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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 vapour pressure of a liquid in a closed container depends uponA. 1 onlyB. 2 onlyC. 1 and 3 onlyD. 1,2,and3 |
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Answer» Correct Answer - a |
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| 2. |
Solubility of a solute in water is dependent on temperature as given by `S=Ae^(-DeltaH//RT)`, where `DeltaH`=heat of solution `"Solute"+H_(2)O(l) hArr "Solution", DeltaH= +- x` For given solution, variation of log S with temperature is shown graphically. Hence, solution is A. `CuSO_(4).5H_(2)O`B. NaClC. SucroseD. CaO |
| Answer» Correct Answer - D | |
| 3. |
The `K_(c)` for `H_(2(g)) + I_(2(g))hArr2HI_(g)` is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will beA. `16`B. `32`C. `64`D. `128` |
| Answer» Correct Answer - C | |
| 4. |
The `K_(c)` for `H_(2(g)) + I_(2(g))hArr2HI_(g)` is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will beA. 16B. 32C. 64D. 128 |
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Answer» Correct Answer - C For this reaction there is no change in equilibrium constant by change of volume. |
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| 5. |
Following gaseous reaction is undergoing in a vessel `C_(2)H_(4) + H_(2)hArrC_(2)H_(6)` , `Delta H = - 32.7` Kcal Which will increase the equilibrium concentration of `C_(2)H_(6)` ?A. Increase of temperatureB. By reducing temperatureC. By removing some hydrogenD. By adding some `C_(2)H_(6)` |
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Answer» Correct Answer - B Exaothermic reaction is favoured by low temperature. |
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| 6. |
For the gas phase reaction `C_(2)H_(4)+H_(2) hArr C_(2)H_(6)(DeltaH=-32.7 "kcal")` carried out in a vessel, the equilibrium concentration of `C_(2)H_(4)` can be increased byA. Increasing the temperatureB. increasing concentration of `H_(2)`C. decreasing temperatureD. increasing pressure |
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Answer» Correct Answer - A `C_(2)H_(4) + H_(2)hArrC_(2)H_(6)` , `Delta H = - 32.7` kcal `Delta n = P_("mole") - R_("mole") = 1 - 2 = - 1` `Delta H = -ve` exothermic reaction `darr` `x prop (1)/(T) uarr` Temperature increase reaction goes toward reactant. Thus, concentration of `C_(2)H_(4)` increase. |
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| 7. |
The synthesis of ammonia is given as: `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH^(ɵ)=-92.6 kJ mol^(-1)` given `K_(c)=1.2` and temperature `(T)=375^(@)C` On increasing the temperature, the value of equilibrium constant `K_(c)`A. IncreasesB. DecreasesC. Remain unchangedD. Cannot be predicted |
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Answer» Correct Answer - B It is an exothermic reaction. Hence, the value of `K_(c)` decreases with increase in temperature. |
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| 8. |
In an equilibrium `A+B hArr C+D`, A and B are mixed in vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reaches, concentration of C was thrice the equilibrium concentration of B. Calculate `K_(c)`. |
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Answer» Correct Answer - A::C `{:(A,+,B,hArr,C,+,D,),(2a,,a,,0,,0,"Initial conc"),((2a-x),,(a-x),,x,,x,"Conc at Eq"):}` Given, initially `[A]=2[B]=2a` Let concentration of B initially be a "mole"/litre `:. K_(C)=(x.x)/((2-x)(a-x))` …(i) Given at equilibrium, `[C]=3[B]` `:. x=3(a-x)` `x=(3a)/4` …(ii) `:.` By equations (i) and (ii), `K_(c)=((3a)/4xx(3a)/4)/((2a-(3a)/4)(a-(3a)/4))=1.8` |
