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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.
| 501. |
For a reaction, `aA+bBhArrcC+dD`, the reaction quotient `Q=([C]_(0)^(c)[D]_(0)^(d))/([A]_(0)^(a)[B]_(0)^(b))`, where `[A]_(0)`, `[B]_(0)`, `[C]_(0)`, `[D]_(0)` are initial concentrations. Also `K_(c)=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))` where `[A]`, `[B]`, `[C]`, `[D]` are equilibrium concentrations. The reaction proceeds in forward direction if `Q lt K_(c)` and in backward direction if `Q gt K_(c)`. The variation of `K_(c)` with temperature is given by: `2303log(K_(C_(2)))/(K_(C_(1)))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`. For gaseous phase reactions `K_(p)=K_(c)(RT)^(Deltan)` where `Deltan=` moles of gaseous products `-` moles of gaseous reactants. Also `-DeltaG^(@)=2.303RT log_(10)K_(c)`. The equilibrium constant `K_(c)` for `A_((g))hArrB_((g))` is `1.1`. The gas having concentration `ge 1` is:A. `[B]` if `[A]=0.91`B. `[B]` if `0.9lt[A]le1`C. both `[A]` and `[B]` if `[A]gt1`D. all are correct |
| Answer» `K_(c)=([B])/([A])=1.1` see all are correct. | |
| 502. |
For a reaction, `aA+bBhArrcC+dD`, the reaction quotient `Q=([C]_(0)^(c)[D]_(0)^(d))/([A]_(0)^(a)[B]_(0)^(b))`, where `[A]_(0)`, `[B]_(0)`, `[C]_(0)`, `[D]_(0)` are initial concentrations. Also `K_(c)=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))` where `[A]`, `[B]`, `[C]`, `[D]` are equilibrium concentrations. The reaction proceeds in forward direction if `Q lt K_(c)` and in backward direction if `Q gt K_(c)`. The variation of `K_(c)` with temperature is given by: `2303log(K_(C_(2)))/(K_(C_(1)))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`. For gaseous phase reactions `K_(p)=K_(c)(RT)^(Deltan)` where `Deltan=` moles of gaseous products `-` moles of gaseous reactants. Also `-DeltaG^(@)=2.303RT log_(10)K_(c)`. The equilibrium constants for the reaction, `CaC_(2(s))+5O_(2(g))hArr2CaCO_(3(s))+2CO_(2(g))` is//are given by:A. `K_(c)=([CO_(2)]^(2))/([O_(2)]^(5))`B. `K_(p)=((n_(CO_(2)))^(2))/((n_(O_(2)))^(5))xx[(P)/(sumn]]^(-3)`C. `K_(p)=((p_(CO_(2)))^(2))/((P_(O_(2)))^(5))`D. either of these |
| Answer» Correct Answer - All are correct. | |
| 503. |
For a reaction, `aA+bBhArrcC+dD`, the reaction quotient `Q=([C]_(0)^(c)[D]_(0)^(d))/([A]_(0)^(a)[B]_(0)^(b))`, where `[A]_(0)`, `[B]_(0)`, `[C]_(0)`, `[D]_(0)` are initial concentrations. Also `K_(c)=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))` where `[A]`, `[B]`, `[C]`, `[D]` are equilibrium concentrations. The reaction proceeds in forward direction if `Q lt K_(c)` and in backward direction if `Q gt K_(c)`. The variation of `K_(c)` with temperature is given by: `2303log(K_(C_(2)))/(K_(C_(1)))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`. For gaseous phase reactions `K_(p)=K_(c)(RT)^(Deltan)` where `Deltan=` moles of gaseous products `-` moles of gaseous reactants. Also `-DeltaG^(@)=2.303RT log_(10)K_(c)`. (`B`) The moisture content of a gas is often expressed as dew point, the temperature at which if gas is cooled becomes saturted with vapour pressure of water at that temperture. Dew point of `H_(2)O` is `43^(@)C` having vapour pressure `0.07 "torr"`. `CaCl_(2)*2H_(2)O_((s))hArrCaCl_(2(s))+2H_(2)O_((g))`, The equilibrium constant should not be more than..... `(atm)^(2)` if `CaCl_(2)` is to be used as desiccant.A. `8.5xx10^(-9)`B. `8.5`C. `8.5xx10^(-3)`D. `4.9xx10^(-3)` |
