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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.
| 701. |
At `1400 K, K_( c) = 2.5 xx 10^(-3)` for the reaction `CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` A ` 10 L` reaction vessel at `1400 K` contains `2.0 mol` of `CH_(4)`, `3.0 mol` of `CS_(2)`, `3.0 mol` of `H_(2)S`. In which direction does the reaction proceed to reach equilibrium? |
| Answer» The reaction will proceed in backward direction to reach equilibrium | |
| 702. |
At `1400 K, K_( c) = 2.5 xx 10^(-3)` for the reaction `CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` A ` 10 L` reaction vessel at `1400 K` contains `2.0 mol` of `CH_(4)`, `3.0 mol` of `CS_(2)`, `3.0 mol` of `H_(2)S`. In which direction does the reaction proceed to reach equilibrium?A. ForwardB. BackwardC. May be forward or backwardD. Reaction is in equilibrium |
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Answer» Correct Answer - B `CH_(4)(g)+2H_(2)S(g)overset(K)(hArr)CS_(2)(g)+4H_(2)(g)` or `K=([CS_(2)][H_(2)]^(2))/([CH_(4)][H_(2)S]^(2)) ...(i), [CS_(2)]=(3/10),[H_(2)]=(3/10),[CH_(4)]=(2/10)` and `[H_(2)S]=(4/10)` Substituting the cocentration values in equation (i). `K=((3/10)(3/10)^(4))/((2/10)(4/10))=3^(5)/10^(5)xx10^(3)/(2xx4^(4))xx243/3200=0.075` Therefore, `K gt 2.5xx10^(-3)` (given). Hence, the reaction will proceed in backward direction. |
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| 703. |
For the reaction `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)` Which of the following sketches may represent above equilibrium? Assume equilibrium can be achieved from either side and by taking any one or more components initially. Give `K_(c )` for the reaction `lt 2`?A. B. C. D. |
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Answer» Given: `K_(c)lt2` for `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)` a. From graph, find the vlue of `K_(c )` `K_(c )=(([PCl_(3)][Cl_(2)])/([PCl_(5)]))_(Eq)=(4xx6)/2 gt 2` (a) is incorrect. b. Incorrect: Since the concentration of both `PCl_(5)(g)` and `Cl_(2)(g)` will either increase or decrease, here it is increasing for `Cl_(2)(g)` and decreasing for `PCl_(3)(g)`. c. Note that (from graph) the initial concentration of `PCl_(5)` is zero. So the equilibrium is established as follows: `PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)` From graph, `x=6`. Also note that (from graph), `b lt 6` (i.e., initial concentration of `Cl_(2))`, which means that `xlt 6`. It is not so, hence choice (c ) is incorrect. Obviously this should be the correct choice. However, let us check it. Find `K_(c )`. `K_(c )=(([PCl_(3)][Cl_(2)])/([PCl_(5)]))_(Eq)=(2xx4)/6 lt 2` `:. (d)` is correct. |
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| 704. |
A vessel of `250` litre was filled with `0.01` mole of `Sb_(2)S_(3)` and `0.01` mole of `H_(2)` to attain the equilibrium at `440^(@)C` as `Sb_(2)S_(3)(s)3H_(2)(g)hArr2Sb(s)+3H_(2)D(g)` After equilibrium, the `H_(2)S` formed was analysed was analysed by dissloved it in water and treating with execedd of `Pb^(20+)` to give `1.19` g of PbS as precipitate. What is the value of `K_(c)` at `440^(@)C` ?A. `1`B. `2`C. `4`D. `8` |
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Answer» Correct Answer - A |
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| 705. |
A vessel of `2.50 litre` was filled with `0.01` mole of `Sb_(2)S_(3)` and `0.01` mole of `H_(2)` to attain the equilibrium at `440^(@)C` as: `Sb_(2)S_(3(s))+3H_(2(g))hArr2Sb_((s))+3H_(2)S_((g))` After equilibrium the `H_(2)S` formed was analysed by dissolving it in water and treating with excess of `Pb^(2+)` to give `1.029g` of `PbS` as precipitate. What is value of `K_(c)` of the reaction at `440^(@)C`? (At weight of `Pb=206`) |
| Answer» Correct Answer - `4.3xx10^(-1)` , | |
| 706. |
