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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.
| 301. |
Which of the following reactions will not be affected by increasing the pressure?A. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`B. `N_(2)(g)+O_(2)(g) hArr 2NO(g)`C. `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)`D. `CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)` |
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Answer» Correct Answer - B::D a. `Deltan=1+1-1=1` b. `Deltan=2-(1+1)=0` c. `Deltan=1-0=1` d. `Deltan=1+1-(1+1)=0` `Deltan` is zero in (b) and (d), so (b) and (d) are not efected by pressure. |
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| 302. |
The equilibrium constant of a reaction is `1xx10^(20)` at `300 K`. The standard free energy change for this reaction is :A. `-115kJ`B. `+115 kJ`C. `+166 kJ`D. `-166 kJ` |
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Answer» `DeltaG^(@)=-2.303RT log K_(p)` `=-2.303xx8.314xx300xxlog10^(20)J` `=-114.88 kJ` |
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| 303. |
For the equilibrium `AB(g) hArr A(g)+B(g)` at a given temperature, the pressure at which one-third of AB is dissociated is numerically equal toA. `8` times `K_(p)`B. `16` times `K_(p)`C. `4` times `K_(p)`D. `9` times `K_(p)` |
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Answer» Correct Answer - A `{:(,AB(g),hArr,A(g),+,B(g)),(underset(("moles"))(t=0),a,,0,,0),(t=t_("eq"),a-x,,x,,x):}` `rArr K_(p)=(p_(A).p_(B))/(p_(AB))=((x/(a+x)P)(x/(a+x)P))/(((a-x)/(a+x)P))` `rArr K_(p)=(x^(2)P)/(a^(2)-x^(2))` and `alpha=x/a=1/3rArr P=8 K_(p)` |
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| 304. |
In the reaction `A+B hArr AB`, if the concentration of A and B is increased by a factor of `2`, it will cause the equilibrium concentration of AB to change toA. Two times to original valueB. Three times to original valueC. SameD. Zero |
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Answer» Correct Answer - A For `A+B hArr AB` `K=([AB])/([A][B])` When `2A+2B hArr 2AB` `K=2AB` |
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| 305. |
Manufacture of ammonia from the elements is represented by `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)+22.4 kcal` The maximum yield of ammonia will be obtained when the process is made to take placeA. At low pressure and high temperatureB. At low pressure and low temperatureC. At high pressure and high temperatureD. At high pressure and low temperature |
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Answer» Correct Answer - D Because the reaction is exothermic and on product side the number of moles of moles of substance is less. |
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| 306. |
Select the incorrect statements:A. `K_(p)` or `K_(c)` are dimenensionless if pressure or concentration are expressed in standard state.B. The neumerical value of `K_(p)` changes with experimental conditions, i.e., P, T, and C at which equilibrium is attained.C. Active mass of reactant = concentration of reactantD. Dissolution of `NH_(3)` in water increases with increasing pressure. |
| Answer» Correct Answer - A::B::C::D | |
| 307. |
`K_(p)` and `K_(c)` are inter related as `K_(p)=K_(c)(RT)^(Deltan)` Answer the following questions: In which of the following equilibria `K_(p)` is less than `K_(c)`?A. `H_(2) + I_(2) hArr2HI`B. `N_(2) + 3H_(2)hArr2NH_(3)`C. `N_(2) + O_(2)hArr2NO`D. `CO + H_(2)OhArrCO_(2) + H_(2)` |
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Answer» Correct Answer - B `K_(p) = K_(c)(RT)^(Delta n)`, When `Delta n =2 - (2 + 1) = - 1`, i.e., negative, `K_(p) lt K_(c)`. |
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| 308. |
Derive the best conditions for the formation of `NH_(3)`. Given, `2NH_(3)hArrN_(2)+3H_(2)`, (`Delta H= +91.94 kJ)` |
