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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.
| 51. |
For the reaction `I_(2)(g) hArr 2I(g)`, `K_(c) 37.6 xx 10^(-6)` at 1000K . If 1.0 mole of `I_(2)` is introduced into a 1.0 litre flask at 1000K, at equilibriumA. `[I_(2)]gt gt [I]`B. `[I_(2)] = [I]`C. `[I_(2)] lt [I]`D. Unpredictable |
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Answer» Correct Answer - A As `K_(c)` is `~~ 10^(-6)` therefore though the equilibrium shall move in forward direction yet `I_(2) gt gt gt I` |
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| 52. |
In the following equilibria `I:A+2B hArr C, K_(eq)=K_(1)` `II: C+DhArr 3A, K_(eq)=K_(2)` `III, 6B+D hArr 2C, K_(eq)=K_(3)` Hence,A. `3K_(1)+K_(2)=K_(3)`B. `K_(1)^(3).K_(2)^(2)=K_(3)`C. `3K_(1)+K_(2)^(2)=K_(3)`D. `K_(1)^(3).K_(2)=K_(3)` |
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Answer» Correct Answer - D To obtain `K_(3)` multiply eqn. I by 3 and add eqn. II `3A+6B hArr 3C, K_(eq)=K_(1)^(3)` `C+D hArr 3A , K_(eq)=K_(2)` Adding, we get `6B+D hArr 2C, K^(eq)=K_(3)=K_(1)^(3).K_(2)` |
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| 53. |
The values of `K_(p_(1))` and `K_(p_(2))` for the reactions `X hArr Y+Z` ….(i) and `A hArr 2B` …(ii) are in ratio of 9 : 1. If degree of dissociation of X and A be equal, then total presure at equilibrium (i) and (ii) are in the ratio.A. `3:1`B. `1:9`C. `36:1`D. `1:1` |
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Answer» Correct Answer - C Suppose the total pressure at equilibrium for the reactions (1) and (2) are `P_(1)` and `P_(2)` respectively. Then `{:(,X,hArr,Y,+,Z),("Initial",1 mol,,0,,0),("At eqm.",1-alpha,,alpha,,alpha):}` Total no. of moles at eqm. `=1+alpha` `P_(X)=(1-alpha)/(1+alpha)P_(1),P_(Y)=(alpha)/(1+alpha)P_(1),P_(Z)=(alpha)/(1+alpha)P_(1)` `K_(P)=(((a)/(1+a)P_(1))^(2))/(((1-a)/(1+a))P_(1))=(alpha^(2)P_(1))/(1-alpha^(2))=alpha^(2)P_(1)` `{:(,A,hArr,2B,),("At eqm.",1-alpha,,2alpha,):}` Total no. of moles at eqm. `=1+alpha` `P_(A)=(1-alpha)/(1+alpha)P_(2),P_(B)=(2alpha)/(1+alpha)P_(2)` `K_(p_(2))=(((2alpha)/(1+alpha)P_(2))^(2))/(((1-alpha)/(1+alpha))P_(2))=(4alpha^(2))/(1-alpha^(2))P_(2)~~P_(2)` `(K_(p_(1)))/(K_(p_(2)))=(alpha^(2)P_(1))/(4alpha^(2)P_(2))=(P_(1))/(4P_(2))=(9)/(1)`Given or `(P_(1))/(P_(2))=(36)/(1)` or ` P_(1) : P_(2) =36 :1` |
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| 54. |
For the reversible reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` at `500^(@)C`, the value of `K_(p)` is `1.44xx10^(-5)` when the partial pressure is measured in atmosphere. The corresponding value of `K_(c)` with concentration in mol `L^(-1)` isA. `1.44xx10^(-5)//(0.082xx500)^(-2)`B. `1.44xx10^(-5)//(8.314xx773)^(-2)`C. `1.44xx10^(-5)//(0.0832xx500)^(2)`D. `1.44xx10^(-5)//(0.0832xx773)^(-2)` |
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Answer» Correct Answer - D `K_(p)=K_(c)(RT)^(Deltan)` or `K_(c)=K_(p)//(RT)^(Deltan)` Here, `K_(c)=1.44xx10^(-5), R=0.082, T=273` Hence `K_(c)=1.44 xx 10^(-5)//(0.082xx773)^(-2)` |
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| 55. |
The yield of `NH_(3)` in the reaction `N_(2)+3H_(2) hArr 2NH_(3) , Delta H =-22.08 `kcal is affected byA. change in pressure and temperatureB. change in temperature and concentration of `N_(2)`C. change in pressure and concentration of `N_(2)`D. change in pressure, temperature and concentration of `N_(2)` |
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Answer» Correct Answer - D Factual question. |
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| 56. |
For the following equilibrium `:` `NH_(2)CO_(2)NH_(4)(s)hArr 2NH_(3)(g)+CO_(2)(s)` `K_(p)` is found to be 0.5 at 4400 K . Hence, partial pressure of `NH_(3)` and `CO_(2)` are respectively `:`A. 2.0,1.0 atmB. 1.0,2.0 atmC. 1.0,0.5 atmD. 0.5,1.0 atm |
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Answer» Correct Answer - C `NH_(2)CONH_(4)(s)hArr 2NH_(3)(g)+CO_(2)(g)` `K_(p)=(p_(NH_(3)))^(2)(p_(CO_(2)))` Let `p_(CO_(2))=x:. p_(NH_(3))=2x` `:. (2x)^(2).x=0.5` `4x^(3)=0.5` `x^(3)=(5)/(40)=(1)/(8)` `x=(1)/(2)=0.5` `:. P_(CO_(2))=0.5,p_(NH_(3))=2xx0.5=1.0` `p_(NH_(3))=1.0` and `p_(CO_(2))=0.5` |