A sample of `CaCO_(3)(s)` is introduced into a sealed container of volume `0.654 L` and heated to `1000 K` until equilibrium is reached. The equilibrium constant for the reaction `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g),` is `3.9xx10^(-2)` atm at this temperature. Calculate the mass of `CaO` present at equilibrium. |
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Answer» `CaCO_(3)(s)(s)hArrCaCO_(2)(g)` `K_(p)=p_(CO_(2))=3.9xx10^(-2)` Let the number of "moles" of `CO_(2)` be formed `=n` `n=(p_(CO_(2))xxV)/(RT)=(3.9xx10^(-2)xx0.654)/(0.082xx1000)=3.11xx10^(-4) "mol"` The amount of `CaO(s)` formed will also be `=3.11xx10^(-4) "mol"` Hence, the mass of `CaO` formed `=3.11xx10^(-4)xx56` `=0.0174 g` |
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| 707. |
Calculate the equilibrium constant `K_(p)` and `K_(c )` for the reaction: `CO(g)+1//2O_(2)(g) hArr CO_(2)(g)`. Given that the partial pressure at equilibrium in a vessel at `3000 K` are `p_(CO)=0.4 atm, p_(CO_(2))=0.2 atm` |
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Answer» Correct Answer - B `CO(g)+1/2O_(2)(g)hArr CO_(2)(g)` `K_(p)=([p_(CO_(2))(g)])/([p_(CO)(g)][p_(O_(2))(g)]^(1//2))` `=0.6/(0.4xx(0.2)^(1//2))=0.6/(0.4xx0.4472)=3.356 "atm"^(-1//2)` `K_(p)=K_(c )(RT)^(Deltan)` `Deltan=1-(1+1/2)=-1/2` `K_(c )=K_(p)/((RT)^(Deltan))` `=3.356/((0.082xx3000)^(-1//2))` `=3.356xx(0.082xx3000)^(1//2)` `=3.356xx15.684=52.67` |
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| 708. |
4.4 grams of `CO_(2)` are introduced into a 0.82 L flask containing excess solid carbon at `627^(circ)C`, so that the equilibrium: The density of equilibrium gaseous mixture corresponds to an average molecular weight of 36. `K_(p)=P_(CO)^(2)/P_(CO_2)and K_(C)([CO]^(2))/([CO_(2)])` `[R=0.082"Lt-atm/mol-K",C=12,0=16]` Total number of moles of equilibrium gaseous mixture is :A. `(1)/(30)`B. `(2)/(15)`C. `(1)/(15)`D. `(1)/(10)` |
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Answer» Correct Answer - D |
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| 709. |
Consider the reaction `X(g) hArr Y(g)+Z(g)` When the system is at equilibrium at `100^(@)`, the concentrations are found to be `[X]=0.2 M, [Y]=[Z]=0.4 M` a. If the pressure of the container is suddenly halved at `100^(@)C`, find equilibrium concentration. b. If the pressure of the container is suddenly doubled at `100^(@)C`, find the equilibrium concentration |
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Answer» `{:(X,hArr,Y,+,Z,),(0.2,,0.4,,0.4,"At equilibrium"):}` `K_(c )=(0.4xx0.4)/0.2=0.8` a. When pressure is halved, `(i.e. P rarr P/2)`, concentration is halved. `:. [X]=0.1 M, [Y]=[Z]=0.2 M` `Q_(c )=(0.2xx0.2)/(0.1)=0.4` `:. Q_(c ) lt K_(c )` (Reaction goes forward) `K_(c )=0.8=((0.2+x)(0.2+x))/((0.1-x))` solving for `x`: `x=(-0.4+sqrt(0.16+4xx0.04))/2=0.08` `:.` Equilibrium concentration: `[X]=0.1-x=0.1-0.08=0.02 M` `[Y]=[Z]=0.2+x=0.2+0.08=0.28 M` b. When pressure is doubled (i.e., `P rarr 2P)`, concentration is doubled. `:. [X]=0.4 M [Y]=[Z]=0.8 M` `Q_(c)=(0.8xx0.8)/0.4=1.6` `:. Q_(c) gt K_(c)` (Reaction goes backward) `K_(c)=0.8=((0.8-x)(0.8-x))/((0.4+x))` Solving for x: `x^(2)-0.4x+0.32=0` `x=(-(-0.4)+sqrt(0.16-1.28))/(2)~~0.73` `:.` Equilibrium concentration: `[X]=0.4+x=0.4+0.73=1.13 M` `[Y]=[Z]=0.8-x=0.8-0.73=0.07 M` |
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| 710. |
For the reaction `A+B hArr C+D`, the initial concentrations of A and B are equal. The equilibrium concentration of C is two times the equilibrium concentration of A. The value of equilibrium constant is ………..A. 4B. 2C. `1//4`D. `1//2` |
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Answer» Correct Answer - A `{:(,A,+,B,hArr,C+D),("At equilibrium",a,+,b,2a,2a):}` `K = (2axx2a)/(axxa) = 4` |
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| 711. |
For the reaction `N_(2(g))+O_(2(g))rArrNO_((g))`, the value of `K_(c)` at `800^(@)` C is 0.1 . What is the value of `K_(p)` at this temperature ?A. 0.5B. 0.01C. 0.05D. 0.1 |
| Answer» Correct Answer - 4 | |
| 712. |