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Answer» High pressure favours reactions showing decrease in moles i.e., forward reaction. Low temperature favours reaction showing exothermic nature i.e., forward reaction. More concentration of `N_(2)` and `H_(2)` also favours forward reaction. |
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| 309. |
`K_(p)` for the following reaction at 700 K is `1.3xx10^(-3) atm^(-1)` . The `K_(c)` at same temperature for the reaction `2SO_(2) + O_(2) rarr 2SO_(3)` will beA. `1.1xx10^(-2)`B. `3.1xx10^(-2)`C. `5.2xx10^(-2)`D. `7.4xx10^(-2)` |
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Answer» Correct Answer - D `K_(p) = K_(c)(RT)^(Delta n)` R = Gas constant `K_(c) = (K_(p))/((RT)^(Delta n)) = (1.3xx10^(-3))/((0.0821xx700)^(-1)) = 7.4xx10^(-2)` |
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| 310. |
The equilibrium constant expression for the equilibrium `2NH_(3)(g) + 2O_(2)(g)hArrN_(2)O(g) + 3H_(2)O(g)` isA. `K_(C) =([N_(2)O][H_(2)O]^(3))/([NH_(3)][O_(2)])`B. `K_(C) =([H_(2)O]^(3) [N_(2)O])/([NH_(3)]^(2) [O_(2)]^(2))`C. `K_(C) =([NH_(3)]^(2))/([N_(2)O][H_(2)O]^(3))`D. `K_(C) =([NH_(3)][O_(2)])/([N_(2)O][H_(2)O])` |
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Answer» Correct Answer - B `2NH_(3)(g) + 2O_(2)(g)hArrN_(2)O(g) + 3H_(2)O(g)` `K_(C) = ((N_(2)O)xx(H_(2)O)^(3))/((NH_(3))^(2)xx(O_(2))^(2))` |
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| 311. |
A reaction `CaF_(2)hArrCa^(2+)+2F^(-)` is at equilibrium. If the concentration of `Ca^(2+)` is increased four times, what will be the change in `F^(-)` concentration as compared to the initial concentration of `F^(-)`?A. `1//4` of the initial valueB. `1//2` of the initial valueC. `2` times of the initial valueD. none of these |
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Answer» `K_(c)=[Ca^(2+)][F^(-)]^(2)`, If `[Ca^(2+)]=4xx[Ca^(2+)]`, To have `K_(c)` constant `[F^(-)]` should be `([F^(-)])/(2)`. |
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| 312. |
For the reaction `2A(g)+B(g)hArr C(g)+D(g), K_c=10^(12)`.if initially 4,2,6,2 moles of A,B,C,D respectively are taken in a 1 litre vessel, then the equilibrium concentration of A is :A. `4xx10^(-4)`B. `2xx10^(-4)`C. `10^(-4)`D. `8xx10^(-4)` |
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Answer» Correct Answer - A |
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| 313. |
28 g of `N_(2)` and 6 g of `H_(2)` were mixed. At equilibrium 17 g `NH_(3)` was produced. The weight of `N_(2)` and `H_(2)` at equilibrium are respectivelyA. 11 g, 0 gB. 1 g, 3 gC. 14 g, 3 gD. 11 g, 3 g |
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Answer» Correct Answer - C `{:(,N_(2),+,3H_(2),rarr,2NH_(3)),(t=0,,,,,),(Initial,(28)/(28),,(6)/(2),,0),(,1,,3,,0),(,"mole",,"mole",,"mole"),("At equilibrium",,,,,),(t=?,1-x,,3-3x,,2x),(,2x=(17)/(17),,2x=1,,x=1//2),("Final",1//2,,3//2,,1),("mole",,,,,),("weight"-14g,,3g,,,):}` |
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| 314. |
28 g of `N_(2)` and 6 g of `H_(2)` were mixed. At equilibrium 17 g `NH_(3)` was produced. The weight of `N_(2)` and `H_(2)` at equilibrium are respectivelyA. `11 g`, zeroB. `1g`, `3`C. `14g`, `3g`D. `11g`, `3g` |
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Answer» `{:(,,N_(2),+,3H_(2),hArr,2NH_(3)),(,,28/28=1,,6/2=3,,0"mole before reaction"),(,,1-1/2,,3-3/2,,17/17=1"mole after reaction"):}` `:. "Mole of" N_(2)=(1)/(2)` `Wt. of N_(2)=14 g` Mole of `H_(2)=(3)/(2)` Wt. of `H_(2)=(3)/(2)xx2=3g` |
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| 315. |
An equilibrium system for the reaction between hydrogen and iodien to give hydrogen iodide at 765 K in a 5 litre volume contains `0.4` mole of hydrogen iodide. The equilibrium constant for the reaction `H_(2) + I_(2)hArr 2HI` isA. `36.0`B. `15.0`C. `0.067`D. `0.028` |
