InterviewSolution
Saved Bookmarks
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.
| 651. |
For the following mechanism, `P + Qoverset(K_(A))underset(K_(B))hArr PQ` `overset(K_(C))underset(K_(D))hArr` R at equilibrium `([R])/([P][Q])` is: [K represents rate constant]A. `(K_(A).K_(B))/(K_(C).K_(D))`B. `(K_(A).K_(D))/(K_(B).K_(C))`C. `(K_(B).K_(D))/(K_(A).K_(C))`D. `(K_(A).K_(C))/(K_(B)K_(D))` |
|
Answer» Correct Answer - D `(K_(A))/(K_(B))=([PQ])/([P][Q])` ....(i) ` (K_(C))/(K_(D))=([R])/([PQ])`.....(ii) On multiply equation (i) and (ii) we get `(K_(A).K_(C))/(K_(B).K_(D))=([R])/([P][Q])` |
|
| 652. |
A mixture of `0.3` mole of `H_(2)` and `0.3` mole of `I_(2)` is allowed to react in a 10 litre evacuated flask at `500^(@)`C. The reaction is `H_(2) + I_(2) hArr 2H`, the K is found to be 64. The amount of undreacted `I_(2)` at equilibrium isA. `0.15` moleB. `0.06` moleC. `0.03` moleD. `0.2` mole |
|
Answer» Correct Answer - B `K_(c) = ([HI]^(2))/([H_(2)][I_(2)]), 64 = (x^(2))/(0.03xx0.03)` `x^(2) = 64xx9xx10^(-4)` `x = 8xx3xx10^(-2) = 0.24` x is the amount of HI at equilibrium amount of `I_(2)` at equilibrium will be `0.30 - 0.24 = 0.06` |
|
| 653. |
The vapour density of fully dissociated `NH_(4)Cl` would beA. Slightly less than half of that of ammonium chloride.B. Half of that of ammonium chloride.C. Double that of ammonium chlorideD. Determined by the amount of solid ammonium chloride used in the experment. |
|
Answer» Correct Answer - B `("normal mol. mass")/("exp.mol.mass")=1+alpha` `NH_(4)CIhArrNH_(3)+HCI alpha=1` `"exp Mol. Mass"=("Normal mol mass")/(2)` |
|
| 654. |
The vapour density of completely dissociated `NH_(4)C1` would beA. Slight less than half that `NH_(4)C1`B. Half that of `NH_(4)C1`C. Double that of `NH_(4)C1`D. Determined by the amount of solid `NH_(4)C1` in the experiment |
|
Answer» Correct Answer - B `("Normal molecular weight")/("experimental molecular wt.") =1+alpha` `NH_(4)C1hArrNH_(3) + HC1` `.: alpha = 1` `:.` "Experimental Molecular wt". `= ("nor.mol.wt.")/(2)` |
|
| 655. |
For the following mechanism, `P + Qoverset(K_(A))underset(K_(B))hArr PQ` `overset(K_(C))underset(K_(D))hArr` R at equilibrium `([R])/([P][Q])` is: [K represents rate constant]A. `(K_(A).K_(B))/(K_(C).K_(D))`B. `(K_(A).K_(D))/(K_(B).K_(C))`C. `(K_(B).K_(D))/(K_(A).K_(C))`D. `(K_(A).K_(C))/(K_(B).K_(D))` |
|
Answer» Correct Answer - D `(K_(A))/(K_(B)) =([PQ])/([P][Q])` ….(i) `(K_(C))/(K_(D)) =([R])/([P][Q])` ……(ii) On multipiy equation (i) and (ii) we get `(K_(A).K_(C))/(K_(B).K_(D)) =([R])/([P][Q])` |
|
| 656. |
The vapour density of fully dissociated `NH_(4)Cl` would beA. Less than half of the vapour density of pure `NH_(4)Cl`B. Double of the vapour density of pure `NH_(4)Cl`C. Half of the vapour density of pure `NH_(4)Cl`D. One-third of the vapour density of pure `NH_(4)Cl` |
| Answer» Correct Answer - C | |
| 657. |
For which of the following reaction does the equilibrium constant depend on the units of concentration?A. `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))` |
|
Answer» Correct Answer - D `Delta n = 1` for this change So the equilibrium constant depends on the unit of concentration. |
|
| 658. |
`100 g` of `NaCl` is stirred in `100 mL` of water at `20^(@)C` till the equilibrium is attained: (`a`) How much `NaCl` goes into the solution and how much of it is left undissolved at equilibrium? The solubility of `NaCl` at `20^(@)C` is `6.15 mol//litre`. (`b`) What will be the amount of `NaCl` left undissolved, if the solution is diluted to `200 mL`? |