Bromine monochloride, `(BrCl)` decomposes into bromine and chlorine and reaches the equilibrium. `2BrCl_((g))hArrBr_(2(g))+Cl_(2(g))` For which `K_(c)=32` at `500 K`. If initially pure `BrCl` is present at a concentration of `3.30xx10^(-3) mol litre^(-1)`, what is its molar concentration in the mixture at equilibrium? |
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Answer» `{:(,2BrCl_((g)),hArr,Br_(2(g)),+,Cl_(2(g))),("Initial conc.",0.0033,,0,,0),("Conc. at equilibrium",(0.0033-a),,a//2,,a//2):}` `K_(c)=([Br_(2)][Cl_(2)])/([BrCl]^(2))=32` `:. ((a)/(2)xx(a)/(2))/((0.0033-a)^(2))=32` or `(a)/(2xx(0.0033-a))=5.66` `:. a=3xx10^(-3)` `:. [BrCl]=3.3xx10^(-3)-3xx10^(-3)` `=3xx10^(-4) mol litre^(-1)` |
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| 713. |
The equilibrium constant at `278K` for `Cu(s)+2Ag^(o+)(aq) hArr Cu^(2+)(aq)+2Ag(s)` is `2.0xx10^(15)`. In a solution in which copper has displaced, some silver ions from the solution, the concentration of `Cu^(2+)` ions from the solution, the concentration of `Cu^(2+)` ions is `1.8xx10^(-2) mol L^(-1)` and the concentration of `Ag^(o+)` ions is `3.0xx10^(-9) mol L^(-1)`. Is the system at equilibrium? |
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Answer» Correct Answer - A::B Applying the law of chemical equilibrium to the given reaction, we have `K=([Cu^(2+)(aq)][Ag(s)]^(2))/([Cu(s)][Ag^(o+)(aq)]^(2))` By convention, putting `[Ag(s)]=1` and `[Cu(s)]=1` `:. K=([Cu^(2+)(aq)])/([Ag^(o+)(aq)]^(2))` Putting `[Cu^(2+)]=1.8xx10^(-2) "mol" L^(-1)` and `[Ag^(o+)]=3.0xx10^(-9) "mol" L^(-1)` `K=(1.8xx10^(-2))/((3.0xx10^(-9))^(2))=2xx10^(15)` Which is same as for the reaction in equilibrium. Hence, the given system is in equilibrium. |
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| 714. |
For the reaction `Cu(s)+2Ag^(o+)(aq)rarr Cu^(2+)(aq)+2Ag(s)` Fill in the blanks in the following table for the three solution at equilibrium. `{:("Solution",[Cu^(2+)(aq)],[Ag^(o+)(aq)],K L^(-1)),(,mol L^(-1),mol L^(-1),mol L^(-1)),(1.,(a),1.0xx10^(-9),2.0xx10^(15)),(2.,2.0xx10^(-7),1.0xx10^(-11),(b)),(3.,2.0xx10^(-2),(c),2.0xx10^(15)):}` |
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Answer» Correct Answer - A::B::C a. `K=([Cu^(2+)])/([Ag^(o+)]^(2))` `2.0xx10^(15)=([Cu^(2+)])/((1.0xx10^(-9))^(2))` `[Cu^(2+)]=2.0xx10^(15)xx(1.0xx10^(-9))^(2)=2.0xx10^(-3)` `:. a=2.0xx10^(-3) "mol" L^(-1)` b. `K=([Cu^(2+)])/([Ag^(o+)]^(2)) =(2.0xx10^(-7))/((1.0xx10^(-11))^(2))=2.0xx10^(15)` `b=2.0xx10^(15) "mol" L^(-1)` c. `K=([Cu^(2+)])/([Ag^(o+)]^(2))` `2.0xx10^(15)=(2.0xx10^(-2))/([Ag^(o+)]^(2))` ltbr or `[Ag^(o+)]=([2.0xx10^(-2)])/([2.0xx10^(15)])=3.16xx10^(-9)` |
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| 715. |
0.6 moles of `PCl_(5)`, 0.3 mole of `PCl_(3)` and 0.5 mole of `Cl_(2)` are taken in a 1 L flask to obtain the following equilibrium , `PCl_(5(g))rArrPCl_(3(g))+Cl_(2(g))` If the equilibrium constant `K_(c)` for the reaction is 0.2 Predict the direction of the reaction.A. Forward directionB. Backward directionC. Direction of the reaction cannot be predictedD. Reaction does not move in any direction |
| Answer» Correct Answer - 2 | |
| 716. |
For the reaction `a+brArrc+d`, initially concentrations of a and b are equal and at equilibrium the concentration of a . What will be the equilibrium constant for the reaction ?A. 2B. 9C. 4D. 3 |
| Answer» Correct Answer - 3 | |
| 717. |
When heated , ammonium carbamate decomate decompoes as follows : NH_(4)COOH_(2)(s) hArr2NH_(3)(g)+CO_(2)(g)` `At a certain temperature , the equilibrium pressure of the system is 0.318atm `K_(p) ` for the reaction is:A. `0.128`B. `0.426`C. `4.76xx10^(-3)`D. none of these |
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Answer» Correct Answer - c |
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| 718. |
When a product is removed from the system which is at equilibrium ……….. reaction is favoured. |
| Answer» Correct Answer - Forward | |
| 719. |