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| 57. |
`CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` For the equilibrium , it is found that `log K_(p)=8-(6400(K))/(T)` `T=527^(@)C` if `:`A. `K_(p)=7.2 xx 10^(-5)`B. `K_(p)=1`C. `K_(p)=13.04 xx 10^(3)`D. `K_(p)=10`. |
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Answer» Correct Answer - B `=T=527^(@)C=527+273=800K` `:. logK_(p)=8-(6400K)/(T)` `=8-(6400)/(800)=8-8=0` `log K_(p)=0` `:. K_(p)=1` |
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| 58. |
`CaCO_(3) hArr CaO + CO_(2)` reaction in a lime kiln goes to completion becauseA. CaO does not react with `CO_(2)` to give `CaCO_(3)`B. backward reaction is very slowC. `CO_(2)` formed escpaes outD. None of these |
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Answer» Correct Answer - C The reaction, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g) uArr` goes to completion in a lime kiln because `CO_(2)` formed is allowed to escape out. |
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| 59. |
The equilibrium: `P_(4)(g)+6Cl_(2)(g) hArr 4PCl_(3)(g)` is attained by mixing equal moles of `P_(4)` and `Cl_(2)` in an evacuated vessel. Then at equilibrium:A. `[Cl_(2)]gt [PCl_(3)]`B. `[Cl_(2)] gt [P_(4)]`C. `[P_(4)] gt [Cl_(2)]`D. `[PCl_(3)] gt [P_(4)]` |
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Answer» Correct Answer - C `{:(,P_(4)(g),+,6Cl_(2)(g),hArr,4PCl_(3)(g)),("Initial",c"mol",,c "mol" ,,),("At eqm.",C(1-alpha),,C(1-6alpha),,4C alpha):}` Clearly, `[P_(4)] gt [Cl_(2)]` at equilibrium |
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| 60. |
Consider the reaction `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. IncreaseB. DecreasesC. Data is not sufficient to predict itD. Remains unaffected |
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Answer» Correct Answer - D K for the reation is given by `K=[CO_(2)].` Because concentration of solids taken to be unity. Since `[CaCO_(3)]` and `[CaO]` do not figure in the expression. There will be no effect of addition of `CaCO_(3)`. |
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| 61. |
`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. 4B. 1C. `1//2`D. `1//4` |
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Answer» Correct Answer - B `K=(2xx2)/(2xx2)=1` |
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| 62. |
For the reaction `2NO_(2)(g) hArr 2NO(g)+O_(2)(g)` `K_(c)=1.8xx10^(-6)` at `184^(@)C, R=0.00831 kJ //` ( mol.K) when `K_(p)` and `K_(c)` are compared at `184^(@)C`, it is foundA. `K_(p)` is greater than `K_(c)`B. `K_(p)` is less than `K_(c)`C. `K_(p)=K_(c)`D. whether `K_(p)` is greater than, less than or equal to `K_(c)` depends upon the total gas pressure. |
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Answer» Correct Answer - A `2NO_(2)(g) hArr 2NO(g)+O_(2)(g)` `Deltan_(g)=2+1-1=1,K_(c )=1.8xx10^(-6)` `T=184^(@)C =184+273=457K` `R=0.00831kJ //(mol K)` `=8.31J K^(-1) mol^(-1)` `=0.0821 L atm K^(-1) mol^(-1)` `K_(p)=K_(c)(RT)^(Deltan_(g))` `K_(p)=1.8xx10^(-6)(0.0821)xx457)` `=1.8xx10^(-6)xx37.52` `=67.53xx10^(-6) gt K_(c )` |
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| 63. |
For the equilibrium `2NO(g)+Cl_(2)(g)hArr 2NOCl(g)` `K_(p)` is related to `K_(c)` by the reactionA. `K_(p)=K_(c)(RT)`B. `K_(p)=K_(c)(RT)^(2)`C. `K_(p)=K_(c)//RT`D. `K_(p)=K_(c)//(RT)^(2)` |
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Answer» Correct Answer - C 3 moles of gaseous reactants change to 2 moles of gaseous products. |
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| 64. |
If `K_(1)=` Rate constant at temperature `T_(1)` and `k_(2)` rate constant at temperature `T_(2)` for a first order reaction, then which of the following relation is correct ?A. `log. (k_(2))/(k_(1))=(2.303)/(RT)[(T_(2)-T_(1))/(T_(1)T_(2))]`B. `log. (k_(2))/(k_(1))=(E_(a))/(2.303RT)[(T_(2)-T_(1))/(T_(1)T_(2))]`C. `log. (k_(2))/(k_(1))=(E_(a))/(2.303RT)[(T_(1)T_(2))/(T_(2)-T_(1))]`D. `log. (k_(1))/(k_(2))=(E_(a))/(2.303RT)[(T_(2)T_(2))/(T_(2)-T_(1))]` |