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| 9. |
For the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g),DeltaH=-93.6 kJ mol^(-1)`. The concentration of `H_(2)` at equilibrium will increase ifA. the temperature is loweredB. the volume of the system is decreasedC. `N_(2)` is added at constant volumeD. `NH_(3)` is added |
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Answer» Correct Answer - 4 From Le- chatelir principle |
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| 10. |
A tenfold increase in pressure on the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` at equilibrium result in ……….. in `K_(p)`. |
| Answer» Correct Answer - No change | |
| 11. |
A `10`-fold increase in pressure on the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`at equilibrium results in …….. in `K_(p)`. |
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Answer» A `10`-fold increase in pressure on the reaction, `N_(2)(g)^(o+)+3H_(2)(g) hArr 2NH_(3)(g)` at equilibrium, result is no change in `K_(p)`. The equilibrium constant depends only upon temperature change. |
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| 12. |
For a gaseous reaction `2B rarr A`, the equilibrium constant `K_(p)` is ……… to`//`than `K_(c)`. |
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Answer» For a gaseous reaction `2B rarr A`, the equilibrium constant `K_(p)` is less than `K_(c)` `K_(p)=K_(c)(RT)^(Deltan)` `Deltan-1-2=-1` `K_(p)=K_(c)(RT)^(-1)` `K_(p)` lt `K_(c)` |
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| 13. |
For the following three reaction 1, 2 and 3, equilibrium constants are given: (1) `CO_((g)) + H_(2)O_((g))hArrCO_(2(g)) + H_(2(g)), K_(1)` (2) `CH_(4(g)) + H_(2)O_((g))hArrCO_((g)) + 3H_(2(g)), K_(2)` (3) `CH_(4(g)) + 2H_(2)O_((g))hArrCO_(2(g)) + 4H_(2(g)), K_(3)` Which of the following relations is correct ?A. `K_(1)sqrt(K_(2))=K_(3)`B. `K_(2)K_(3)=K_(1)`C. `K_(3)=K_(1)K_(2)`D. `K_(3)K_(2)^(3)K_(1)^(2)` |
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Answer» `{:(CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g)),K_(1).....(i)),(CH_(4)(g)+H_(2)O_((g))hArrCO_((g))+3H_(2(g)),K_(2).....(ii)),(bar(CH_(4)+2H_(2)O_((g))hArrCO_(2(g))+4H_(2(g)),K_(3)......(iii))):}` Eq. (`iii`) `=` eq. (`i`) `+` eq. (`ii`) Thus, `K_(3)=K_(1)*K_(2)` |
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| 14. |
For the reaction, `CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g))` at a given temperature, the equilibrium amount of `CO_(2(g))` can be increased by:A. adding a suitable catalystB. adding an inert gasC. decreasing the volume of the containerD. increasing the amount of `CO_((g))` |
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Answer» (d) Any factor which favours the forward reaction will increase the equilibrium concentration of `CO`, and addition of `CO_(2)` will favour the forward reaction. |
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| 15. |
For the equilibrium: `CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g))` the standard enthalpy and entropy changes at `300 K` and `1200 K` for the forward reaction are as follows: `{:(DeltaH_(300 K)^(@)=-41.16kJmol^(-1),), (DeltaS_(300 K)^(@)=-0.0424kJmol^(-1)), (DeltaH_(1200 K)^(@)=-32.93kJmol^(-1),), (DeltaS_(1200K)^(@)=-0.0296kJmol^(-1)):}` In which direction will the reaction be spontaneous? (`a`) At `300 K`, (`b`) At `1200 K`, when at equilibrium `P_(CO)=P_(CO_(2))=P_(H_(2))=P_(H_(2)O)=1atm` Also calculate `K_(p)` for the reaction at each temperature. |
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Answer» (`a`) `K_(p)=8.94xx10^(4)` at `300K`, (`b`) `K_(p)=0.773` at `1200K` , |
