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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.
| 51. |
In the given reaction, `Na(s)toNa(g)` The enthalpy of atomisation is same as theA. enthalpy of dissociationB. enthalpy of sublimationC. enthalpy of associationD. enthalpy of vaporisation |
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Answer» Correct Answer - B `Na(s)toNa(g),Delta_(g)H^(ө)=108.4kJmol^(-1)` the enthalpy of atomisation is same as the enthalpy of sublimation. |
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| 52. |
In an irreversible process taking place at constant T and p and in which only pressure volumee work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria.A. `(dS)_(V,E) lt 0, (dG)_(T,p) lt0`B. `(dS)_(V,E) gt 0,(dG)_(T,p) lt 0`C. `(dS)_(V,E)=0,(dG)_(T,p)=0`D. `(dS)_(V,E)=0,(dG)_(T,p) gt 0` |
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Answer» Correct Answer - B It is an irreversible and spontaneous process. `(dS)_((V,S))=+ve` `(dG)_(T,p)=-ve` |
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| 53. |
Thermodynamics is not concerned about….A. energy changes involvedin a chemical reactionB. the extent to which a chemical reaction proceedsC. the rate at which a reaction proceedsD. the feasibility of a chemical reaction |
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Answer» Correct Answer - C Thermodynamics deals with the energy changes, feasibility and extent of a reaction, but not with the rate and mechanism of a process. |
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| 54. |
The second law of thermodynamics says that in a cyclic processA. work cannot be converted into heatB. heat cannot be convertd into workC. work cannot be completely converted into heatD. heat cannot be completely converted into work |
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Answer» Correct Answer - D According to second law, work can always be completely converted into heat bhut heat cannot be completely conoverted into work. |
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| 55. |
The decreasing order of bond dissociation energies of C-C,C-H and H-H bonds isA. `H-H gt -C-H gt -C-C-`B. `-C-C-gt-C-HgtH-H`C. `-C-Hgt-C-C-gtH-H`D. `-C-C-gtH-Hgt-C-H` |
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Answer» Correct Answer - A The decreasing order of bond dissociation energies of C-C, C-H and H-H bonds is as follows `H-H gt-C-H gt-C-C-` |
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| 56. |
Select the incorrect expression from the following expressions.A. `DeltaS_("total")=DeltaS_("system")+DeltaS_("surr")`B. `DeltaS_("surr")=(DeltaH_("surr"))/(T)=-(DeltaH_("sys"))/(T)`C. `DeltaS_("total") lt 0` (spontaneous process)D. `DeltaG=Delta-TDeltaS` |
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Answer» Correct Answer - C We know that, `DeltaS_("total")=DeltaS_("system")+DeltaS_("surr")` Also, entropy change of surroundings `DeltsS_("surr")=(DeltaS_("surr"))/(T)=-(DeltaH_("sys"))/(T)` For spontaneous process, `DeltaS_("total")gt0` and `DeltaG=Delta-TDeltaS` |
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| 57. |
The standard molar entropy of `H_(2)O(l)` is 70 `JK^(-1)mol^-1`. Standard molar entropy of `H_(2)O(s)` isA. more than 70 `JK^(-1)mol^(-1)`B. less than 70 `JK^(-1)mol^(-1)`C. equal to 70 `JK^(-1) mol^(-1)`D. none of these |
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Answer» Correct Answer - B Less than 70 `JK^(-1)mol^(-1)`, because ice is more ordered than `H_(2)O(l)`. |
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| 58. |
The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from 1kg of coke is (assume coke to be 100% carbon). (Given: Enthalpies of combustion of `CO_(2),CO and H_(2)` are 393.5 kJ, 285 kJ, 285 kJ respectively all at 298K)A. `0.79:1`B. `0.69:1`C. `0.86:1`D. `0.96:1` |
