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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.
| 1. |
The exothermic formation of `ClF_(3)` is represented by thr equation: `Cl_(2)(g)+3F_(2)(g) hArr 2ClF_(3)(g), DeltaH=-329 kJ` Which of the following will increase the quantity of `ClF_(3)` in an equilibrium mixture of `Cl_(2), F_(2)`, and `ClF_(3)`?A. increasing the temperatureB. removing `Cl_(2)`C. increasing the volume of the containerD. adding `F_(2)` |
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Answer» Correct Answer - D Adding `F_(2)` will shift the equilibrium in the forward direction and as such quantity of `CiF_(3)` in the equilibrium mixture will increase. |
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| 2. |
The equilibrium constant for the reaction `CaSO_(4).5H_(2)H_(2)O(s)hArr CaSO_(4).3H_(2)O(s)+2H_(2)O(g)` is equal toA. `([CaSO_(4).3H_(2)O][H_(2)O]^(2))/([CaSO_(4).5H_(2)O])`B. `([CaSO_(4).3H_(2)O])/([CaSO_(4).5H_(2)O])`C. `[H_(2)O]^(2)`D. `[H_(2)O].` |
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Answer» Correct Answer - C Concentration of solids is taken to be unity. Thus `[CaSO_(4).5H_(2)O(s)]` `[CaSO_(4).3H_(2)O(s)]=1` and `K=[H_(2)O]^(2)` |
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| 3. |
Which of the following equilibrium will shift left on increasing temperature. `(i) " " H_(2)O(g) hArr +1//2 O_(2)(g)` `(ii) " " CO_(2)(g) +C(s) hArr CO(g)` `(iii) " " C(s, "diamond") hArr C(s, "graphite")`A. Only (i)B. Both (i) and (ii)C. Only (iii)D. None |
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Answer» Correct Answer - C Out of diamond and graphite, at a given temperature graphite has lesser energy and the process `[Delta H^(@) ._(f) C ("graphite" )=0]`. `C(s,"diamond") hArr C(s, "graphite")` is an exothermic process. As such an increase in temperature shifts the equilibrium in the backward direction. |
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| 4. |
Consider the following reversible reactionat equilibrium: `2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g), DeltaH=+24.7 kJ` Which one of the following changes in conditions will lead to maximum decomposition of `H_(2)O(g)`?A. Increasing both temeprature and pressureB. Decreasing temperature and increasing pressureC. Increasing temperature and decreasing pressureD. Increasing temperature at constant pressure |
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Answer» Correct Answer - C `2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g) , DeltaH =241.7 kJ` Here, the forward reaction is endothermic and proceeds with an increase in number of moles of gaseous constituents, therefore, high temperature and low pressure will favour the forward reaction leading to maximum decomposition of `H_(2)O` |
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| 5. |
Given the equilibrium system `:` `NH_(4)Cl(g) hArr NH_(44)^(+)(aq)+Cl^(-)(aq), (DeltaH = + 35 kcal //` mol ). What change will shift the equilibrium to the right ?A. decreasing the temperatureB. increasing the temperatureC. dissolving NaCl crystals in the equilibrium mixtureD. dissolving `NH_(4)NO_(3)` crystals in the equlibrium mixture |
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Answer» Correct Answer - B As the given dissociation is endothermic, increase of temperature will shift the equilibrium in the forward direction. |
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| 6. |
There is 50% dimer formation of benzoic acid `(C_(6)H_(5)COOH)_(2)` in benzene solution. `2C_(6)H_(5)COOHhArr (C_(6)H_(5)COOH)_(2)`. Hence abnormal molecular weight of benzoic acid (theorectical value `=122g mol^(-1))` isA. 61 g `mol^(-1)`B. 244 g `mol^(-1)`C. 163 g `mol^(-1)`D. 81 g `mol^(-1)` |