|
Answer» Solubility of `NaCl=6.15mol//litre` `=6.15xx58.5g//litre` `=(6.15xx58.5)/(10)g//100mL` `=36g//100mL` (`a`) Thus amount of `NaCl` in `100mL` at `20^(@)C` get dissolved `=36g` and amount of `NaCl` in `100mL` at `20^(@)C` remained undissolved `=100-36=64g` (`b`) If volume of solution is diluted to `200mL`, `36g` more of `NaCl` will be dissolved leaving only `28g NaCl` dissolved in `200 mL`. |
|
| 659. |
For which of the following reaction does the equilibrium constant depend on the units of concentration?A. `NO(g) hArr 1/2 N_(2)(g)+1/2 O_(2)(g)`B. `C_(2)H_(5)OH(l)+CH_(3)COOH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)`C. `2HI(g) hArr H_(2)(g)+I_(2)(g)`D. `COCl_(2)(g) hArr CO(g)+Cl_(2)(g)` |
| Answer» Correct Answer - D | |
| 660. |
The yield of product in the reaction, `A_(2)(g)+2B(g) hArr C(g)+Q KJ` would be higher at:A. Low temperature and high pressureB. High temperature and high pressureC. Low temperature and low pressureD. High temperature and low pressure |
| Answer» Correct Answer - A | |
| 661. |
Assertion: Water boils at higher temperature in pressure cooker. Reason: Increase in pressure leads to an increase in boiling temperature.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» Boiling of a liquid occurs, when its vapour pressure becomes equal to atmospheric pressure. | |
| 662. |
Unit of equilibrium constant for the reversible reaction `H_(2) + I_(2)hArr 2HI` isA. `mol^(-1)` litreB. `mol^(-2)` litreC. mol `litre^(-1)`D. None of these |
|
Answer» Correct Answer - D Unit of `K_(p) = (atm)^(Delta n)` Unit of `K_(c) = ("mole"//"litre")^(Delta n)` `=["mole"//"litre"]^(0) = 0` |
|
| 663. |
The vapour density of `N_(2)O_(4)` at a certain temperature is `30`. Calculate the percentage dissociation of `N_(2)O_(4)` this temperature.A. `53.3%`B. `106.6%`C. `26.7%`D. None of these |
|
Answer» Correct Answer - A `(VD_(i))/(VD_(e)) =(n_(e))/(n_(i))` `(46)/(30) = (1+alpha)` `1.533 = 1+ alpha` `alpha =0.533` `%` dissociation `=53.3%` |
|
| 664. |
The equation `alpha=(D-d)/((n-1)d)` is correctly matched for: (`alpha` is the degree of dissociation, D and d are the vapour densities before and after dissociation, respectively).A. B. C. D. |
|
Answer» Correct Answer - B `alpha=(D-d)/((n-1)d)rArr(n-1)alpha=D/d-1` `rArr D/d=1+(n-1) alpha` `rArr d/D=1/(1+(n-1)alpha)=1/((1-alpha+nalpha))` `rArr` Any equation in which the total number of moles on product side =n and on reactant side=`1`. Clearly, (b) satisfies. |
|
| 665. |
In the dissociation of `N_(2)O_(4)` into `NO_(2), (1+ alpha)` values with the vapour densities ratio `((D)/(d))` is given by: [`alpha` degree of dissociation, D-vapour density before dissociation, d-vapour density after dissociation]A. B. C. D. |
| Answer» Correct Answer - A | |
| 666. |
The vapour density of `N_(2)O_(4)` at a certain temperature is `30`. Calculate the percentage dissociation of `N_(2)O_(4)` this temperature. |
| Answer» Correct Answer - 50 | |
| 667. |
To the system, `LaCl_(3)(s)+H_(2)O(g) hArr LaClO(s)+2HCL(g)-"Heat"` already at equilibrium, more water vapour is added without altering temperature or volume of the system. When equilibrium is re-established, the pressure of water vapour is doubled. The pressure of `HCl` present in the system increases by a factor ofA. `2`B. `2^(1//2)`C. `2^(1//3)`D. `2^(2)` |
