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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.
| 351. |
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. `200g of CaCO_(3)(g)` taken in 4Ltr container at a certain temperature. `K_(c)` for the dissociation of `CaCO_(3)` at this temperature is found to be `1//4` mole `Ltr^(-1)`then the concentration of CaO in mole/litre is : [Given :`rho_(CaO)=1.12gcm^(-3)][Ca=40,O=16]`A. `(1)/(2)`B. `(1)/(4)`C. `0.02`D. `20` |
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Answer» Correct Answer - C |
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| 352. |
In a reversible chemical reaction, equilibrium is said to have been established when theA. Concentrations of reactants and products are equalB. Opposing reactions ceaseC. Speeds of opposing reactions become equalD. Temperature of opposing reactions are equal |
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Answer» Correct Answer - C At equilibrium, the rate of forward reaction becomes equal the rate of opposing reaction or backward reaction. |
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| 353. |
What is the active mass of `5.6` litres of `O_(2)` at S.T.P.? |
| Answer» Correct Answer - `0.044M`. | |
| 354. |
Indicate the correct answer out of the following for the reaction: `NH_(4)Cl+H_(2)OhArrNH_(4)OH+HCl`A. the reaction is retarded by the addition of `KOH`B. the reaction is favoured by the addition of `NH_(4)OH`C. the reaction is retarded by the addition of hydrogen ionD. none of these |
| Answer» An increase `H^(+)` favours the backward reaction. | |
| 355. |
Hydrogen and oxygen were heated together in a closed vessel. The equilibrium constant is found to decrease after `2000^(@)C`. Which is responsible for this?A. Backward reaction predominatesB. Forward reaction predominatesC. Both forward and backward reaction have same rateD. It is a property of the system, hence no reason for lower value |
| Answer» It is the characteristic of given reaction. | |
| 356. |
Calculate `DeltaG` for the reaction at `25^(@)C` when partial pressure of reactants `H_(2)`, `CO_(2)`, `H_(2)O` and `CO` are `10`, `20`, `0.02` and `0.01atm` respectively. `H_(2(g))+CO_(2(g))hArrH_(2)O_((g))+CO_((g))` Given `G_(f)^(@)` for `H_(2(g))`, `CO_(2(g))`, `H_(2)O_((g))` and `CO_((g))` are `0`, `-394.37`, `-228.58`, `-137.15 KJ` respectively. |
| Answer» Correct Answer - `5.61 kJ` , | |
| 357. |
Consider these reactions and their corresponding `K_s`. `{:(1/2N_2+O_2toNO_2,K_1),(2NO_2to2NO+O_2,K_2),(NOBrtoNO+1/2Br_2,K_3):}` Express the K value for the reaction below in terms of `K_1,K_2 and K_3` `1/2N_2+1/2O_2+1/2Br_2toNOBrK=?`A. `K_1+K_2/2-K_3`B. `K_1+(K_2)^(1//2)-K_3`C. `(K_1K_2)/(2K_3)`D. `K_1(K_2)^(1//2)//K_3` |
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Answer» Correct Answer - D |
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| 358. |
In reaction: `CH_(3)COCH_(3)(g)hArrCH_(3)CH_(3)(g)+CO(g),` if the initial pressure of `CH_(3)COCH_(3)(g)` is `150 mm` and at equilibrium the mole fraction of `CO(g)` is `1/3`, then the value `K_(P)` isA. `50 mm`B. `100 mm`C. `33.3 mm`D. `75 mm` |
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Answer» Correct Answer - A `1` mol of `CH_(3)COCH_(3) =1` mol" of `CH_(3)CH_(3)` `1` mol of `CO` = mol fraction of `CO=1/3` Pressure of gas = mole fraction xx Initial pressure `=1/3xx150=50 mm` `:. K_(p)=((P_(CH_(3)CH_(3)))(P_(CO)))/((P_(CH_(3)COCH_(3))))=(50xx50)/50=50 mm` Alternative method: `CH_(3)COCH_(3)hArrCH_(3)CH_(3)+CO` `{:("Initial",,1,0,0),("At equilibrium",,1-x,x,x):}` Total "moles" `=1-x+x+x=1+x` `:.` mole fraction of `CO=(x)/(1+x)=1/3 :. x=1/2` `:.` mole of each gas `=1/2` `:.` mole fraction of each gas `=(1/2)/(1/2+1/2+1/2)=1/3` `:.` Pressure of each = mole fraction xx Initial pressure `=1/3xx150=50 mm` `:. K_(p)=((P_(CH_(3)CH_(3)))(P_(CO)))/((P_(CH_(3)COCH_(3))))=(50xx50)/50=50 mm` |