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Answer» Correct Answer - C `M_(f)=122,M_(0)=?` `y=(1)/(2)` `x=50%=(1)/(2)` `x=(M_(f)-M_(0))/(M_(0)(y-1))` `(1)/(2)=(122-M_(0))/(M_(0)((1)/(2)-1))` `(1)/(2)=((M_(0)-122)^(2))/(M_(0))` `M_(0)=4(M_(0)-122)` `M_(0)=4M_(0)-122xx4` `3M_(0)=122xx4` `M_(0)=(122xx4)/(3)=162.67` `M_(0)~~163` |
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| 7. |
At equilibrium of the reaction , `N_(2)O_(4)(g) hArr 2NO_(2)(g)` the observed molecular weight of `N_92)O_(4)` is 80g `mol^(-1)` at 350K. The percentage dissociation of `N_(2)O_(4)(g)` at 350K isA. `10%`B. `15%`C. `20%`D. `18%` |
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Answer» Correct Answer - B `alpha=(D-d)/(d)=(M_(f)-M_(0))/(M_(0))=(92-80)/(80)xx100=15%` |
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| 8. |
Before equilibrium is set-up for the chemical reaction, `N_(2)O_(4)hArr 2NO_(2)`, vapour density of the gaseous mixture was measured. If D is the theoretical value of vapour density, variation of x with `D//d` is by the graph. |
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Answer» Correct Answer - C See hint to Q. 46 `x=(D)/(d)-1` At `A,x=0` `:. (D)/(d)=1` |
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| 9. |
In the dissociation of `N_(2)O_(4)` into `NO_(2). (1+x)` values with the vapour densities ratio `((D)/(d))` is given by `:`A. B. C. D. |
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Answer» Correct Answer - A `N_(2)O_(4)(g)hArr 2NO_(2)(g)` Here `y=2` Degree of dissociation, `x=(D-d)/(d(y-1))` `x=(D-d)/(d)` `x=(D)/(d)-1` `1+x=(D)/(d)` A graph between `1+x` ( along Y-axis) and `(D)/(d)` (along X-axis) , is a straight line passing through origin. `(y-mx` type ) with slope, `m=1`. |
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| 10. |
The vapour density of `N_(2)O_(4)` at a certain temperature is `30`. Calculate the percentage dissociation of `N_(2)O_(4)` this temperature.A. `53.3%`B. `106.6%`C. `26.7%`D. None |
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Answer» Correct Answer - A Mol. Wt. of `N_(2)O_(4)=92` Vapour density, `D=(92)/(2)=46` `N_(2)O_(4)(g) hArr 2NO_(2)(g)` Here `y=2` `d=30` `x=(D-d)/(d(y-1))=0.533=53.3%` |
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| 11. |
The factor which changes equilibrium constant of the reaction `A_(2)(g)+B_(2)(g) rarr 2AB(g)` isA. Total pressureB. Amounts of `A_(2)` and `B_(2)`C. TemperatureD. Catalyst. |
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Answer» Correct Answer - C K changes with temperature. |
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| 12. |
If the equilibrium constant for the reaction `2AB hArr A_(2)+B_(2)` is 49, what is the value of equilibrium constant for `ABhArr (1)/(2) A_(2)+(1)/(2)B_(2)`A. 49B. 2401C. 7D. 0.02 |
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Answer» Correct Answer - C `K_(2)=sqrt(K_(1))=sqrt(49)=7`. |
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| 13. |
When 3 moles of A and 1 mole of B are mixed in 1 litre vessel, the following reaction takes place `A_((g)) + B_((g))hArr2C_((g)). 1.5` moles of C are formed. The equilibrium constant for the reaction isA. 0.12B. 0.5C. 0.25D. 4 |
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Answer» Correct Answer - D `{:(,A,+,B,hArr, 2C),("Initial moles",3,,1,,1.5),("Eqm. moles",2.25,,0.25,,1.5):}` `K_(c)=((1.5)^(2))/(2.25xx0.25)=4.00` |
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| 14. |
