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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. |
Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: `2NO_((g))+Br_(2(g))hArr2NOBr_((g))`. When `0.087 "mole"` of `NO` and `0.0437 "mole"` of `Br_(2)` are mixed in a closed container at constant temperature, `0.0518 "mole"` of `NOBr` is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine. |
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Answer» `{:(,,2NO_((g)),+,Br_(2(g)),hArr,2NOBr_((g))),(,"Initial mole",0.0870,,0.0437,,-),(,"Equilibrium",(0.087-2a),,(0.0437-a),,2a):}` Given: `2a=0.0518` `:. a=0.0259` At equilibrium, `NO=0.0870-0.0518` `=0.0352 "mole"` `:. Br_(2)=0.0437-0.0259=0.0178 "mole"` |
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| 52. |
For the reaction is equilibrium : `2NOBr_((g))hArr2NO_((g))+Br_(2(g))` If `P_(Br_(2))` is `(P)/(9)` at equilibrium and `P` is total pressure, prove that `(K_(p))/(P)` is equal to `(1)/(81)`.A. `1//9`B. `1//81`C. `1//27`D. `1//3` |
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Answer» `K_(p)=((P_(NO))^(2)(P_(Br)))/((P_(NOBr))^(2))`, `underset(P-(3P)/(9)=(6P)/(9))(2NOBr)hArrunderset((2P)/(9))(2NO)+underset((P)/(9))(Br_(2))` Total pressure `=(6P)/(9)+(2P)/(9)+(P)/(9)=P` `K_(P)=(((2P)/(9))^(2)xx((P)/(9)))/(((6P)/(9))^(2))` `K_(p)=(P)/(81)` |
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| 53. |
For the reaction is equilibrium : `2NOBr_((g))hArr2NO_((g))+Br_(2(g))` If `P_(Br_(2))` is `(P)/(9)` at equilibrium and `P` is total pressure, prove that `(K_(p))/(P)` is equal to `(1)/(81)`. |
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Answer» `{:(,2NOBr_((g)),hArr,2NO_((g)),+,Br_(2(g))),("P at equilibrium",p-(3P)/(9)=(6P)/(9),,(2P)/(9),,(P)/(9)):}` `:. P_(T)=(6P)/(9)+(2P)/(9)+(P)/(9)=P` `K_(p)=(((2P)/(9))^(2)((P)/(9)))/(((6P)/9)^(2))=(P)/(81)` |
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| 54. |
A given sample of `N_(2)O_(4)` in a closed vessel shows `20%` dissociation in `NO_(2)` at `27^(@)C` and `760 "torr"`. The sample is now heated upto `127^(@)C` and the analysis of mixture shows `60%` dissociation at `127^(@)C`. The molecular weight of mixture at `27^(@)C` is :A. `76.66`B. `78.69`C. `66.52`D. `80.24` |
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Answer» `0.8axx92+0.4axx46=1.2axxM` at `300K` `:. M=76.66` |
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| 55. |
A given sample of `N_(2)O_(4)` in a closed vessel shows `20%` dissociation in `NO_(2)` at `27^(@)C` and `760 "torr"`. The sample is now heated upto `127^(@)C` and the analysis of mixture shows `60%` dissociation at `127^(@)C`. The density of the equilibrium mixture (in `g//litre`) at `27^(@)C` is :A. `3.1`B. `4.2`C. `2.1`D. `3.7` |
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Answer» At `300 K`, `K_(p)=K_(c)(RT)^(Deltan)` `:. K_(c)=(0.165)/(0.0821xx300)=6.7xx10^(-3)` Also `K_(c)=(((0.4a)/(V))^(2))/(((0.8a)/(V)))=6.7xx10^(-3)` `:. (a)/(V)=(6.7xx10^(-3))/(0.2)=0.0335` Also density of mixture at `300 K` `=(1.2axx76.67)/(V)` (`d=(w)/(V)`) `=1.2xx0.0335xx76.67` `=3.082 g//litre` |
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| 56. |
A given sample of `N_(2)O_(4)` in a closed vessel shows `20%` dissociation in `NO_(2)` at `27^(@)C` and `760 "torr"`. The sample is now heated upto `127^(@)C` and the analysis of mixture shows `60%` dissociation at `127^(@)C`. The total pressure of equilibrium mixture (in atm) at `127^(@)C` is :A. `1.78 atm`B. `2.01 atm`C. `3.18 atm`D. `1.33 atm` |