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Answer» `K_(p)=(P_(H_(2)O))^(2)` `=(0.07)^(2)=4.9xx10^(-3)("torr")^(2)` `=(4.9xx10^(-3))/((760)^(2)=8.5xx10^(-9)atm^(2)` If `CaCl_(2)` is to be used as desiccant `K_(p)` should be lesser than `8.5xx10^(-9)atm^(-2)`. |
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| 504. |
If `K_(c)/(K_(p))-"log"1/(RT)=0` then above is ture for the following equilibrium reaction :A. `NH_(3)(g)hArr(1)/(2)N_(2)+(3)/(2)H_(2)(g)`B. `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)`C. `2NO_2(g)hArrN_2O_4(g)`D. `H_2(g)+I_2(g)hArr2HI(g)` |
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Answer» Correct Answer - A::B |
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| 505. |
For a reversible reaction `alphaA+betaBhArr^.cC+^.dD`,the variation of K with temperature is given by log`(K_(2))/(K_(1))=(-DeltaH^circ)/(2.303R)[(1)/T_(2)-(1)/T_(1)]`then,A. `K_(2) gt K_1if T_2 gt T_1` for an endothermic changeB. `K_(2) lt K_1if T_2 gt T_1` for an endothermic changeC. `K_(2) gt K_1if T_2 gt T_1` for an endothermic changeD. `K_(2) lt K_1if T_2 gt T_1` for an endothermic change |
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Answer» Correct Answer - A::D |
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| 506. |
For the raction: `PCI_5(g)hArrPCI_3(g)+CI_2(g)` The forward reaction at constant temperature is favouraved by:A. introduncing chlorine gas at constant volumeB. introducing an inert gas at constant pressureC. increasing the volune of the conatainerD. introducing `PCI_5` at constant volume |
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Answer» Correct Answer - B::C::D |
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| 507. |
For the reaction, `A+BhArr2C` `K_(C)=1`. If the initial concentration of A,B and C are 1m, 1m and 2m respectively then, at equilibrium.A. `[A]=[B]=[C]`B. `[A]=(4)/(3)M`C. `[B]=(2)/(3)M`D. `[A]=(1)/(2)[C]` |
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Answer» Correct Answer - A::B |
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| 508. |
The average person can see the red colour imparted by the complex `[Fe(SCN)]^(2+)` to an aqueous solution if the concentration of the complex is `6xx10^(-6)` M or greater.What minimum concentration of KSCN would be required to make it possible to detect 1 ppm (part per million) of Fe(III) in a natural water sample ? The instability constant for `Fe(SCN)^(2+)hArrFe^(3+)+SCN^(-) "is"7.142xx10^(-3)`A. `0.0036 M`B. `0.0037 M`C. `0.0035 M`D. None of these |
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Answer» Correct Answer - A 0.0036M must be the correct answer
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| 509. |
One mole of a compound AB reacts with 1 mole of a compound CD according to the equation `AB + CD hArr AD + CB`. When equilibrium had been established it was found that `(3)/(4)` mole each of reactant AB and CD has been converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction isA. `(9)/(16)`B. `(1)/(9)`C. `(16)/(9)`D. 9 |
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Answer» Correct Answer - D `AB + CDhArr AD + CD` mole at t = 0 ,1,1,0,0 Mole at equilibrium `(1-3/4)(1-3/4)hArr(3/4)(3/4)` `0.25` `0.25` `0.75` `0.75` `K_(c) = (0.75xx0.75)/(0.25xx0.25) = (0.5625)/(0.0625) = 9` |
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| 510. |
`K_(p)`has the value of `10^(-6) atm^(3) and 10^(-4)atm^(3)` at 298 K and 323 K respectiely for the reaction `CuSO_(4).3H_(2)O(s)hArrCuSO_(4)(s)+3H_(2)O(g)``Delta_(r)H^(@)` for the reaction is :A. `7.7KJ//mol`B. `-147.41KJ//mol`C. `147.41KJ//mol`D. none of these |
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Answer» Correct Answer - c |
