InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Calculate the workdone, and change in internal enegry taking place when an ideal gas undergos isothermal expansion. |
| Answer» a. `DeltaU` for an ideal gas, `U` depends on temperature. Since temperature is constant<br> `rArr DeltaU = 0`<br> b. According to the first law<br> `DeltaU = dq +dW`<br> Since `DeltaU =0`<br> `:. dq=- dw` | |
| 2. |
Internal enegry of a system of molecules is determined by taking into consideration itsA. Kinetic enegryB. Vibrational enegryC. Rotantional energyD. All kinds of energies present in the molecules |
| Answer» All kinds of energies present in the molecules. | |
| 3. |
The enthalpies of neutralisation of a string acid `HA` and a weaker acid `HB` by `NaOH` are `-13.7` and `-12.7 kcal Eq^(-1)`, respectively. When one equivalent of `NaOH` is added to a mixture containing one equivalent of `HA` and `HB`, the enthalpy change was `-13.5kcal`. In what ratio is the base distributed between `HA` and `HB`? |
| Answer» Let `x` equivalent of `HA` and `y` equivalent of `HB` are taken in the mixture<br> `x +y = 1 …(i)`<br> Ethalpy change `=x xx13.7 +y xx12.7 = 13.5 …(ii)`<br> Solving equation (i) and (ii), we get<br> `x = 0.8, y = 0.2`<br> `x:y = 4:1` | |
| 4. |
For the percipitation reaction of `Ag^(o+)` ions with `NaCI`, which of the following statements is true?A. `DeltaH` is zero for the reaction.B. `DeltaG` is zero for the reaction.C. `DeltaG` is negative for the reaction.D. `DeltaG` should be equal to `DeltaH`. |
| Answer» `DeltaG=-ve` for spontaneous process | |
| 5. |
The lattice energy of solid `NaCI` is `180 kcal mol^(-1)`. The dissolution of the solid in `H_(2)O` is endothermic to the extent of `1.0 kcal mol^(-1)`. If the hydration energies of `Na^(o+)` and `CI^(Theta)` ions are in the ratio of `6:5` what is the enthalpy of hydration of sodium ion?A. `-85 kcal mol^(-1)`B. `-98 kcal mol^(-1)`C. `+82 kcal mol^(-1)`D. `+100 kcal mol^(-1)` |
| Answer» We know that<br> Ethalpy of solution =Lattice enegry + Hydration enthalpy<br> or `Delta_(sol)H^(Theta) = Delta_("lattice") H^(Theta) +Delta_("hyd")H^(Theta)`<br> `:. Delta_("hyd")H^(Theta) = Delta_("sol")H^(Theta) - Delta_("lattice")H^(Theta) = 1.0 - 180`<br> `=- 179.0 kcal mol^(-1)`<br> The ratio of hydration energies of `Na^(o+)` and `CI^(Theta)` is `6:5`.<br> `:.` Hydration enegry of `Na^(o+) = (6)/(11) xx (-179.0)`<br> `=- 97.63 kcal ~~ 98.0 kcal` | |
| 6. |
For the reaction,` 2Co +O_(2)rarr 2CO_(2), DeltaH = - 560 kJ `. Two moles of CO and one mole of`O_(2)` are taken in a container of volume 1 L. They completely reacts to form two moles of `CO_(2)`. The gases deviate appreciably from ideal behaviour. Ifthe pressure in the vessel changesfrom 70 to 40 atm,find the magnitude ( absolute value) of `DeltaU` at 500 K `( 1 L atm = 0.1 kJ )`. |
| Answer» `H =U +PV`<br>`DeltaH+ DeltaU + Delta(PV) ` or ` DeltaH = DeltaU + VDeltaP`<br> or `DeltaU = DeltaH -VDeltaP = - 560kJ - 1L ( 40 atm -70 atm)`<br> `=- 560 kJ + 30 Latm = - 560 kJ + 3-0 xx 0.1kJ = - 557 kJ `<br> Absolute value `=557 kJ . ` | |
| 7. |
The enthalpy of neutralization of oxalic acid by a strong base is `- 25.4 kcal mol^(-1)`. The enthalpy of neutralization of strong acid and strong base is `-13.7 kcal equiv^(-1)` . The enthalpy of dissociation of `H_(2)C_(2)O_(4) hArr 2H^(+) C_(2)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)` |
| Answer» Correct Answer - B<br>Enthalpy of neutralization of oxalic acid by strong base `= - 254 kcalmol^(-1) = - 12.7 kcale quiv^(-1)`<br> `:. `Enthalpy of dissociation of oxalic acid<br> `= 13.7 - 12.7 kcal e quiv^(-1)`<br> `=1.0 kcal e quiv^(-1) = 2.0 kcal mol^(-1)` | |
