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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.
| 551. |
In a reaction mixture containing `H_(2),N_(2)` and `NH_(3)` at partial pressure of 2 atm, 1 atm and 3 atm respectively, the value of `K_(p)` at `725K` is `4.28xx10^(-5)atm^(-2)` . In which direction the net reaction will go ? `N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`A. ForwardB. BackwardC. No net reactionD. Direction cannot be predicted |
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Answer» Correct Answer - 2 `Q=(3^(2))/(2^(3)xx1)=(9)/(8)` Since `QgtK_(c)` reaction will proceed in backward direction. |
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| 552. |
In a reversible reaction, the catalystA. Increases the activation energy of the backward reactionB. Increases the activation energy of the forward reactionC. Decrases the activation energy of both forward and backward reactionD. Decreases the activation energy of forward reaction |
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Answer» Correct Answer - C Dearease the activation energy of both forward and backward reaction. |
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| 553. |
Statement: The equilibrium constant for a reaction having positive `DeltaH^(@)` increases with increase of temperature. Explanation: The temperature dependence of the equilibrium constant is related to `DeltaH^(@)` and not `DeltaS^(@)` for the reaction.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
| Answer» The temperature dependence of the equilibrium constant is related to `DeltaS^(@)` as well as `DeltaH^(@)` for the reaction. | |
| 554. |
On the velocity in a reversible reaction, the correct explanation of the effect of catalyst is.A. It provides a new reaction path of low activation energyB. It increases the kinetic energy of reacting moleculesC. It displaces the equilibrium state on right sideD. It decreases the velocity of backward reaction |
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Answer» Correct Answer - A Effect of catalyst is that it attains equilibrium quickly by providing a new reaction path of low activation energy. It does not alter the state of equilibrium. |
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| 555. |
For the reaction `C_(2)H_(6)(g)hArrC_(2)H_(4)(g)+H_(2)(g)` `K_(p)` is `5xx10^(-2)` atm. Calculate the mole per cent of `C_(2)H_(6)(g)` at equilibruium if pure `C_(2)H_(6)` at `1` atm is passed over a suitable catalyt at `900 K` :A. `20`B. `33.33`C. `66.66`D. None of these |
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Answer» Correct Answer - C |
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| 556. |
In the gas phase reaction, `C_(2H_(5) + H_(2)hArrC_(2)H_(6)`, the equilibrioum constant can be expressessed in units ofA. `"litre"^(-1)"mole"^(-1)`B. `"litre mole"^(-1)`C. `"mole"^(2)"litre"^(-2)`D. `"mole litre"^(-1)` |
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Answer» Correct Answer - B `K = ([C_(2)H_(6)])/([C_(2)H_(4)][H_(2)]) = (["mole"//"litre"])/(["mole"//"litre"]["mole"//"litre"])` `= "litre"//"mole"` or litre `"mole"^(-1)` |
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| 557. |
Assertion: The equilibrium in physical system is also dynamic in nature. Reason: The equilibrium ice `hArr` water is static is static in nature.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C Ice `hArr` Water shows dynamic equilibrium. |
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| 558. |
Which one of the following statements is not corrrect ?A. The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equiliberiumB. Enzymes catalyse mainly bio-chemical reactionsC. Coenzymes increase the catalytic activity of enzymeD. Catalyst does not intiate any reaction |