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| 511. |
An equilibrium mixture in a vessel of capacity `100` litre contain `1 "mol" N_(2).2"mol"O_(2) "and" 3 "mol" NO`. Number of moles of `O_(2)` to be added so that at new equilibrium the conc. Of `NO` is found to be `0.04` mol//lt.A. `(101//18)`B. `(101//9)`C. `(202//9)`D. None of these |
| Answer» Correct Answer - A | |
| 512. |
5 moles of `SO_(2)` and 5 moles of `O_(2)` react in a closed vessel. At equilibrium 60% of the `SO_(2)` is consumed . The total number of gaseous moles`(SO_(2),O_(2)andSO_(3))` in the vessel is :-A. 5.1B. 3.9C. 10.5D. 8.5 |
| Answer» Correct Answer - 4 | |
| 513. |
The densities of graphite and diamond are `22.5` and `3.51` gm `cm^(-3)`. The `Delta_(f)G^(ɵ)` values are `0 J mol^(-1)` and `2900 J mol^(-1)` for graphite and diamond, respectively. Calculate the equilibrium pressure for the conversion of graphite into diamond at `298 K`. |
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Answer» We have `C_(g)rarr C_(d)` `DeltaG^(ɵ)=DeltaG_(("diamond"))-DeltaG_(("graphite"))` `rArr (2900-0)=2900 J "mol"^(-1) :. DeltaG^(ɵ)=2900 J "mol"^(-1)` `Volume=("Mass")/("density")` `:. DeltaV=(V_(d)-V_(g))=(12/3.51-12/2.25)xx10^(-6) m^(3) "mol"^(-1)` `=-1.91xx10^(-6) m^(3) "mol"^(-1)` We know that, `(DeltaG)_(T)=DeltaV del P` `:. int_(g)^(d)del (DeltaG)=int_(P_(1))^(P_(2)) DeltaV del P` `DeltaG=deltaV(P_(2)-P_(1))` `:. P_(2)=(DeltaG)/(DeltaV)+P_(1)` or `P_(2)=(2900 J "mol"^(-1))/(-1.91xx10^(-6) m^(3) "mol"^(-1))+1013225 Pa` `=1.52xx10^(9) Pa` `:.` Pressure `=1.52xx10^(9) Pa` |
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| 514. |
Densities of diamond and graphite are `3.5` and `2.3 g mL^(-1)`, respectively. The increase of pressure on the equilibrium `C_("diamond") hArr C_("graphite")`A. favours backward reactionB. favours forward reactionC. have no effectD. nothing can be predicted |
| Answer» Le Chatelier principle is not valid for solid-solid systems. | |
| 515. |
If a system at equilibrium is subjected to a change of any one of the factors such as concentration , pressure or temperature, the system adjusts itself in such a way so as to minimise the effect of that change. For the reaction `N_(2)(g)+O_(2)(g)hArr2NO(g)` If pressure id increased by reducing the volume of the container then :A. Total pressure at equilibrium will change.B. Concentration of all the component at equilibrium will change.C. Concentration of all the component at equilibrium will remain sameD. Equilibrium will shift in the forward direction |
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Answer» Correct Answer - A::B Number of moles will remain unchanged but due to decreased volume pressure will get increased and also the concentrations. |
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| 516. |
For reaction A(g) + B(g) we start with 2 moles of A and B each. At equilibrium `0.8` moles of AB is formed. Then how much of A changes to ABA. `20%`B. `40%`C. `60%`D. `4%` |
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Answer» Correct Answer - B `{:(,,A(g),+,B(g),hArr,AB(g)),(,"Initial mole",2,,2,,0),(,"At equilibrium",2-x,,2-x,,x=0.8):}` `:. %` of A changed to AB = `(0.8)/(2)xx100=40%` |
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| 517. |