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Answer» At `300 K`, `underset(=0.8a)underset(a(1-0.6))underset(a)(N_(2)O_(4))hArrunderset(=0.4a)underset(2xx0.2a)underset(0)(2NO_(2))` At `400 K`, `underset(0.4a)underset(a(1-0.6))underset(a)(N_(2)O_(4))hArrunderset(1.2a)underset(0.6xx2a)underset(0)(2NO_(2))` `PV=nRT` `1xxV=1.2axxRxx300` .....(`1`) `PV=1.6axxRxx300` `:.P=1.78 atm` at `400K` |
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| 57. |
One mole of `H_(2)` and one mole of `CO` are taken in a `10 litre` vessel and heated to `725 K`. At equilibrium, `40 per cent` of water (by mass) reacts with carbon monoxide according to the equation, `H_(2)O_((g))+CO_((g))hArrH_(2(g))+CO_(2(g))` Calculate the equilibrium constant for the reaction. |
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Answer» `{:(,H_(2)O_((g)),+,CO_((g)),hArr,H_(2(g)),+,CO_(2(g))),("Initial moles",1,,1,,0,,0), ("Mole at eqm.",(1-.04),,(1-0.4),,0.4,,0.4):}` (`40%` of `H_(2)O` reacts at eqm.) `K_(c)=([CO_(2)][H_(2)])/([H_(2)O][CO])=(0.4xx0.4)/(0.6xx0.6)=0.44` |
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| 58. |
Solid `Ca(HCO_(3))_(2)` decomposes as `Ca(HCO_(3))_(2)(s)hArrCaCO_(3)(s)+CO_(2)(g)+H_(2)O(g)` If the total pressure is `0.2` bat at `420 K,` what is the standard free energy change for the given reaction `(Delta_(r)G^(@))` ?A. `840 kJ//mol`B. `3.86 kJ//mol`C. `6.98 kJ//mol`D. `16.083 kJ//mol` |
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Answer» Correct Answer - D |
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| 59. |
For the reation at `300 K` `A(g)hArrV(g)+S(g)` `Delta_(r)H^(@)=-kJ//mol, Delta_(r)S^(@)=-0.1K^(-1).mol^(-1)` What is the value of equilibrium constant ?A. `0`B. `1`C. `10`D. None of these |
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Answer» Correct Answer - B |
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| 60. |
Each question contains STATEMENT-1 (Assertion) and STATEMENT-2( Reason). Examine the statements carefully and mark the correct answer according to the instruction given below: STATEMENT-1: The catalyst does not alter the equilibrium constant. STATEMENT-2: Because for the catalysed reaction and uncatalysed reaction `DeltaH` reamains same and equilibrium constant depends of `Delta H`.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - A |
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| 61. |
Calculate the pressure for `CO_(2)` at equilibrium if `DeltaG^(@)=31.1kcal` at `300 K` for `CaCO_(3(s))hArrCaO_((s))+CO_(2(g))` |
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Answer» `CaCO_(3(g))hArrCaO_((s))+CO_(2(g))` `:. K_(p)=P_(CO_(2))` Also `-DeltaG^(@)=2.303RT log K_(p)` `-31.1xx10^(3)=2.303xx2xx300 log K_(p)` `:. K_(p)=P_(CO_(2))=3.11xx10^(-23)` |
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| 62. |
statement-1 : A reaction with `K_(P)=(1)/(1.005)atm^2` is expected to be spobtaneous at standard conditions. statement-2: Reactions with negative `Delta^circ` will be spontaneous at standard condition.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statment-2is True ,Statement-2is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A |
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| 63. |
`2NH_3(g) hArr N_2(g)+3H_2(g)` If degree of dissociation of ammonia at equilibrium is 0.7 then observed molecular weight of reaction mixture at equilibrium :A. 12B. 10C. 15D. 17 |
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Answer» Correct Answer - B |
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| 64. |