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| 16. |
If for `2A_2B(g) hArr 2A_2(g)+B_2(g), K_p`=TOTAL PRESSURE (at equilibrium ) and starting the dissociation from 4 mol of `A_2B` then:A. degree of dissociation of `A_2B` will be (2/3)B. total no of moles at equilibrium will be (14/3)C. at equilibrium the no of moles of `A_2B` are not equal to the no of moles of `B_2`D. at equilibrium the no of moles of `A_2B` are equal to the no of moles of `A_2` |
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Answer» Correct Answer - A |
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| 17. |
For the reaction `:` `" "2A+B hArr 3C` at `298 K, K_(c)=49` A `3L` vessel contains 2,1 and 3 molesof `A, B` and `C` respectively. The reaction at the same temperatureA. (A) must proceed in forward directionB. (B) must proceed in backward directionC. (C) must be equilibriumD. (D) can not be predicted |
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Answer» Correct Answer - A `Q=([C])^(3)/([A]^(2)[B])=(3//3)^(3)/((2//3)^(2)(1//3))=6.75` `QltK_(C)` The reaction will proceed in forward direction to attain equilibrium. |
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| 18. |
For the reaction, `C_2H_6(g) hArr C_2H_6(g) + H_2(g).K_p` is 0.05 at 900 K. If an initial mixture comprising 20 mol of `C_2H_6` and 80 mol of inert gas is passed over a dehydrogenation catalyst at 900 K, what is the equilibrium mole percentage of `C_2H_6` in the gas mixture ? The total pressure is kept at 0.5 bar :A. 4.3B. 9.67C. 8.76D. 72.5 |
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Answer» Correct Answer - C |
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| 19. |
`CH_3OH`, methanol can be prepared from CO and `H_2`. `CO(g)+2H_2(g)hArr CH_3OH(g)` The value of `K_p` at 500 K is `6.23xx10^(-3)` When total pressure (in atm) at equilibrium is required to convert 25% of CO to `CH_3OH` at 500 K if CO and `H_2` comes from `CH_4(g)+H_2O(g) to CO(g)+3H_2(g)` Given your answer excluding decimal places. |
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Answer» Correct Answer - 10 |
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| 20. |
Consider the decomposition of solid `NH_(4)HS` in a flask containing `NH_(3)(g)` at a pressure of `2` atm. What will be the partial pressure of `NH_(3)(g) "and" H_(2)S(g)` after the equilibrium has been attained? `K_(p)` for the reaction is `3`.A. `P_(NH_(3))=4 "atm", P_(H_(2))s=2` atmB. `P_(MH_(3))=1.732 "atm", P_(h_2)s=1.732` atmC. `P_(NH_(3))=3 "atm", P_(H_(2))s=1` atmD. `P_(NH_(3))=1 "atm", P_(H_(2))s=1` atm |
| Answer» Correct Answer - C | |
| 21. |
Solid ammonium cabamate dissociae to give ammonia and carbon dioxide as folows `NH_(2)COOMH_(4)(s)hArr2NH_(2)(g)+CO_(2)(g)` which of the following graph correctly represents the equilibrium.A. B. C. D. |
| Answer» Correct Answer - C | |
| 22. |
A 2.0 L container is charged with a mixture of 6.0 moles of CO(g) and 6.0 moles of `H_2O(g)` and the following reaction takes place : `CO(g)+H_2O(g)hArrCO_2(g)+H_2(g)` When equilibrium is reached the `[CO_2]`=2.4 M.What is the value of `K_c` for the reaction ?A. 16B. `4.0`C. `0.25`D. `0.063` |
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Answer» Correct Answer - A |
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| 23. |
For the equilibrium system : `CO(g)+2H_2(g)hArr CH_3OH(l)` what is `K_c`?A. `K_c=([CH_3OH])/(2[CO][H_2])`B. `K_c=([CH_3OH])/([CO][H_2]^2)`C. `K_c=1/(2[CO][H_2])`D. `K_c=1/([CO][H_2]^2)` |
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Answer» Correct Answer - D |