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Answer» Correct Answer - A `H_(2) + I_(2)hArr2HI` At equilibrium `(0.4)/(5) (0.4)/(5) (2.4)/(5)` `K_(c) = (((2.4)/(5))^(2))/(((0.4)/(5))((0.4)/(5)))` `K_(c) =(2.4xx2.4)/(0.4xx0.4)` `K_(c) = (5.76)/(0.16)` `K_(c) = 36` |
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| 316. |
In the preceding problem, if `[A^+] and [AB_(2)^(-)]` are y and x respectively under equilibrium produced by addings the substance AB to the solvents than `K_(1)//K_(2)` is equal to :A. `(y)/(x)(y-x)^(2)`B. `((y)^2(x+y))/(x)`C. `(y^(2)(x+y)^2)/(x)`D. `(y)/(x)(x-y)` |
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Answer» Correct Answer - A |
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| 317. |
Consider the equilibrium reaction: `4NH_(3((g)))+3O_(2((g)))hArr2N_(2((g)))+6H_(2)O_((g))` `(DeltaH=-1268KJ)` Which change will cause the reaction to shift to the right?A. Increase the temperatureB. Decrease the volume of the container.C. Add a catalyst to speed up the reaction.D. Remove the gaseous water by allowing it to react and be absorbed by `KOH`. |
| Answer» Correct Answer - D | |
| 318. |
At Assuming complete decomposition of `NH_(3)` and `N(2)H_(4)` `P=0.3 atm, P=2.7 atm` `T=300K, T=200K` `VL, VL` mole `%` of `NH_(3)` in original mixture is (assume both concentration same volume)A. `25%`B. `20%`C. `75%`D. `37.5%` |
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Answer» Correct Answer - C Using equation `PV=nRT` `n_(1)`= moles of `NH_(3)` `n_(2)`= moles of `N_(2)H_(4)` `0.3xxV=(n_(1)+n_(2))xxRxx300` `2.7xxV=(2n_(1)+3n_(2))xxRxx1200` `(2n_(1)+3n_(2))/(n_(1)+n_(2))=(9/4)` `n_(1)/n_(2)=(1/3)` `n_(1)/n_(2)=75%` |
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| 319. |
Physical and chemical equilibrium can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium we have a principle named Le Chatelier principle. According to this principle, even if we make some changes in equilibrium, then also the system even re-establishes the equilibrium by undoing the effect. In the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`. If we increase the pressure of the system, the equilibrium isA. Shifts in the product sideB. Shift un reactant sideC. Remains unchangedD. Cannot be predicted |
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Answer» Correct Answer - A Sift in product side because on increasing the pressure of the system, the equilibrium shift to lesser number of molecules side, i.e., the product side. |
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| 320. |
Two gases X and Y, one being the dimer of other are at equilibrium. Increase of volume at constant temperature or increase of temperature at constant pressure favours the formation of more Y. The reaction could not be represented by:A. `2XhArrY+Q` caloriesB. `2XhArrY-Q` caloriesC. `2YhArrX-Q` caloriesD. `2YhArrX+Q` calories |
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Answer» Correct Answer - A::B::C |
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| 321. |
Physical and chemical equilibrium can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium we have a principle named Le Chatelier principle. According to this principle, even if we make some changes in equilibrium, then also the system even re-establishes the equilibrium by undoing the effect. Consider the following equilibrium: `2NO_(2) hArr 2NO_(3), DeltaH=-ve`, If `O_(2)` is added and volume of the reaction vessel is reduced, the equilibriumA. Shift in the product sideB. Shifts in the reactant sideC. Cannot be predictedD. Remains unchanged |
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Answer» Correct Answer - A Shift in product side. |
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| 322. |