`NH_(4)COONH_(2)(s)hArr2NH_(3)(g)+CO_(2)(g)` If equilibrium pressure is 3 atm for the above reaction, then `K_(p)` for the reaction isA. 27B. 4C. 3D. 9 |
| Answer» Correct Answer - 2 | |
| 720. |
What weight of solid ammonium carbamate `(NH_(2)COONH_(4))`, when vaporised at `200^(@)C` will have a volume of `8.96 litre` at `1.0 atm` pressure. Assume that the solid completely decomposes into `CO_(2)` and `NH_(3)` at `200^(@)C` and `1.0 atm` :A. `4 g`B. `6 g`C. `5 g`D. `10 g` |
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Answer» `{:(,NH_(2)COONH_(4),hArr,2NH_(3),+,CO_(2),),("Before dissociation",a,,0,,0),("After dissociation",a,,2a,,a):}` Total mole after decomposition `=3a` `PV=nRT` `:. 1xx8.96=3axx0.0821xx473` `:. A=0.077` `:. Wt of NH_(2)COONH_(4)=0.077xx78=6.0 g` |
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| 721. |
In an aqueous solution of volume 500 ml, the the reaction of `2Ag^(+) + CuhArrCu^(2+) + 2Ag` reached equilibrium the `[Cu^(2+)]` was x M. When 500 ml of water is further added, at the equilibrium `[Cu^(2+)]` will beA. `2x M`B. `x M`C. between x M and `x//2 M`D. less than `x//2 M` |
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Answer» Correct Answer - D `{:(2Ag^(+)+,Cu,hArr,Cu^(2+),+,2Ag),(aM,,,xM,,if V is 500ml):}` If volume is dobled, then for `K_(c) = (x)/(a^(2))` to have same value `[Cu^(2+)]` will be less than `x//2` |
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| 722. |
Solid Ammonium carbamate dissociates as: `NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g).` In a closed vessel, solid ammonium carbonate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of `NH_(3)` at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure. Also find the partial pressure of ammonia gas added. |
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Answer» `NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)` Let `P`= original equilibrium pressure. From the mole ratio of `NH_(3)` and `CO_(2)` at equilibrium, we have `p_(NH_(3))=2/3 P` and `p_(CO_(2))=P/3` `rArr K_(p)=(p_(NH_(3)))^(2).p_(CO_(2))=(2/3P)^(2) (P/3)=4/27 P^(3)` Now `NH_(3)` is added such that `p_(NH_(3))=P` Find the pressure of `CO_(2)` `rArr 4/27 P^(3)=P^(2) p_(CO_(2)) rArr p_(CO_(2))=4/27 P` Total new pressure `=P_("new")=p_(NH_(3))+p_(CO_(2))` `rArr P_("new")=P+4/27 P=31/27 P` `rArr` Ratio =`P_("new")/P_("original")=(31/27P)/P=31/27` Let `x` be the partial pressure of `NH_(3)` added at original equilibrium. `NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)` `{:("At equilibrium",,2/3 P,,1/3 P),("When" NH_(2) " is added",,2/3P+x,,1/3 P),("At new equilibrium",,2/3P+x-2y,,1/3P-y):}` `rArr 2/3 P+x-2y=p_(NH_(3))` and `1/3 P-y=P_(CO_(2))=4/27P` `rArr` Solve to get: `x=19/27 P` |
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| 723. |
In which of the following equilibrium system the rate of the backward reaction is favoured by increase of pressureA. `PC1_(5)hArrPC1_(3) + C1_(2)`B. `2SO_(2) + O_(2)hArr2SO_(3)`C. `N_(2) + 3H_(2)hArr2NH_(3)`D. `N_(2) + O_(2)hArr2NO` |
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Answer» Correct Answer - A The rate of backward reaction is favoured by increase of pressure in the reaction as `Delta n` is positive `PC1_(5)hArrPC1_(3) + C1_(2)` |
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| 724. |
For which of the following `K_(p)` may be equal to `0.5` atmA. `2HIhArrH_(2) + I_(2)`B. `PC1_(5(g))hArrPC1_(3) + C1_(2)`C. `N_(2) + 3H_(2)hArr2NH_(3)`D. `2NO_(2)hArrN_(2) O_(4)` |
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Answer» Correct Answer - B For `K_(p) = 0.5` atm `Delta n = 1` (since the unit is atm) and `PC1_(5)hArrPC1_(3) + C1_(2)` `Delta n = 1` |
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| 725. |
A vessel at `1000 K` contains carbon dioxide with a pressure of `0.5 atm`. Some of the carbon dioxide is converted to carbon monoxide on addition of graphite. Calculate the value of `K_(p)` if total pressure at equilibrium is `0.8 atm`.A. `1.8` atmB. 3 atmC. `0.3` atmD. `0.18` atm |