statement-1 : In dilute aqueous solution, water is present is such large excess such that its concentration remains essentially constant during any reaction involving water. statement-2 : The term `[H_(2)O]` does not appear in any equilibrium constant expression for a reaction taking place in dilute aqueous solution.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statment-2is True ,Statement-2is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - B |
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| 65. |
For the following isomerisation reaction `ci`s-butene-`2hArrtrans-`butene-2 `K_(p)=732` Which of the following statement is true at point `A`?A. `Qgt(p)`B. `QltK_(p)`C. `Q=K=1`D. `Q=K=1.732` |
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Answer» Correct Answer - D AT point `A, Q=tan60^(@)C` `Q=1.732` `therefore Q=K=1.732` |
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| 66. |
The value of `DeltaG^(@)`for the phosphorylation of glucose in glycolysis is `15KJ//mol`. Find the value of `K_(eq) "at" 300 K`.A. `e^(6)`B. `._(10)-(6)/(2.303)`C. `(1)/(e^(-6))`D. `._(10)(2.303)/(6)` |
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Answer» Correct Answer - B `DeltaG^(@)=-RT"in"K_(eq) InK_(eq)=-6` `15000=-(25)/(3)xx300"In"K_(eq) K_(eq)=e^(-6)` ltbr. `InK_(eq)=-(15000)/(2500)=6` |
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| 67. |
statement-1 : A net reaction can occur only if a system is not a equilibrium . statement-2 : All reversible reactions occur to reach a state of equilibrium.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statment-2is True ,Statement-2is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A |
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| 68. |
Which of the following statements is correct for a reversible process in a state of equilibrium ?A. `DeltaG=2.30RTlogK`B. `DeltaG^(@)=-2.30RTlogK`C. `DeltaG^(@)=2.30RTlogK`D. `DeltaG=-2.30RTlogK` |
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Answer» Correct Answer - B `DeltaG^(@)=-2.303RTlogK` |
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| 69. |
Which of the following statements is correct for a reversible process in a state of equilibrium ?A. `Delta G^(@) = - 2.30` RT log KB. `Delta G^(@) = - 2.30` RT log KC. `Delta G = - 2.30` RT log KD. `Delta G = 2.30` RT log K |
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Answer» Correct Answer - A `Delta G^(@) = - 2.30` RT log K because at equilibrium `Delta G = 0` |
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| 70. |
For the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)K_(c) = 66.9` at `350^(@)C` and `K_(c) = 50.0` at `448^(@)C` . The reaction hasA. `Delta H = + ve`B. `Delta H = - ve`C. `Delta H = zero`D. `Delta H =` Not found the signs |
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Answer» Correct Answer - B K decreases as increase in temp. Hence exothermic `Delta H lt 0` |
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| 71. |
If the value of equilibrium constant for a particular reaction is`1.6xx10^(12)` , then art equilibrium the system will containA. mostly productsB. similar amounts of reactions and productsC. all reactionsD. mostly reactants |
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Answer» Correct Answer - A `K = 1.6xx10^(12) = ("Product")/("Reactant")` The value of K is high which means reaction proceeds almost to completion, i.e., the system will contain mostly products. |
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| 72. |
In the reaction, `H_(2)(g)+I_(2)(g)hArr2HI(g)` The concentration of `H_(2), I_(2)`, and `HI` at equilibrium are `8.0, 3.0` and `28.0` mol per `L` respectively. Determine the equilibrium constant. |