The persentage of ammonia produced from nitrogen and hydrogen under conditions of temperature and pressure is given in the graph Use the graph answering the following questions: What happens to the percentage of ammonia produced when the pressure is increased?A. The `%` is decreasedB. The `%` is increasedC. No effectD. Cannot be predicted |
| Answer» Correct Answer - B | |
| 323. |
Select the correct statemens for following equilibrium:A. On increasing pressure, melting point decreases while boiling point of `CO_2` increases.B. On increasing pressure, melting point decreases while boiling point increases for `H_2O`C. On increasing pressure,sublimation temperature increases for both `CO_2 and H_2O`D. On increasing pressure,sublimation temperature increases for `CO_2` but decreases for `H_2O`. |
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Answer» Correct Answer - A::B::C |
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| 324. |
The triple point of `CO_(2)` occurs at `5.1` atm and `-56^(@)C`. Its critical temperature is `31^(@)C`. Solid `CO_(2)` is more dense than liquid `CO_(2)`. Under which combination of pressure and temperature is liquid `CO_(2)` stable at equilibrium?A. 10 atm and `-25^(@)C`B. `5.1` atm and `-25^(@)C`C. 10 atm `33^(@)C`D. `5.1` atm and `-100^(@)C` |
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Answer» Correct Answer - A |
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| 325. |
Which of the following is/are correct about chemical equilibrium?A. Equilibrium conditions is most stable condition under given conditonsB. Equilibrium can be achieved from both reactants as well as prducts sideC. Catalyst does not affect the equilibrium constant and equilibrium compostionD. For any given reaction equilibrium constant dopends on temperature only |
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Answer» Correct Answer - A::B::C::D |
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| 326. |
The persentage of ammonia produced from nitrogen and hydrogen under conditions of temperature and pressure is given in the graph Use the graph answering the following questions: What conditions of pressure produce the highest percentage of ammonia?A. At least `50` atmB. At least `150` atmC. At least `300` atmD. At least `100` atm |
| Answer» Correct Answer - C | |
| 327. |
Which statement about the triple point of a substance is/are incorrect?A. The triple point for a substance varies with pressureB. The three phases (solid, liquid, gas) have same densityC. The three phases (solid, liquid,gas) are in equilibriumD. The three phases(solid, liquid, gas) are indistinguishable in appearance. |
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Answer» Correct Answer - A::B::D |
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| 328. |
For which of the following reaction does the equilibrium constant depend on the units of concentrationA. `NO_((g))hArr(1)/(2)N_(2(g)) + (1)/(2)O_(2(g))`B. `Zn_((s)) + Cu_((aq))^(2+)hArrCu_((s)) + Zn_((aq))^(2+)`C. `C_(2)H_(2)OH_((l)) + CH_(3)COOH_((l))hArrCH_(3)COOC_(2)H_(5(l)) + H_(2)O_((l))` (Reaction carried in an inert solvent)D. `COCI_(2(g))hArr(CO_((g)) + CI_(2(g)` |
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Answer» Correct Answer - D `Delta n = 1` for this change So the equilibrium constant depends on the unit of concentration. |
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| 329. |
Which among the following equilibrium, `K_(P)` does not depend upon the initial pressure of ractants?A. `H_2(g)+I_2(s)hArr2HI(g)`B. `H_2(g)+CI_2(g)hArr2HCI(g)`C. `N_2(g)+O_2(g)hArr2NO(g)`D. `N_2(g)+3H_2(g)hArr2NH_3(g)` |
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Answer» Correct Answer - A::B::C::D |
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| 330. |