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Answer» Correct Answer - A Equilibrium constant is not affected by pressence of catalyst hence statement (1) is incorrect. It is only affected by temperature. |
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| 559. |
In Haber process 30 litre of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only`50%` of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end ?A. 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogenB. 20 litres ammonia, 20 litres nitrogen, 15 litres hydrogenC. 10 litres ammonia, 25 litres nitrogen, 15 litrers hydrogenD. 20 litres ammoina, 10 litres nitrogen, 30 litres hydrogen |
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Answer» Correct Answer - C `{:(N_(2)+3H_(2),hArr,2NH_(3)),(30,30,0),(30-x,30-x,2x):}` 2x = 10, `x = (10)/(2) = 5` `N_(2) = 30 - 5 = 25 L` `H_(2) = 30 - 3xx5 = 15 L` |
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| 560. |
The `K_(p)` for the reaction `N_(2)O_(4)hArr2NO_(2)` is `640 mm` at `775 K`. Calculate the percentage dissociation of `N_(2)O_(4)` at equilibrium pressure of `160 mm`. At what pressure, the dissociation will be `50%`? |
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Answer» `{:(,N_(2)O_(4),hArr,2NO_(2)),("Mole before equilibrium",1,,0),("Mole at equilibrium",(1-x),,2x):}` `K_(p)=(4x^(2))/((1-x))xx[(P)/(sumn)]^(Deltan)` `640=(4x^(2))/((1-x))(160)/((1+x))` `4=(4x^(2))/(1-x^(2))` or `1-x^(2)=x^(2)` or `2x^(2)=1` ` :. x^(2)=1//2` or `x=0.707=70.7%` Also `640=(4xx(0.5)^(2))/(0.5)xx(P)/(1.5)` (if `x=0.5`) `:. P=480 mm` |
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| 561. |
For the dissociation reaction `N_(2)O_($) (g)hArr 2NO_(2)(g)`, the degree of dissociation `(alpha)`interms of `K_(p)` and total equilibrium pressure P is:A. `alpha=sqrt((4P+K_(p))/(K_(p)))`B. `alpha=sqrt((K_(p))/(4P+K_(p)))`C. `alpha=sqrt((K_(P))/(4P))`D. None of these |
| Answer» Correct Answer - 2 | |
| 562. |
Equilibirum constants `K_(1)` and `K_(2)` for the following equilibria `NO(g) + (1)/(2)O_(2)hArrNO_(2)(g)` and `2NO_(2)(g)hArr` `2NO(g) + O_(2)(g)` are related asA. `K_(2)= (1)/(K_(1)^(2))`B. `K_(2) = (1)/(K_(1))`C. `K_(2) = K_(1)^(2)`D. `K_(2) = (K_(1))/(2)` |
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Answer» Correct Answer - A `K_(1) = ([NO_(2)])/([NO][O_(2)])` and `K_(2) = ([NO]^(2)[O_(2)])/([NO_(2)]^(2))` If eqn. is multiplied by n, then new equilibrium constant is `K_(1)n` and if equ. Is reversed, then new equilibrium constant is reciprocal of pervious. `:. K_(2) = (1)/(K_(1)^(2))` |
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| 563. |
Using moler concentrations, what is the unit of `K_(c)` for the reaction ? `CH_(3)OH(g)hArrCO(g)+2H_(2)(g)`A. (A) `M^(-1)`B. (B) `M^(2)`C. (C) `M^(-1)`D. (D) `M` |
| Answer» Correct Answer - B | |
| 564. |
An alloy consists of rubidium and one of the other alkali metals. A sample of `4.6g` of the alloy when allowed to react with water, liberates `2.241dm^(3)` of hydrogen at `STP`. Relative atomic masses: `A_(T)(Li)=7,A_(T)(Na)=23,A_(T)(K)=39,A_(T)(Rb)=85.5,A_(T)(Cs)=1.33` Which alkali metal is the component of the alloy? |
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Answer» Correct Answer - `M-alkali metal Reaction:`2M+2H_(2)Orarr2MOH+H_(2)` `n(H_(2)):0.1`mol `n(M):0.2`mol Mean molar mass: `M=(4.6g)/(0.2mol)=23gmol^(-1)` |
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| 565. |