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Answer» `H_(2)(g)+I_(2)(g)hArr2HI(g)` Applying the law of mass action `K_(c )=([HI]^(2))/([H_(2)][I_(2)])` Given `[H_(2)]=8.0 "mol" L^(-1)` `[I_(2)]=3.0 "mol" L^(-1)` `[HI]=2.8 "mol" L^(-1)` So, `K_(c )=((28.0)^(2))/((8.0)xx(3.0))=32.66` |
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| 73. |
For the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)` at 721 K the value of equilibrium constant `(K_(c))` is 50. When the equilibrium concentration of both is `0.5` M, the value of `K_(p)` under the same condtions will beA. 40B. 60C. 50D. 30 |
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Answer» Correct Answer - C The equilibrium constant does not change when concentration of reactant is changed as the concentration of product also get changed accordingly. |
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| 74. |
For the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)K_(c) = 66.9` at `350^(@)C` and `K_(c) = 50.0` at `448^(@)C` . The reaction hasA. `DeltaH= +ve`B. `DeltaH=-ve`C. `DeltaH=zero`D. `DeltaH` sign can not be determined |
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Answer» Correct Answer - B |
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| 75. |
For the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)` at 721 K the value of equilibrium constant `(K_(c))` is 50. When the equilibrium concentration of both is `0.5` M, the value of `K_(p)` under the same condtions will beA. `0.002`B. `0.2`C. `50.0`D. `50//RT` |
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Answer» Correct Answer - C For the reaction `H_(2) + I_(2)hArr 2HI` `Delta n = 0` So `K_(p) = K_(c) :. 50.0` |
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| 76. |
a. For which of the following reactions, `K_(p)` is equal to `K_(c)` ? i. `H_(2)+I_(2)hArr2HI` ii. `N_(2)+3H_(2)hArr2NH_(3)` iii. `PCl_(5)hArrPCl_(3)+Cl_(2)` b. For which of the following cases does the reaction go garthest to completion: `K=1, K=10^(10), K=10^(-10)`? |
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Answer» `H_(2)+I_(2)hArr2HI` only `Deltan=2-2=0` i.e., equation (iii) reduces to `K_(p)=K_(c )`. b. The reaction having `K=10^(10)` will go fathest to complection because the ratio [product]`//`[reactant] is maximum in this case [Eqn. (1)]. |
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| 77. |
For the reaction `H_(2)(g)+I_(2)(g)hArr2HI(g)` the equilibrium constant `K_(p)` changes withA. total pressureB. catalystC. concentration of`H_(2)and I_(2)`D. temperature |
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Answer» Correct Answer - d |
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| 78. |
Consider the reaction :- `2CO(g)+2H_(2)O_((g))hArr2CO_(2(g))+2H_(2(g))eq.const=K_(1)` `CH_(4(g))+H_(2)O_((g))hArrCO_((g))+3H_(2(g)),eq.const=K_(2)` `CH_(4(g))+2H_(2)O_((g))hArrCO_(2(g))+4H_(2(g)),eq.const=K_(3)` Which of the following ralation is correct ?A. `K_(3)=(K_(1))/(K_(2))`B. `K_(3)=(K_(1)^(2))/(K_(2)^(2))`C. `K_(3)=K_(1)K_(2)`D. `K_(3)=sqrt(K_(1)).K_(2)` |
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Answer» Correct Answer - d |
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| 79. |
Which of the following statements is//are correct:A. At equilibrium, vapour pressure of solution and refractive index of eq. mixture becomes constant.B. Equilibrium can be attained in both homogenous and heterogenous reaction.C. Approach to the equilibrium is fast in initial state but gradually it decreases.D. Equilibrium is dynamic in nature |
| Answer» Correct Answer - A::B::C::D | |
| 80. |