Effect of temperature on the equilibrium process analysed by using the thermodynamics From the thermodynamics reaction `DeltaG^(@)=-2.30RTlogk` `DeltaG^(@):` Standing free energy change `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`…(ii) `DeltaH^(@) :`Standard heat of the reaction gt From eqns.(i) and(ii) `-2RTlogk=DeltaH^(@)=TDeltaS^(@)` `DeltaS^(@)` : standard entropy change `implies" "logK=-(DeltaH^(@))/(2.3RT)+(DeltaS^(@))/(2.3R)` Clearly, if a plot of k vs 1/T is made then it is a straight lone having slope `=(-DeltaH^(@))/(2.3R)` amd y intercept `=(DeltaS^(@))/(2.3R)` If at temperature `T_(1)` equilibrium constant be `k_(1)` and at temperature `T_(2)`equilibrium constant be `k_(2)` then : `implies" "logK_(1)=-(DeltaH^(@))/(2.3RT_(1))+(DeltaS^(@))/(2.3R)`..(iv) `implies" "logK_(2)=-(DeltaH^(@))/(2.3RT_(2))+(DeltaS^(@))/(2.3R)`...(v) Substracting e.q (iv) from (v), we get from the relation we can conclude that the of equilibrium constant increase in temperature for endothermic reaction eith but value of equilibrium constant decrease with the increase in temperature for exothermic reaction For exothermic reaction if `DeltaS^(@)lt0` then the sketch of log k vs `(1)/(T)` may beA. B. C. D. |
| Answer» Correct Answer - B | |
| 518. |
Effect of temperature on the equilibrium process analysed by using the thermodynamics From the thermodynamics reaction `DeltaG^(@)=-2.30RTlogk` `DeltaG^(@):` Standing free energy change `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`…(ii) `DeltaH^(@) :`Standard heat of the reaction gt From eqns.(i) and(ii) `-2RTlogk=DeltaH^(@)=TDeltaS^(@)` `DeltaS^(@)` : standard entropy change `implies" "logK=-(DeltaH^(@))/(2.3RT)+(DeltaS^(@))/(2.3R)` Clearly, if a plot of k vs 1/T is made then it is a straight lone having slope `=(-DeltaH^(@))/(2.3R)` amd y intercept `=(DeltaS^(@))/(2.3R)` If at temperature `T_(1)` equilibrium constant be `k_(1)` and at temperature `T_(2)`equilibrium constant be `k_(2)` then : `implies" "logK_(1)=-(DeltaH^(@))/(2.3RT_(1))+(DeltaS^(@))/(2.3R)`..(iv) `implies" "logK_(2)=-(DeltaH^(@))/(2.3RT_(2))+(DeltaS^(@))/(2.3R)`...(v) Substracting e.q (iv) from (v), we get from the relation we can conclude that the of equilibrium constant increase in temperature for endothermic reaction eith but value of equilibrium constant decrease with the increase in temperature for exothermic reaction If statndard heat of dissociation of `PCl_(5)` is 230 cal then slope of the graph of log vs `(1)/(T)` is :A. `+50`B. `-50`C. `10`D. None |
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Answer» Correct Answer - B Slop`=(-DeltaH^(@))/(2.3R)=-(230)/(2.3xx2)=-50` |
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| 519. |
In an equilibrium reaction for which `Delta G^(@) = 0` , the equilibrium constant K = |
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Answer» Correct Answer - B If `Delta G^(@) = 0` `Delta G^(@) = - 2.303` RT log `K_(p)` log `K_(p) = 0` `(.: log 1 = 0)` `K_(p) = 1` . |
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| 520. |
The partial pressure of `CH_(3)OH_((g))`, `CO_((g))` and `H_(2(g))` in equilibrium mixture for the reaction, `CO_((g))+2H_(2(g))hArrCH_(3)OH_((g))` are `2.0`, `1.0` and `0.1` atm respectively at `427^(@)C`. The value of `K_(P)` for deomposition of `CH_(3)OH` to `CO` and `H_(2)` is:A. `10^(2)` atmB. `2xx10^(2)atm^(-1)`C. `50 atm^(2)`D. `5xx10^(-3)atm^(2)` |
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Answer» Correct Answer - D `K_(p) = P_(CH_(3)OH)/(P_(H_(2))^(2)xxP_(CO)) = (2)/(1xx(0.1)^(2))= 200` `K_(p)` for reverse reaction will be `K_(p)^(1) = (1)/(K_(p)) = (1)/(200) = 5xx10^(-3) atm^(2)` |
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| 521. |
For the reaction `NH_(4)HS(s)rArrNH_(3)(g)+H_(2)S(g) ,K_(p)=0.09`. The total pressure at equilibrum is :-A. 0.3 atmB. 0.09 atmC. 0.6 atmD. 0.36 atm |