In a reversible reaction `Aunderset(K_2)overset(K_1)hArrB` the initial concentration of A and B are a and b in moles per litre and the equilibrium concentrations are (a-x) and (b+x) respectively, Express x in terms of `K_1, K_2, a and b`.A. `(K_1a-K_2b)/(K_1+K_2)`B. `(K_1a-K_2b)/(K_1-K_2)`C. `(K_1a-K_2b)/(K_1K_2)`D. `(K_1a+K_2b)/(K_1+K_2)` |
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Answer» Correct Answer - A |
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| 566. |
For the reaction `3A(g)+B(g) hArr 2C(g)` at a given temperature , `K_c=9.0`.What must be the volume of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium ?A. 6 LB. 9 LC. 36 LD. None of these |
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Answer» Correct Answer - A |
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| 567. |
For the reaction `3A(g)+B(g) hArr 2C(g)` at a given temperature , `K_c=9.0`.What must be the volume of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium ? |
| Answer» Correct Answer - `6L` | |
| 568. |
For the reaction `3A(g)+B(g) hArr 2C(g)` at a given temperature, `K_c`=9.0 what must be the volume (in L) of the flask , if a mixture of 2.0 mol each of A,B and C exist in equilibrium ? |
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Answer» Correct Answer - 6 |
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| 569. |
`1` mole of a gas `A` is taken in a vessel of volume `1L`. It dissociates according to the reaction `A(g)hArrB(g)+C(g) "at" 27^(@)C`. Forward and backward reaction rate constants for the reaction are `1.5xx10_(-2) "and" 3xx10_(-2)` respectively. Find the concentrations of `A,B "and" C` at equilibrium. |
| Answer» Correct Answer - `[A]_(eq)=[B]_(eq)=[C]_(eq)=1//2M,Kp=12.3 "atm", K_(c)=0.5` m(Unitless). | |
| 570. |
`K_(c)` for `A+BhArrC+D` is `10` at `25^(@)C`. If a container contains `1`, `2`, `3`, `4` `mol//litre` of `A`, `B`, `C` and `D` respectively at `25^(@)C`, the reaction shell proceed:A. from left to rightB. from right to leftC. equilibriumD. either of these |
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Answer» `A+BhArrC+D` `Q=([C][D])/([A][B])=(3xx4)/(1xx2)=6` `K_(c)=10`, Thus to increase the value of `Q` to `K_(c)`, forward reaction occurs. |
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| 571. |
For the reaction `A(aq)hArrB(aq)+2C(aq) K_C` at `25^@` is `4xx10^(-19)` Concentration of B in a solution that had originally C and B concentration of 0.1 M and 0.03 M respectively is :A. 0.03 MB. `7.5xx10^(-12) M`C. `7.5xx10^(-15) M`D. `7.5xx10^(-18) M` |
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Answer» Correct Answer - D |
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| 572. |
Two solids `A` and `D` dissociates into gaseous products as follows `C(s)hArrB(g)+C(g),K_(P_(1))=300 , D(s)hArrE(g)+C(g)K_(P_(2)=600` at `27^(@)C`, then find the total pressure of the solid mixture. |
| Answer» Correct Answer - 60 atm | |
| 573. |
Two solid `X` and `Y` dissociate into gaseous products at a certain temperature as followas: `X(s)hArrA(g)+C(g)` , and `Y(s)hArrB(g)+C(g)` At a given temperature, the pressure over excess solid `X` is `40 mm` and total pressure over solid `Y` is `80 mm`. Calculate a. The value of `K_(p)` for two reactions. b. The ratio of moles of A and B in the vapour state over a mixture of `X` and `Y`. c. The total pressure of gases over a mixture of `X` and `Y`. |