For the reaction, `ZnCO_(3)(s) hArr ZnO(s)+CO_(2)(g)` expression for the partial pressure constant for the above reaction would beA. `K_(p)=([ZnO][CO_(2)])/([ZnCO_(3)])`B. `K_(p)=(p_((ZnO))xxp_(CO_(2)))/(p_(ZnCO_(3)))`C. `K_(p)=p_((ZnO))^(2)xxp_(CO_(2))`D. `K_(p)=p_(CO_(2))` |
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Answer» Correct Answer - D For reaction, `ZnCO_(3)(s) hArr ZnO(s)+CO_(2)(g)` `K_(p)=(p_((ZnO))xxp_(CO_(2)))/(p_((ZnCO_(3))))` Now, `p_((ZnCO_(3)))=p_((ZnO))=1` as per conventions `:. K_(p)=p_(CO_(2))` |
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| 15. |
The equilibrium constant for a reaction `A+B hArr C+D` is `1xx10^(-2)` at `298 K` and is `2` at `273 K`. The chemical process resulting in the formation of `C` and `D` isA. exothermicB. EndothermicC. unpredictableD. There is no relationship between K and `Delta H`. |
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Answer» Correct Answer - B On raising temperature K increase. It means the forward reaction is favoured on increasing temeprature. Thus, forward reaction should be endothermic. |
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| 16. |
The value of equilibrium constant of the reaction, HI `hArr(1)/(2)H_(2) + (1)/(2)I_(2)` is `8` . The equilibrium constant of the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)` will beA. `1./16`B. `1//64`C. 16D. `1//8` |
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Answer» Correct Answer - B For `HI(g) hArr (1)/(2)H_(2)(g)+(1)/(2) I_(2)(g) ,K=8` For `(1)/(2)H_(2)(g)+(1)/(2)I_(2)(g)hArr HI(g), K=(1)/(8)` For `H_(2)(g)+I_(2)(g) hArr 2HI(g)` `K=((1)/(8))^(2)=(1)/(64)` |
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| 17. |
Equilivalent amounts of `H_(2)` and `I_(2)` are heated in a closed vessel till equilibrium is obtained. If `80%` of the hydrogen is converted to `HI`, the `K_(c)` at this temperature isA. 64B. 16C. 0.25D. 14 |
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Answer» Correct Answer - A `{:(,H_(2),+,I_(2),hArr,2HI),("Initial",a" mol",,a " mol",,),("At eqm.",a-0.8a=2a,,0-0.8=0.2a,,2xx0.8a=1.6a),("Molar conc.",(0.2a)/(V),,(0.2a)/(V),,(1.6a)/(V)):}` `K_(c)=([HI]^(2))/([H_(2)][I_(2)])=((1.6a//V)^(2))/((0.2a//V)(0.20//V))` `=((16)^(2))/((0.2)^(2))=(2.56)/(0.04)=64` |
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| 18. |
Naphthalene, a white solid used to make mothballs, has a vapour pressure of 0.10 mm Hg at `27^(@)C`. Hence `K_(p)` and `K_(c)` for the equilibruim are `: C_(10)Hg(s)hArr C_(10)Hg(g)`A. 0.10,1.10B. `0.10,4.1xx10^(-3)`C. `1.32xx10^(-4)xx10^(-3)`D. `5.34xx10^(-6),1.32xx10^(-4)` |
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Answer» Correct Answer - C `C_(10)H_(8)(s)hArr C_(10)H_(8)(g)` `K_(p)=P_(C_(10)H_(8))=(0.10)/(760)atm` `=1.315xx10^(-4)atm` `K_(p)=K_(c)(RT)^(Deltan)` Here, `R=0.0821 L " atm "K^(-1)moll^(-1),` `T=27+273=300K` `Delta n =1-0=1` `K_(c)=(K_(p))/((RT)^(Deltan))=(1.315xx10^(-4))/((0.0821xx300))` `=(1.315xx10^(-4))/(8.21xx3)=(131.5xx10^(-6))/(24.63)` `=5.339xx10^(-6)` |
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| 19. |
For a reaction, `H_2+I_2hArr 2HI` at 721 K , the value of equilibrium constant is 50. If 0.5 moles each of `H_2` and `I_2` is added to the system the value of equilibrium constant will be :A. 0.02B. 0.2C. 50D. 25 |
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Answer» Correct Answer - C `K=([HI]^(2))/([H_(2)][I_(2)])=50=(4alpha^(2))/((1-alpha)^(2))` `=sqrt(50)=(2a)/(1-a)=7.07` or `2alpha=7.07-7.07 alpha` `9.07alpha=7.07 ` or `alpha=(7.07)/(9.07)=0.78` If 0.5 moles each of `H_(2)` and `I_(2)` are taken than Equilibrium conc. `[H_(2)]=0.5-0.39, ` `[I_(2)]=0.5-0.39,HI=0.78` `K=(0.78xx0.78)/(0.11xx0.11)=50` |