Select correct statements:A. Low pressure is favourable for evaporation of `H_2O(I)`.B. The degree of dissociation of `CaCO_3(s)` decreases with increase in pressure.C. If the equilibrium constant of `A_2(g)+B_2(s)` to 2AB(g) is 25, then equilibrium constant for `AB(g)to(1)/(2)(g)+(1)/(2)B_2(g)is0.2.`D. If solid product is added to an equilibrium mixture,then equilibrium will be unaffected. |
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Answer» Correct Answer - A::B::C::D |
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| 331. |
2 moles each of `SO_(3)`, CO, `SO_(2)` and `CO_(2)` is taken in a 1 L vessel. If `K_(C)` for `SO_(3) + COhArrSO_(2) + CO_(2)` is `1//9` then:A. (A) total no. of moles at equilibrium are less than `8`B. (B) `n(SO_(#))+n(CO_(2))=4`C. (C) `[n(SO_(2))//n(CO)]lt1`D. (D) both (B) and (C). |
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Answer» Correct Answer - D `SO_(3)(g)+CO(g)hArrSO_(2)(g)+CO_(2)(g)` `g=((2+x)/(2-x))^(2)` `{:(2,2,2,2,),(2+x,2+x,2-x,2-x,x=1):}` `n_(eeq)=3+3+1+1=8" " n(SO_(3))+n(CO_(2))=4` `(n(SO_(2)))/(n(CO))=(1)/(2)lt1` Therefore, (D) option is correct. |
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| 332. |
2 moles each of `SO_(3)`, CO, `SO_(2)` and `CO_(2)` is taken in a 1 L vessel. If `K_(C)` for `SO_(3) + COhArrSO_(2) + CO_(2)` is `1//9` then:A. total no. of moles at equilibrium are less than 8B. `n(SO_(3)) + n(CO_(2))=4`C. `[n(SO_(2))//n(CO)]lt 1`D. both (b) and (c). |
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Answer» Correct Answer - D `{:(,SO_(3),+CO,hArr,SO_(2),+,CO_(2)),(t=0,2M,2M,,2M,,2M),(t=oo,2+x,2+x,,2-x,,2-x):}` |
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| 333. |
For which of the following reactions `K_(p) = K_(c)`A. `2NOC1(g)hArr2NO(g) + C1_(2)(g)`B. `N_(2)(g) + 3H_(2)(g)hArr2NH_(3)(g)`C. `H_(2)(g) + C1_(2)(g)hArr2HC1(g)`D. `N_(2)O_(4)(g)hArr2NO_(2)(g)` |
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Answer» Correct Answer - C `K_(p) = K_(c)(R_(T))^(Delta n)` , `Delta n= 2-2 =0` |
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| 334. |
Two solid `A "and" B` are present in two different container having same volume and same temperature following equilibrium are established: In container `(1)A(s)hArrD(g)+C(g)P_(T)=40 "atm at equilibrium"` In container `(2)B(s)hArrE(g)+F(g)P_(T)=60 "atm at equilibrium"` If excess of `A "and" B` are added to a third container having double the volume and at same temperature then, the total pressure of this container at equilibrium is:A. `50` atmB. `100` atm`C. `200` atmD. `70` atm |
| Answer» Correct Answer - B | |
| 335. |
Which of the following equilibrium reactions would be affected by change in pressure?A. `N_(2)+O_(2)hArr2NO`B. `2SO_(2)+O_(2)hArr2SO_(3)`C. `PCl_(5)hArrPCl_(3)+Cl_(2)`D. `H_(2)+Cl_(2)hArr2HCl` |
| Answer» Correct Answer - (b,c) | |
| 336. |
`K_(p)` for the reaction given below is `1.36 "at" 499K`. Which of the following equations can be used to calculate `K_(c)` for this reaction? `N_(2)O_(5(g))rarrNO_(2(g)+NO_(3))`A. `K_(C)=([(0.0821)xx(499)])/([1.36])`B. `K_(C)=([(1.36)xx(0.0821)])/([499])`C. `K_(C)=([1.36])/([(0.0821)xx(499)])`D. `K_(C)=([(1.36)xx(499)])/([0.0821])` |
| Answer» Correct Answer - C | |
| 337. |
At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass: `C_((s))+CO_(2(s))hArr2CO_((g))` Calculate `K_(c)` for the reaction at the above temperature. |
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Answer» `{:(,,C_((s)),+,CO_(2(g)),hArr,2CO_((g))),(,"At equilibrium",9.45g ,,90.55g,,),(,,=(9.45)/(44)"mole",,=(90.55)/(28)"mole",,),(,,=0.21 "mole",,=3.23 "mole",,):}` `K_(p)=((n_(CO))^(2))/(n_(CO_(2)))xx[(P)/(sumn)]^(1)` `=(3.23xx3.23)/(0.21)xx(1)/(3.44)` Now, `K_(c)=(K_(p))/((RT)^(Deltan))=(14.44)/([0.0821xx1127]^(1))` `=0.156 mol litre^(-1)` |