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Answer» `X(S)hArrA(g)+C(g)` At wquilibrium, `A` and `C` are in equal proportions, so their pressures will br same. `p_(A)=p_(C )` Also `p_(A)+p_(C )=(20)^(2)=400 mm^(2)` `Y(s)hArrB(g)+C(g)` `p_(B)=p_(C )=40 mm(p_(B)+p_(C )=80)` rArr `K_(p)=p_(B).p_(C )=40^(2)=1600 mm^(2)` b. Now for a mixture of `X` and `Y`, we will have to consider both the equilibrium simultaneously. `X(s)hArrA(g)+C(g)` `Y(s)hArrB(g)+C(g)` Let `p_(A)=a mm, p_(B)=b mm` Note that the pressure of `C` due to dissociation of `X` will also be a mm and similarly the pressure of `C` due to dissociation of `Y` will also be `b mm`. `rArr p_(C )=(a+b) mm` `K_(p)(for X)=p_(A).p_(C )=a(a+b)=400 ...(i)` `K_(p)(for Y)=p_(B).p_(C )=(a+b)=1600 ...(ii)` From (i) and (ii), we get: `a/b=1/4` as volume and temperature are constant, the "mole" ratio will be same as the pressure ratio. c. The total pressure `=P_(T)=p_(A)+p_(B)+p_(C )` `=a+b+(a+b)=2(a+b)` Adding (i) and (ii) `a+b=sqrt(K_(PX)+K_(PY))=sqrt(2000)=20sqrt(5)mm` rArr Total pressure `=2(a+b)=89.44 mm` |
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| 574. |
`A(s)hArrB(g)+C(g) K_(p_(1))=36 atm^(2)` `E(s)hArrB(g)+D(g) K_(p_(2))=64atm^(2)` Both solids `A "&" E` were taken in a container of constant volume at a give temperature. Total pressure in the container after equilibrium isA. `6` atmB. `5` atnC. `10` atmD. `20` atm |
| Answer» Correct Answer - D | |
| 575. |
`CuSO_(4),5H_(2)O(s)hArrCuSO_(4)(s)+5H_(2)O(g)K_(P)=10^(-10) "moles of" CuSO_(4).5H_(2)O(s)` is taken in a `2.5L` container at `27^(@)C` then at equilibrium [Take: `R=(1)/(12)` litre atm `mol^(-1)K^(-1)`]A. Moles of `CuSO_(4).5H_(2)O` left in the container is `9xx10^(-3)`B. Moles of `CuSO_(4).5H_(2)O` left in the container is `9.8xx10^(-3)`C. Moles of `CuSO_(4)` Left in the container is `10^(-3)`D. Moles of `CuSO_(4)` left in the container is `2xx10^(-4)` |
| Answer» Correct Answer - B::D | |
| 576. |
The molar concentration of `A` and `B` are `0.80 mol//litre` each . On mixing them, the reaction starts to proceed as : `A+BhArrC+D`, and attain equilibrium. At equilibrium molar concentration of `C` is `0.60 mol//litre`. Find the value of `K_(c)` of the reaction. |
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Answer» Given, `[C]` at equilibrium `=0.6 mol//litre` `[D]=0.6 mol//litre` Thus, concentration of `A` and `B`, which reacts, are `0.6 mol//litre` at equilibrium. `{:(,A,+,B,hArr,C,+,D),("Initial conc.",0.8,,0.8,,0,,0),("Equilibrium conc.",(0.8-0.6),,(0.8-0.6),,0.6,,0.6),(,=0.2,,=0.2,,0.6,,0.6):}` `:. K_(c)=([C][D])/([A][B])=(0.6xx0.6)/(0.2xx0.2)=9` |
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| 577. |
At a certain temperature `T`, a compound `AB_(4)(g)` dissociates as `2AB_(4)(g)hArrA_(2)(g)+4B_(2)(g)` with a degree of dissociation `alpha`, which compared to unity. The expressio of `K_(P)` in terms of `alpha` and total pressure `P` is:A. `256 P^(3) alpha^(5)`B. `4Palpha^(2)`C. `8P^(3)alpha^(5)`D. None of these |
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Answer» Correct Answer - C `2AB_(4)(g)hArrA_(2)(g)+4B_(2)(g)` `{:(underset(("moles"))(t=0),a,,0,0),(t=t_(Eq),a-aalpha,,(aalpha)/2,2aalpha):}` `K_(p)=(p_(A_(2)).p_(B_(2))^(4))/p_(AB_(4))^(2)` `p_(A_(2))=(aalpha//2)/(a+3/2aalpha).p~~(alphap)/2, p_(B_(2))=(2aalpha)/(a+3/2aalpha).P~~2alphap` `p_(AB_(4))=(a-aalpha)/(a+3/2aalpha).p~~p` `K_(p)=((alphap)/2.(2alphap)^(4))/(p^(2))=8alpha^(5).p^(3)` |
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| 578. |
consider the following gaseous equilibrium with equilibrium constant `K_(1)` and `K_(2)` respectively `SO_(2)(g)+ 1//2O_(2)hArrSO_(3)(g)` `2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)` The equilibrium constants are related asA. `K_(2)=(K_(1))^(-1)`B. `K_(2)=sqrt((1)/(K_(1))`C. `K_(2)=((1)/(K_(1)))^(2)`D. `sqrtK_(1)` |
| Answer» Correct Answer - B | |
| 579. |
`2A(S)hArrB(g)+2C(g)+3D(g)` Total pressure developed in closed conatainer by decomposition of A at equilibrium is 12 atm at `727^(@)C`. Calculate `DeltaG^(@)` (in cal)of the reaction at `727^(@)C`. (R=2cal/mol-K,In2=0.7,In3=1.1) [Fill your answer as `|DeltaG^@|/(100)]`. |