| Answer» Correct Answer - B | |
| 668. |
i. The initial pressure of `PCl_(5)` present in one litre vessel at `200 K` is `2` atm. At equilibrium the pressure increases to `3` atm with temperature increasing to `250`. The percentage dissociation of `PCl_(5)` at equilibrium isA. `30%`B. `60%`C. `0.2%`D. `20%` |
|
Answer» Correct Answer - D i. First method: Pressure at `200 K=2 "atm"` Pressure at `250 K=P "atm"` Using relation, `P_(1)/T_(1)=P_(2)/T_(2)rArr P/250=2/200` `:. P=2.5 "atm"` `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",a,,0,,0),("At equilibrium",a-alpha,,alpha,,alpha),("Total moles" =a+alpha,,,,,):}` `("Intial mole")/("moles at equilibrium")= (("Intial pressure after"),("change of temperature"))/("Equilibrium pressure")` `(a)/(a+alpha)=2.5/3 rArr alpha=1/5 a` Therefore, `%` of `PCl_(5)` dissociated `=alpha/axx100` `=a/(5xxa)xx100=20%` Secondethod: Using relation: `alpha=(T_(2)P_(2)-T_(2)P_(1))/(T_(2)P_(1))` `=(200xx3-250xx2)/(250xx2)=(600-500)/(500)` `alpha=1/5` `% "of" alpha=1/5xx100=20%` |
|
| 669. |
To the system, `LaCl_(3)(s)+H_(2)O(g) hArr LaClO(s)+2HCL(g)-"Heat"` already at equilibrium, more water vapour is added without altering temperature or volume of the system. When equilibrium is re-established, the pressure of water vapour is doubled. The pressure of `HCl` present in the system increases by a factor ofA. 2B. `sqrt2`C. `sqrt3`D. `sqrt5` |
|
Answer» Correct Answer - B |
|
| 670. |
During thermal dissociation of a gas, the vapour density.A. Remains the sameB. IncreasesC. DecreasesD. Increases in some cases and decreases in others |
| Answer» Correct Answer - C | |
| 671. |
If the pressure of `N_(2)//H_(2)` mixture in a closed vessel is 100 atmospheres and `20%` of the mixture then reacts the pressure at the same temperature would be .A. `100`B. `90`C. `85`D. `80` |
| Answer» `20 %` mixture reacts to form `10 % NH_(3)` , Thus `80%` mixture and `10 % NH_(3)` left or total pressure left `=90 atm` since `100%` mixture has `100 atm`. | |
| 672. |
In a chemical equilibrium, `K_(c)=K_(p)` whenA. The number of molecules entering into a reaction is more than the number of molecules produced.B. The number of molecules entering into the reaction is equal to the number of molecules produced.C. the number of molecules entering into the reaction is less to the number of moleculed produced.D. None of the above |
| Answer» Correct Answer - B | |
| 673. |
If `340 g` of a mixture of `N_(2)` and `H_(2)` in the correct ratio gas a `20%` yield of `NH_(3)`. The mass produced would be:A. `16 g`B. `17 g`C. `20 g`D. `68 g` |
|
Answer» `20%` yield of `NH_(3)` and thus `20%` of `340 g` `=(20xx340)/(100)=68 g` |
|
| 674. |
The chemical equilibrium is ………… in nature. |
| Answer» Correct Answer - Dynamic | |
| 675. |
For the chemical reaction `3X(g)+Y(g) hArr X_(3)Y(g)`, the amount of `X_(3)Y` at equilibrium is affected byA. Temperature and pressureB. Temperature onlyC. Pressure onlyD. Temperature, pressure, and catalyst |
|
Answer» Correct Answer - A The given reaction will be exothermic in nature due to the formation of three X-Y bonds from the gaseous atoms. The reaction is also accompanied with the decrease in the gaseous species. Hence, the reaction will be affected by both temperature and pressure. The use of catalyst does not affect the equilibrium concentration of the species in the chemical reaction. |
|
| 676. |