| Answer» Correct Answer - B | |
| 65. |
For which of the following reaction `K_(p)=K_(c)` ?A. `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`B. `2NOCl(g) hArr 2NO(g) +Cl_(2)(g)`C. `H_(2)(g)+I_(2)(g) hArr 2HI(g)`D. `H_(2)(g)+Cl_(2)(g) hArr HCl(g)`. |
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Answer» Correct Answer - D When `Delta n =0, K_(p)=K_(c)` |
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| 66. |
In which of the following equilibrium, change in the volume of the system does not alter the number of moles?A. `N_(2)(g)+O_(2)(g) hArr 2NO(g)`B. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`C. `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`D. `SO_(2)Cl_(2)(g) hArr SO_(2)(g)+Cl_(2)(g)` |
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Answer» Correct Answer - A A reaction which does not involve a change in the number of moles of gaseous reactants and products (i.e., `Deltan_(g)=0` or `n_(p)=n_(r ) )` is not affected by change in pressure. |
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| 67. |
For reaction : `H_(2)(g)+I_(2)(g) hArr 2HI(g)` at certain temperature, the value of equilibrium constant is `50`. If the volume of the vessel is reduced to half of its original volume, the value of new equilibrium constant will beA. 25B. 50C. 100D. Unpredictable |
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Answer» Correct Answer - B There is no change in no. of gas moles, therefore the expression for K is independent of volume. Hence K will remain same. |
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| 68. |
What are the most favourable conditions for the reaction `SO_(2)+//2 O_(2) hArr SO_(3), Delta H =-ve` to occurA. low T, high PB. low T and low PC. high T, low PD. high T, high P. |
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Answer» Correct Answer - A Since the reaction is exothermic , it is favoured by low temperature . Since reaction is accompanied by decrease in the number of moles, it is favoured by high pressure. |
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| 69. |
For the reaction `2HI hArr H_(2)+I_(2)`A. `K_(p) gt K_(c)`B. `K_(c) gt K_(p)`C. `K_(p) =K_(c)`D. None of these |
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Answer» Correct Answer - C `Delta n _(g)=0` Since `K_(p)=K_(c)(RT)^(Deltan)`hence `K_(p)=K_(c).` |
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| 70. |
For the reaction `CO(g)+(1)/(2) O_(2)(g) hArr CO_(2)(g),K_(p)//K_(c)` isA. RTB. `(RT)^(-1)`C. `(RT)^(-1//2)`D. `(RT)^(1//2)` |
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Answer» Correct Answer - C `Deltan_(g)=1-(1+(1)/(2))=-(1)/(2)` `K_(p)=K_(c ) (RT)^(Deltan)` or `K_(p)//K_(c )=(RT)^(Deltan)=(RT)^(-1//2)` |
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| 71. |
When ethanol and acetic acid are mixed together in equimolar proportions, equilibrium is attained when `2//3rd` of acid and alcohol are consumed. The equilibrium constant for the reaction isA. 0.4B. 40C. `4xx10^(2)`D. `4xx10^(@)` |
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Answer» Correct Answer - D `{:(CH_(3)COOH,+,C_(2)H_(5)OH,hArr,CH_(3)COOC_(2)H_(5),+,H_(2)O),(1 "mol",,1"mol",,-,,-),(1//3"mol",,1//3"mol",,2//3"mol",,2//3"mol"):}` `K=(2//3xx2//3)/(1//3xx1//3)=4` |
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| 72. |
1 mol of hydrogen and 2 mol of iodine are taken initially in a 2L vessel. The number of moles of hydrogen at equilibrium is `0.2`. Then the number of moles of iodine and hydrogen iodide at equilibrium areA. 1.2,1.6B. 1.8,1.0C. 0.4,2.4D. 0.8,2.0 |
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Answer» Correct Answer - A `{:(,H_(2),+,I_(2),hArr , 2HI),("Initial ",1 "mol",,2"mol",,),("At eq.",0.2"mol",,2-0.8,,2xx0.8):}` `(0.8` mole of `H_(2)` have reacted ). |
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| 73. |