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| 338. |
The formation constant of `Ni(NH_3)_6^(2+)` is `6xx10^8` at `25^@C`.If 50 ml of 2.0 M `NH_3` is added to 50 ml of 0.20 M solution of `Ni^(2+)` , the concentration of `Ni^(2+)` ion will be nearly equal to :A. `3xx10^(-10)` mole `litre^(-1)`B. `2xx10^(-10)` mole `litre^(-1)`C. `2xx10^(-9)` mole `litre^(-1)`D. `4xx10^(-8)` mole `litre^(-1)` |
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Answer» Correct Answer - D ` {:(,Ni^(2+)+,6NH_(3), to,[Ni(NH)_(3)]^(6+),K_(f)=6xx10^(8)),(t=0,0.01 "mole",0.01 "mole",,0):}` `K_(c) = ([Ni(NH_(3))_(6)^(+6)])/([Ni^(+2)][NH_(3)]^(6))` `=((0.1))/([Ni^(2+)](0.4)^(6)) =6xx10^(8)` `[Ni^(2+)]=4xx10^(-8)` |
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| 339. |
At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass: `C_((s))+CO_(2(g))hArr2CO_((g))` Calculate `K_(c)` for the reaction at the above temperature. |
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Answer» Given: If total mass of the mixture of `CO "and" CO_(2) "is" 100g,` then `CO=90.55 "and" CO_(2)=100-90.55=9.45g` Asked: `K_(c)=?` Formulae: `K_(P)=(2_(CO)^(2))/P_(CO_(2))` Substitution & Calculation: Number of moles of `CO=90.55//28=3.234` Number of moles of `CO_(2)=9.45//44=0.215` `therefore P_(CO)=(3.234)/(3.234+0.215)xx1"atm"=0.938`atm `P_(CO_(2))=(0.215)/(3.234+0.215)xx1"atm"=0.062` atm `K_(P)=P_(CO)^(2)/(P_(CO_2))=(0.938)^(2)/(0.062)=14.19` `Deltan_(g)=2-1=1 thereforeK_(P)=K_(c)(RT)"or" K_(c)=(K_(P))/(RT)=(14.19)/(0.0821xx1127)=0.153` |
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| 340. |
`H_(2)(g) + I_(2)(g)hArr2HI(g)` When 46 g of `I_(2)` and 1 g of `H_(2)` gas heated at equilibrium at `450^(@)C`, the equilibrium mixture contained `1.9` g of `I_(2)` . How many moles of `I_(2)` and HI are present at equilibrium ?A. `0.0075` and `0.147` molesB. `0.005` and `0.147` molesC. `0.0075` and `0.347` molesD. `0.0052` and `0.347` moles |
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Answer» Correct Answer - C Moles of `I_(2)` taken `=(46)/(254) =0.181` Moles of `H_(2)` taken `=(1)/(2) =0.5` Moles of `I_(2)` remaining `=(1.9)/(254)=0.0075` Mols of `I_(2)` used `=0.81 - 0.0075 = 0.1735` Moles of `H_(2)` used `=0.1735` Moles of `H_(2)` reamining `=0.5 - 0.1735 = 0.3265` Moles of HI formed `=0.1735xx2=0.347` At equilibrium, moles of `I_(2) =0.0075` moles Moles of HI `=0.347` moles |
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| 341. |
A mixture of one mole of `CO_(2)` and "mole" of `H_(2)` attains equilibrium at a temperature of `250^(@)C` and a total pressure of `0.1` atm for the change `CO_(2)(g)+H_(2)(g) hArr CO(g)+H_(2)O(g)`. Calculate `K_(p)` if the analysis of final reaction mixture shows `0.16` volume percent of CO. |
| Answer» Correct Answer - `1.03xx10^(-5)` , | |
| 342. |
In a chemical equilibrium, the rate constant for the backward reaction is `7.5xx10^(-4)` and the equilibrium constant is `1.5` the rate constant for the forward reaction is:A. `5xx10^(-4)`B. `2xx10^(-3)`C. `1.125xx10^(-3)`D. `9.0xx10^(-4)` |
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Answer» Correct Answer - C `K_(c) = (K_(f))/(K_(b))` `K_(f) = K_(c)xxK_(b) = 1.5xx7.5xx10^(-4) = 1.125xx10^(-3)` |
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| 343. |
The low of mass action applicable to heterogenous equilibria. |
| Answer» Correct Answer - True | |
| 344. |
For a gaseous reaction : `A(g)to3B(g)+C(g),DeltaH` is positive and the reaction attains equilibrium at 1 bar total pressure and 400 K. Identify the incorrect statement(s) regarding the above reaction:A. On increase of temperature,equilibrium will be shifted in forward direction.B. When inert gas is introduced into a rigid container containing above equilibria equilibrium shifts towards left.C. `Delta_(400)^(circ)=0`,for the above reaction.D. If volume of vessel containing the above equilibria is increased without change in temperature then partial pressure kof B decreases as compared to original equilibrium partial pressure of B. |
