_(6)(s)+6O_(2)(g)to2CO_(2)(g)+6H_(2)(l)` `Delta_(f)^(@)=DeltaCH^(@)=[6DeltafH^(@)(CO_(2))+6DeltaH_(f)^(@)H(H_(2)O)_(6)]-[Delta^(@)(C_(6)H_(12)O_(6))+6DeltafH^(@)*O_(2)]` `=6xx(-400)+6xx(-300)-(1300)` `=-2900 kJ//mol` `=-(2900)/(180)kJ//g=16.11 kJ//g` |
|
| 441. |
`2H_(2)(g)+O_(2)(g)to2H_(2)O(l)+xkJ` In the above reactionA. `zkJ mol^(-1)` is the value of `DeltaH_(f)` for waterB. `DeltaH_(f)` of `H_(2)O(l)` is `-xkJ`C. `DeltaH_(f)` of `H_(2)O(l)` is `-x//2kJ mol^(-1)`D. `DeltaH_("comb")` of `H_(2)` is `-x//2kJ mol^(-1)` |
|
Answer» Correct Answer - C::D Both `(C )` and `(D)` statements about the reaction are correct. |
|
| 442. |
One mole of monoatomic ideal gas at `T(K)` is exapanded from `1L` to `2L` adiabatically under constant external pressure of `1` atm . The final tempreture of gas in kelvin isA. `T`B. `(T)/(2^(5//3-1))`C. `T-(2)/(3xx0.0821)`D. `T+(2)/(3xx0.0821)` |
|
Answer» Correct Answer - C Here `P_("ext")=1"atm"` `V_(1)=1L ` and `V_(2)=2L` `R=0.0821 L "atm" K^(-1) mol^(-1)` `gamma=5//3` (For a monoatomic gas) `T_(1)=T K` and `T_(2) = ?` Now `C_(P)-C_(V)=R` or `(C_(P))/(C_(V))-1=(R )/(C_(V))` `gamma-1=(R )/(C_(V))` `C_(V)=(R )/(gamma-1)` `C_(V)(T_(2)-T_(1))=P_("ext")(V_(1)-V_(2))` `(R )/(gamma-1)(T_(2)-T_(1))=P_("ext")(V_(1)-V_(2))` Substiuting the values `(0.0821)/(5//3-1)(T_(2)-T_(1))=1xx(!-2)` `T_(2)-T=(2//3)/(0.0821)` `T_(2)=T-(2)/(3xx0.0821)` |
|
| 443. |
The free energy for a reaction having `DeltaH=31400` cal, `DeltaS=32" cal "K^(-1)mol^(-1)` at `1000^(@)C` is:A. `-9336cal`B. `-7386cal`C. `-1936cal`D. `+9336cal` |
|
Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS=31400-1273xx32=-9336kcal` |
|
| 444. |
Melting point of a solid is x K and its latent heat of fusion is 600 cal `mol^(-1)`. The entropy change for fusion of 1 mol solid is 2 cal `mol^(-1)K^(-1)`. The value of x will be:A. 100KB. 200KC. 300KD. 400K |
|
Answer» Correct Answer - C `(DeltaH_("fusion"))/(T_("mp"))=DeltaS_(fusion)` `(600)/(T)=2` `T=300K` |
|
| 445. |
5 moles of oxygen are heated at constant volume from `10^(@)C` to `20^(@)C`. The change in internal energy of a gas. `[C_(P)=7.03 cal mol^(-1) deg^(-1)` and `R=8.3J mol^(-1) deg^(-1)]` |
|
Answer» `R=8.31J" "mol^(-1)deg^(-1) and R=8.31" J "mol^(-1)` `=1.99" cal "mol^(-1)deg^(-1)` We know that `C_(P)-C_(V)=R` or `C_(V)=C_(P)-R=7.03-1.99=5.04cal" "mol^(-1)deg^(-1)` ltBrgt Heat absorbed by 5 mole of oxygen in heating from `10^(@)C` to `20^(@)C` `=5xxC_(V)xxDeltaT=5xx5.04xx10=252`cal Since, the gas is heated at constant volume, no external work is done, i.e., So, change in internal energy will be equal to heat absorbed, `DeltaU=q+w=252+0=252`cal |
|
| 446. |
A constant pressure calorimeter consists of an insulated beaker of mass `92g` made up of glass with heat capacity `0.75 J K^(-1) g^(-1)`. The beaker contains `100 mL` of `1M HCI` at `22.6^(@)C` to which `100 mL` of `1M NaOH` at `23.4^(@)C` is added. The final temperature after the complete reactions is `29.3^(@)C`, What is `DeltaH` per mole for this neutralization reaction? Assume that the heat capacities of all solutions are equal that of same volumes of water. |
|
Answer» Initial average temperature of the acid and base `=(22.6+23.4)/(2)=23.0^(@)C` Rise in temperature `=(29.3-23.0)=6.3^(@)C` Total heat produced `=(92xx0.75+200xx4.184)xx6.3` `=(905.8)xx6.3=5706.54J` Enthalpy of neutralisation`=-(5706.54)/(100)xx1000xx1` `=57065.4J=-57J` |
|
| 447. |
To what pressure must a certain ideal gas `(gamma=1.4)` at 373 K and 1 atmospheric pressure be compressed adiabatically in order to raise its temperature to 773K? |
| Answer» Correct Answer - 7.89 atm | |
| 448. |
Calculate the work done when 50 g of iron is dissolved in HCl at `25^(@)C` in (i) a cloed vessel and (ii) an open beaker when the atmospheric pressure is 1atm. |
|
Answer» (i) When the reactiion is carried in a closed vessel, the change in volume is zero. Hence, the work done by the system will be zero. (ii) when iron dissolves in HCl, hydrogen is produced. `undersetunderset(50g)(56g)(Fe)+2HCltoFeCl_(2)+undersetunderset((1)/(56)xx50"Mole")(1" mole")(H_(2))` Volume of hdyrogen produced at `25^(@)C` `=(nRT)/(P)=(50)/(56)xx(0.0821xx298)/(1)=21.84L` this is equal to volme change when the reaction is carried in open beaker. Work done by the system `=-PDeltaV=-1.0xx21.84` `=-21.84"Litre"-atm =-2212.89J` |
|
| 449. |
Calculate the work done when 65.38 g of zinc dissolves in hydrochloric acid in an open beaker at 300K. (At. Mass of Zn=65.38) |
| Answer» `-2494.2J` , i.e.,w ork is done by the system | |
| 450. |
Given the bond energies `N=N` , `H` and `H-H` bond are `945, 436` and `391KJmol^(-1)` respectively, the enthalpy change of the reaction `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` isA. `93KJ`B. `102KJ`C. `90KJ`D. `105KJ` |
|
Answer» Correct Answer - A `DeltaH=(945+3xx439)-(2xx3xx391)` `=2253-2346` `=-93KJ` |
|