When `H_(2)` and `I_(2)` are mixed and equilibrium is attained,thenA. amount of HI formed is equal to the amount of `H_(2)` dissociatedB. HI dissociation stopsC. the reaction stops completelyD. None of these |
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Answer» Correct Answer - D `(a) H_(2)+I_(2) rarr 2HI_(2)` `:. ` 1 mole of `H_(2)` gives 2 moles of HI. `(b)` At equilibrium the backward and forward reactions do not stop. (c ) The reaction does not stop but `R_(f) =R_(b)`. |
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| 74. |
In the case of gaseous homogeneous reaction, the active mass of the reaction is obtained by the expression.A. `(PV)/(RT)`B. `(P)/(RT)`C. `(RT)/(P)`D. `(n)/(V) RT`. |
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Answer» Correct Answer - B Active mass `=` Molar conc. `=(n)/(V)` `PV=nRT` `(n)/(V)=(P)/(RT)` |
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| 75. |
Which of the following favours the reverse reaction in chemical equilibrium ?A. Increasing the concentration of the reactantB. Removal of at least one of the products at regular intervalsC. Increasing the concentrations of one or more of the productsD. increasing the pressure. |
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Answer» Correct Answer - C If the concentration of a product is increased, by Le-Chaterlier principle, equilibrium shifts in the backward direction. |
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| 76. |
The equilibrium constant for a reacton `N_(2)(g)+O_(2)(g)=2NO(g)` is `4xx10^(-4)` at `2000 K`. In the presence of catalyst, the equilibrium constant is attained `10` times faster. The equilibrium constant in the presence of catalyst, at `2000 K` isA. `40 xx 10^(-4)`B. `4xx10^(-4)`C. `4xx10^(-2)`D. Difficult to compute without more data |
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Answer» Correct Answer - B Equilibrium constant (K) values is independent of the catalyst. |
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| 77. |
For the reaction `CO(g)+Cl_(2)(g) harr COCl_(2)(g), K_(p)//K_(c)` is equal to:A. `(1)/(RT)`B. `1.0`C. `sqrt(RT)`D. RT |
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Answer» Correct Answer - A `K_(p)=K_(c ) (RT)^(Deltan_(g))` or `(K_(p))/(K_(c))=(RT)^(Deltan_(g))` `CO(g)+Cl_(2)(g) hArr COCl_(2)(g)` Here, `Deltan_(g)=n_(p)-n_(r)=1-2=-1` `:. (K_(p))/(K_(c))=(RT)^(-1)` or `(K_(p))/(K_(c))=(1)/(RT)` |
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| 78. |
What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?A. `K_(c)=([P_(4)O_(10)])/([P_(4)][O_(2)]^(5))`B. `K_(c)=(1)/([O_(2)]^(5))`C. `K_(c)=[O_(2)]^(5)`D. `K_(c)=([P_(4)O_(10)])/(5[P_(4)][O_(2)])` |
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Answer» Correct Answer - B In the heterogeneous equilibrium `P_(4)(s)+5O_(2)(g) hArr P_(4) O_(10)(s)` `K_(c)=([P_(4)O_(10)(s)])/([P_(4)(s)][O_(2)(g)]^(5))` Here `[P_(4)O_(10)(s)]=[P_(4)(s)]=1` ( By convention ) `:. K_(c ) =(1)/([O_(2)]^(5))` |
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| 79. |
Consider the equilibrium reactions, `H_(3)PO_(4)overset(K_(1))(hArr ) H^(+)+H_(2)PO_(4)^(-)` `H_(3)PO_(4)^(-)overset(K_(2))(hArr ) H^(+)+HPO_(4)^(-2)` `HPO_(4)^(-2)overset(K_(3))(hArr)H^(+)+PO_(4)^(-3)` The equilibrium constant K for the following dissociation `H_(3)PO_(4)hArr 3H^(+)+PO_(4)^(-)` isA. `K_(1) // K_(2)K_(3)`B. `K_(1) K_(2) K_(3)`C. `K_(2) // K_(1)K_(3)`D. `K_(1)+K_(2)+K_(3)` |
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Answer» Correct Answer - B In multistage equilibrium , `K=K_(1). K_(2).K_(3)` |
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| 80. |
`I_(2)+I^(ɵ)hArrI_(3)^(ɵ)` This reaction is set-up in aqueous medium. We start with `1` mol of `I_(2)` and `0.5` mol of `I^(ɵ)` in `1 L` flask. After equilibrium reached, excess of `AgNO_(3)` gave `0.25` mol of yellow precipitate. Equilibrium constant isA. 1.33B. 2.66C. 2D. 3 |
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Answer» Correct Answer - A `{:(,I_(2),+,I^(-),hArr,I_(3)^(-)),("Initial",1M,,0.5M,,-),("At eqm,. " ,(1-x)M,,0.5M,,x "mol"):}` Moles of `AgI=` Moles of `I^(-)` in solution `:. [I^(-)]_("eqn")=0.25M` `0.5-x=0.2:. x=0.25` `K=(x)/((1-x)(0.5-x))` `=(0.25)/(0.75xx0.25)=1.33` |