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Answer» Correct Answer - B Number of moles in 1 kg coke`=(1000)/(12)=83.33mol` For the combustion of 1 kg of coke `C+O_(2) to CO_(2)" "DeltaH=393.5kJ` `implies`Heat liberated from 1 mole of coke=393.5 kJ `therefore`Heat liberated from 83.33 mole of coke `H_(1)=(393.5xx83.33)kJ` Also, for the burning of water gas `C+H_(2)O to underset("Water gas")ubrace(CO+H_(2))` `CO+H_(2)+O_(2)toCO_(2)+H_(2)O,DeltaH=285+285=570kJ` `therefore`Ratio of heat liberated from burning of water gas obtained from 1 mole of coke `H_(2)=(570xx83.33)kJ` `therefore`Required ratio `=H_(1):H_(2)=393.5:570=0.69:1` |
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| 59. |
What is the equilibrium constant, K for the following reaction at 400 K? `2NOCl (g) hArr 2NO(g)+Cl_(2)(g)` `DeltaH=77.2" kJ "mol^(-1) and DeltaS=122JK^(-1)mol^(-1)` at 400K.A. `-3.708`B. `1.95xx10^(-4)`C. `2.8xx10^(4)`D. `1.67xx10^(-5)` |
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Answer» Correct Answer - B `DeltaG=DeltaH-TDeltaS` `=77.2-400xx122xx10^(-3)` `=+28.4kJ" "mol^(-1)=28400" J "mol^(-1)` Further, `DeltaG=-2.303RTlogK` or `28400=-2.303xx8.314xx400xxlogK` `logK=-(28400)/(2.303xx8.314xx400)` `logK=-3.708` `K=`Antilog (-3.708)`=1.95xx10^(-4)` |
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| 60. |
A cylinder of gas is assumed to contain 11.2 kg of butane `(C_(4)H_(10))`. If a normal family needs 20000 kJ of energy per day, the cylinder will last in (given that `DeltaH` for combustion of butane is -2658 kJ)A. 20 daysB. 25 daysC. 26 daysD. 24 days |
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Answer» Correct Answer - C Cylinder contains 11.2 kg or 193.10 mole of butane. [`because` moleclar mass of butane=58] `because`Energy released by 1 mole of butanne=-2658 kJ `therefore` Energy released by 193.10 mole of butane `=-2658xx193.10` `=5.13xx10^(5)kJ` `therefore`Cylinder will last in `(5.13xx10^(5))/(20000)=25.66` or 26 days. |
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| 61. |
How many joules of heat are absorbed when 70.0 g of water is completely vaporised at its boiling point?A. 23352 JB. 7000 JC. 15813 JD. 158200 J |
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Answer» Correct Answer - D Since, the process involves phase change, heat absorbed, Q=mass`xx`latent heat of vaporisation Given mass=70.0g=0.07 kg, `L_(V)=2260kJ//kg` `therefore Q=0.07xx2260kJ=158.2kJ=158200J` |
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| 62. |
The equilibrium constannt of a reaction is 0.008 at 298 K. the standard energy change of the reaction at the same temperature isA. `-11.96kJ`B. `-5.43kJ`C. `-8.46kJ`D. `+11.96kJ` |
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Answer» Correct Answer - D Standard free energy change, `DeltaG^(@)=-2.303RTlogK_(c)` `=-2.303xx8.314xx298xxlog` 0.008 `=+11965.16J=11.96kJ` |
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| 63. |
The gaseous mixture containing 2 moles of each of two ideal gases `A(C_(V),m=(3)/(2)R)` and `B(C_(V),m=(5)/(2)R)`. Find out the average molar heat capacity at constant volume.A. 8RB. 3RC. 2RD. R |
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Answer» Correct Answer - C Average `C_(V),m=(n_(1)C_(V_(1))m_(1)+n_(2)C_(V_(1))m_(2))/(n_(1)+n_(2))` `=(2xx(3)/(2)R+2xx(5)/(2)R)/(2+2)=2R` |
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| 64. |
`DeltaU^(@)` for combustion of methane is -X kJ `mol^(-1)`. The value of `DeltaH^(@)` isA. `=DeltaU^(@)`B. `gtDeltaU^(@)`C. `lt DeltaU^(@)`D. `=0` |
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Answer» Correct Answer - C `CH_(4)(g)+2O_(2)(g) to CO_(2)(g)+2H_(2)O(l)` `Deltan_(g)=(n_(p)-n_(r))=1-3=-2` `DeltaH^(@)=DeltaU^(@)+Deltan_(g)RT=-X-2RT` Hence, `DeltaH^(@) lt Delta U^(@)` |
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| 65. |
In a process, `701 J` of heat is absorbed by a system and `394 J` of work is done by the system. What is the change in internal energy for the process?A. `-464J`B. `+464J`C. `+307J`D. `-307J` |
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Answer» Correct Answer - C We know, internal energy change, `DeltaE=q+W` `=+701J+(-394J)=+307J`. Hence, internal energy of the system increases by 307 J. |
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| 66. |
A system absorb `600 J` of heat and work equivalent to `300 J` on its surroundings. The change in internal energyA. 300 JB. 400 JC. 500 JD. 600 J |