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| 359. |
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 physical equilibrium is not static but dynamic in nature. STATEMENT-2: The pysical equilibrium is a state in which two opposing process are proceeding at the same rate.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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| 360. |
Assertion (A) : Equilibrium constant of a reaction increases if temperature is increased Reason (R) : The forward reaction becomes faster with increase of temperature.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 Forward reaction increase with temperature only for endothermic reaction, while for exothermic it is reverse. |
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| 361. |
The reaction having higher value of equilibrium constant is faster than the reaction having lower value of equilibrium constant. |
| Answer» Correct Answer - True | |
| 362. |
Assertion: The equilibrium constant may show higher or lower values with increase in temperature. Reason: The change depends on the heat of reaction at equilibrium.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`. |
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Answer» `2.303 log (K_(c_(2)))/(K_(c_(1)))=(DeltaH)/(R )[(T_(2)-T_(1))/(T_(1)T_(2))]` If `DeltaH= -ve`, `K_(c_(2)) lt K_(c_(1))` If `DeltaH= +ve`, `K_(c_(2)) gt K_(c_(1))` |
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| 363. |
Assertion: The equilibrium constant may show higher or lower values with increase in temperature. Reason: The change depends on the heat of reaction at equilibrium.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 `2.303 log (K_(c2))/(K_(c1)) =(Delta H)/(R) [T_(2) - T_(1))/(T_(1)T_(2))` , if `Delta H = -ve`, `K_(c_(2)) lt K_(C_(1))` if `Delta H = +ve`, `K_(C_(2)) gt K_(C_(1)` . |
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| 364. |
In a reaction `A + B hArrC + D`, the concentration of A, B, C and D (in moles/litre) are `0.5, 0.8, 0.4` and `1.0` respectively. The equilibrium constant isA. `0.1`B. `1.0`C. `10`D. `oo` |
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Answer» Correct Answer - B For A + B `hArr C + D ` `K = ([C][D])/([A][B]) = (0.4xx1)/(0.5xx0.8) = 1` |
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| 365. |
For the hypothetical reaction : `A+B hArrC+D`, the equilibrium constant,K, is less than 1.0 at `25^(circ)C` and decreased by `35%` on changing the temperature to `45^(circ)C`. What must be ture according to this information?A. The `DeltaH^circ` for the reaction is negative .B. The `DeltaS^circ` for the reaction is positive.C. The `DeltaG^circ` for the reaction at `25^cricC` is negative.D. The `DeltaG^circ` for the reaction at `45^circC` is zero. |
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Answer» Correct Answer - A |
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| 366. |
`Au(s) hArr Au (l)` above mentioned equilibrium is fovoured atA. High pressure, lowtemperatureB. High pressure, high temperatureC. Low pressure, high temperatureD. Low pressure, low temperature |
