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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. |
Under what conditions of temperatue and pressure the formation of atomic hydrogen from molecular hydrogen will be favoured mostA. High temperatuer and high pressureB. Low temperatue and low pressureC. High temperatue and low pressureD. Low temperatue and high pressure |
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Answer» Correct Answer - C High temperature and low pressure. |
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| 2. |
Unit of equilibrium constant for the reversible reaction `H_(2)+I_(2)hArr2HI` isA. `mol^(-1)` litreB. `mol^(-2)` litreC. mol `litre^(-1)`D. None of these |
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Answer» Correct Answer - D Unit of `K_(p)=(atm)^(Deltan)` Unit of `K_(c)=("mole/litre")^(Deltan)` `=["mole/litre"]^(0)=0` |
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| 3. |
4 moles each of `SO_(2) and O_(2)` gases are allowed to react to from `SO_(3)` in a closed vassel. At equlibrium `25%` of `O_(2)` is used up. The total number of moles of all the gases at equlibrium isA. `6.5`B. `7.0`C. `8.0`D. `2.0` |
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Answer» Correct Answer - A `2SO+O_(2)hArr2SO_(3)` `{:("Initial moles",4,4,0),("Moles at eqm.",4(1-0.25),4(1-0.25),0.25xx2),("Total number of moles",=41,(1-0.25)+4,(1-0.25)+0.5):}` `=4-1+4-1+0.5=6.5.` |
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| 4. |
Three moles of `PCl_(5).` three moles of `PCl_(3)` and two moles of `Cl_(2)` are taken in a closed vessel. IF at equlibrium the vessel has 1.5 moles of `PCl_(5),` the number of moles of `PCl_(3)` present in it isA. 6B. `4.5`C. 5D. 3 |
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Answer» Correct Answer - B `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))` `{:("3",3,2,("Initially")),("(3-x)",(3+x),(2+x),("Aqulibrium")):}` Given that, `(3-x)=1.5, thereforex=1.5,` Therefore no. of moles of `PCl_((g))Presentis =3+1.5=4.5` |
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| 5. |
The equlibrium concentration of X,Y and `YX_(2),` are 4,2 and 2 moles respectively for the equlibrium `2X+Y hArr YX_(2).` The value of `K_(c)` isA. `0.625`B. `2.0625`C. `6.25`D. `0.00625` |
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Answer» Correct Answer - B `K_(c)=([YX_(2)])/([X]^(2)[Y])=(2)/(4xx4xx2)=1/16=0.0625.` |
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| 6. |
Assertion : ON cooling a frezing maxture, colour of the mixture turns to pink from deep blue for a reaction `Co(H_(2)P)_(6(aq))^(2+)+4Cl_(aq)^(-)hArrCoCl_(4(aq))^(2-)+6H_(2)O_((1)).` Reason: Reaction is endothermic so on cooling, the reacton moves to backward direction.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - C | |
| 7. |
Suppose the reaction `PCl_(5(g))hArr PCl_(3(s))+Cl_(2(g))` is a closed vessel tat equlibrium stage. What is the effect on equlibrium concentration of `Cl_(2(g))` byadding `PCl_(5)` at constant temperatureA. DecreasesB. IncreasesC. UnaffectedD. Cannot be described without the value of `k_(p)` |
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Answer» Correct Answer - B On adding more `PCVl_(5),` equlibrium shifts forward. |
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| 8. |
In the equlibrium `AB hArr A+B,` if the equlibrium concentration of A is doubled, the equlibrium concentratio of B would becomeA. TwiceB. HalfC. `1//4^(th)`D. `1//8^(th)` |
| Answer» Correct Answer - B | |
| 9. |
Ina chemical equilibrium `A+B hArrC+D,` when one mole each ofhte two reactant are mixed, `0.6` male each of the products are formed. The equilibrium constant calculated isA. 1B. `0.36`C. `2.25`D. `4//9` |
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Answer» Correct Answer - C `A+BhArrC+D` initial `1" "1" "0" "0` Remaining at equlibrium `0.4" "0.4" "0.6" "0.6` `K=([C][D])/([A][B])=(0.6xx0.6)/(0.4xx0.4)=36/16=2.25.` |
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| 10. |
In te gas phase reaction, `C_(2)H_(4)+H_(2)hArr C_(2)H_(6),` the equlibrium constant can be expressed in units ofA. `"litre"^(-2)"mole"^(-1)`B. litre `"mole"^(-1)`C. `"mole"^(2)"litre"^(2)`D. mole `"litre"^(-1)` |
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Answer» Correct Answer - B `K=([C_(2)H_(6)])/([C_(2)H_(4)][H_(2)])=(["mole/litre"])/(["mole/litre][mole/litre"])` `="litre/mole. or litre mole"^(-1)` |
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| 11. |