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Answer» Correct Answer - A `DeltaE=Q+W=600+(-30)=300J` [`because` work is done by the system] |
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| 67. |
1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C (graphite)`+O_(2)(g)rarrCO_(2)(g)` During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?A. `-2.48xx10^(2)kJ" "mol^(-1)`B. `2.48xx10^(2)"kJ "mol^(-1)`C. `-5.46xx10^(2)"kJ "mol^(-1)`D. `5.46xx10^(2)"kJ "mol^(-1)` |
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Answer» Correct Answer - A Given, in bom calorimeter, volume remains the constant, thus, the heat involved is internal energy i.e., `DeltaU`. We know that, Since, heat is lost be the system. `thereforeq_(V)=-C_(V)DeltaT=-20.7kJ//Kxx(299-298)K=-20.7kJ` (Here, negative sign indicates the exothermic nature of the reaction.) Thus, `DeltaU` for the combustion of the 1 g of graphite`=-20.7kJ" "K^(-1)` For cumbusion of 1 mole (12.0 g) of graphite `=(12.0" g "mol^(-1)xx(-20.7kJ))/(1g)=-2.48xx10^(2)" kJ "mol^(-1)` Since, `Deltan_(g)=0,DeltaH=DeltaU=-2.48xx10^(2)"kJ "mol^(-1)` |
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| 68. |
In view of the signs of `DeltaG^(ө)` for the following reactions, `PbO_(2)+Pb to 2PbO,Delta_(r)G^(ө) lt 0`, `SnO_(2)+Snto 2SnO,Delta_(r)G^(ө) gt 0` Which oxidation states are more characteristic for lead and tin?A. For lead +4, for tin +2B. For lead +2, ffor tin +2C. For lead +4, for tin +4D. For lead +2, for tin +4 |
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Answer» Correct Answer - D `overset(+4)(Pb)O_(2)+overset(0)(Pb) to 2 overset(+2)(Pb)O` Since, `Delta_(r)G^(@) lt 0`, hence +2 state of lead is favourable. `overset(+4)+(Sn)O_(2)+overset(0)(Sn) to 2 overset(0)(Sn)O` Since, `Delta_(r)G^(@)gt0` it means forward reaction is not spontaneous. `2overset(+2)(Sn)O+overset(+4)(Sn)O_(2) to 2overset(0)(S)n` `Delta_(r)G^(@) lt0`, thus +4 state of tin is favourable. |
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| 69. |
One mole of which o the following has the highest entropy?A. Liquid nitrogenB. Hydrogen gasC. MercuryD. Diamond |
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Answer» Correct Answer - B Greater the disorder in a system, the higher is the entropy. Hence, entropy is highest for hydrogen gas. |
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| 70. |
The heat of neutralisation is highest for the reaction betweenA. `NH_(4)OH-CH_(3)COOH`B. `HNO_(3)-NH_(4)OH`C. `NaOH-CH_(3)COOH`D. `HCl-NaOH` |
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Answer» Correct Answer - D Heat of neutralisation of strong acid by strong base is always greater than the heat of neutralisation of weak acid or weak base because some energy is used to ionise weak acid or weak base. (a) `underset("weak base-weak acid")(NH_(4)OH-CH_(3)COOH)` (b) `underset("Strong acid-weak base")(HNO_(3)-NH_(4)OH)` (c) `underset("strong base-weak acid")(NaOH-CH_(3)COOH)` (d) `underset("strong acid-strong base")(HCl-NaOH)` `because`HCl is a strong acid and NaOH is a strong base. `therefore` Their heat of neutralisation will be highest among the given. |
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| 71. |
Calculate the difference between `DeltaE and DeltaH` for the following reaction at `27^(@)C` (in kcal) `C("graphite")+2H_(2)(g)toCH_(4)(g)`A. `-0.6`B. `-1.2`C. `+0.6`D. `+1.2` |
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Answer» Correct Answer - C `C("graphite")+2H_(2)(g) to CH_(4)(g)` `DeltaH=DeltaE+Deltan_(g)RT` `DeltaE-DeltaH=-Deltan_(g)RT` `DeltaE-DeltaH=-(-1)xx2xx300` `=600cal=0.6kcal` |
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| 72. |
The enthalpy of vaporisation of liquid water using the data `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.77kJ//mol` `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g),DeltaH=-241.84kJ//mol` isA. `+43.93kJ//mol`B. `-43.93kJ//mol`C. `527.61kJ//mol`D. `-527.61kJ//mol` |