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Answer» Correct Answer - 178 |
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| 580. |
Two solid compounds `A "and" C` dissociate into gaseous product at temperature `T` as follows: (i) `A(s)hArrB(g)+D(g) K_(P_(1))=25(atm)^(2)` (ii) `C(s)hArrE(g)+D(g) K_(P_(2))=975(atm)^(2)` Both solid are present in same container then calculate total pressure over the solid mixture. |
| Answer» Correct Answer - 40 atm | |
| 581. |
At `727^(@)C` and `1.2 atm` of total equilibrium pressure, `SO_(3)` is partially dissociated into `SO_(2)` and `O_(2)` as: `SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g)` The density of equilibrium mixture is `0.9 g//L`. The degree of dissociation is:, `[Use R=0.08 atm L mol^(-1) K^(-1)]`A. `1/3`B. `2/3`C. `1/4`D. `1/5` |
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Answer» Correct Answer - A |
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| 582. |
The equilibrium constant of the reactions `SO_(2)(g)+1/2 O_(2)(g)hArrSO_(3)(g)` and `2SO_(2)(g)+O_(2)hArr2SO_(3)(g)` are `K_(1)` and `K_(2)` respectively. The relationship between `K_(1)` and `K_(2)` isA. `K_(1)=K_(2)`B. `K_(2)=K_(1)`C. `K_(1)=sqrt(K_(2))`D. `K_(2)=sqrt(K_(1))` |
| Answer» Correct Answer - (b,c) | |
| 583. |
Given reaction is `2X_((gas)) + Y_((gas))hArr2Z_((gas)) + 80` Kcal Which combination of pressure and temperature gives the highest yield of Z at equilibrium ?A. 1000 atm and `500^(@)C`B. 500 atm and `500^(@)C`C. 1000 atm and `100^(@)C`D. 500 atm and `100^(@)C` |
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Answer» Correct Answer - C The reaction takes place with a reduction in number of moles (volume) and is exothermic. So high pressure and low temperature will favour the reaction in forward direction. |
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| 584. |
Which among the following reactions is favoured in forward direction by increase of temperature?A. `N_(2)(g)3H_(2)(g) hArr 2NH_(3)(g)+22.9 kcal`B. `N_(2)(g)+O_(2)(g) hArr 2NO(g)-42.8 kcal`C. `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)+45.3 kcal`D. `H_(2)(g)+Cl_(2)(g)-44 kcal hArr 2HCl(g)` |
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Answer» Correct Answer - B It is an endothermic reaction, hence the rise in temperature will favour forward direction. |
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| 585. |
At certain temperature compound `AB_(2)(g)` dissociates accoring to the reacation `2AB_(2)(g) hArr2AB (g)+B_(2)(g)`With degree of dissociation `alpha` Which is small compared with unity, the expression of `K_(p)` in terms of `alpha` and initial pressure P is :A. `p(alpha^(3))/(2)`B. `(Palpha^(2))/(3)`C. `P(alpha^(3))/(3)`D. `(Palpha^(2))/(2)` |
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Answer» Correct Answer - a |
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| 586. |
At a certain temperature, the equilibrium constant `(K_(c ))` is `16` for the reaction: `SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)` If we take one mole of each of the equilibrium concentration of `NO` and `NO_(2)`? |
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Answer» `SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)` `{:("Initial conc".,1,1,1,1),("Equilibrium conc".,1-x,1-x,1+x,1+x):}` Applying the law of mass action, `K_(c)=([SO_(3)][NO])/([SO_(2)][NO_(2)])=((1+x)(1+x))/((1-x)(1-x))=16` `(1+x)/(1-x)=4` or `1+x=4-4x` or `5x=3`, i.e., `x=3/5=0.6` Concentratio of `NO_(2)` at equilibrium `=(1-0.6)=0.4 "mol"` Concentration of `NO` at equilibrium`=(1+0.6)=1.6 "mol"` |