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| 20. |
In which of the following cases, the reaction goes farthest to completion?A. `K=10^(3)`B. `K=10^(-2)`C. `K=10`D. `K=10^(0)` |
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Answer» Correct Answer - A The reaction , in which value of K is more, goes farthest to completion. |
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| 21. |
In an exothermic reaction, a `10^(@)` rise in temperature willA. decrease the value of equilibrium constantB. double the value of `K_(c)`C. not produce any change in `K_(c)`D. produce some increase in `K_(c)` |
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Answer» Correct Answer - A In exothermic reaction, increase in temperature favours the backward reaction i.e., decreases the value of equilibrium constant value. |
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| 22. |
The equilibrium constant `(K_(p))` for the reaction, `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))` is `16`. If the volume of the container is reduced to half of its original volume, the value of `K_(p)` for the reaction at the same temperature will be:A. 32B. 64C. 16D. 4 |
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Answer» Correct Answer - C The value of `K_(p)` depends only upon temperature. |
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| 23. |
In a reversible reaction, two substances are in equilibrium. If the concentration of each one is reduced to half, the equilibrium constant will beA. reduced to half of its original valueB. doubledC. sameD. reduced to one fourth its original value. |
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Answer» Correct Answer - C Equilibrium constant depends only upon temperature. It is independent of concentration of reactants or products. |
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| 24. |
In which case `K_(p)` is less than `K_(c)` ?A. `PCl_(5)hArr PCl_(3)+Cl_(2)`B. `H_(2)+Cl_(2) hArr 2HCl`C. `2SO_(2)+O_(2) hArr 2SO_(3)`D. All of above |
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Answer» Correct Answer - C `K_(p)=K_(c) (RT)^(Delta n)` For `K_(p)` to be less than `K_(c), Delta n` has to be negative `(RT gt 1)`. For `2SO_(2)+O_(2) hArr 2SO_(3)` `Delta n =2-(2+1)= - 1` Therefore for this reaction `K_(p) lt K_(c)`. |
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| 25. |
The reaction `3Fe(s)+4H_(2)O hArr Fe_(3)O_(4)(s) +4H_(2)(g)` is reversible if it is carried outA. at constant pressureB. at constant temperatureC. in an open vesselD. in a closed vessel. |
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Answer» Correct Answer - D Factual question. |
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| 26. |
At what relative humidity with `Na_(2)SO_(4)` be deliquescent (absorb moisture) when exposed to the air at `0^(@ )C` ? Given `Na_(2)SO_(4).10H_(2)O(s) hArr Na_(2)SO_(4)(s) +10H_(2)O(g), K_(p)=4.08xx10^(-25)` and vapour pressure of water at `0^(@)C =4.58` Torr ?A. below 60.5%B. above 60.5%C. above 39.5%D. below 39.5% |
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Answer» Correct Answer - B `Na_(2)SO_(4).10H_(2)O(s)hArr Na_(2)SO_(4)(s)+10H_(2)O(g)` `K_(p)=4.08xx10^(-25)` `:. (p_(H_(2)O))^(10)=4.08xx10^(-25)` `p_(H_(2)O)=(4.08xx10^(-25))^((1)/(10))=x("say")` `log x =(1)/(10)log (4.08xx10^(-25))` `=(1)/(10)[log 4.08+log 10^(-25)]` `=(1)/(10)[log 4.08-25]` `=(1)/(10)[0.611-25]` `=0.0611-2.5=bar(3).5611` `x=` antilog `bar(3).5611=3.64xx10^(-3)` `Na_(2)SO_(4)(s)` will absorb water if `R_(H)` is more than `=(3.64xx10^(-3))/(4.58//760)xx100` `=(0.364xx760)/(4.58)=60.4%` |