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Answer» Correct Answer - B::C |
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| 345. |
In a chemical equilibrium, the rate constant for the backward reaction is `7.5xx10^(-4)` and the equilibrium constant is `1.5` the rate constant for the forward reaction is:A. `2xx10^(-3)`B. `5xx10^(-4)`C. `1.12xx10^(-3)`D. `9.0xx10^(-4)` |
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Answer» Correct Answer - C `K = (r_(f))/(r_(b)) implies 1.5 = (r_(f))/(7.5xx10^(-4)) implies r_(f) = 1.12xx10^(-3)` |
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| 346. |
The equilibrium constant in a reversible reaction at given temperatureA. Depends on the initial concentration of the reactantsB. Depends on the concentration of the products at equilibriumC. Does not depend on the initial concetrationsD. It is not characteristic of the reaction |
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Answer» Correct Answer - C Equilibrium constant is independent of original concentration of reactant. |
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| 347. |
In a general reaction `A+B hArr AB`, which value of equilibrium constant most favours the production of AB?A. `9.0xx10^(-3)`B. `3.0xx10^(-3)`C. `9.0xx10^(-7)`D. `9.0xx10^(-12)` |
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Answer» Correct Answer - A::B::C Because higher the value of equilibrium constant more favourable would be the product formation. |
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| 348. |
In a chemical reaction, equilibrium is said to have been established when theA. Concentrations of reactants and products are equalB. Opposing reactions ceasesC. Velocities of opposing reaction become equalD. Temperature of opposing reactions are equal |
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Answer» Correct Answer - C Rate of forward reaction = Rate of backward reaction i.e., when `R_(f)=R_(b)`, equilibrium is established. |
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| 349. |
In a chemical reaction, equilibrium is said to have been established when theA. the concentrations of the reactants and product are equalB. the rate of the opposing reactions become equalC. the temperature of the opposing reactions are equalD. there is no change in the concentration of either the product or the rectant with time. |
| Answer» Correct Answer - (b,d) | |
| 350. |
If more than one phase is present in the reversible reaction then it is said to be heterogenous system. Example: `CaO(s)+CO_(2)(g)` Expression of equilibrium constant for the above reaction can be taken as : `K=([CaO(s)][CO_(2)(g)])/([CaO(s)])`." ".....(i) Now concentration of `CaO(s)=[CaO(s)]` `=("moles of CaO")/("volume of CaO")` as density of `CaO[rho_(CaO(s))]` and molar mass of `CaO[M_(CaO(s))]`are a fixed quantity therefore concentration of pure solid and liquid term is uncharge with respect to time. Hence, equilibrium constant for the equation (i) can be written as : `K_(C)=[CO_(2)(g))]` `K_(P)=P_(CO_2)` As `K_(p) and K_(c)` is not containing solid terms therefore, addition or removel of pure solid and pure liquid has no effect on the equilibrium process. `K_(p)` for the reaction `NH_(4)I(s)hArrNH_(3)(g)+HI(g)`is `1//4 at 300K`.If above equilibrium is established by taking 4 moles of `NH_(4)I(s)` in 100 litre contanier, then moles of `NH_(4)I(s)` left in the container at equilibrium is `["Taken R=1/12Lt.atm mol"^(-1)K^(-1)]`.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B |
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