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| 24. |
For `NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)` reaction started only with `NH_(4)HS(s)`, the observed pressure for reaction mixture in equilibrium is `1.2` atm at `106^(@)C`. What is the value of `K_(p)` for the reaction?A. `1.44atm^(2)`B. `0.36atm^(2)`C. `0.16atm^(2)`D. `3.6atm^(2)` |
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Answer» Correct Answer - B `NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)` `P P` `2P=1.2` `P=0.6` `K_(p)=P^(2)=(0.6)^(2)=0.36 atm^(2)` |
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| 25. |
Consider the heterogeneous equilibrium in a closed container `NH_(4)HS (s)hArrNH_(3) (g) + H_(2)S(g)` If more `NH_(4)Hs` is added to the equilibriumA. Partial pressure of `NH_(3)` increasesB. Partial pressure of `H_(2)S` increasesC. Total pressure in the container increasesD. No effect on partial pressure of `NH_(3)` and `H_(2)S` |
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Answer» Correct Answer - D `NH_(4)HS (s)hArrNH_(3)(g) + H_(2)S(g)` `K_(p)= NH_(3)(g)xxH_(2)S(g)` Partial pressure of `NH_(3)` and `H_(2)O` do not affect `NH_(4)HS(s)` because `NH_(4)HS(s)` is solid for solid active mass unity. |
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| 26. |
On decomposition of `NH_(4)HS` , the following equilibrium is estabilished: `NH_(4) HS(s)hArrNH_(3)(g) + H_(2)S(g)` If the total pressure is P atm, then the equilibrium constant `K_(p)` is equal toA. P atmB. `P^(2) atm^(2)`C. `P^(2) // 4 atm^(2)`D. 2P atm |
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Answer» Correct Answer - C `P_(NH_(3)) = P_(H_(2)S) = (P)/(2)` Hence `K_(p) = P_(NH_(3))xxP_(H_(2)S)` `=(P)/(2)xx(P)/(2) = (P^(2))/(4)` |
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| 27. |
The equilibrium constant of the reaction `H_(2)(g)+I_(2)(g)hArr2HI(g)` at `426^(@)C` is `55.3`, what will be the value of equilibrium constant a. if the reaction is reversed and b. if the given reaction is represented as `3H_(2)+3I_(2)hArr6HI`? |
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Answer» The reverse reaction of the given reaction is `2HIhArrH_(2)+I_(2)` :. Equilibrium constant `=1/55.3` b. The reaction `3H_(2)+3I_(2)hArr6HI` has been obtained by multiplying the reaction, `H_(2)I_(2)hArr2HI` by `3`, hence, `K=(55.3)^(3)` |
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| 28. |
At equilibrium, the concentrations of `N_(2)=3.0xx10^(-3)M, O_(2)=4.2xx10^(-3) M,` and `NO=2.8xx10^(-3) M` in a sealed vessel at `800K`. What will be `K_(c)` for the reaction `N_(2)(g)+O_(2)(g)N_(2)(g)+O_(2)(g)hArr2NO(g)2NO(g)` |
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Answer» For the reaction equilibrium constant `K_(c )` can be written as `K_(c )=([NO]^(2))/([N_(2)][O_(2)])=((2.8xx10^(-3)M)^(2))/((3.0xx10^(-3)M)(4.2xx10^(-3)M))=0.622` |
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| 29. |
In the system `A_((s))hArr2B_((g))+3C_((g))`, if the concentration of `C` at equilibrium is increased by a factor of `2`, it will cause the equilibrium concentration of `B` to change to:A. two times the original valueB. one half of its original valueC. `2sqrt(2) times` its original valueD. `(1)/(2sqrt(2))times` its original value |
| Answer» `K_(c)=[C]^(3)[B]^(2)` if `[C]=[2C]`, then to have `K_(c)` constant `[B]` should be reduced to `([B])/(2sqrt(2))`. | |
| 30. |
In the system `A_((s))hArr2B_((g))+3C_((g))`, if the concentration of `C` at equilibrium is increased by a factor of `2`, it will cause the equilibrium concentration of `B` to change to:A. Two times original valueB. One half of its original valueC. `2sqrt(2)`times to the original valueD. `(1)/(2sqrt(2))` times the original value |