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Answer» Correct Answer - C `Au(s) + "heat" hArr Au(l)` (i) There is an increase in volume in the forward direction, hence the melting is favoured at low pressure. (ii) Since melting is an endothermic process, it is favoured at high temperature. |
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| 367. |
What is the direction of a reversoble reaction when one of the products of the reaction is removed?A. The reaction moves towards right hand side.B. The reaction moves towards left hand sideC. The reaction moves towards both hand sideD. The reaction stops. |
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Answer» Correct Answer - A Removal of one product will result will result in decrease in the concentration of products and hence the reaction shifts to forward direction. |
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| 368. |
For a reversible gaseous reaction `N_(2)+3H_(2)hArr2NH_(3)` at equilibrium , if some moles of `H_(2)` are replaced by same number of moles of `T_(2)` (T is tritium , isotope of H and assume isotopes do not have different chemical properties ) without affecting other parameters , then:A. the sample of ammonia obtained after something will be radioactive .B. moles of `N_(2)` after the change will be different as compared to moles of `N)(2)` present before the changeC. the volue of `K_(p) "or" K_(c)` will changeD. the average molecular mass of new equilibrium will be same as that of old equilibrium |
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Answer» Correct Answer - a |
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| 369. |
For `NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)`, the observed, pressure for reaction mixture in equilibrium is `1.12` atm at `106^(@)C`. What is the value of `K_(p)` for the reaction? |
| Answer» Correct Answer - `0.3136 atm^(2)` , | |
| 370. |
The equilibrium constant for : `PClrArrPCl_(3)+Cl_(2)` is 0.5 at 500 K. What is the state of vessel shown below ? A. The contents of the vessel are not in equilibrium , with an excess of `PCl_(5)`.B. The contents of vessel are in equilibrium.C. The contents of the excess of `PCl_(3)and Cl_(20`.D. Nothing can be predicted redicted regarding the equilibrium just from the diagram given above. |
| Answer» Correct Answer - 2 | |
| 371. |
For the reaction, `N_(2)+3H_(2)hArr2NH_(3)`, in a vessel equal moles of `N_(2)` and `H_(2)` are mixed to attain equilibrium. At equilibrium:A. `[N_(2)]=[H_(2)]`B. `[N_(2)]gt[H_(2)]`C. `[N_(2)]lt[H_(2)]`D. `[H_(2)]gt[NH_(3)]` |
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Answer» `{:(N_(2),+,3H_(2),hArr,2NH_(3)),(a,,a,,0),((a-x),,(a-3x),,):}` ` :. (a-x) gt (a-3x)` |
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| 372. |
At 473 K,`K_(c)` for the reaction `OCl_(5(g))rArrPCl_(3(g))Cl_(2(g))` is `8.3xx10^(-3)`. What will be the value of `K_(c)` for the formation of `PCl_(5)` at the same temperature ?A. `8.3xx10^(3)`B. 120.48C. `8.3xx10^(-3)`D. 240 .8 |
| Answer» Correct Answer - 2 | |
| 373. |
Consider the reactions (i) `PCI_(5)(g)hArrPCI_(3)(g)+CI_(2)(g)` (ii) `N_(2)O_(4)(g)hArr2NO_(2)(g)` The addition of an inert gas at constant volumeA. will increase the dissociation of `PCI_(5)` aswell as `N_(2)O_(4)`B. will reduce the dissociation of `PCI_(5)` aswell as `N_(2)O_(4)`C. will increase the dissociation of `PCI_(5)` and step up the formation of `NO_(2)`D. will not disturb the equilibrium of the reactions |
| Answer» Correct Answer - D | |
| 374. |
Addition of water to which of the following equilibria causes it to shift in the backward direction?A. `CH_(3)NH_(2)(aq)+H_(2)O(l)hArrCH_(3)NH_(3)(aq)+OH^(-)(aq)`B. `AgCI(s)hArrAg^(+)(aq)+CI^(-)(aq)`C. `HCN(aq)+H_(2)O(l)hArrH_(3)O^(+)(aq)+CN^(-)(aq)`D. `[Cr(dien)_(2)]^(3+)(aq)+3H_(2)O(l)+3CI^(-)(aq)hArr[Cr(H_(2)O)_(3)CI_(3)](aq)+2dien(aq)` |