A mixture of `2` moles of `CH_(4)` and `34 g` of `H_(2)S` was placed in an evacuated chamber, which was then heated to an maintained at `727^(@)C`. When equilibrium was established in the gaseous phase reaction: `CH_(4)+2H_(2)ShArrCS_(2)+4H_(2)`, the total pressure in the container was `0.92 atm` and the partial pressure of hydrogen was `0.2 atm`. Calculate the volume of container. |
| Answer» Correct Answer - `300.38 litre` , | |
| 81. |
`64 gm` of `CH_(4) "and" 68gm "of" H_(2)S` was placed in an close container and heated up to `727^(@)C` following equilibrium is established in gaseous phase reaction is: `CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` The total pressure at equilibrium is `1.6` atm and partial pressure of `H_(2) "is" 0.8` atm. ThenA. Total moles at equilibrium `4.8`B. `K_(P)=K_(C)(RT)^(2)`C. Mole fraction `H_(2)` at equilibrium`=0.5`D. On increasing moles of `H_(2)S` equilibrium constant increases. |
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Answer» Correct Answer - A::B::C::D `CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` `{:(t=0,64gm,68gm),("Moles",4"mole",2"mole"):}` `4-x 2-2x x 4x` Total moles at equilibrium`=4-x+2-2x+x+4x=(6-2x)` Acc. To `PV=nRT,T=727+273=1000K` `1.84xxV=(6-2x)xxRxx1000 `....(1) For `H_(2) "gas" V,P_(H_(2))=n_(H_(2))RTto0.8xxV=4xxRxx1000` ....(2) From eq (1) and(2) `(1.6)/(0.8)=((6-2x))/(4x)` `x=0.6` |
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| 82. |
For a reaction `A+BhArr2C` the equilibrium concentration of `(A)` and `(B)` are `20` mole//L when volume is doubled the new equilibrium concentration of `(A)` was found to be `15` mol//L thenA. Ratio of concentration of `A "and" B` at new equilibrium is 3//4B. Value of equilibrium constant for both cases are remain sameC. Concentration of `C` at new equilibrium become halfD. Equilibrium concentration of `C` at new equilibrium `(10sqrt20)/(sqrt20-sqrt15)` |
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Answer» Correct Answer - A::B::D `{:(,A,+,2B,hArr,2C),("at vol=v lit",20,,20,,a),("at vol=2v lit",(20)/(2),,(20)/(2),,(a)/(2)):}` at new equilibrium `((20)/(2)+(x)/(2))((20)/(2)+x)((a)/(2)-x)` given that `(A)_(new)=15=((20)/(2)+(x)/(2)rArrx=(15-10)xxL` `x=101` for Ist equilibrium `K_(C)=((a)/(2))^(2)/(((20)/(2)(20)/(2))^(2))=((a)/(2))^(2)/(10+100)=(((a)/(2))^(2)/(1000))` for IInd equilibrium `K_(C)=((a)/(2)-10)^(2)/((20/(2)+(x)/(2))((20)/(2)-x)^(2))=((a)/(2)-10)^(2)/(15xx25)=((a)/(2)-10)^(2)/(275)` `(K_(C) "for" I=K_(C) "for II")` `((a)/(2))^(2)/(1000)=((a)/(2)-10)^(2)/(375)` `((a)/(2))/(31.62)=((a)/(2)-10)/(19.365)` `19.365xx(a)/(2)=31.62xx(a)/(2)-31.62xx10` `9.6825xx0=15.81xx0-316.2` `6.1275xxa=316.2` `a=51.60M` `K_(C)=(C)^(2)/((20)(20)^(2))=(51.6xx51.6)/(20xx20xx20)=332.82xx10^(-3)` `K_(C)=0.333` approxly. |
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| 83. |
In which of the follolwing, the reaction proceeds towards completionA. `K = 10^(3)`B. `K = 10^(-2)`C. `K = 10`D. `K = 1` |
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Answer» Correct Answer - A Those reactions which have more value of K proceeds towards completion. |
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| 84. |
In line kilns, the following reaction, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` proceeds to completion because ofA. of high temperatureB. `CO_(2)` escapes outC. `CaO` is removedD. of low temperature |
| Answer» Removal of product always favours forward reaction. | |
| 85. |
In which of the following reaction, the value of `K_(p)` will be equal to `K_(c)` ?A. `H_(2) + I_(2) hArr 2Hl`B. `PCI_(5) hArrPCI_(3) + CI_(2)`C. `2NH_(3)hArrN_(2) + 3H_(2)`D. `2SO_(2) + O_(2)hArr2SO_(3)` |