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Answer» Correct Answer - d |
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| 31. |
For the reaction, `2HCl_((g))hArrH_(2(g))+Cl_(2(g))`, the equilibrium constant is `1.0xx10^(-5)`. What is the concentration of `HCl` if the equilibrium concentrations of `H_(2)` and `Cl_(2)` are `1.2xx10^(-8)M` and `1.2xx10^(-9)M` respectively?A. `1.2xx10^(-3)M`B. `1.2xx10^(-7)M`C. `1.2xx10^(-4)M`D. `1.2xx10^(-6)M` |
| Answer» Correct Answer - (b,d) | |
| 32. |
The value of `K_(p)` at `298 K` for the reaction `1/2 N_(2)+ 3/2 H_(2) hArr 2NH_(3)` is found to be `826.0`, partial pressure being measured atmospheric units. Calculate `DeltaG^(ɵ)` at `298 K`. |
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Answer» `DeltaG^(ɵ)=-2.303RT log K_(p)` `=-2.303xx1.98xx298xxlog 826` `=-3980 cal` |
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| 33. |
For which of the following reaction is product formation favoured by law pressure and high temperature?A. `H_(2)(g)+I_(2)(g)hArr2HI(g),DeltaH^(@)=-9.4KJ`B. `CO_(2)(g)+C(s)hArr2CO(g),DeltaH^(@)=172.5KJ`C. `CO(g)+2H_(2)(g)hArrCH_(3)OH,DeltaH^(@)=-21.7KJ`D. 3O_(2)(g)hArr2O_(3)(g),DeltaH^(@)=285KJ` |
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Answer» Correct Answer - b |
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| 34. |
For which of the following reaction is product formation favoured by law pressure and high temperature?A. `CO_(2)(g)+C(s)hArr2CO(g), DeltaH^(@)=172KJ`B. `CO(g)+2H_(2)(g)hArrCH_(3)OH, DeltaH^(@)=-21.7KJ`C. `2O_(3)(g) hArr3O_(2)(g), DeltaH^(@)=-285Kj`D. `H_(2)(g)+F_(2)(g)hArr2HF(g),DeltaH^(@)=-541Kj` |
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Answer» Correct Answer - c |
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| 35. |
`2` mol of `N_(2)` is mixed with `6` mol of `H_(2)` in a closed vessel of one litre capacity. If `50% N_(2)` is converted into `NH_(3)` at equilibrium, the value of `K_(c)` for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`A. `4//27`B. `27//4`C. `2//27`D. `20` |
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Answer» Correct Answer - A::B::C `K=([NH_(3)]^(2))/([H_(2)]^(3)[N_(2)])` for `50% N_(2)` `K=4/27` |
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| 36. |
`2` mol of `N_(2)` is mixed with `6` mol of `H_(2)` in a closed vessel of one litre capacity. If `50% N_(2)` is converted into `NH_(3)` at equilibrium, the value of `K_(c)` for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`A. `4//27`B. `27//4`C. `1//27`D. `24` |
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Answer» Correct Answer - A `underset((a-x))oversetaN_(2)+underset((b-3x))overset3H_(2)harrunderset((2x))overset02NH_(3)` `50%` Dissociation of `N_(2)` take palce so, At equilibrium `(2xx50)/(100) = 1` , value of x = 1 `K_(c) = ([2]^(2))/([1][3]^(3)) = (4)/(27)` So `K_(c) = (4)/(27)`c |
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| 37. |
28 g of `N_(2)` and 6 g of `H_(2)` were kept at `400^(@)C` in 1 litre vessel, the equilibrium mixture contained `27.54` g of `NH_(3)`. The approximate value of `K_(c)` for the above reaction can be `("in mole"^(-2) "litre"^(2))`A. 75B. 50C. 25D. 100 |
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Answer» Correct Answer - A `{:(,N_(2)+3H_(2),hArr,2NH_(3)),("Initial conc".,1,3,0),("at equilibrium",1-0.81,3-2.43,1.62),(," "0.19,,0.57):}` No. of moles `N_(2) = (28)/(28) = 1` mole No. of moles of `H_(2) = (6)/(2) = 3` mole No. of moles of `NH_(3) = (27.54)/(17) = 1.62` moles `K_(c) = ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3)) = ([1.62]^(2))/([0.19][0.57]^(3)) = 75` |