`CO+2H_(2) rarr CH_(3)OH` (all gases). An equilibrium mixture consists of `2.0` atm `CH_(3)OH, 1` atm CO and `0.1` atm `H_(2)`. The volume, at same T. Find new equilibrium pressures. |
|
Answer» Correct Answer - A `{:(CO(g),+,2H_(2)(g),hArr,CH_(3)OH(g)),(0.2 "mol",,-,,-),((0.2-0.1),,x,,0.1 "mol" larr "at equilibrium"):}` Total moles of equilibrium `=0.1+x+0.1=0.2+x` Also total moles `=(PV)/(RT)=(4.92xx5)/(0.0821xx600)=0.5` `rArr 0.5=0.2+x` `rArr x=0.3`="mol" of `H_(2)` at equilibrium `K_(c)=([CH_(3)OH])/([CO][H_(2)]^(2))=(0.1//5)/((0.1//5)(0.3//5)^(2))=277.8` If there is no catalyst, no reaction occurs. `n=n_(CO)+n_(H_(2))=0.2+0.5=0.7` `P=n(RT)/V=6.88 "atm"` |
|
| 677. |
Calculate the volume percent of chlorine gas at equilibrium in the dissociation of `PCl_(5)(g)` under a total pressure of `1.5` atm. The `K_(p)` for its dissociation `=0.3`. |
|
Answer» Correct Answer - B::C `{:(,PCl_(5)(g),hArr,Pcl_(3)(g),+,Cl_(2)(g)),(underset(("moles"))(t=0),a,,0,,0),(t=t_("eq"),a-x,,x,,x):}` `K_(p)=(P_(PCl_(3)).P_(Cl_(2)))/(P_(PCl_(5)))=([x/(a+x)P][x/(a+x)P])/((a-x)/(a+x)P)` `rArr K_(p)=(x^(2)P)/(a^(2)-x^(2))=(alpha^(2)P)/(1-alpha^(2))=0.3[alpha=x/a]rArr alpha=sqrt(1/6)` `%Cl_(2)=x/(a+x)=alpha/(1+alpha)xx100%=28.6%` |
|
| 678. |
At a certain temperature , `K_(p)` for dissociation of solid `CaCO_(3)` is `4xx10^(-2)` atm and for the reaction, `C(s)+CO_(2) hArr 2CO` is `2.0` atm, respectively. Calculate the pressure of CO at this temperature when solid `C, CaO, CaCO_(3)` are mixed and allowed to attain equilibrium. |
| Answer» Correct Answer - `0.28` , | |
| 679. |
The following pictures represents the equilibrium state for three different reactions of the type `A_(2)+X_(2)hArr2AX` `(X=B,C "or" D)` Which reaction has the largest equilibrium constant?A. `A_(2)+B_(2)hArr2AB`B. `A_(2)+C_(2)hArr2AC`C. `A_(2)+D_(2)hArr2AD`D. None of these |
| Answer» Correct Answer - B | |
| 680. |
A catalyst …………… the equilibrium state but helps to attain in lesser time. |
| Answer» Correct Answer - Does not change | |
| 681. |
The dissociation of `CaCO_(3)` is suppressed at high pressure |
| Answer» Correct Answer - True | |
| 682. |
More of `SO_(3)` decompose at loe temperature. |
| Answer» Correct Answer - True | |
| 683. |
The equilibrium constant `(K)` for the reaction. `A+2BhArr2C+D` is:A. `([C]^(2)[D])/([A][2B])`,B. `([2C][D])/([A][2B])`C. `([C][D])/([A][B])`D. `([C]^(2)[D])/([A][B]^(2))` |
| Answer» Correct Answer - D | |
| 684. |
In the gaseous phase reaction `C_(2)H_(4) +H_(2) hArr C_(2)H_(6),` the equilibrium constant can be expressed in the units to :A. `mol^(2)dm^(-3)`B. `dm^(3)mol^(-1)`C. `dm^(3)mol^(-1)`D. `mol.dm^(-3)` |
| Answer» Correct Answer - B | |
| 685. |
NO and `Br_(2)` at initial pressures of `98.4` and `41.3` torr respectively were allowed react at `300 K`. At equilibrium the total pressure was `110.5` torr. Calculate the value of equilibrium constant, `K_(p)` and the standard free energy change at `300 K` for the reaction: `2NO(g)+Br_(2)(g) hArr 2NOBr(g)` |
|
Answer» Correct Answer - A::B::C::D `{:(NO,rarr,Br_(2),hArr,2NOBr),(98.4,,41.3,,),((98.4-2x),,(41.3-x),,2x):}` `P_(t)` (at equilibrium)`=(98.4-2x)+(41.3-x)+2x` `=110.5` (given) `rArr x=29.2` `{:(P_(NO),=98.4-2x=40,),(P_(Br_(2)),=41.3-x=12.1,),(,P_(NOBr)=2x=58.4,):}] rArr K_(P)=(P_(NOBr)^(2))/(P_(NO)^(2)P_(Br))` `=0.1762 "torr"^(-1)` `K_(p)`= should be in atm units for calculation of `DeltaG^(ɵ)`. `K_(p)=0.1762xx760=133.9 "atm"^(-1)` `DeltaG^(ɵ)=-2.303 RT log K_(p)=-12216.26 J "mol"^(-1)` `=-12.22 kJ "mol"^(-1)` |