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| 81. |
For the equilibrium `2SO_(2)+O_(2)hArr 2SO_(3)` we start with 2 moles of `SO_(2)` and 1 mole of `O_(2)` at 3 atm. When equilibrium is attained, pressure changes to 2.5atm. Hence, `K_(p)` is `:`A. `3 atm^(-1)`B. `2.5 atm^(-1)`C. `2 atm ^(-1)`D. `0.5 atm^(-1)` |
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Answer» Correct Answer - C Initially , `P=3 atm` Total no. of moles `=2+1=3` `PV=nRT` `3xx V =3RT` `V=RT` `{:(,2SO_(2),+,O_(2),hArr,2SO_(3)),("Initial",2,,1,,-),("At eqm.",2-2x,,1-x,,2x):}` Total no. of moles at eqm. `=3-x` `PV=nRT` `2.5xxV=(3-x)RT` `2.5 xx RT =(3-x)RT` `3-x=2.5` `x=0.5` No. of moles at eqm. `SO_(2)=2(1-x)=2(1-0.5)=1` `O_(2)=1-x=1-0.5=0.5` `SO_(3)=2x=2xx0.5=1` We know `p_(A)=x_(A)P` `p_(SO_(2))=(1)/(2.5)xx2.5=1 "atm"` `p_(O_(2))=(0.5)/(2.5)xx2.5=0.5 "atm"` `p_(SO_(3))=(1)/(2.5)xx2.5=1"atm"` `K_(p)=((p_(SO_(3)))^(2))/((p_(O_(2)))(p_(SO_(2)))^(2))` `=((1)^(2))/((0.5)(1)^(2))=2 atm^(-1)` |
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| 82. |
For the reaction `A+B hArr C+D` taking place in a 1 L vessel, equilibrium concentration of `[C]=[D]=0.5M` if we start with 1 mole each A and B. Percentage of A converted into C if we start with 2 moles of A and 1 mole of B, isA. `25%`B. `40%`C. `66.66%`D. `33.33%` |
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Answer» Correct Answer - D `{:("Initial conc. ",A,+,B,hArr,C,+,D),("Eqm. conc.",(1-0.5)M,,(1-0.5)M,,0.5M,,0.5M):}` `[A]=[B]=[C]=[D]=0.5` `:. K=([C][D])/([A][B])` In second case, `{:(,A,+,B,hArr,C,+,D),("Initial eonc. ",2M,,1M,,,,),("Eqm. conc.",(2-x)M,,(1-x)M,,xM,,xM):}` `K=([C][D])/([A][B])` `:. (x^(2))/((2-x)(1-x))=1` `x^(2)=(2-x)(1-x)` `x^(2)=-2x-x+2+x^(2)` `:. x=(2)/(3)` `:. %` of A converted to `C=(x)/(2)xx100` `=(2xx100)/(3xx2)=33.33%` |
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| 83. |
Volume of the flask in which species are transferred is double of the earlier flask. In which of the following cases, equilibrium will be shifted in the forward direction ?A. `N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`B. `N_(2)(g)+O_(2)(g)hArr 2NO(g)`C. `PCl_(5)(g) hArr PCl(g)+Cl_(2)(g)`D. `2NO(g)hArrN_(2)(g)+O_(2)(g)` |
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Answer» Correct Answer - C As volume is doubled, the pressure is reduced to one-half. Under these conditions, the equilibrium involving increase in the number of moles of gaseous components in the forward direction will be shifted forward. Thus, eqm. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` will be shifted in the forward direction. |
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| 84. |
One mole each of A and B and 3 moles each of C and D are placed in 11L flask, if equilibrium constant is 2.25 for `A+B hArr C+D` equilibrium concentration of A and C will be in the ratio `:`A. `2:3`B. `3:2`C. `1:2`D. `2:1` |
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Answer» Correct Answer - A `{:(,A,+,B,hArr,C,+,D),("Initial conc. ",1,,1,,3,,3),("At eqm. ",1-x,,1-x,,3+x,,3+x):}` `K=([C][D])/([A][B])=((3+x)^(2))/((1-x)^(2))` `2.25=((3+x)^(2))/((1-x^(2)))` `:. (3+x)/(1-x)=1.5` `3+x=1.5-1.5x` `2.5x=1.5` `x=(-15)/(25)=(-3)/(5)` `[A]=1-x=1+(3)/(5)=(8)/(5)` ltbgt `[C]=3+x=3-(3)/(5)=(12)/(5)` `[A]:[C]=8:12` or `[A]:[C]=2:3` |
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| 85. |
For the reaction `C(s) +CO_(2)(g) rarr 2CO(g), k_(p)=63` atm at 100 K. If at equilibrium `p_(CO)=10p_(CO_(2)) ` then the total pressure of the gases at equilibrium isA. 6.3 atmB. 6.93 atmC. 0.63 atmD. 0.693 atm |
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Answer» Correct Answer - B `C(s) +CO_(2)(g) hArr 2CO(g)` `k_(p)=(p_(CO^(2)))/(p_(CO))` or 63=((10p_(CO_(2)))^(2))/(p_(CO_(2)))` `p_(CO_(2))=(63)/(100)=0.63 atm` `p_(CO)=10p_(CO_(2))=10xx0.63atm =6.3` `:. p_("total")=p_(CO_(2))+p_(CO)=0.63+6.3` `=6.93` |