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| 38. |
Which one is the correct representation for the reaction `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`A. `K_(p)=([p_(SO_(3))]^(2))/([p_(SO_(2))]^(2)[p_(O_(2))])`B. `K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])`C. `K_(p)=([n_(SO_(3))]^(2))/([n_(SO_(2))]^(2)[n_(O_(2))])xx[(P)/("Total mole")]^(-1)`D. All the above |
| Answer» Correct Answer - D | |
| 39. |
For the reaction `CuSO_(4).5H_(2)O(s) hArr CuSO_(4).3H_(2)O(s)+2H_(2)O(g)` Which one is the correct representation?A. `K_(p)=[p_(H_(2)O)]^(2)`B. `K_(c)=[H_(2)O]^(2)`C. `K_(p)=K_(c)(RT)^(2)`D. All |
| Answer» Correct Answer - D | |
| 40. |
`CuSO_(4).5H_(2)O(s)CuSO_(4).3H_(2)O(s)+2H_(2)O(s) K_(p)=0.4xx10^(-3)atm^(2)` Which of following statement are correc:A. `DeltaG^(@)=-RTlnP_(H_(2)O)` where `P_(H_(2)O)="partial pressure of" H_(2)O` at equilibrium.B. At vapour pressure of `H_(2)O=15.2` torr relative humidity of `CuSO_(4).5H_(2)O "is"100%`C. In pressure of aqueous tension of `24` torr, `CuSO_(4).5H_(2)O` can not loss moisture.D. In presence of dry atmosphere in open container `CuSO_(4).5H_(2)O` will completely convert into `CuSO_(4).3H_(2)O` |
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Answer» Correct Answer - B::C::D (A) `K_(p)=(P_(H_(2)O))^(2)` `=0.4xx10^(-3)=4xx10^(-4)atm^(2)` `(P_(H_(2^O)))_(equi)=2xx10^(-2)atm` `=2xx10^(-2)atm` `=2xx10^(-2)xx760=15.2` torr (B) Relative humidity`=(15.2)/(15.2)xx100=100%`(IfV.P.`=15.2` torr) (C) When `P_(H_(2)O)gt(P_(H_(2)O))_(equi)` backward shift so at `24` torr pressure reaction shift backward. (D) In dry atmosphere and open container reaction shift completely in forward. |
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| 41. |
Consider the equilibrium `N_(2)(g) + 3H_(2)(g)hArr2NH_(3)(g)` , `Delta H = - 93.6 kJ` . The maximum yield of ammonia obtained byA. Decrease of temperature and increase of pressureB. Increase of tempreature and decrease of pressureC. Decrease of both the temperature and pressureD. Increase of both the temperature and pressure |
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Answer» Correct Answer - A For high yield of ammonia, low temperature and high pressure will result in high concentration of the reactant molecule. |
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| 42. |
Assertion : The equilibrium constant for the reaction `CaSO_(4).5H_(2)O(s)hArrCaSO_(4).3H_(2)O(s) + 2H_(2)O(g)` is `K_(C) = ([CaSO_(4).3H_(2)O][H_(2)O]^(2))/([CaSO_(4).5H_(2)O])` Reason : Equalibrium constant is the ration of the product of molar concentration of the substance produced to the product of the molar concentrations of reactants with each concentrations term raised to the power equal to the respective stoichiometric constant.A. If both assertion and reason are true and reason is the correct explanation of the assertion.B. If both assertion and reason are true and reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D Assertion is false but reason is true. `K_(c) = [H_(2)O]^(2)`, because concentration of solids is taken to be unity. |
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| 43. |
The value of `1og_(10)` K for a reaction `AhArrB` is: (Given, `Delta_(r)H_(298K)^(@)=-54.07kJ" "mol^(-1),Delta_(r)S_(298K)^(@)=10JK^(-1)" "mol^(-1) and R=8.314JK^(-1)" "mol^(-1),2.303xx8.314xx298=5705`)A. `5`B. `10`C. `95`D. `100` |