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| 27. |
For a reaction if `K_(p) gt K_(c)` , the forward reaction is favoured by `(T gt 15K)`A. High temperatureB. Low temperatureC. Low pressureD. High temperature |
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Answer» Correct Answer - C `K_(p)=K_(c )(RT)^(Deltan_(g))` or `(K_(p))/(K_(c ))=(RT)^(Deltan_(g))` Now, `K_(p) gt K_(c) ( `given) `implies (K_(p))/(K_(c)) gt 1` `(RT)^(Deltan_(g)) gt1` At `T gt 15K` `RT gt 0.0821 xx 15 L ` atm `mol^(-1)` `RT gt 1.22 L` atm `mol^(-1)` `implies Delta n_(g) gt 0` `implies ` No. of moles of gaseous products `gt `No. of moles of gaseous reatants Such a reaction in the forward direction will favoured by low pressure. |
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| 28. |
In a lime kiln, to get higher yield of `CO_(2)` the measure that can be taken is `:`A. to maintain high temperatureB. to pump out `CO_(2)`C. to remove CaOD. to add more `CaCO_(3)` |
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Answer» Correct Answer - B The reaction taking place in a lime kiln is `CaCO_(3) overset(Delta) (hArr) CaO(s) +CO_(2)(g)` Removing one of the product will shift the equilibrium in the forward direction. As such removing `CO_(2)` as and when it forms, will shift the equilibrium in the forward direction. This will increase the yield of `CO_(2)`. |
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| 29. |
For the reaction of XO with `O_(2)` to form `XO_(2)` the equilibrium constant at `398 K` is `1.0 xx 10^(-4) mol^(-1)`. If 1.0 mol of XO and 2,0 mol of `O_(2)` are placed in a 1.0 L flask and allowed to come to equilibrium, the equilibrium concentration of `XO_(2)` will beA. `1.4 xx 10^(-2) mol L^(-1)`B. `2.8 xx10^(-2) mol L^(-1)`C. `2.8 xx 10^(-3) mol L^(-1)`D. none of these |
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Answer» Correct Answer - A Units of K indicate that the reaction involved is , `{:(,2XO,+,O_(2),hArr ,2XO_(2)),("Initial conc. " (mol L^(-1)),1.0,,2.0,,-),("At equilibrium " (mol L^(-1)),(1.0-2x),,(2.0-x),,2x):}` `K=([XO_(2)]^(2))/([XO]^(2)[O_(2)])=1.0xx10^(-4)mol^(-1)` As value of K is very small , x is also very small `1.0 -2x ~~ 1.0` and `2.0-x =2.0` `1.0 xx10^(-4)=((2x)^(2))/((1.0)^(2)(2.0))impliesx=7.1xx10^(-3)` `:. [XO_(2)]=2x=2xx7.1xx10^(-3)` `=1.4xx10^(-2) mol L^(-1)` |
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| 30. |
For a hypothetical reaction `4A+5B hArr 4P +6Q` . The equilibrium constant `K_(c )` has units.A. `mol L^(-1)`B. `mol^(-1)L`C. `(molL^(-1))^(-2)`D. unit less. |
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Answer» Correct Answer - A `K_(c)=([P]^(4)[Q]^(6))/([A]^(4)[B]^(5))` `=([mol L^(-1)]^(4)[mol L^(-1)]^(6))/([molL^(-1)]^(4)[mol L^(-1)]^(5))=mol L^(-1).` |
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| 31. |
The equilibrium constant, K for the reaction `:` `2HI (g) hArr H_(2)(g)+I_(2)(g)` at room temperature is 2.85 and that of 698 K is `1.4 xx 10^(-2)` . This implies that the forward reaction isA. ExothermicB. EndothermicC. ExergonicD. Unpredictable |
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Answer» Correct Answer - A On increasing temperature, the value of equilibrium constant decreases indicating backward shift. Since increase in temperature shifts the equilibrium towards endothermic side. Therefore, forward reaction is exothermic. |
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| 32. |