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| 86. |
For a gaseous reaction `xA+yB hArr lC +mD`A. `K_(p)=Kc`B. `K_(p)=(K_(c))^(l+m)`C. `K_(p)=K_(c)(RT)^((l+m)-(x+y))`D. `K_(p)=1//K_(c).` |
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Answer» Correct Answer - C `K_(p)=K_(c)(RT)^(Deltan)` where `Delta n=(l+m)-(x+y)` |
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| 87. |
One mole of `N_(20O_(4)(g)` at 300K is kept in a closed container under one atmosphere. It is heated to 600K when 20% of `N_(2)O_(4)(g)` is converted to `NO_(2)(g)` `N_(2)O_(4)hArr 2NO_(2)(g)` Hence resultant pressure is `:`A. 1.2 atmB. 2.4 atmC. 2.0 atmD. 1.0 atm |
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Answer» Correct Answer - B `{:(,N_(2)O_(4),hArr,2NO_(2),),("Initial",1 mol,,,),("Equilibrium",0.8 mol,,0.4 mol,):}` Total moles at equilibrium `=0.8 +0.4 =1.2` mol Pressure due to 1 mol when temperature is doubled `(300-600K) ` is 2 atm. |
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| 88. |
For the equilibrium system `2HX(g) hArr H_(2)(g)+X_(2)(g)` the equilibrium constant is `1.0xx 10^(-5)`. What is the equilibrium concentration of HX if the equilibrium concentrations of `H_(2)` and `X_(2)` are `1.2xx10^(-3) M` and `1.2 xx 10^(-4) M` respectively ?A. `12 xx 10^(-4) M`B. `12 xx 10^(-3) M`C. `12 xx 10^(-2) M`D. `12 xx 10^(-6) M` |
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Answer» Correct Answer - C `[HX]^(2)=([H_(2)][X_(2)])/(K_(c))` |
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| 89. |
In the reaction `BrO^(-3)(aq) + 5Br^(-) (aq) + 6H^(+) rarr 3Br_(2)(1) + 3H_(2)O(1)` The rate of appearance of bromine `(Br_(2))` is related to rate of disapperance of bromide ions as folllwoing :A. `(d[Br_(2)])/(dt)=(3)/(5)(d[Br^(-)])/(dt)`B. `(d[Br_(2)])/(dt)=(-3)/(5)(d[Br^(-)])/(dt)`C. `(d[Br_(2)])/(dt)=(-5)/(3)(d[Br^(-)])/(dt)`D. `(d[Br_(2)])/(dt)=(-5d)/(3)(d[Br^(-)])/(dt)` |
| Answer» Correct Answer - B | |
| 90. |
The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient `Q_(c)` in place of equilibrium constant `(K_(c))` by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of `Q_(c)` . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of `Q_(c)` and `K_(c)`. The three possible cases are shown as comparison of `K_(c)` and `Q_(c)` in the following figures. Change in Gibbs free energy, i.e., `Delta G` is the driving force of any reaction. For spontaneous reaction , `Delta G =-ve` For non-spontaneous reaction , `Delta G=+ve` For reaction at equilibrium , `Delta G =0` Thermodynamically, we know that `Delta G= Delta G^(@)+ RT ln Q`, where `Q` is reaction quotient and `Delta G^(@)=` change in Gibbs energy at standard condition. For equilibrium `A(g) hArr B(g) (K_(eq) =1.732)` If the pressure of the system [varied by introducing a stream of `A (g)` and B (g) is represented by the curve at constant temperature T. Suppose the equilibrium system `N_(2)O_(4)(g) hArr 2NO_(2)(g)` `N_(2)O_(4)(g)` is in a cylinder fitted with a movable piston . Which of the following statements is correct ?A. If piston is pushed downwards at constant temperature, `Q_(c) gt K_(c)` and the direction shifts in the left direction.B. If pistaon is pushed downwards at constant temperature `Q_(c) gt K_(c)` and the reaction shifts in the right direction.C. If piston is released at constant temperature , `Q _(c) gt K_(c)` and the reaction shifts in the left direction.D. If piston is released at a constant temperature, and the reaction shifts in the right direction. |
| Answer» Correct Answer - A | |
| 91. |