|
| 686. |
If `log(K_(C))/(K_(P))-log(1)/(RT)=0`, then above is true for the following equilibrium reactionA. `NH_(2)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`B. `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)`C. `2NO_(2)(g)hArrN_(2)O_(4)(g)`D. `H_(2)(g)+I_(2)(g)hArr2HI(g)` |
|
Answer» Correct Answer - A::B `"log"(K_(C))/(K_(p))="log"(1)/((RT))` `rArrK_(p)=K_(c)(RT) rArr Deltan=1` |
|
| 687. |
The equilibrium concentration of x, y and z are `4, 2` and `2` mol `L^(-1)`, respectively, at equilibrium of the reaction `2x+y hArr z`. The value of `K_(c)` is …………….. |
| Answer» `1/16 "mol"^(-2) L^(2)` | |
| 688. |
The `K_(P)` values for three reactions are `10^(-5), 20` and `300` then what will be the correct order of the percentage composition of the products. |
| Answer» Since `K_(P)` order is `10^(-5)lt20lt300` so the percentage composition of products will be greatest for `K_(P)=300.` | |
| 689. |
The equilibrium constant for the reaction `H_(2)+Br_(2)hArr2HBr` is `67.8` at `300(@)K`. The equilibrium constant for the dissociation of `HBr` is:A. `0.0147`B. `67.80`C. `33.90`D. `8.349` |
| Answer» Correct Answer - A | |
| 690. |
The reaction for which `K_(P)=K_(C)` is satisfiedA. `A(g)+2B(g)hArr3C(g)`B. `A(s)hArrB(g)`C. `2A(g)hArrB(g)+C(g)`D. `A(s)+B(g)toC(s)+2D(g)` |
| Answer» Correct Answer - A::C | |
| 691. |
For the reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)`, the value of `K_(p)` is `1.7xx10^(3)` at `500K` and `1.7xx10^(4)` at `600K`. Which of the following is/are correct ?A. The proportions of `NO_(2)` in the equilibrium mixture is increased by decrease in pressure.B. The standard enthalpy change for the forward reaction is negativeC. Units of `K_(p)` are `atm^(-1)`D. At `500 K` the degree of dissociation of `N_(2)O_(4)` decreases by `50%` by increasing the pressure by `100%` |
|
Answer» Correct Answer - A `Deltan=2-1=1` a. That is, with the decrease of pressure, reaction shifts towards right i.e. proportions of `NO_(2)` increases. Statement (a) is correct. b. Value of K increases with increase of temperature and hence reaction is endothermic, i.e., `DeltaH=+ve` Hence statement (b) is incorrect. c. `K_(p)=([p_(NO_(2))]^(2))/([p_(N_(2)O_(4))])=("atm"^(2))/("atm")="atm"` Hence statement (c) is incorrect. d. `{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial",1 "atm",,0),("At eq.",1-alpha,,2alpha):}` `K_(p)=((2alpha)^(2))/(1-alpha)=1.7xx10^(3)` at `500 K` `[(1-alpha~~1), "since" alpha "is small"]` `4alpha^(2)=1.7xx10^(3)` `alpha_(1)=sqrt((1.7xx10^(3))/4)` `alpha=0.206xx10^(2)` When the pressure was `1` atm, now let it be `2` atm `(100% "increase")`. `{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial",2,,0),("At eq.",2-alpha,,2alpha):}` `K_(p)=((2alpha)^(2))/(2-alpha)=1.7xx10^(3)` at `500 K` `[2-alpha~~2, "since" alpha "is small"]` `(4alpha^(2))/2=1.7xx10^(3)` `alpha_(2)=sqrt((1.8xx10^(3))/2)` `alpha_(2)/alpha_(1)=sqrt((1.7xx10^(3)xx4)/(2xx1.7xx10^(3)))=sqrt(2)=1.4` `:.` When the pressure is increased `100%` the decrease in `alpha` is `1.4` times which is not `50%`. Hence statement is wrong. e. `K_(p)` at `600 K=1.78xx10^(4)` `N_(2)O_(4)(g) hArr 2NO_(2)(g)` `Deltan=2-1=1` Since, by decrease of pressure reaction goes forward, i.e., more of `N_(2)O_(4)` will dissociate. It means by decreasing pressure dissociation of `N_(2)O_(4)` increases. Hence, the statement is wrong. |