| 8. |
The lattice energy of NaCl is 180 kcal `//` mol . The dissociation of the solid in water in the form of ioins is endothermicto the extent of 1 kcal` //` mol.If the solvation energies of`Na^(+)` and `Cl^(-)` ions are in theratesof`6:5` , calculate the enthalpy of hydration of `Na^(+)` ions. |
| Answer» (i) `NaCl(s) rarr Na^(+) (g) + Cl^(-) (g), DeltaH =180kcal mol^(-1)`<br> (ii) `Na^(+) Cl^(-)(s) overset( + aq)(rarr) Na^(+)(aq), Cl^(-)(aq) , Delta H = 1 kcalmol^(-1)`<br> (iii) `Na^(+)(g)+aq rarr Na^(+)(aq) , Delta H = 6 x ( `Aim)<br> (iv) `Cl^(-)(g) +aq rarrCl^(-) (aq), DeltaH = 5x`<br> Eqn. (ii) - Eqn. (iv) gives Enq. (i)<br> `:. 180 = 1-5x ` or ` - 11x = 179 `or ` x= ( -179)/( 11)`<br> `:. ` Enthalpy of hydration of`Na^(+)`ion `= 6x = ( -179)/( 11) xx 6 =-97.6 kcal mol^(-1)` | |
| 9. |
Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporation of `90.0g` of water at `100^(@)C`. Assume that water behaves as an ideal gas and heat of evaporation of water is `540 cal g^(-1) (R = 2.0 cal mol^(-1)K^(-1))`. |
| Answer» Total heat change<br> `DeltaH = 90.0 xx 540 = 48600cal`<br> Now, `DeltaH = DeltaU +P DeltaV`<br> Here, `DeltaV =(V_(vapour) -V_(liquid)) = V_(vapour)`<br> (Volume of liquid is negligible as compared to volume of vapor)<br> `DeltaH = DeltaU +PV_(vapour) = DeltaU +nRT`<br> `n = (90)/(18) = 5mol, R = 2 cal K^(-1) mol^(-1), T = 373 K`<br> `:. DeltaH = DeltaU +5 xx2 xx 373`<br> `DeltaH = DeltaU +3730`<br> or `DeltaU = DeltaH - 3730 =48600 - 3730 = 44870cal` | |
| 10. |
What is the criteriafor spontaneity in terms of free energy change ? |
| Answer» If`DeltaG`is negative , process is spontaneous , if `DeltaG` is positive , direct process is non-spontaneous and if `DeltaG=0`, the processis in equilibrium | |
| 11. |
Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporationof 90.0 g of water at `100^(@)C` . Assume that water vapour behave as an ideal gas and heat of evaporation of water is 540 cal `g^(-1) ` `( R = 2.0 cal mol^(-1)K^(-1))` |
| Answer» Correct Answer - `Delta H = 48600 cal, Delta U = 44870cal`<br>`DeltaH = 540 xx 90 cal = 48600 cal, DeltaU = DeltaH - Deltan_(g)RT`,<br> `90 g H_(2) O = 5 ` moles , `Delta n_(g) = 5 ` for moles of `H_(2)O(l) rarr 5` moles of `H_(2)O(g)` | |
| 12. |
Same mass of diamond and graphite ( both being carbon) are burnt in oxygen. Will be heat produced be same or different ? Why ? |
| Answer» Heat evolved will be difficult . This is because they have different crystal structure. | |
| 13. |
In an exothermic reaction, heat is evolved and system loses heat to the surrounding. For such system.A. `q_(p)` will be negativeB. `Delta_(r)H` will be negativeC. `_(p)`will be positiveD. `Delta_(r)H` will be positive |
| Answer» Correct Answer - a,b<br>`q_(p)` and `Delta_(r)H` are `-ve` for exothermic reaction. | |
| 14. |
`NH_(3)(g)+3Cl_(2)toNCl_(3)(g)+3HCl(g),DeltaH_(1)` `N_(2)(g)+3H_(2)(g)to2NH_(3)(g),DeltaH_(2)` `H_(2)(g)+Cl_(2)(g)to 2HCl(g),DeltaH_(3)` The heat of formation of `NCl_(3)`(g) in the terms of `DeltaH_(1),DeltaH_(2),DeltaH_(3)` is :A. `DeltaH_(f)=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`B. `DeltaH_(f)=DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`C. `DeltaH_(f)=DeltaH_(1)-(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`D. none of these |
| Answer» Correct Answer - b<br> | |
| 15. |