The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient `Q_(c)` in place of equilibrium constant `(K_(c))` by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of `Q_(c)` . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of `Q_(c)` and `K_(c)`. The three possible cases are shown as comparison of `K_(c)` and `Q_(c)` in the following figures. Change in Gibbs free energy, i.e., `Delta G` is the driving force of any reaction. For spontaneous reaction , `Delta G =-ve` For non-spontaneous reaction , `Delta G=+ve` For reaction at equilibrium , `Delta G =0` Thermodynamically, we know that `Delta G= Delta G^(@)+ RT ln Q`, where `Q` is reaction quotient and `Delta G^(@)=` change in Gibbs energy at standard condition. For equilibrium `A(g) hArr B(g) (K_(eq) =1.732)` If the pressure of the system [varied by introducing a stream of `A (g)` and B (g) is represented by the curve at constant temperature T. Suppose `N_(2)O_(4)(g)` is enclosed in a cylinder fitted with a movable piston which attains the following equilibrium `N_(2)O_4)(g) hArr 2NO_(2)(g)` Given that for the 10 atmosphere pressure of the equilibrium mixture, the content of `NO_(2)` is `8xx1^(5) ` ppm. if the piston of cylinder is moved upward in such a manner so that the volume of the gaseous mixture becomes double, then what will be new ppm of `NO_(2)(g)` in the cylinder ? ( Assuming that the temperature of the cylinder remains constant )A. `8.2 xx 10^(5) `ppmB. `10^(5)` ppmC. `8.72 xx 10^(5)` ppmD. `7.4 xx 10^(5)` ppm |
| Answer» Correct Answer - C | |
| 92. |
The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient `Q_(c)` in place of equilibrium constant `(K_(c))` by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of `Q_(c)` . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of `Q_(c)` and `K_(c)`. The three possible cases are shown as comparison of `K_(c)` and `Q_(c)` in the following figures. Change in Gibbs free energy, i.e., `Delta G` is the driving force of any reaction. For spontaneous reaction , `Delta G =-ve` For non-spontaneous reaction , `Delta G=+ve` For reaction at equilibrium , `Delta G =0` Thermodynamically, we know that `Delta G= Delta G^(@)+ RT ln Q`, where `Q` is reaction quotient and `Delta G^(@)=` change in Gibbs energy at standard condition. For equilibrium `A(g) hArr B(g) (K_(eq) =1.732)` If the pressure of the system [varied by introducing a stream of `A (g)` and B (g) is represented by the curve at constant temperature T. If A and B are enclosed in the cylinder and piston of the cylinder be moved downward so that volume of cylinder becomes half, then what will be the effect in `K_(c)` at constant temeprature ?A. `K_(c)` will increaseB. `K_(c)` will decreaseC. `K_(c)` has no relation with `K_(p)`D. No effect in `K_(c)` |
| Answer» Correct Answer - D | |
| 93. |
The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient `Q_(c)` in place of equilibrium constant `(K_(c))` by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of `Q_(c)` . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of `Q_(c)` and `K_(c)`. The three possible cases are shown as comparison of `K_(c)` and `Q_(c)` in the following figures. Change in Gibbs free energy, i.e., `Delta G` is the driving force of any reaction. For spontaneous reaction , `Delta G =-ve` For non-spontaneous reaction , `Delta G=+ve` For reaction at equilibrium , `Delta G =0` Thermodynamically, we know that `Delta G= Delta G^(@)+ RT ln Q`, where `Q` is reaction quotient and `Delta G^(@)=` change in Gibbs energy at standard condition. For equilibrium `A(g) hArr B(g) (K_(eq) =1.732)` If the pressure of the system [varied by introducing a stream of `A (g)` and B (g) is represented by the curve at constant temperature T. For the equilibrium `n` Butane `(g) hArr ` Isobutane `(g),(K_(eq)=1.732)` If the pressure of the system (varied by introducing a stream of n butane and isobutane ) is represented by the curve at constant temperature T. At a particular point Q, which of the following statements holds good ?A. The reaction moves in the backward direction.B. The reaction movess in the forward direction.C. The reaction is at equilibrium .D. The data is insufficient to prodict |
| Answer» Correct Answer - C | |
| 94. |