| Answer» Correct Answer - D | |
| 375. |
`CaCI_(2).6H_(2)O(s)hArrCaCI_(2)(s)+6H_(2)O(s)` `K_(p)=6.4xx10^(-17) atm^(6)` Excess solid `CaCI_(2).6H_(2)O "&" CaCI_(2) "are taken in a container containing some water vapours at a pressure of "1.14 "torr at a particular temp"`.A. `CaCI_(2)(s)` acts as drying agent under given condition.B. `CaCI_(2)(s)` acts as hydrosceopic substance given condition.C. `CaCI_(2)6H_(2)O(s)` acts as effuoroscent substance.D. Mass of `CaCI_(2).6H_(2)O(s)` increases dueto some reaction. |
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Answer» Correct Answer - C Equilibrium partial pressure of `H_(2)O(g)=(24xx10^(-18))=2xx10^(-3)atm=1.52` torr |
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| 376. |
`A(g)+2B(s) hArr 2C(g)` Initially 2 mole A (g), 4 mole of B(s) and 1 mole of an inert gas are present in a closed container. After equilibrium has established, total pressure of container becomes 9 atm.If A(g) is 50% consumed at equilibrium, then , calculate `K_p` for the :A. 9 atmB. `36/5` atmC. 12 atmD. `2/3` atm |
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Answer» Correct Answer - A |
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| 377. |
The state of equilibrium refers toA. State of restB. Dynamic stateC. Stationary stateD. State of inertness |
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Answer» Correct Answer - B Dynamic state |
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| 378. |
For the reaction, `A+2B hArr C`, the expession for equilibrium constant isA. `([A][B]^(2))/([C])`B. `([A][B])/([C])`C. `([C])/([A][B]^(2))`D. `([C])/([2B][A])` |
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Answer» Correct Answer - C `A+2B overset(K)(hArr) C` According to law of mass action `K=([C])/([A][B]^(2))` |
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| 379. |
Which is/are correct?A. `2.303 log K=-DeltaH^(ɵ)//RT+DeltaS^(ɵ)//R`B. `DeltaG^(ɵ)=-2.303RT log K`C. `-2.303 log K=-DeltaH^(ɵ)//RT^(2)+DeltaS^(ɵ)//R`D. `2.303 log K=(1//RT)(DeltaH^(ɵ)+DeltaS^(ɵ))` |
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Answer» Correct Answer - A::B `DeltaG^(ɵ)=-nRT ln K` and `2.303 log K=(-DeltaH^(ɵ))/(RT)+(DeltaS^(ɵ))/(R )` |
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| 380. |
A saturated aqueous dolution of `Mg(OH)_(2)` has a vapour pressure of `759.5 mm` at `373 K`. Calculate the solubility product of `Mg(OH)_(2)`. (Assume molarity equals molality). |
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Answer» `p_(s)=759.5 mm, P_(H_(2)O)=760 mm` at `373 K` `:. DeltaP=p_(s)-P_(H_(2)O)=0.5` `Deltap//p=0.5//760=x_(2)=6.5xx10^(-4)` We know, molality `(m)=("moles of solute")/("Volume of solvent (kg)")` `:. M=(x_(2)xx1000)/(x_(1)xxMw_(1))=((6.5xx10^(-4))xx1000)/((1-6.5xx10^(-4))xx18)=0.036` `Mg(OH)_(2) hArr Mg^(2+)+2OH^(ɵ)` `i=3` (Assuming `100%` ionisation) Solubility `(S)=m//i` `S=0.036//3=0.012 M` `K_(SP)=4S^(3)=4xx(0.012)^(3)=6.8xx10^(-6)` |
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| 381. |
At constant temperature, the equilibrium constant `(K_(p))` for the decomposition reaction `N_(2)O_(4) hArr 2NO_(2)` is expressed by `K_(p)=4x^(2)p//(1-x^(2))`, where p=pressure x= extent of decomposition. Which of the following statements is true?A. `K_(p)` increases with increase of P.B. `K_(p)` increases with increases of x.C. `K_(p)` increase with decrease of x.D.`K_(p)`remains constant with change in p and x |