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Answer» Correct Answer - B `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)=-54.07xx1000-298xx10=-54070-2980=-57050` `DeltaG^(@)=-2.303RT log_(10)K` `-57050=-2.303xx298xx8.314 log_(10)K=-5705log_(10)K` `log_(10)K=10` |
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| 44. |
For the following reaction: `K=1.7xx10^(7)` at `25^(@)C` `Ag^(o+)(aq)+2NH_(3)(aq) hArr [Ag(NH_(3))_(2)]^(o+)` What is the value of `DeltaG^(ɵ)` in kJ? |
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Answer» `DeltaG^(ɵ)=-2.303RT log K` `=-2.303xx7.314xx298.15 log (1.7xx10^(7))` `=-41.2 kJ` |
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| 45. |
`K_(1)` and `K_(2)` are equilibrium constants for reaction (i) and (ii) `N_(2)(g)+O_(2)(g) hArr 2NO(g)` …(i) `NO(g) hArr 1//2 N_(2)(g)+1//2O_(2)(g)` …(ii) then,A. `K_(1)=(1//K_(2))^(2)`B. `K_(1)=K_(2)^(2)`C. `K_(1)=1//K_(2)`D. `K_(1)=(K_(2))^(@)` |
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Answer» Correct Answer - A `N_(2)+O_(2) hArr 2NO` …(i) `NO hArr 1/2 N_(2)+1/2O_(2)` …(ii) Equation (ii) is obtained by reversing equation (i) and dividing by `2`. `:. K_(2)=1/((K_(1))^(1//2))` `rArr (K_(2))^(2)=1/K_(1)` `rArr K_(1)=1/((K_(2))^(2))=(1/K_(2))^(2)` |
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| 46. |
The equilibrium constant `K_(p)` for the homogeneous reaction is `10^(-3)`. The standard Gibbs free energy change `DeltaG^(ɵ)` for the reaction at `27^(@)C ("using" R=2 cal K^(-1) mol^(-1))` isA. ZeroB. `-1.8 kcal`C. `-4.154 kal`D. `+4.154 kcal` |
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Answer» Correct Answer - D `DeltaG^(ɵ)=-RT "In" K_(p)` `=-2xx300xxIn (10^(-3))` `=+4.154 kcal` |
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| 47. |
What is `DeltaG^(ɵ)` for the following reaction? `1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g), K_(p)=4.42xx10^(4)` at `25^(@)C`A. `-26.5 kJ mol^(-1)`B. `11.5 kJ mol^(-1)`C. `-2.2 kJ mol^(-1)`D. `-0.97 kJ mol^(-1)` |
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Answer» Correct Answer - A `DeltaG^(ɵ)=-2.303RT log K_(p)` `=-2.303xx8.314xx298.15xxlog (4.42xx10^(4))` `=-26.5 kJ "mol"^(-1)` |
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| 48. |
For `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g), K_(c)` is equal to …………..A. `K_(c)=(1)/([CO_(2)])`B. `K_(c)=[CO_(2)]`C. `K_(c)=([CaO][CO_(2)])/([CaCO_(3)])`D. `K_(c)=([CaCO_(3)])/([CaO][CO_(2)])` |
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Answer» Correct Answer - B `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` According the to of law ,ass action, `K_(c )=[CO_(2)]`, because the concentration of solid substance is taken as unity. |
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| 49. |
The equilibrium constant `K_(p)` for a homogeneous gaseous reaction is `10^(-8)`. The standard Gibbs free energy change `DeltaG^(ɵ)` for the reaction `("using" R=2 cal K^(-1) mol^(-1))` isA. `10.98 kcal`B. `-1.9 kcal`C. `-4.1454 kcal`D. `+4.1454 kcal` |
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Answer» Correct Answer - A `DeltaG^(ɵ)=-2.303 RT log K` `=-2.303xx2xx298 log 10^(-8)` `=-2.303xx2xx298xx-8 cal` `=10980 cal=10.98 kcal` |
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| 50. |
For the reaction `C(s)+CO_(2)(g) hArr 2CO(g)`, the partial pressure of `CO_(2)` and `CO` is `2.0` and `4.0` atm, respectively, at equilibrium. The `K_(p)` of the reaction isA. `0.5`B. `5.0`C. `30.0`D. `8.0` |
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Answer» Correct Answer - D As `K_(p)=(P_(CO))^(2)/P_(CO_(2))=(4)^(2)/(2)=16/2=8` |
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