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Answer» Correct Answer - A `{:(,CO_(2)(g),+,C(s),hArr,2CO(g)),("Initial",,,0.5atm,,,__),("At equilibrium",,,0.5-p,,2patm):}` This is a case of heterogeneous equilibrium. C(s) being solid is not considered. Total pressure of `CO_(2)` and CO gases. `P_(CO_(2)) + P_(CC) = P_(total)` `0.5- (p+2p) = 0.8, p=0.3` atm `:.` `P_(CO_(2)) = 0.5 - 0.3 = 0.2` atm `P_(CO) = 2p = 0.6` atm `K_(p) = (P^(2)CO)/(PCO_(2)) = (0.6xx0.6)/(0.2) = 1.8` atm |
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| 726. |
A vessel at `1000 K` contains carbon dioxide with a pressure of `0.5 atm`. Some of the carbon dioxide is converted to carbon monoxide on addition of graphite. Calculate the value of `K_(p)` if total pressure at equilibrium is `0.8 atm`. |
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Answer» `CO_(2)(g)+C(s)hArr2CO(g)` `{:(Initial,,0.5 "atm",,0 "atm"),(At equilibrium,,(0.5-x),,2x "atm"):}` At equilibrium, the total pressure is `0.8 "atm"`. `p_("total")=p_(CO_(2))+p_(CO)` `0.8=(0.5-x)+2x=0.5+x` or `x=0.3 "atm"` Applying the law of mass action `K_(p)=((p_(CO))^(2))/(p_(CO_(2)))=((2xx0.3)^(3))/0.2=0.36/0.2=1.8 "atm"` |
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| 727. |
Two moles of HI were heated in a sealed tube at `440^@ C` till the equilibrium was reached. HI was found to be 22% decomposed.The equilibrium constant for disssociation is :A. 0.282B. 0.0796C. 0.0199D. 1.99 |
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Answer» Correct Answer - C |
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| 728. |
In the following reaction started only with `A_(B), 2A_(B)(g) hArr3A_(2)(g)+A_(4)(g)` mole fraction of `A_(2)` is found to `0.36` at a total pressure of `100 atm` at equilibrium. The mole fraction of `A_(8)(g)` at equlibrium is `:`A. `0.28`B. `0.72`C. `0.18`D. None of these |
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Answer» Correct Answer - 1 `{:(,2A_(B),hArr,2A_(3),+,3A_(2),+,A_(4)),(t=0,2,,0,,0,,0),(t=t_(aq),2-2alpha,,2alpha,,3alpha,,alpha):}` `n_(r)=2+4alpha` given mole fraction of `A_(2)` is `=0.36`. `0.36=(3alpha)/(2+4alpha)` `alpha=0.46` Mole fraction of `A_(B)=(2-2alpha)/(2+4alpha)=(2-2xx0.46)/(2+4xx0.46)=0.28` |
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| 729. |
`PCl_5` decomposes as `PCl_5(g) hArr PCl_3(g)+Cl_2(g)`.If at equilibrium , total pressure is P and density of gaseous mixture is d at temperature T then degree of dissociation `(alpha)` is : (Molecular wt. of `PCl_5=M`)A. `alpha=1-(PM)/(dRT)`B. `alpha=1-(dRT)/(PM)`C. `alpha=(PM)/(dRT)-1`D. `alpha=(dRT)/(PM)-1` |
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Answer» Correct Answer - C |
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| 730. |
For which of the reaction, the ratio `(K_(P))/(K_(C))` is maximum and minimum respectively. (a) `X(g)hArr2Y(g)` (b) `2X(g)hArrY(g)` (c) `X(g)hArr2Y(g)+Z(g)` (d) `X(s)hArrY(g)`A. c,bB. a,bC. c,dD. a,c |
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Answer» Correct Answer - A For `(K_(P))/(K_(C))` to be maximum, `(Deltan)` has to be maximum & or `(K_(P))/(K_(C))` to be minimum, `(Deltan)` has to be minimum |
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| 731. |
Phosphorus pentachloride dissociates as follows in a closed reaction vessel. `PCl_5(g) hArr PCl_3(g) +Cl_2(g)` If total pressure at equilibrium of the reactions mixture is P and degree of dissociation of `PCl_5`is x, the partial pressure of `PCl_3` will be:A. `(x/(x+1))P`B. `((2x)/(1-x))P`C. `(x/(x-1))P`D. `(x/(1-x))P` |
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Answer» Correct Answer - A |
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| 732. |