Thermodynamics mainly deals withA. interrelation of various forms of energy and their transformation from one formto another.B. energy changes in the processes which depend only on initial and final statesof the microscopic system containing a few molecules.C. how and at what rate these energy transformations are carried out.D. the system in equilibrium state or moving from one equilibrium state to another equilibrium state. |
| Answer» Correct Answer - a,d<br>Thermodynamics deals with (a) and (b)and not with (b) and(c ) . | |
| 16. |
Which of the following is not correct?A. `DeltaG` is zero for a reversible reactionB. `DeltaG` is positive for a spontaneous reactionC. `DeltaG` is negative for a spontaneous reactionD. `DeltaG` is positive for a non- spontaneous reaction |
| Answer» Correct Answer - B<br>`DeltaG` is `-ve` and not `+ve` for a spontaneous reaction. `DeltaG` is zero for reaction in equilibrium . `DeltaG` is `+ve` for a non-spontaneous reaction. | |
| 17. |
In `C_(2)H_(4)` , formation of `(C= C)` and`(C-C)` is `- 590 kJ // `mole and `- 331 kJ //`mole respectively. What is enthalpy change when ethylene polymerizes to form polythene ?A. `+259 kJ mol^(-1)`B. `+ 72 kJ mol^(-1)`C. `- 259 kJ mol^(-1)`D. `- 72 kJ mol^(-1)` |
| Answer» Correct Answer - D<br> | |
| 18. |
An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg.K and B = 2xx 10^-2 cal//kg.K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion for water = `8xx 10^4 cal//kg`, specific heat of water `=10^3 cal//kg.K`). |
| Answer» Correct Answer - D<br>Heat lost by container<br> `Q=int_(500)^(300)(A+BT)dt =-21600`m cal.<br> Heat gained by ice in melting `Q_(1)=0.1xx80000=8000` cal<br> Heat gained by water when its temp rises from `0^(@)C` to `27^(@)C`<br> `Q_(2)=0.1xx1000xx27=2700` cal<br> Total heat gained `=10700` cal<br> `21600m=10700 :.m=0.49kg` | |
| 19. |
Work is the mose of transfrence of energy. If the system involves gaseous substance and there is difference of pressure between system and surroundings, such a work is referred to as pressure - volume work `(W_(PV)= -P_(ext)DeltaV)`. It has been observed that reversible work done by the system is the maximum obtainable work. `w_(rev) gt w_(ir r)` The works of isothermal and adiabatic processes are different from each other. for isothermal reversible proces, `W_("isothermal reversible")=2.303 nRT log_(10) (V_(2)/V_(1))` `W_("adiabatic reversible")=C_(V) (T_(1)-T_(2))` A thermodynamic system goes in acyclic process as represented in the following `P-V` diagram: 1. For an ideal gas, consider only `P-V` work in going from an initial state The net work done during the complete cycle is given by the area:A. Cycle ACBDAB. `A A_(1)B_(1)BDA`C. `A A_(2)B_(2)B`D. Half of area bonded by curve. |
| Answer» Correct Answer - A | |
| 20. |
Work is the mose of transfrence of energy. If the system involves gaseous substance and there is difference of pressure between system and surroundings, such a work is referred to as pressure - volume work `(W_(PV)= -P_(ext)DeltaV)`. It has been observed that reversible work done by the system is the maximum obtainable work. `w_(rev) gt w_(ir r)` The works of isothermal and adiabatic processes are different from each other. for isothermal reversible proces, `W_("isothermal reversible")=2.303 nRT log_(10) (V_(2)/V_(1))` `W_("adiabatic reversible")=C_(V) (T_(1)-T_(2))` Calculate work done when 1 mole of an ideal gas is expended reversibly from 20 L to 40 L at a constant temperature of 300 K.A. `7.78 kJ`B. `-1.73 kJ`C. `11.73 kJ`D. `-4.87 kJ` |
| Answer» Correct Answer - B | |