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Answer» Correct Answer - D `K_(p)` depends upon temperature only. |
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| 382. |
At constant temperature, the equilibrium constant `(K_(p))` for the decomposition reaction `N_(2)O_(4) hArr 2NO_(2)` is expressed by `K_(p)=4x^(2)p//(1-x^(2))`, where p=pressure x= extent of decomposition. Which of the following statements is true?A. `K_(p)` increases with increase in p.B. `K_(p)` increases with increase in x.C. `K_(p)` increases with decrease in x.D. `K_(p)` remains constant with change in p and x |
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Answer» Correct Answer - D The equilibrium constant does not change at all with changes in concentrations, volume, pressure and presence of catalyst. It changes only with changes in temperature of the system. For endothermic reaction, the value of K increases with increase in temperature and vice versa. For exothermic reaction, the value of K decreases with increase in temperature and vice versa. |
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| 383. |
For the reaction `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, which of the following factors will have no effect on the value of equilibrium constant?A. TemperatureB. Initial concentration of `N_(2)O_(4)`C. Pressure of catalystD. Pressure |
| Answer» Correct Answer - B::C::D | |
| 384. |
The graph relates. In `K_(eq.) vs.(1)/(T)` for a reaction. The reaction must be: A. exothermicB. endothermicC. `DeltaH` is negligibleD. high spontaneous at ordinary temperature |
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Answer» In `(K_(2))/(K_(1))=(DeltaH)/(R )[(T_(2)-T_(1))/(T_(1)T_(2))]` Since `K` increase with decrease of temperature and thus `DeltaH=-ve`. |
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| 385. |
Consider the following equilibrium in a closed container `N_(2)O_(4)(g) hArr 2NO_(2)(g)` At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statement holds true regarding the equilibrium constant `(K_(p))` and the degree of diffociation `(alpha)`?A. Neither `K_(p)` nor `alpha` changesB. Both `K_(p)` and `alpha` changeC. `K_(p)` changes but `alpha` does not changeD. `K_(p)` does not change but `alpha` changes |
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Answer» Correct Answer - D The equilibrium constant depends only upon temperature. For endothermic reaction, the value of K increases with increase in temperature and vice-versa. The degree of dissociation must change to adjust for the new concentrations. |
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| 386. |
The equilibrium constant for the given reaction: `SO_(3(g))hArrSO_(2(g))+1//2O_(2(g))`, `(K_(c)=4.9xx10^(-2))` The value of `K_(c)` for the reaction: `2SO_(2(g))+O_(2(g))hArr2SO_(3(g))`, will be :A. `416`B. `2.40xx10^(-3)`C. `9.8xx10^(-2)`D. `4.9xx10^(-2)` |
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Answer» `K_(c_(1))=([SO_(2)][O_(2)]^(1//2))/([SO_(3)])=4.9xx10^(-2)` `K_(c_(2))=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=[(1)/(K_(c_(1)))]^(2)=[(1)/(4.9xx10^(-2))]^(2)` `=416.5` |
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| 387. |
For a given reversible reaction at a fixed temperature, equilibrium constants `K_(p)` and `K_(c)` are related by ……… |
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Answer» `K_(p)=K_(c )(RT)^(Deltan)` `Deltan=`number of moles of gaseous products numbers of moles of gesous reactants |
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| 388. |