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| 587. |
For the reaction `2A(g)hArrB(g)+3C(g),` at a given temperature ,`K_(c)=16.` What must be the volume of the flask , if a mixture of 2 mole rach of A,B and C exist in equilibrium ?A. `(1)/(4)`B. `(1)/(2)`C. 1D. none of these |
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Answer» Correct Answer - b |
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| 588. |
At a certain temperature and `2` atm pressure equilibrium constant `(K_(p))` is `25` for the reaction `SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)` Initially if we take `2` moles of each of the four gases and `2` moles of inert gas, what would be the equilibrium pparital pressure of `NO_(2)`?A. `1.33` atmB. `0.1665` atmC. `0.133` atmD. None of these |
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Answer» Correct Answer - C |
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| 589. |
Equilibrium constant for the reactions, `2 NO+O_(2)hArr2 NO_(2) "is" K_(c_(1)`, `NO_(2)+SO_(2)hArrSO_(3)+NO "is" L_(C_(2)` and `2 SO_(3)hArr2 SO_(2)+O_(2) "is" K_(c_(3)` then correct reaction is:A. `K_(C_(3))=K_(C_(1))xxK_(C)(2)`B. `K_(C_(3))xxK_(C_1)xxK_(C_(2))^(2)=1`C. `K_(C_(3))xxK_(C_(1))xxK_(C_2)=1`D. `K_(C_(3))xxK_(C_(1))^(2)xxK_(C_(2)=1` |
| Answer» Correct Answer - B | |
| 590. |
In the gaseous equilibrium `A+2B hArr C+"Heat"`, the forward reaction is favoured:A. Low P, High TB. Low P, Low TC. High P, Low TD. High P, High T |
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Answer» Correct Answer - C `A+2B hArr C+ "Heat"` The equation shows, that it is exothermic reaction. Since the heat is released in the reaction, so the reaction is favoured in forward direction at low temperature. `Deltan=1-(2+1)=-2` Since the number of moles decreases, `:.` the forward reaction is favoured at high pressure. |
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| 591. |
`8` mol of gas `AB_(3)` are introduced into a `1.0 dm^(3)` vessel. It dissociates as `2AB_(3)(g) hArr A_(2)(g)+3B_(2)(g)` At equilibrium, 2 mol of `A_(2)` is found to be present. The equilibrium constant for the reaction isA. `72 mol^(2)L^(-2)`B. `36 mol^(2)L^(-2)`C. `3 mol^(2)L^(-2)`D. `27 mol^(2)L^(-2)` |
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Answer» `{:(,2AB_(3(g)),hArr,A_(2(g)),+,3B_(2(g))),("At t=0",8,,0,,0),("At eqm.",8-a,,a//2,,3a//2):}` Thus `K_(c)=([A_(2)][B_(2)]^(3))/([AB_(3)]^(2))`, Also `(a)/(2)=2` `:. A=4` and `[AB_(3)]=(4)/(1)`, `[A_(2)]=(2)/(1)` and `[B_(2)]=(6)/(1)` `=(2xx6^(3))/(4^(2))=27 mol^(2)litre^(2-)` |
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| 592. |
The equilibrium `K_(c)`for the reaction `SO_(2)(g)NO_(2)(g)hArrSO_(3)(g)+NO(g)is 16` 1 mole of rach of all the four gases is taken in `1dm^(3)` vessel , the equilibrium concentration of NO would be:A. `0.4M`B. `0.6M`C. `1.4M`D. `1.6M` |
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Answer» Correct Answer - d |
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| 593. |
`AB_(2)` dissociates as `AB_(2)(g) hArr AB(g)+B(g)`. If the initial pressure is `500` mm of Hg and the total pressure at equilibrium is `700` mm of Hg. Calculate `K_(p)` for the reaction. |
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Answer» Correct Answer - A::C After dissociation, suppose the decrease in the pressure of `AB_(2)` at equilibrium is `p` mm. `{:(,AB_(2)(g),hArr,AB(g),+,B(g)),("Initial pressure",500 mm,,0,,0),("Pressure at equilibrium" ,(500-p) mm,,p mm,,p mm):}` `:.` Total pressure at equilibrium `=500-p+p+p` `=500+p mm` `500+p=700` (Given) or `p=200 mm` Hence, at equilibrium, `p_(AB_(2))=500-200=300 mm` `p_(AB)=200 mm, p_(B)=200 mm` `:. K_(p)=(p_(AB)xxp_(B))/p_(AB_(2))=(200xx200)/300=133.3 mm` |