| Answer» Correct Answer - 3 | |
| 522. |
Consider the following gas - phase reaction : `2A(g)+B(g)rArrC(g)+D(g)` An equilibrium mixture of reactants and products is subjected to the following changes :- (I) A decrease in volume (II) An increase in temperature (III) Addition of reactants (IV) Addition of an inert gas at constant volume . Which of these changes affect the composition of the equilibrium mixture but leaves the value of `K_(c)` unchanged : -A. I, IIB. I, II, IIIC. II, III, IVD. I, III |
| Answer» Correct Answer - 4 | |
| 523. |
The conversion of ozone into oxygen is exothermic . Under what conditions is ozone the most stable ? `2O_(3)(g)rArr3O_(2)(g)`A. At low pressure and low pressure and low temperatureB. At high pressure and high temperatureC. At high pressure and low temperatureD. At low pressure and high temperature |
| Answer» Correct Answer - 2 | |
| 524. |
The conversion of ozone into oxygen is exothermic under what conditions is ozone is most stable? `2O_(3)(g)hArr3O_(2)(g)`A. At low pressure and low temperatureB. At high pressure and high temperatureC. At high pressure and low temperatureD. At low pressure and high temperature |
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Answer» Correct Answer - b |
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| 525. |
What will be the effect of the equilibrium constant on increasing temperature. If the reaction neither absorbs heat nor releases heat?A. Equililbrium constant will remain constant.B. Equilibrium constant will decrease .C. Equilibrium constant will increase.D. Can not be predicted. |
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Answer» Correct Answer - a |
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| 526. |
pure nitrosyl chloride (NOCl) gas was heated to `240^(@)`C` in a 1.0 container .At equilibrium the total pressure was 1.0 atm and the NOCl pressure was 0.64 atm . What would be the value of `K_(P)` ?A. `1.02 atm`B. `16.875xx10^(-3)`atmC. `16xx10^(-2)`atmD. none of these |
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Answer» Correct Answer - b |
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| 527. |
`5.1g` of solid `NH_(4)HS` is introduced in a `16.4` lit. vessel & heated upto `500 K` `K_(B)` for equilibrium `NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)` is `0.16`. The maximum pressure developed in the vessel will be `:`A. `0.8 atm`B. `0.40 atm`C. `0.5 atm`D. None of these |
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Answer» Correct Answer - 3 `{:(,NH_(4)HS(s),hArr,NH_(3)(g),+,H_(2)S(g),),(t=0,0.1"mole",,,,,),("for max. pressure",-,,0.1,0.1,,0.20"gaseous mole"),(Pxx16.4=0.2xx0.82xx500,,rArr,,P=0.5,,):}` |
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| 528. |
Solid `hArr` liquid equilibrium can be achieved only at melting point of the substance. |
| Answer» Correct Answer - True | |
| 529. |
If pressure is applied to the equilibrium of solid-liquid. The melting point of the solid:A. will not changeB. may increase or decrease depending upon nature of solid speciesC. will always increaseD. will always decrease |
| Answer» `V_((s))gtV_((l))` for ice-water and `V_((s)) lt V_((l))` for others. | |
| 530. |
For the system `3A + 2BhArr` C, the expression for equilibrium constant isA. `([3A][2B])/(C)`B. `([C])/([3A][2B])`C. `([A]^(3) [B]^(2))/([C])`D. `([C])/([A]^(3) [B]^(2))` |
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Answer» Correct Answer - D Equilibrium constant for the reaction, `3A + 2Bh C` is `K= ([C])/([A]^(3) [B]^(2))` |