Which of the following statements is false ?A. The greater the concentration of the substances involved in a reaction, the lower the speed of the reactionB. The point of dynamic equilibrium is reached when the reaction rate in one direction just balances the reaction rate in the opposite directionC. The dissociation of weak electrolyte is a reversible reactionD. The presence of free ions facilitates chemical changes |
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Answer» Correct Answer - B At equilibrium, the rate of rorward and backward reactions become equal. |
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| 389. |
A chemical reaction is catalyst X. Hence XA. reduce ethalpy of the reactionB. decreases rate constant of the reactionC. increases activeation energy of the reactionD. does not affect equilibrium constant of reaction |
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Answer» Correct Answer - D Catalyst does not affect equilibrium constant. |
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| 390. |
In the reaction `A+3BrArr2C`, initially 10 mole each of A and B were taken . If equilibrium pressure is 8 atm and`alpha " of "A=20%`then `K_(p)` will be :-A. `(1)/(4)`B. `(1)/(8)`C. `(1)/(16)`D. `(1)/(2)` |
| Answer» Correct Answer - 2 | |
| 391. |
Equilibrium constants are given (in atm) for the following reactions at `0^(@)C` : `SrCl_2 . 6H_2O(s) hArr SrCl_2 . 2H_2O(s)+4H_2O(g) K_p=5xx10^(-12)` `Na_2HPO_4 . 7H_2O(s)+5H_2O(g) K_p=2.43xx10^(-13)` `Na_2SO_4 . 10H_2O(s) hArr Na_2SO_4(s)+10H_2O(g) K_p=1.024xx10^(-27)` The vapour pressure of water at `0^@C` is 4.56 torr. At what relative humidities will `Na_2SO_4` be deliquescent (i.e. absorb moisture) when exposed to the air at `0^@C` ?A. above 33.33%B. below 33.33%C. above 66.66%D. below 66.66% |
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Answer» Correct Answer - A |
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| 392. |
In a 7.0 L evacuated chamber, 0.50 mole `H_(2)` and 0.50 mole `I_(2)` react at `427^(@)` C. According to reaction `H_(2)(g)+I_(2)(g)hArr2HI(g)` At the given temperature, `K_(c)=49` for the reaction. What is the value of `K_(p)`?A. 7B. 49C. 24.35D. None of these |
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Answer» Correct Answer - B |
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| 393. |
Calculate the equilibrium constant `(K_(c))` for the reaction given below , if at equilibrium maxture conyains 5.0 mole of `A_(2),3 mole of `B_(2)` and 2 mole of `AB_(2)` at 8.21 atm and 300K `A_(2)(g)+2B_(2)(g)hArr2AB_(2)(g)+Heat`A. `1.333`B. `2.66`C. `20`D. none of these |
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Answer» Correct Answer - b |
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| 394. |
The equilibrium constant `K_(c)` for the following reaction at `842^(@)C`is `7.90xx10^(-3)` .What is `K_(p)`at same temperature ? `(1)/(2)f_(2)(g)hArrF(g)`A. `8.64xx10^(-5)`B. `8.26xx10^(-4)`C. `7.90xx10^(-2)`D. `7.56xx10^(-2)` |
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Answer» Correct Answer - d |
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| 395. |
Equilibrium constants are given (in atm) for the following reactions at `0^(@)C` : `SrCl_2 . 6H_2O(s) hArr SrCl_2 . 2H_2O(s)+4H_2O(g) K_p=5xx10^(-12)` `Na_2HPO_4 . 7H_2O(s)+5H_2O(g) K_p=2.43xx10^(-13)` `Na_2SO_4 . 10H_2O(s) hArr Na_2SO_4(s)+10H_2O(g) K_p=1.024xx10^(-27)` The vapour pressure of water at `0^@C` is 4.56 torr. At what relative humidity will `Na_2SO_4 . 10H_2O` be efflorescent when exposed to air at `0^@ C`?A. above 33.33%B. below 33.33%C. above 66.66%D. below 66.66% |