16 mol of `PCl_(5)(g)` is placed in 4 `dm^(-3)` closed vessel. When the temperature is raised to 500 K, it decompses and at equilibrium, 1.2 mol of `PCl_(5)(g)` remains. What is `K_(c)` value for the decomposition of `PCl_(5)(g)` to `PCl_(3)(g)` and `Cl_(2)(g)` at 500K.A. 0.013B. 0.05C. 0.033D. 0.067 |
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Answer» Correct Answer - C `{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Inital",1.6 "mol",,0,,0),("At eqm.",1.2 "mol",,0.4"mol",,0.4"mol"),("Molar conc.",(1.2)/(4)"mol"dm^(-3),,(0.4)/(4)"mol"dm^(-3),,(0.4)/(4)"mol"dm^(-3)),(,=0.3 mol dm^(-3),,=0.1mol dm^(-3),,=0.1 moldm^(-3)):}` `K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.1xx0.1)/(0.3)=0.033` |
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| 33. |
`NH_(4)ClO_(4)` crystallises in orthohombic structure with unit cell volume of `395 Å^(3)` . The structure of `HClO_(4).H_(2)O` will beA. tetragonal will unit cell volume of `210 Å^(3)`B. cubic with unit cell volume of `78Å^(3)`C. orthorhomibic with unit cell volume of `370 Å^(3)`D. the data given is not sufficient to find the structure. |
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Answer» Correct Answer - C Both substances are ionic with lattice sites occupied by ions. In `HClO_(4).H_(2)O` cation is `H_(3)O^(+)` and there are no `H_(2)O` molecules of hydration. The anions in both are same i.e., `ClO_(4)^(-)` ions. `H_(3)O^(+)` and `NH_(4)^(+)` are nearly isoelectronic ( have nearly same number of electrons ) and as such occupy nearly equal space. |
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| 34. |
The theoretically computed equilibrium constant for the polymerisation of formaldehyde to glucose in aqueous solution : `6HCHO hArr C_6H_12O_6` is `1.0xx10^24` If 1 M-solution of glucose was taken, what should be the equilibrium concentration of formaldehyde ?A. `1.6 xx 10^(22) M`B. `1.6 xx 10^(-4) M`C. ` 6 sqrt(6xx10^(4)) xx 10^(3) M`D. None of these |
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Answer» Correct Answer - B A very high value of K indicates that amount of glucose dissociated into formaldehyde is negligible. Therefore , at equilibrium `[C_(6)H_(12)O_(6)]=1.0 mol L^(-1)` `K([C_(2)H_(12)O_(6)])/([HCHO]^(6))` `6xx10^(22)=(1)/([HCHO]^(6))` `[HCHO]^(6)=(1)/(6xx10^(22))` `implies [HCHO]=1.6 xx 10^(-4) M` |
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| 35. |
A 550 K, the `K_(c)` for the following reaction is `10^(4) mol^(-1) L` `X(g)+Y(g) hArr Z (g)` At equilibrium, it was observed that `[X]= (1)/(2)[Y]=(1)/(2)[Z]` What is the value of [Z] ( in mol `L^(-1))` at equilibrium ?A. `2xx10^(-4)`B. `10^(-4)`C. `2xx10^(4)`D. `10^(4)` |
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Answer» Correct Answer - A `K_(c)=([Z])/([X][Y])` `[X]=(1)/(2)[Y]=(1)/(2)[Z]=a` (say) `:. [Z]=2a,[Y]=2a,[X]=a` `:. 10^(4)=(2a)/(a.2a)` or `a=10^(-4)` `:. [Z]=2a=2xx10^(-4)` |
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| 36. |
`A + B hArrC + D`. If finally the concentrations of A an d B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.A. `4//9`B. `9//4`C. `1//9`D. 4 |
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Answer» Correct Answer - D `{:(,A+B,hArr,C+D,),("Initial",a,,a,),("At eqm.",a-x,a-x,x" "x,):}` But we are given that `x=2(a-x)` or `3x=2a` or `a=(3)/(2)x=1.5 x` `:. K=(x xx x)/((a-x)^(2))=(x^(2))/((1.5x-x)^(2))` `=(x^(2))/(0.25x^(2))=4` |
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| 37. |