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Answer» Correct Answer - A (I) `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.77kJ//mol` (II) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-241.84kJ//mol` `H_(2)O(l)toH_(2)O(g),dDeltaH=?` Subtracting the Eqs. (I) from (II), we get `H_(2)O(l)toH_(2)O(g)` `DeltaH=-241.84-(-285.77)` `DeltaH=+43.93kJ//mol` |
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| 73. |
The `DeltaH_(f)^(@)` of `O_(3),CO_(3),NH_(3) and HI` are 142.2, -393.3, -46.2 and +25.9 kJ per mol, respectively. The order of their increasing stabilities will beA. `O_(3),CO_(2),NH_(3),HI`B. `CO_(2),NH_(3),HI,O_(3)`C. `O_(3),HI,NH_(3),O_(2)`D. `NH_(3),HI,CO_(2),O_(3)` |
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Answer» Correct Answer - C Energy absorbed `prop` stability of compound Energy released `prop` stability of compound `142.2 gt 25.9 gt -46.2 gt -393.2` i.e. `O_(3) gt HI gt NH_(3) gt CO_(2)` |
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| 74. |
The enthalpy of reaction, `H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(g)` is `DeltaH_(1)` and that of `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)` is `DeltaH_(2)`. ThenA. `DeltaH_(1) lt DeltaH_(2)`B. `DeltaH_(1)+DeltaH_(2)=0`C. `DeltaH_(1) gt DeltaH_(2)`D. `DeltaH_(1)=DeltaH_(2)` |
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Answer» Correct Answer - A The enthalpy of `H_(2)O(l)` is less than that of `H_(2)O(g)`, hence, more energy will be released when `H_(2)O(l)` is formed, therefore `DeltaH_(1) lt DeltaH_(2)`. |
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| 75. |
Formation of a solution from two components can be considered as (1). pure solvent`to`separated solvennt molecules, `DeltaH_(1)` (2). Pure solute `to`separated solute molecules, `DeltaH_(2)` (3). Separated solvent and solute molecules`to`solution, `DeltaH_(3)` Solution so formed will be ideal ifA. `DeltaH_("soln")=DeltaH_(1)-DeltaH_(2)-DeltaH_(3)`B. `DeltaH_("soln")=DeltaH_(3)-DeltaH_(1)-DeltaH_(2)`C. `DeltaH_("soln")=DeltaH_(1)+DeltaH_(2)+DeltaH_(3)`D. `DeltaH_("soln")=DeltaH_(1)+DeltaH_(2)+DeltaH-DeltaH_(3)` |
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Answer» Correct Answer - C For ideal solution, `DeltaH_("solution")=DeltaH_(1)+DeltaH_(2)+DeltaH_(3)` |
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| 76. |
The heat energy required to ionise the following molecules is given as follows `N_(2)(g) overset(DeltaH_(1))to H_(2)^(o+)(g),O_(2)(g) overset(DeltaH_(2))to O_(2)^(+)(g)` `Li_(2)(g) overset(DeltaH_(3)) to Li_(2)^(o+)(g),C_(2)(g) overset(DeltaH_(4))to C_(2)^(o+)(g)` The correct decreasing order of energy in terms of heatA. `DeltaH_(1) gt DeltaH_(3) gt DeltaH_(2) gt DeltaH_(4)`B. `DeltaH_(2) gt DeltaH_(3) gt DeltaH_(1) gt DeltaH_(4)`C. `DeltaH_(3) gt DeltaH_(4)gt DeltaH_(1) gt DeltaH_(2)`D. `DeltaH_(3) gt DeltaH_(1) gt DeltaH_(4) gt DeltaH_(2)` |
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Answer» Correct Answer - C Enthalpy of ionisation depends upon the size of molecules. Larger the size lesser will be ionisation energy. |
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| 77. |
According to second law of thermodynamics, a process (reaction) is spontaneous, if during the processA. `DeltaS_("universe")gt0`B. `DeltaS_("universe")=0`C. `DeltaH_("system")=0`D. `DeltaS_("universe")=DeltaS_("system")` |
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Answer» Correct Answer - A When a spontaneous (naturallyh occuring) process takes place, it is accompanied by increase in entropy, i.e., `DeltaS_("universe")=DeltaS_("system")+DeltaS_("surroundings")gt0`. |
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| 78. |
Calculate `DeltaG^(Theta)` for the conversion of oxygen to ozone, `((3)/(2)) O_(2)(g) hArr O_(3)(g) at 298 K`, of `K_(p)` for this conversion is `2.47 xx 10^(-29)`.A. `1.63kJmol^(-1)`B. `164kJmol^(-1)`C. `16.3kJmol^(-1)`D. `1630kJmol^(-1)` |
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Answer» Correct Answer - B Given, temperature, `T=298K and K_(p)=2.47xx10^(-29)` We know, that `Delta_(r)G^(ө)=-2.303RTlogK_p` `thereforeDelta_(r)G^(ө)=-2.303(8.314"JK"^(-1)mol^(-1))xx(298K)(log2.47xx10^(-29))` `=163000" J "mol^(-1)=163" kJ "mol^(-1)` |
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