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| 594. |
`XY_(2)` dissociates `XY_(2)(g) hArr XY(g)+Y(g)`. When the initial pressure of `XY_(2)` is `600` mm Hg, the total equilibrium pressure is `800` mm Hg. Calculate K for the reaction Assuming that the volume of the system remains unchanged.A. `50.0`B. `100.0`C. `166.6`D. `400.0` |
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Answer» Correct Answer - B `{:(,XY_(2),hArr,XY,+,Y),("Initial",P,,0,,0),("Final",P-x,,x,,x):}` Initial `P=600` mm Hg, Final `P=800` mm Hg Total "moles" `prop` final P. `:. P-x+x+x prop 800` `600-x+x+xprop 800` `x=200` mmHg `K=(x xx x)/(P-x)=(200xx200)/(400)=100 Hg` |
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| 595. |
The value of `K_(c)` for the reaction: `A_(2)(g)+B_(2)(g) hArr 2AB(g)` at `100^(@)C` is `49`. If `1.0 L` flask containing one mole of `A_(2)` is connected with a `2.0 L` flask containing one mole of `B_(2)`, how many moles of AB will be formed at `100^(@)C`? |
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Answer» Correct Answer - A `{:(,A_(2),+,B_(2),hArr,2AB),("Initial",1,,2,,),("Total",V=1+2=3,,,,),("At equlibrium",(1-x)/3,,(2-x)/3,,(2x)/3):}` `K=50=((2x//3)^(2))/(((1-x)/3)((2-x)/3))` `x=7/9 "mol"` Number of moles of `AB` formed at equilibrium `=14/9=1.55 "moles"` |
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| 596. |
Assertion: Melting of ice at `10^(@)C` is a reversible process. Reason: Evaportation of water at `100^(@)C` and 1 atm pressure is a reversible priocess.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D Water `hArr` Vapour at b. pt. V.P. `= 760` mm |
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| 597. |
Assertion: A dynamic equilibrium means a balance between the tendency towards minimum and mazimum ethalpies. Reason: The reaction having `Delta H = -ve` occurs form high ethalpy side to low enthalpy side and the reaction `Delta H = +ve` occurs form low enthalpy side to high enthalpy side.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A A+B `hArr` C+D, `Delta H = +ve`, then C+D `hArr` A+B, `Delta H = -ve`, i.e., a balance in enthalpies is noticed at equilibrium. |
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| 598. |
If the equilibrium constant for the reaction `A_(2)+B_(2) hArr 2AB` is K, then the backward reaction, `AB hArr 1/2 A_(2)+1/2 B_(2)` the equilibrium constant is `1//K`. |
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Answer» False. If the equilibrium constant for the reaction `AB hArr 1/2 A_(2)+1/2 B_(2)` the equilibrium constant is `sqrt(1//K)` because the reaction is reversed as well as divided by two. |
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| 599. |
For the reaction equilibrium, `N_(2)O_(4(g))hArr2NO_(2(g))`, the concentration of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8xx10^(-2)` and `1.2xx10^(-2)` mol/L respectively. The value of `K_(c)` for the reaction is:A. `3xx10^(-3)`MB. `3xx10^(3)`MC. `3.3xx10^(2)`MD. `3xx10^(-1)`M |
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Answer» Correct Answer - A `K_(c) = ([NO_(2)]^(2))/([N_(2)O_(4)]) = ((1.2xx10^(-2))^(2))/(4.8xx10^(-2)) = 3.0xx10^(-3)` M |
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| 600. |
For the reaction , `A+2B hArr 2C`, the rate constant for the forward and the backward reactions are `1xx10^(-4)` and `2.5xx10^(-2)` respectively.The value of equilibrium constant, K, for the reaction would be :A. `1xx10^(-4)`B. `2.5xx10^(-2)`C. `4xx10^(-3)`D. `2.5xx10^(2)` |
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Answer» Correct Answer - C |
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