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Answer» Correct Answer - A `n_(p) = n_(r)` then `K_(p) = K_(c)` where `n_(p)` = number of moles of product `n_(r)` = number of moles of reactant |
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| 86. |
A change in the free energy of a system at constant temperature and pressure will be: `Delta_(sys)G = Delta_(sys)H -T Delta_(sys)S` At constant temperature and pressure `Delta_(sys) G lt 0` (spontaneous) `Delta_(sys)G = 0` (equilibrium) `Delta_(sys)G gt 0` (non-spontaneous) For a system in equilibrium, `DeltaG = 0`, under conditions of constantA. temperature and pressureB. temperature and volumeC. energy and volumeD. pressure and volume |
| Answer» `(DeltaG)_(T.P.)=0`, at equilibrium state. | |
| 87. |
For homogeneous gas reaction `4NH_(3) + 5O_(2)hArr4NO + 6H_(2)O`. The equilibrium constant `K_(c)` has the unit ofA. `("concentration")^(1)`B. `("concentration")^(-1)`C. `("concentration")^(9)`D. `("concentration")^(10)` |
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Answer» Correct Answer - A `4NH_(3)(g) + 5O_(2)(g)hArr4NO(g) + 6H_(2)O(g)` at equilibrium x x x x `K_(c) =((x)^(4)xx(x)^(6))/((x)^(4)xx(x)^(5))(x) = "concentration"` `K_(c) =(x)` |
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| 88. |
The `K_(p)//K_(c)` ratio for the reaction: `4NH_(3)(g)+7O_(2)(g)hArr4NO(g)+6H_(2)O(g)`, at,`127^(@)C` isA. `0.0301`B. `0.0831`C. `1.0001`D. `33.26 |
| Answer» Correct Answer - A | |
| 89. |
One "mole" of `N_(2)` is mixed with three moles of `H_(2)` in a `4L` vessel. If `0.25% N_(2)` is coverted into `NH_(3)` by the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`, calculate `K_(c)`. Also report `K_(c)` for `1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)` |
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Answer» `{:(,N_(2)(g),+,3H_(2)(g),hArr,2NH_(3)(g)),("moles at t=0",1,,3,,0),("moles at equilibrium",(1-x),,(3-3x),,2x):}` `:. K_(c )=((2x//V)^(2))/((1-x//V)(3-3x//V)^(3))=(4x^(2)V^(2))/((1-x)(3-3x)) ...(i)` Now given `x=0.25//100 and V=4 L` `:.` By equation (i), `K_(c )=1.49xx10^(-5) L^(2) "mol"^(-2)` Now for `1//2 N_(2)+3//2 H_(2) hArr NH_(3)` `K_(c_(1))=([NH_(3)])/([N_(2)]^(1//2)[H_(2)]^(3//2))` `:. K_(c_(1))=sqrt((K_(c )))=3.86xx10^(-3) L "mol"^(-1)` |
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| 90. |
For the equilibrium at `298 K, N_(2)O_(4)(g) hArr 2NO_(2)(g), G_(N_(2)O_(4))^(ɵ)=100 kJ mol^(-1)` and `G_(NO_(2))^(ɵ)=50 kJ mol^(-1)`. If `5` mol of `N_(2)O_(4)` and `2` moles of `NO_(2)` are taken initially in one litre container than which statement are correct.A. reaction proceeds in forward directionB. `K_(c)=1`C. `DeltaG=-0.55 KJ, DeltaG^(ɵ)=0`D. At equilibrium `[N_(2)O_(4)]=4.84 M` and `[NO_(2)]=0.212 M` |
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Answer» Correct Answer - A::B::C::D `DeltaG=DeltaG^(ɵ)+2.303 RT log Q` `DeltaG=2xxG_(NO_(2))^(ɵ)-G_(N_(2)O_(4))^(ɵ)=2xx50-100=0` `:. DeltaG=0+2.303xx8.314xx10^(-3)xx298 log 22/5` `=0-0.55 kJ` `:. DeltaG=-0.55 kJ`, i.e., reaction proceed in forward direction Also `DeltaG^(ɵ)=0=2.303 RT log K :. K=1` `{:("Now,",N_(2)O_(4),=,2NO_(2)),(,5,,2),(,5-x,,2+2x):}` `:. K_(p)=((P_(NO_(2))))/((P_(N_(2)O_(4))))=1=((2+2x)^(2))/(5-x)` or `x=0.106` |
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| 91. |
Calculate the equilibrium concentration ratio of C to A if equimolar ratio of A and B were allowed to come to equilibrium at 300K. `A(g)+B(g) iff C(g) +D(g) , DeltaG^(@)=`-830 cal. |
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Answer» Correct Answer - 2 |
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| 92. |