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 +CD hArr AD +CB` Original concentration of `[AB]=[CD]=(1)/(V)` mol `L^(-1)` where V is the volume of the vessel in L. `K_(c )=([AD][CB])/([AB][CD])` At equilibrium `[AB]=[CD]=(1-3//4)/(V)=(1)/(4V)"mol" L^(-1)` `[AD]=[CB] =(3)/(4V) "mol" L^(-1)` `K_(c )=(((3)/(4V))((3)/(4V)))/(((1)/(4V))((1)/(4V)))=9` |
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| 38. |
For the reaction `2NH_(3)(g) hArr N_(2)(g) +3H_(2)(g)` the units of `K_(p)` will beA. atmB. `(atm)^(3)`C. `(atm)^(-2)`D. `(atm)^(2)` |
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Answer» Correct Answer - D `K_(p)=((p_(H_(2)))^(3)(p_(N_(2))))/((p_(NH_(3)))^(2))=((atm)^(3)(atm))/((atm)^(2))=(atm)^(2)` |
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| 39. |
The equilibrium constant for the reaction `H_(3)BO_(3) +` glycerin `hArr (H_(3)BO_(3)` glycerine ) is 0.90. Glycerine present per litre of 0.1 M `H_(3)BO_(3)` to convert `60%` of `H_(3)BO_(3)` into `(H_(3)BO_(3)` glycerine ) is `:`A. 0.167 MB. `1.73 M`C. 0.0167 MD. 10.67 M |
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Answer» Correct Answer - B `{:(,H_(3)BO_(3),+"glycerine",hArr,H_(3)BO_(3)."glycerine"),("Initial",0.1M,,," xM"),("At eqm.",0.1(1-0.6)M,,,(x-0.06)M " "0.06M):}` `K=([H_(3)BO_(3)."glycerine"])/([H_(3)BO_(3)]["glycerine"])` `=0.9=(0.06)/((0.04)(x-0.06))` `(9)/(10)xx(4)/(6)=(1)/(x-0.06)` `x-0.06=1.67` `x=1.67+0.06` `x=1.73M` |
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| 40. |
Assertion (A) `:` The equilibrium is not static but a dynamic one. Reason (R ) `:` The chemical equilibrium is an apparent state of rest in which two opposing reactions are proceeding at the same rate.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 | |
| 41. |
Assertion (A) `:` Addition of an inert gas to the equilibrium mixture has no effect on the state of equilibrium at constant volume or at constant pressure. Reason (R ) `:` The addition of inert gas at constant volume will not alter the concentrations of the reactants as well as products of a reaction mixture.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 - d | |
| 42. |
The state of equilibrium refers toA. State of restB. Dynamic stateC. Stationary stateD. State of inertness. |
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Answer» Correct Answer - B Question based on facts |
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| 43. |
For the following equilibrium `N_(2)O_(4)(g)hArr 2NO_(2)(g)` `K_(p)` is found to be equal to `K_(c)`. This is attained when `:`A. `T=1K`B. `T=12.18K`C. `T=27.3K`D. `T=273K` |
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Answer» Correct Answer - B `K_(p)=K_(c)(RT)^(Deltan)` Here, `K_(p)=K_(c),Deltan=2-1=1` `:. (RT)^(Deltan)=1` `(0.0821xxT)=1` `T=(1)/(0.0821)=(10000)/(821)=12.18K` |
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| 44. |
A gas bulb is filled with `NO_(2)` gas and immersed in an ice bath at `0^(@)C`, which becomes colourless after sometime. This colourless gas will be:A. `NO_(2)`B. `N_(2)O`C. `N_(2)O_(4)`D. `N_(2)O_(5)` |
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Answer» Correct Answer - C `2NO_(2)underset("High temp")overset(0^(@)C)(hArr)N_(2)O_(4)` ( colourless) |
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| 45. |
For the equilibrium `:` `I: C_(6)H_(5)COOH+H_(2)OhArr C_(6)H_(5)COO^(-)+H_(3)O^(+),K_(1)=6.30xx10^(-5)` `II: C_(6)H_(5)COOH+OH^(-)hArrC_(6)H_(5)COO^(-)+H_(2)O, K_(2)=6.30xx10^(9)` `III: H_(2)O+H_(2)OhArr H_(3)O^(+)+OH^(-), K_(3)=?` `K_(3)` using above equilibria is` :`A. `(6.30)^(2)xx10^(4)`B. `1.0xx10^(-14)`C. `1xx10^(14)`D. `(6.30)^(-2)xx10^(-4)` |