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| 531. |
For the hypothetical reactions, the equilibrium constant `(K)` value are given `AhArrB,K_(1)=2,BhArrC,K_(2)=4,` `ChArrD,K_(3)=3` The equilibrium constant (K) for the reaction `AhArrD` isA. `48`B. `6`C. `27`D. `24` |
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Answer» Correct Answer - D The reaction `AhArrD` is obtained by adding the three given reactions `:. K=K_(1)xxK_(2)xxK_(3)=2xx4xx3=24` or `K_(1)=([B])/([A]), K_(2)([C])/([B]), K_(3)=([D])/([C])` On multiplying `K_(1),K_(2)` and `K_(3)`, we get `K=K_(1)xxK_(2)xxK_(3)=([D])/([A])=24` |
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| 532. |
For the reactions, `{:(AhArrB, K_(c)=2), (BhArrC, K_(c)=4), (ChArrD, K_(c)=6):}` `K_(c)` for the reaction, `AhArrD` is:A. `(2+4+6)`B. `(2xx4)//6`C. `(4xx6)//2`D. `2xx4xx6` |
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Answer» `2=([B])/([A])`, `4=([C])/([B])` and `6=([D])/([C])` Thus on multiplying, `2xx4xx6=([D])/([A])`, i.e., `K_(c)` for the reaction. |
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| 533. |
The following reaction occurs at `700 K`. Arrange them in the order of increasing tendency to proceed to completion. I. A. `II lt I lt IV lt III`B. ` III lt IV lt I lt II`C. `I lt III lt IV lt II`D. `IV lt III lt I lt II` |
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Answer» Correct Answer - B Higher the equilibrium constant, more will the reaction go to completion. `III lt IVlt I lt II` |
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| 534. |
For the equilibrium `MgCO_3(g)oversetDeltahArrMgO(s)CO_2(s)` which of the following expressions is correct ?A. `K_p=pco_2`B. `K_p=([MgO][CO_2])/([MgCO_3])`C. `K_p=(p_(Mgo).p_(CO_2))/(p_(MgCO_3))`D. `K_p=(p_(Mgo)+p_(CO_2))/(p_(MgCO_3))` |
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Answer» Correct Answer - A In hetrogeneous system. `K_(c)and K_(p)` are not depend upon the concentration of pressure of solid substence Hence. At equilibrium their concentratiaon or pressure are assumed as one. `MgCO_3(s)hArrMgO(s)+CO_2(g)` `therefore K_p=Pco_(2)` |
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| 535. |
Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:A. increased mass of `BaO_2`B. increased mass of `BaO`C. increased temperature of equilibriumD. increased mass of `BaO_2` and `BaO` both |
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Answer» Correct Answer - C `BaO_2(s)overset(r_1)underset(r_2)hArrBaO(s)+O_2(g),DeltaH=+ve` According to law of mass action, the rate of forward reaction `=r_1` `r_1prop[BaO_2]` or `r_1=k_1[BaO_2]` `BaO_2` is solid substence in pure state concentration `=1m`. then, `r_=k_1` Similarly the rate of backward reaction `=r_2` `r_2prop[BaO][O_2]` or `r_2=k_2[BaO][O_2]` `because` Concentration of solid `[BaO]=1" "[O_2(g)]` `therefore r_2=k_2[O_2]` At reuilibrium , `r_1=r_2` `K_1=K_2[O_2]` or `K_1=K_2.p_(o.2)` where `(o ._2)` =partial pressure of `O_2` or `(K_1)/(K_2)=p_(o ._2)" "("equilibrium constant" )` `because (K_1)/(K_2)=K` or `K=p_(o.2)` |
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| 536. |