The activation energy of `H_(2)+I_(2) hArr 2HI` in equilibrium for the forward reaction is `167 kJ mol^(-1)` whereas for the reverse reaction is `180 kJ mol^(-1)`. The presence if catalyst lowers the activation energy by `80 kJ mol^(-1)`. Assuming that the reaction are made at `27^(@)C` and the frequency factorr for the forward and backward reactions are `6xx10^(-4)` and `3xx10^(-3)`, respectively, calculate `K_(c )`. |
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Answer» The lowering of activation energy by a catalyst occurs for forward reaction as well as for backward reaction. Thus, in presence of catalyst, Energy of activation for forward reaction `(DeltaH_(1))` `=167-80=37 kJ "mole"^(-1)` Energy of activation for backward reaction `(DeltaH_(2))` `=180-80=100 kJ "mole"^(-1)` For forward reaction, `K_(1)=A_(1) e^(-DeltaH_(1)//RT)` For backward reaction `K_(2)=A_(2) e^(-DeltaH_(2)//RT)` where `A_(1)` and `A_(2)` are frequency factors and `DeltaH_(1)` and `DeltaH_(2)` are energies of activation. `:. K_(c )=K_(1)/K_(2)=A_(1)/A_(2). e^([(-DeltaH_(1)//RT)+(DeltaH_(2)//RT)])` `=(6xx10^(-4))/(3xx10^(-3))xxe^([(-87+100)//(8.314xx10^(-3)xx300)])` `K_(c )=2xx10^(-1) e^(+13//(8.314xx300xx10^(3)))=36.8` |
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| 733. |
For the reaction `2HIhArrH_(2)(g)+l_(2)(g)`A. `K_(p)=K_(c)`B. `K_(p)gtK_(c)`C. `K_(p)ltL_(c)`D. `L_(c)=sqrtK_(P)` |
| Answer» Correct Answer - A | |
| 734. |
If for `2A_(2)B(g)hArr2A_(2)(g)+B_(2)(g),K_(P)="TOTAL PRESSURE" ("at equilibrium") " and starting the dissociation from" 4 "mol of" A_(2)B` then:A. degree of dissociation of `A_(2)B "will be" (2//3)`B. total no. of moles at equilibrium will be `(14//3)`.C. at equilibrium the no. of moles of `A_(2)B` are no equal to the no. of moles of `B_(2)`D. at equilibrium the no. of moles of `A_(2)B` are equal to the no. of moles of `A_(2)` |
| Answer» Correct Answer - A | |
| 735. |
In an evaculated closed isolated chamber at `250^@C` , 0.02 mole `PCl_5` and 0.1 mole `Cl_2` are mixed. `(PCl_5 ltimplies PCl_3+Cl_2)`.At equilibrium density of mixture was 2.48 g/L and pressure was 1 atm. The number of total moles of equilibrium will be approximately :A. 0.012B. 0.022C. 0.039D. 0.045 |
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Answer» Correct Answer - C |
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| 736. |
The activation energy of `H_(2)+I_(2) hArr 2HI(g)` in equilibrium for the forward reaction is `167 kJ mol^(-1)` whereas for the reverse reaction is `180 kJ mol^(-1)`. The presence of catalyst lowers the activation energy by `80 kJ mol^(-1)`. Assuming that the reactions are made at `27^(@)C` and the frequency factor for forwatd and backward reactions are `4xx10^(-4)` and `2xx10^(-3)` respectively, calculate `K_(c)`. |
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Answer» A catalyst lowers the activation energy for forward reaction as well as for backward reaction by equal amount. Thus, in presence of catalyst, Energy of activation for forward reaction `(E_(a1))=167-80=87 kJ "mol"^(-1)` Energy of activation for backward reaction `(E_(a2))=180-80=100 kJ "mol"^(-1)` `:.` For forward reaction, `K_(1)=A_(1)e^(-Ea//RT)` For backward reaction, `K_(2)=A_(2)e^(-Ea//RT)` where `A_(1)` and `A_(2)` are frequency factors and `E_(a1)` and `E_(a2)` are energies of activation after addition of catalyst. `:. K_(c )=K_(1)/K_(2)=A_(1)/A_(2) e^([(-E_(a1)//RT)+(-E_(a2)//RT)])` `=A_(1)/A_(2)e^(([E_(a2)-E_(a1)])/(RT))=(4xx10^(-4))/(2xx10^(-3)) e^(((100-87))/((8.314xx10^(-3)xx300)))` `:. K_(c )=2xx10^(-1) e^([13//8.314xx10^(-3)xx300])` `K_(c )=36.8` |
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| 737. |
`0.96 g` of `HI` were heated to attain equilibrium `2HIhArrH_(2)+I_(2)`. The reaction mixture on titration requires `15.7mL` of `N//10` hypo solution. Calculate degree of dissociation of `HI`. |
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Answer» Correct Answer - `20.9%` meq of `l_(2)=(W_(l_2))/(127)xx1000=1.57` `W_(l_(2))=(127xx1.57)/(1000)` `n_(l_(2))=(1.57)/(2000)``2HI(g)harrH_(2)(g)+l_(2)(g)` `a-aalpha , (aalpha)/(2) , (aalpha)/(2) , (0.96)/(256)xxalpha=(1.57)/(2000)` `alpha=0.209` |