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Answer» Correct Answer - B |
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| 396. |
Equilibrium constants are given (in atm) for the following reactions at `0^(@)C` : `SrCl_2 . 6H_2O(s) hArr SrCl_2 . 2H_2O(s)+4H_2O(g) K_p=5xx10^(-12)` `Na_2HPO_4 . 7H_2O(s)+5H_2O(g) K_p=2.43xx10^(-13)` `Na_2SO_4 . 10H_2O(s) hArr Na_2SO_4(s)+10H_2O(g) K_p=1.024xx10^(-27)` The vapour pressure of water at `0^@C` is 4.56 torr. Which is the most effective drying agent at `0^@C` ?A. `SrCl_2 . 2H_2O`B. `Na_HPO_4 . 7H_2O`C. `Na_2SO_4`D. All equally |
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Answer» Correct Answer - A |
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| 397. |
For the dissociation reaction `N_(2)OhArr2NO_(2)(g),` the equilibrium constant `K_(P)` is `0.120` atm at `298 K` and total pressure of system is `2` atm. Calculate the degree of dissociation of `N_(2)O_(4)`. |
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Answer» For the reaction `N_(2)O(g)hArr2NO_(2)(g)` `{:("At equilibrium",,1-alpha,,2alpha),("moles",,,,):}` Let `apha` be the degree of dissociation and P is the total pressure then Total number of moles `=2alpha+1-alpha=1+alpha` `:. P_(N_(2)O)= "moles"` of `N_(2)OxxP_(total)` `=((1-alpha)/(1+alpha))P` and `P_(NO_(2))=((2alpha)/(1+apha))P` `:. K_(P)=((P_(NO_(2))))/(P_(N_(2)O))=([((2alpha)/(1+alpha))P]^(2))/(((1-alpha)/(1+alpha))P)` `=(4alpha^(2)P^(2))/(1+alpha)xx(1+alpha)/((1-alpha)P)=(4alpha^(2)P)/(1-alpha^(2)) ...(1)` Given `K_(P)=0.120 "atm", P=2 "atm"` Subtituting all the values in equation (i), we get `0.120=(4alpha^(2)(2))/((1-alpha^(2)))=(8alpha^(2))/((1-alpha^(2)))` `rArr 0.120(1-alpha^(2))=8alpha^(2)` `:.` Degree of dissociation, `alpha=(0.120/8.12)^(1//2)=0.121` |
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| 398. |
A sample of air consisting of `N_(2)` and `O_(2)` was heated to `2500 K` until the equilibrium `N_(2)(g)+O_(2)(g)hArr2NO(g)` was established with an equlibrium constant, `K_(c )=2.1xx10^(-3)`. At equilibrium, the `"mole"%` of `NO` was `1.8`. Eatimate the initial composition of air in mole fraction of `N_(2)` and `O_(2)`. |
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Answer» Let the total number of mole of `N_(2)` and `O_(2)` be `100` containing a mole of `N_(2)` initially. `N_(2)+O_(2)hArr2NO` `{:(a,(100-a),,0,"Initial mol"),((a-x),(100-a-x),,2x,"mole at equilibrium"):}` mole `%` of NO at equilibrium `=(2x)/((a-x)+(100-a-x)+2x)xx100=1.8` `:. x=0.9` Thus, at equilibrium, mole of `N_(2)=(a-0.9)` mole of `O_(2)=(100-a-0.9)=(99.1-a)` "mole" of `NO=2x=2xx0.9=1.8` `K_(c )=((2x//V)^(2))/(((a-0.9)/V)((99.1-a)/V))=((2x)^(2))/((a-0.9)(99.1-a))` `2.1xx10^(-3)=((1.8)^(2))/((a-0.9)(99.1-a))` `a=74.46%` and `20.54%` As we know that `%` of `N_(2)` in the air is more than that of `O_(2)`. |
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| 399. |
In one experiment, certain amount of `NH_(4)I(s)` was heated rapidly in a closed container at `375^(@)C`. The following equilibrium was established: `NH_(4)I(s)hArrNH_(3)(g)+HI(g)` Excess of `NH_(4)I(s)` remained unreacted in the flask and equilibrium pressure was `304mm` of Hg. After some time, the pressure started increasing further owing to the dissociation of `HI`. 2HI(g)hArrH_(2)(g)+l_(2)(g) K_(C)=0.010` calculate final pressure.A. `331mm "of" Hg`B. `335mm "of" Hg`C. `662mm "of" Hg`D. `151 mm "of" Hg` |
| Answer» Correct Answer - A | |
| 400. |
For any reaction, greater the value of equilibrium constant greater is the extent of reaction. |
| Answer» Correct Answer - True | |