Consider the chemical reaction: `underset(("Green solution"))(Ni^(2+)(aq))+6NH_(3)(aq) hArr underset(("Blue solution"))([Ni(NH_(3))_(6)]^(2+)) (aq)` When `H^(o+)(aq)` is added, the colour green is favoured. Use one or more of the following interpretations to answer the questions: i. Some unreacted `Ni^(2+)(aq)` is present in the solution at equilibrium ii. Some unreacted `NH_(3)(aq)` is present in the solution at equilibrium iii. The colour change indicates new equilibrium conditions with reduced `[Ni(NH_(3))_(6)]^(2+)(aq)` iv. The colour change indicates new equilibrium conditions with increased `[Ni(NH_(3))_(6)]^(2+)(aq)`. The deepening of blue colour on dissolving more `Ni(NO_(3))_(2)` supports interpretation (s).A. i onlyB. i and iv onlyC. ii and iv onlyD. i and ii only |
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Answer» Correct Answer - B Addition of `Ni^(2+)` causing more blue colour means reaction moving in the forward direction. `rArr [Ni(NH_(3))_(6)]^(2+)` has increased and `NH_(3)` (unreacted) was present initially. |
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| 537. |
Consider the chemical reaction: `underset(("Green solution"))(Ni^(2+)(aq))+6NH_(3)(aq) hArr underset(("Blue solution"))([Ni(NH_(3))_(6)]^(2+)) (aq)` When `H^(o+)(aq)` is added, the colour green is favoured. Use one or more of the following interpretations to answer the questions: i. Some unreacted `Ni^(2+)(aq)` is present in the solution at equilibrium ii. Some unreacted `NH_(3)(aq)` is present in the solution at equilibrium iii. The colour change indicates new equilibrium conditions with reduced `[Ni(NH_(3))_(6)]^(2+)(aq)` iv. The colour change indicates new equilibrium conditions with increased `[Ni(NH_(3))_(6)]^(2+)(aq)`. The deepening of blue colour on addition of more `NH_(3)(aq)` supports interpretation(s).A. i onlyB. i and iv onlyC. i and ii onlyD. ii and iv only |
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Answer» Correct Answer - D Addition of `NH_(3)` causing more blue colour means reaction moving in the forward direction. `rArr [Ni(NH_(3))_(6)]^(2+)` has increased and `Ni^(2+)` (unreacted) was present initially. |
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| 538. |
Equilibrium constants `(K)` for the reaction `2NO(g)+Cl_(2)(g)hArr2NOCl(g)` is correctly given by the expressionA. `([NOCl]^(2))/([NO]^(2)[Cl_(2)])`B. `([2NOCl])/([2NO][Cl_(2)])`C. `([NO]^(2)+[Cl_(2)])/([NOCl])`D. `([NO]^(2)[Cl_(2)])/([NOCl]^(2))` |
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Answer» Correct Answer - A `2NO(g)+Cl_(2)(g)overset(K)(hArr)2NOCl(g)` According the law of mass action `K=([NOCl]^(2))/([NO]^(2)[Cl_(2)])` |
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| 539. |
For a system, `a+2BhArrC` , the equilibrium comcentrations are `[A]=0.06, [B]=0.12`, and `[C]=0.216`. The `K_(c )` for the reaction isA. `120`B. `400`C. `4xx10^(-3)`D. `250` |
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Answer» Correct Answer - D `A+2BhArrC` `K=([C])/([A][B]^(2))rArr ((0.216))/((0.06)(0.012)^(2))=250` |
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| 540. |
Which oxide is most stable?A. AOB. BOC. COD. DO |
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Answer» Correct Answer - A Because minimum the value of equilibrium constant, slower will be the reaction and hence maximum stable will be the reactant (here `AO` oxide) |
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| 541. |
If the value of equilibrium constant is large, ……….. Are more stable. |
| Answer» Correct Answer - Products | |
| 542. |
The reaction which proceeds in the forward direction isA. `SnCl_(4)+Hg_(2)Cl_(2)hArrSnCl_(2)+2HgCl_(2)`B. `NH_(4)Cl+NaOHhArrH_(2)O+NH_(3)+NaCl`C. `Mn^(2+)+2H_(2)O+Cl_(2)hArrMnO_(2)+4H^(+)+2Cl`D. `S_(2)O_(6)^(2-)+3I^(-)hArr2S_(2)O_(3)^(2-)+I_(2)` |
| Answer» Rest all occur in backward direction. | |
| 543. |