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| 738. |
In a container of constant volume at a particular temparature `N_(2)` and `H_(2)` are mixed in the molar ratio of `9:13`. The following two equilibria are found o be coexisting in the container `N_(2)(g)+3H_(2)(g)harr2NH_(3)(g)` `N_(2)(g)+2H_(2)(g)harr N_(2)H_(4)(g)` The total equiibrium pressure is found to be `305` atm while partial pressure of `NH_(3)(g)` and `H_(2)(g)` are `0.5` atm and `1` atm respectivly. Calculate of equilibrium constants of the two reactions given above. |
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Answer» Let the initial partial pressure of `N_(2)` be `9P` and `13P` respectively Total pressure `=P_(N_(2))+P_(H_(2))+P_(NH_(3))+P_(N_(2)H_(4))=3.5` atm `(9P-x-y)+(13P-3x-2y)+2x+y=3.5`atm .....(1) ` P_(NH_(3))=2x=0.5` atm .....(2) `P_(H_(2))=(13P-3x-2y)= 1` atm ....(3) from (1) `implies(9P-x-y)+1 ` atm `+0.5+y=3.5` `implies(9P-x)=2` atm so `9P=2.25` `P=0.25` atm from (3) equation `2y=1.5` `y=0.75` atm so `P_(N_(2))=9P-x-y=1.25` atm `P_(H_(2))=1` atm `P_(NH_(3))=0.5` atm `P_(N_(2)(H_(4))=0.75` atm So, `K+(P_(1))=(P_(NH_(3))^(2))/P_(H_(2)^(3).P_(N_(2)))=(0.5xx0.5)/(1xx1xx1xx1.25)=0.2atm^(-2)` `K_(P_(2)=P_(N_2H_(4))/P_(N_(2).P_(H_(2))^(2))=(0.75)/(1xx1xx1.25)=0.6 atm^(-2)` |
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| 739. |
The equilibrium constants `K_(p1)` and `K_(p2)` for the reactions X `hArr` 2Y and Z `hArr` P + Q, respectively, are in the ratio of `1:9` . If the degree of dissures at these equilibria is:A. `1:1`B. `1:3`C. `1:9`D. `1:36` |
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Answer» Correct Answer - D |
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| 740. |
The vapour density of `Pcl_(5)` is `104.16` but when heated to `230^(@)C`, its vapour density is reduced to `62`. The degree of dissociation of `PCl_(5)` at `230^(@)C` is …………A. `6.8%`B. `68%`C. `46%`D. `64%` |
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Answer» Correct Answer - B `PC1_(5)(g)hArrPC1_(3)(g) + C1_(2)(g)` `1+alpha =(D)/(d)` `1+alpha =(104.25)/(62)` `1+alpha=1.68` `:. alpha = 0.68` or `68%` |
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| 741. |
The equilibrium constants `K_(p1)` and `K_(p2)` for the reactions X `hArr` 2Y and Z `hArr` P + Q, respectively, are in the ratio of `1:9` . If the degree of dissures at these equilibria is:A. `1:36`B. `1:1`C. `1:3`D. `1:9` |
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Answer» Correct Answer - A `{:(X,hArr,2Y),(1,,0):}` `(1-x) 2x` `implies K_(P1) =((2x)^(2))/((1-x))((P_(1))/(1+x))^(1)` `{:(Z,hArr,P,+,Q),(1,,0,,0),((1-x),,x,,x):}` `K_(P2) = (x^(2))/((1-x))((P_(2))/(1+x))^(1)` . `(4xxP_(1))/(P_(2)) =(1)/(9) = (P_(1))/(P_(2)) =(1)/(36)` |
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| 742. |
For the chemical equilibrium, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` `Delta_(r)H^(ɵ)` can be determined from which one of the following plots?A. B. C. D. |
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Answer» Correct Answer - A `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` `K_(p)=P_(CO_(2))` According to Arrhenius equation: `K=Ae^(-DeltaHɵ_(r )//RT)` `"log" K_(p)= log A-(DeltaH_(r)^(@))/(2.303 RT)` `log P_(CO_(2))=log A-(DeltaH_(r )^(@))/(2.303R)1/T …(i)` `Y=C+MX` Graph (a) represents (i) and its slope will be used to determine the heat of the reaction. |
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| 743. |
For the chemical equilibrium, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` `Delta_(r)H^(ɵ)` can be determined from which one of the following plots?A. B. C. D. |
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Answer» Correct Answer - A `CaCO_(3)(S)harrCaO(S)+CO_(2)(g)` `K_(P)=P_(CO_(2))` `log K_(P)=logA-(DeltaH_(r)^(0))/(2.303RT)` `log P_(CO_(2))=logA-(DeltaH_(r)^(@))/(2.303) (1)/(RT)......(i)` Graph `(a)` represents `(i)` and its slope will be used to determine he heat of the reacion. So, |