Which one is reversible process?A. Melting of ice at `10^(@)C`B. Mixing of two gasesC. Evaporation of water at `100^(@)C` and `1 atm` pressureD. none of these |
| Answer» At `100^(@)C` and `1 atm`, `WaterhArrVapour` | |
| 93. |
Assertion (A) : For `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`, if more `Cl_(2)` is added the equilibrium will shift in backward direction. Hence, equilibrium constant will decrease. Reason (R) : Addition of inert gas to the equilibrium mixture at constant volume does not alter the equilibrium.A. If both (A) and (R) are correct, and (R) is the correct explanation for (A)B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A)C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct. |
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Answer» Correct Answer - C Addition of inert gas shifts the equilibrium to backward direction. |
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| 94. |
`COG_(2(g))` in presence of catalyst at `1000^(@)C` shows the equilibrium: `2COF_(2(g))hArrCO_(2(g))+CF_(4(g))` At equilibrium `500 mL` of the equilibrium mixture at `STP` contains `300 mL` of (`COF_(2)` and `CO_(2)`) at `STP`. If total pressure is `10 atm`, calculate `K_(p)`. |
| Answer» Correct Answer - `4` , | |
| 95. |
In which of the following reactions is `K_(p)ltK_(c)`?A. `CO_(g)+CI_(2)(g)hArrCOCI_(2)(g)`B. `CH_(4)(g)+H_(2)O(g)hArrCO(g)+3H_(2)(g)`C. `2BrCI(g)hArrCI_(2)(g)+Br_(2)(g)`D. `I_(2)(g)hArr2I(g)` |
| Answer» Correct Answer - A | |
| 96. |
A mixture of 3 moles of `SO_(2)`, 4 moles of `NO_(2)`, 1 mole of `SO_(3)` and 4 moles of NO is placed in a 2.0L vessel. `SO_(2)(g)+NO_(2)(g) iff SO_(3)(g)+NO(g)`. At equilibrium, the vessel is found to contain 1 mole of `SO_(2)`. Calculate the value of `K_(C)`. |
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Answer» Correct Answer - 9 |
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| 97. |
When `4 mol` of `A` is mixed with `4 mol` of `B, 4 mol` of `C` and `D` are formed at equilibrium, according to the reaction `A+B hArr C+D` the equilibrium constant isA. `sqrt(2)`B. `2`C. `1`D. `4` |
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Answer» Correct Answer - C `A+B hArr C+D` According to the law of mass action `K=([C][D])/([A][B])=(4xx4)/(4xx4)=1` |
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| 98. |
Choose incorrect option(s) for the given reaction:A. `A(s)+B(g)hArrC(g)`: At equilibrium, if pressure is increased, no effect on equilibrium.B. `X(g)hArrY(g)+Z(g)`,If total pressure of the system is decreased at equilibrium,it will shift in forward direction.C. `H_2O(g)+CO(g)hArrH_2(g)+CO(g)`,inert gas is added at constant volume at equilibrium,so total pressure will increase and no effect on equilibrium.D. `H_2O(g)+CO(g)hArrH_2(g)+CO(g)`,inert gas is added at constant volume at equilibrium,so total pressure will increase and no effect on equilibrium. |
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Answer» Correct Answer - A::D |
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| 99. |
The rate at which a substance reacts, depends on its:A. Active massB. molecular massC. Equivalent massD. Total volume |
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Answer» Correct Answer - A Active mass |
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| 100. |
The rate at which a substance reacts, depends on its:A. Atomic weightB. Molecular weightC. Equivalent weightD. Active mass |
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Answer» Correct Answer - C According to Le-Charelier principle when concentration of reactant increases, the equilibrium shifts in favour of forward reaction. |
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