The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient `Q_(c)` in place of equilibrium constant `(K_(c))` by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of `Q_(c)` . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of `Q_(c)` and `K_(c)`. The three possible cases are shown as comparison of `K_(c)` and `Q_(c)` in the following figures. Change in Gibbs free energy, i.e., `Delta G` is the driving force of any reaction. For spontaneous reaction , `Delta G =-ve` For non-spontaneous reaction , `Delta G=+ve` For reaction at equilibrium , `Delta G =0` Thermodynamically, we know that `Delta G= Delta G^(@)+ RT ln Q`, where `Q` is reaction quotient and `Delta G^(@)=` change in Gibbs energy at standard condition. For equilibrium `A(g) hArr B(g) (K_(eq) =1.732)` If the pressure of the system [varied by introducing a stream of `A (g)` and B (g) is represented by the curve at constant temperature T. If the gaseous substances A,B,C, be in equilibrium as under `:` `{:(B(g),hArr,A(g),,),(10%,,60%,,):}` `{:(B(g),hArr,C(g),,),(10%,,30%,,):}` then which of the following is the correct order of stability of A,B, and C in the equilibrium mixture ?A. `AgtBgtC`B. `CgtBgtA`C. `AgtCgtB`D. `BgtAgtC` |
| Answer» Correct Answer - C | |
| 95. |
Which of the following is a characterisstic of a reversible reaction ?A. it never proceeds by a catalystB. it proceeds onlly in the forward directionC. number of moles of reactants and products are equal.D. |
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Answer» Correct Answer - A A reversible reaction never proceeds to completion. |
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| 96. |
When the rate of formation of reactants is equal to the rate of formation of products, this is known as ,A. Chemical reactionB. Chemical equilibriumC. Chemical kineticsD. None |
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Answer» Correct Answer - B In a chemical reaction at equilibrium, Rate of forward reaction `=` Rate of backward reaction |
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| 97. |
Assertion (A ) `:` The value of equilibrium constant tells us about the extent to which the reactants are converted into the products before the equilibrium is attained at that given temperature. Reason (R ) `:` A small value of K means only small quantities of reactant have undergone change into the products.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true but R is not a correct explanation of A.C. A is true but R is falseD. A is false but R is true. |
| Answer» Correct Answer - a | |
| 98. |
For the reaction `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` the forward reaction at constant temeprature is favoured byA. introducing inert gas at constant pressureB. introducing inert gas at constant volumeC. introducing `PCl_(5)` at constant volumeD. introducing `Cl_(2)` at constant volume. |
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Answer» Correct Answer - A,C The forward reaction is favoured by addition of `PCl_(5)` or by addition inert gas at constant pressure. |
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| 99. |
For reaction, `PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g)` the value of `K_(c)` at `250^(@)C` is 26. The value of `K_(p)` at this temperature will be .A. 0.0006B. 0.57C. 0.83D. 0.46 |
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Answer» Correct Answer - A `K_(p)=K_(c) (RT)^(Deltan)=26(82.05xx523)^(-1)` `=(26)/(82.05xx523)=6.06xx10^(-4)` |
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| 100. |
1 mole of `PCl_(5)` taken at 5 atm, dissociates into `PCl_(3)` and `Cl_(2)` to the extent of 50% `PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)` Thus `K_(p)` is `:`A. 2.5B. 1.67C. 0.5D. 2 |
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Answer» Correct Answer - A `{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Initial",1"mol",,,,),("At eqm.",1-0.5,,0.5,,0.5):}` Total no. of moles at eqm. `=1.5` Moles of , `PCl_(5)=0.5` `PCl_(3)=0.5` `Cl_(2)=0.5` Initially `PV=nRT` `5xxV=1xxRT` `V=(RT)/(5)` At eqm. `PV=1.5 RT` `(PRT)/(5)=1.5RT` `P=7.5 "atm"` `:. p_(PCl_(5))=(0.5)/(1.5)xx7.5=2.5"atm"` `p_(PCl_(3))=2.5 "atm"` `p_(Cl_(2))=2.5"atm"` `K_(p)=(p_(PCl_(3))xxp_(Cl_(2)))/(p_(PCl_(5)))=(2.5xx2.5)/(2.5)=2.5"atm"` |
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