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| 744. |
For the chemical equilibrium, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` `Delta_(r)H^(ɵ)` can be determined from which one of the following plots?A. B. C. D. |
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Answer» Correct Answer - a |
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| 745. |
`4` moles of A are mixed with `4` moles of B, when `2` moles of C are formed at equilibrium according to the reaction `A+B hArr C+D`. The value of equilibrium constant isA. `(1)/(4)`B. `(1)/(2)`C. `1`D. `4` |
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Answer» Correct Answer - C `{:(,,A,+,B,hArr,C,+,D),("Initial conc.",,4,,4,,0,,0),("After T time conc.",,(4-2),,(4-2),,2,,2):}` Equilibrium constant `= ([C][D])/([A][B]) = (2xx2)/(2xx2) = 1` |
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| 746. |
0.5 moles of `N_2` gas is mixed with 0.72 moles of `O_2` gas in a 2 litre tank at 2000 K. The two gases react as : `N_2(g)+O_2(g) hArr 2NO(g), K_p=4.9xx10^(-5)` at 2000 K. The equilibrium concentration of NO(g) will be :A. `4.2xx10^(-3) M`B. `6.3xx10^(-3) M`C. `2.1xx10^(-3) M`D. `7xx10^(-3) M` |
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Answer» Correct Answer - C |
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| 747. |
Two moles of `PCl_(5)` were heated to `327^(@)C` in a closed two-litre vessel, and when equilibrium was achieved, `PCl_(5)` was found to be `40%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant `K_(p)` and `K_(c)` for this reaction.A. `0.530`B. `0.266`C. `0.130`D. `0.170` |
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Answer» Correct Answer - B `{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),(t=0,2,,0,,0),(t=t_("eq"),(2-0.08)/2,,0.8/2,,0.8/2):}` `:. K_(c )=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.8/2xx0.8/2)/(1.2/2)=0.64/2.4=0.266` |
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| 748. |
`CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` in closed container at equilibrium. What would be the effect of addition of `CaCO_(3)` on the equilibrium concentration of `CO_(2)`.A. IncreasesB. DecreasesC. Data is not sufficientD. Remains unaffected |
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Answer» Correct Answer - D `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` `K_(c)=P_(CO_(2))`, therefore, `K_(c)` only depends upon concentration of `CO_(2)` as long as temperature remains constants and both `CaCO_(3)(s)` and `CaO(s)` are present. |
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| 749. |
`4` moles of A are mixed with `4` moles of B, when `2` moles of C are formed at equilibrium according to the reaction `A+B hArr C+D`. The value of equilibrium constant isA. `4`B. `1`C. `1//2`D. `1//4` |
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Answer» Correct Answer - B `{:(,A,+,B,hArr,C,+,D),("Initial",4,,4,,0,,0),("Final",4-x,,4-x,,x,,x),(,x=2,,,,,,):}` `:. [A]=4-2=2, [B]=4-2=2` `[C]=2, [D]=2` `K=([C][D])/([A][B])=(2xx2)/(2xx2)=1` |
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| 750. |
If the concentration of `OH^(-)` ions in the reaction `Fe(OH)_(3)(s)hArrFe^(3+)(aq.)+3OH^(-)(aq.)` is decreased by `1//4` times, then the equilibrium concentration of `Fe^(3+)` will increase byA. 8 timesB. 16 timesC. 64 timesD. 4 times |
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Answer» Correct Answer - C `Fe(OH)_3(s)hArrFe^(3+)(aq)+3OH^(-)(aq)` `K=([Fe^(3+)][OH^(-)]^(3)]/([Fe(OH)_(3)])` To maintain equilibrium constant, let the concentration of `Fe^(3+)` is increased x times, on decreasing the concentration of `OH^(-)` by `(1)/(4)` times `K=([xFe^(3+)][OH^(-)]^(3))/([Fe(OH)_(3)])`.....(ii) By dividing eq. (ii) by (i) we get `(1)/(64)xxX=1rArrX=64` times. |
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