|
| 692. |
`N_(2)O_(4) hArr 2NO_(2), K_(c)=4`. This reversible reaction is studied graphically as shown in the figure. Select the correct statement out of `I, II` and `III`. I : Reaction quotient has maximum value at point `A` II : Reaction proceeds left to right at a point when `[N_(2)O_(2)]=[NO_(2)]=0.1 M` III : `K=Q` when point `D` or `F` is reached: A. I, IIB. II, IIIC. IID. I, II, III |
|
Answer» Correct Answer - B `N_(2)O_(4)(g) hArr 2NO_(2)(g), K_(c )=4` From graph, at point `A, Q=0` When `[N_(2)O_(4)]=[NO(2)]=0.1 M` `rArr Q=([NO_(2)]^(2))/([N_(2)O_(4)])=0.1 lt K_(c ) rArr ` Reaction is movind in forward direction. At point `D` or `F[NO_(2)]` and `[N_(2)O_(4)]` is constant with time `rArr` Achivement of chemical equilibrium. |
|
| 693. |
`N_(2)O_(4)(g)hArr2NO_(2)(g),K_(C)=4`. This reversible reaction is studied graphically as shown in figure. Select the correct statements. A. Reaction quotient has maximum value at point `A`B. Reaction proceeds left to riht at a point when `[N_(2)O_(4)]=[NO_(2)]=0.1M`C. `K_(C)=Q` when point `D` or `F` is reached:D. None of these |
| Answer» Correct Answer - B::C | |
| 694. |
The relation between `K_(p)` and `K_(c)` of a reversible reaction at constant temperature is `K_(p)`=………….. |
| Answer» `K_(p)=K_(c )(RT)^(Deltan)` | |
| 695. |
For a reaction `N_(2) + 3H_(2)hArr2NH_(3)`, the value of `K_(c)` does not depend upon: (A) Initial concentration of the reactans (B) Pressure (C) Temperature (D) catalystA. Only CB. A, B, CC. A, B, DD. A, B, C, D |
|
Answer» Correct Answer - C Equilibrium constant is only temperature dependent. |
|
| 696. |
`10.00` mol of pure, `1-`butyne was taken in a flask with some basic catalyst, sealed and heated to `500K` where it isomerised to produce `2-`butyne and `1,3-`butadiene. At equlibrium `3.0` mole of `2-`butyne and `5.5` mole of `1.3-`butadiene were found to be present in the flask. If `2.0` mole of `1.3-`butadiene wer added further to the above equilibrium mixture. What would be the new equilibrium composition when equilibrium was established again? 1-butyne 2-butyne 1,3-butadieneA. 1.80moles 3.60moles 3.60molesB. 3.60moles 6.60moles 1.80molesC. 1.80moles 6.6moles` 3.60molesD. 3.60moles 1.80moles 6.60moles |
| Answer» Correct Answer - A | |
| 697. |
The relation between `K_(p)` and `K_(c)` of a reversible reaction at constant temperature is `K_(p)`=…………..A. `K_(c)=K_(p)(RT)^(Deltan)`B. `K_(p)=K_(c)(RT)^(-Deltan)`C. `K_(p)=K_(c)(RT)^(Deltan)`D. `K_(c)=K_(p)(RT)^(-Deltan)` |
| Answer» Correct Answer - (c,d) | |
| 698. |
A catalyst accelerates a reaction primarily by stablizing theA. substrateB. productC. intermediateD. transition state |
| Answer» Correct Answer - D | |
| 699. |
Rate of reaction curve for equilibrium can be like: [`r_(f)` = forward rate, `r_(b)` = backward rate]A. B. C. D. |
|
Answer» Correct Answer - A Concentration of reactant and product remains constant w.r.t time. And, rate of [AT EQUILIBRIUM] forward reaction `(r_(f))` = rate of backward reaction `(r_(b))`. |
|
| 700. |
The equilibrium : `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))` shows that `K_(p) (atm)` is double to the value of `K_(c)` (`mol litre^(-1)`) at a particular temperature `T`, then `T` is :A. `2 K`B. `12.18 K`C. `24.36 K`D. `27.3 K` |
|
Answer» `(K_(p))/(K_(c))=(RT)^(Deltan)`, `:. 2=(0.0821xxT)^(1)` `:. T=24.36 K` |
|