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Answer» Correct Answer - B `C_(6)H_(5)COOH+H_(2)O hArr C_(6)H_(5)COO^(-)+H_(3)O^(+),K=K_(1)` `C_(6)H_(5)COO^(-)+H_(2)O hArr C_(6)H_(5)COOH+OH^(-), K=(1)/(K_(2))` Adding, we get `H_(2)O +H_(2)O hArr H_(3)O^(+)+OH^(-), K_(3)=(K_(1))/(K_(2))` `:. K_(3)=(6.30xx10^(-5))/(6.30xx10^(9))=10^(-14)` |
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| 46. |
For an equilibrium reaction, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8 xx 10^(-2)` and `1.2 xx 10^(-2) mol//L` respectively. The value of `K_(c)` for the reaction isA. `3.3 xx 10^(2) mol L^(-1)`B. `3.3xx10^(-1) mol L^(-1)`C. `3.3 xx 10^(-3) mol L^(-1)`D. `3.3 xx 10^(3) mol L^(-1)` |
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Answer» Correct Answer - C `N_(2)O_(4)(g) hArr 2NO_(2)(g)` `K_(c)=([NO_(2)]^(2))/([N_(2)O_(4)])=((1.2xx10^(-2)"mol"L^(-1))^(2))/(4.8xx10^(-2)"mol"L^(-1))` `=(1.44xx10^(-4)mol^(2)L^(-2))/(4.8xx10^(-2)"mol"L^(-1))` `=(14.4)/(4.8)xx10^(-3)"mol" L^(-1)` `=3.3xx10^(-3)"mol" L^(-1)` |
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| 47. |
In the dissociation of `PCl_(5)` as `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` If the degree of dissociation is `alpha` at equilibrium pressure P, then the equilibrium constant for the reaction isA. `K_(p)=(alpha^(2))/(1+alpha^(2)P)`B. `K_(p)=(alpha P^(2))/(1-alpha^(2))`C. `K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`D. `K_(p)=(alpha^(2)P)/(1-alpha^(2))` |
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Answer» Correct Answer - D `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial moles",1,,-,,-),("Equilibrium moles :",1-alpha,,alpha,,alpha):}` Total moles `=1- alpha + alpha + alpha = (1+alpha)` Total pressure `=P` `:. P_(PCl_(5))=((1-alpha)P)/((1+alpha))` `P_(PCl_(3))=(alphaP)/((1+alpha)),P_(Cl_(2))=(alphaP)/((1+alpha))` Now `K_(p)=(P_(Cl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))=(alpha^(2)P^(2))/((1+alpha)(1-alpha)P)` `=(alpha^(2)P)/(1-alpha^(2))` |
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| 48. |
Assertion (A) `:` The values of equilibrium constant of forward and backward reactions are same. Reason (R ) `:` Under particular set of conditions, the values of equilibrium constants of forward and backward reactions are reciprocal to each other.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 - d | |
| 49. |
Equimolar concentrations of `H_(2)` and `I_(2)` are heated to equilibrium in a 2 L flask. At equilibrium, the forward and backward rate constants are found to be equal. What percentage of initial concentration of `H_(2)` has reached at equilibrium ?A. 0.33B. 0.66C. `50%`D. 0.4 |
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Answer» Correct Answer - A `{:(,H_(2)(g),+,I_(2)(g),hArr,2HI(g)),("Initial",1 mol,,1 mol,,0),("At eqm.",1-x,,1-x,,2x),("Molar conc.",(1-x)/(2),,(1-x)/(2),,(2x)/(2)mol L^(-1)):}` `K=((2x)^(2))/((1-x)(1-x))=(4x^(2))/((1-x)^(2))` But `k_(f)//k=1impliesk=1 :. (2x)/(1-x)=1 ` or `x=1//3` Percent dissociation `=100//3=33.33%` |
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| 50. |
If `K_(1)` and `K_(2)` are the equilibrium constants of the equilibria (a) and (b) respectivel, what is the relationship between the two constants ? (a) `SO_(2)+(1)/(2) O_(2)(g)hArr SO_(3)(g),K_(1)` (b) `2SO_(3)(g) hArr 2SO_(2)(g)+O_(2)(g),K_(2)`A. `(K_(1))^(2)=(1)/(K_(2))`B. `K_(2)=(K_(1))^(2)`C. `K_(1)=(1)/(K_(2))`D. `K_(1)=K_(2)` |
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Answer» Correct Answer - A `(a) SO_(2)(g)+(1)/(2)O_(2)hArr SO_(3)(g), K_(1)` `(b) 2SO_(3)(g)hArr 2SO_(2)(g)+O_(2)(g),K_(2)` `(K_(1))^(2)=(1)/(K_(2))` or `K_(1)=sqrt((1)/(K_(2)))` |
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