`K_(p)` and `K_(c)` are inter related as `K_(p)=K_(c)(RT)^(Deltan)` Answer the following questions: Which of the following have `K_(p)-K_(c)`?A. `H_(2)(g)+I_(2)(g) hArr 2HI(g)`B. `N_(2)(g)+O_(2)(g) hArr 2NO(g)`C. `2NO(g)+Cl_(2)(g) harr 2NOcl(g)`D. `2SO_(2)(g)+O_(2)(g) harr 2SO_(3)(g)` |
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Answer» Correct Answer - A::B `K_(p)=K_(c)(RT)^(Deltan)` |
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| 544. |
The equilibrium mixture for `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g)` present in `1 L` vessel at `600^(@)C` contains `0.50, 0.12`, and `5.0` moles of `SO_(2), O_(2)`, and `SO_(3)` respectively. a. Calculate `K_(c)` for the given change at `600^(@)C`. b. Also calculate `K_(p)`. c. How many moles of `O_(2)` must be forced into the equilibrium vessel at `600^(@)C` in order to increase the concentration of `SO_(3)` to `5.2` mol? |
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Answer» Correct Answer - A::B::C `{:(,2SO_(2),+,O_(2),hArr,2SO_(3)),("moles at",0.5,,0.12,,5.0),("equilibrium",,,,,),("Volume"=1 L,,,,,):}` a. `K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=((5)^(2))/((0.5)^(2)(0.12))` `K_(c)=833.33` b. `K_(P)=K_(C)(RT)^(Deltan)=833.33xx(0.0821xx873)^(-1)` `=11.62 ("atm")^(-1)` |
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| 545. |
The reaction which proceeds in the backward direction isA. `Fe_(3)O_(4)+6HCl=2FeCl_(3)+3H_(2)O`B. `NH_(3)+H_(2)O+NaCl=NH_(4)Cl+NaOH`C. `SnCl_(4)+Hg_(2)Cl_(2)=SnCl_(2)+2HgCl_(2)`D. `2CuI+I_(2)+4K^(o+)=2Cu^(2+)+4KI` |
| Answer» Correct Answer - B::C::D | |
| 546. |
For the reaction `CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)` at a given temperature, the equilibrium amount of `CO_(2)(g)` can be increased byA. Adding a suitable catalystB. Adding an inert gasC. Decreasing the volume of the containerD. Increasing the amount of `CO(g)` |
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Answer» Correct Answer - D Since the given equilibrium involves `Deltan_(g)=0`, there will be no effect of adding an inert gas and decreasing the volume of the container. A catalyst can help in achieving the equilibrium rapidly without affecting the equilibrium amount of the species involved in the reaction. Only increasing the amount of a reactant can shift the equilibrium amount of a product. |
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| 547. |
The magnitude of equilibrium constant is a measure of ………. to which the reversible reaction proceeds in a particular direction at a given ……….. |
| Answer» Correct Answer - Extent, temperature | |
| 548. |
In a mixture of `N_(2)` and `H_(2)` in the ratio `1:3` at `30 atm` and `300^(@)C`, the `%` of `NH_(3)` at equilibrium is `17.8`. Calculate `K_(p)` for `N_(2)+3H_(2)hArr2NH_(3)`. |
| Answer» `K_(p)=7.29xx10^(-4)atm^(-2)`, | |
| 549. |
For which of the following reaction, `(K_(p))/(K_(c))` ratio is maximum ?A. `CO(g)+(1)/(2) O_(2)(g) leftrightarrowCO_(2)(g)`B. `H_(2)(g)+I_(2)(g)leftrightarrow2HI(g)`C. `PCl_(5)(g)leftrightarrowPCl_(3)(g)+Cl_(2)(g)`D. `7H_(2)(g)+2NO_(2)(g)leftrightarrow2NH_(2)(g)+4H_(2)O(g)` |
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Answer» Correct Answer - 3 `K_(p)=K_(c)(RT)^(Deltan)` `Deltan` is maximum for `PCl_(5)hArrPCl_(3)+Cl_(2)` |
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| 550. |
If for a particular reversible reaction, `K_(c)=57` at `355^(@)C " " ` and `" " k_(c)=68` at `450^(@)C` then `:`A. `DeltaHlt0`B. `DeltaHgt0`C. `DeltaH=0`D. `DeltaH` whose sign can be determined |
| Answer» Correct Answer - 2 | |