Partial pressue of `O_(2)` in the reaction `2Ag(2)O((s))hArr4A_((s))+O_(2(g))` sA. `K_(p)`B. `sqrt(K_(p))`C. `3sqrt(K_(p))`D. `2K_(p)` |
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Answer» Correct Answer - A `2Ag_(2)O_((s))hArr2Ag_((s))+O_(2(g))` For this reaction, `K_(p)=P_(O2)(Ag_(2)O and Ag "are in solid state")` |
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| 12. |
28 g of `N_(20 and 6g of H_(2)` were keip at `40^(@)C` in 1 litre vesscel the equilibrium mixture contained `24.54g` 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 equlibrium",1-0.81,3-2.43,1.62),(,0.19,0.57,):}` No. of moles of `N_(2)=28/28=3` mole No. of moles of `H_(2)=6/2=3` moles `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=([1.62]^(2))/([0.19][0.57]^(3))=75` |
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| 13. |
One mole o `N_(2)O_(4)` gasat 300 is keip in a closed ctainer at 1 atm. It is heatedto 600 K when `20%` by mass of `N_(2)O_(4)` decomposes to `NO_(2(g)).` The resultant pressure in the container would beA. `1.2` atmB. `2.4` atmC. `2.0` atmD. `1.0` atm |
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Answer» Correct Answer - C `{:("Initial",1,0),("no. of moles",,),("No. of moles",(1-0.2),0.4),("at equlibrium",,):}` Total number of moles at equlibrium `=1.2` Initially, `P_(1)V=n_(1)RTor 1xxV=1xxRT_(1)or V=RT_(1)` At equlibrium, `P_(2)V=n_(2)RT_(2)` `or,P_(2)=(1.2)/(V)RT_(2)=(1.2)/(RT_(1))RT_(2)=(1.2xx600)/(300)=2.4atm.` |
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| 14. |
At a certain temp. `HIhArrH_(2)+I_(2)` only `50%` HI is dissociated at equilibrium. The equiliium constant isA. `0.25`B. `1.0`C. `3.0`D. `0.50` |
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Answer» Correct Answer - A `underset(50)underset(100)(2HI)hArrunderset(25)underset(0)(H_(2))+underset(25)underset(0)(I_(2))` `([H_(2)][I_(2)])/([HI]^(2))=(25xx25)/(50xx50)=0.25.` |
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| 15. |
In the reaction, `A+BhArr2C.` at equilibrium, the concenration of A and B is `0.20 molI^(-1)` each and that of c was found to be `0.6molI^(-1).` tTe equiibrium constant of the reaction isA. `2.4`B. 18C. `4.8`D. 9 |
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Answer» Correct Answer - D `A+Boverset(K_(2))hArr2C` `K_(c)=([C]^(2))/([A][B])=([0.6]^(2))/([0.2][0.2])=9` |
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| 16. |
The reaction which proceeds in the forward direction isA. `Fe_(2)O_(3)+6HCl=2FeCl_(3)+3H_(2)O`B. `NH_(3)+H^(2)O+NaCl=NH_(4)Cl+NaOH`C. `SnCl_(4)+Hg_(2)Cl_(2)=SnCl_(2)=SnCl_(2)+2HgCl_(2)`D. `2CuI+I_(2)+4K^(+)=2Cu^(2+)+4KI` |
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Answer» Correct Answer - A Due to absence of hydrolystion of `FeCl_(3)` backward reaction will not take place. |
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| 17. |
In which of the following, the reaction proceeds towards completionA. `K=10^(3)`B. `K=10^(-2)`C. `K=10^(-2)`D. `K=1` |
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Answer» Correct Answer - A Those reacton which have more value of K proceeds towards completion. |
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| 18. |
In line kilns, the following reaction, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` proceeds to completion because ofA. Of the high temperatureeB. CaO is more stable than `CaCO_(3)`C. CaO is not dissciatedD. `CO_(2)` escapes continuously |
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Answer» Correct Answer - D In make klin `CO_(2)` escapes regularly so reaction proceeds in forward direction. |
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| 19. |
For two different acids with same concentrationA. The relative strngth is expressed as `(alpha_(1))/(a_(2))`B. Relative strength is expressed as `(K_(a_(1)))/(K_(a_(2)))`C. Relative strength is expressed as `sqrt((K_(a_(1)))/(K_(a_(2))))`D. `(pH_(1))/(pH_(2))` |
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Answer» Correct Answer - A::C Let `alpha_(1), and alpha_(2)` be the degree of dissociation of two acids and `alpha_(1),alpha_(2)` are very small when compared to unity. `(calpha_(1)^(2))/(1-alpha_(1))=K_(a_(1)),(calpha_(2)^(2))/(1-alpha_(2))=K_(a_(2))impliessqrt((K_(a_(1)))/(K_(a_(2))))=(alpha_(1))/(alpha_(2))=` relative strength Hence choices (a) and (b) and (c) are correct while (b) and (d) are incorrect. |
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| 20. |
Degree of hydrolysis for a salt of strong acid and weak base isA. Independent of dilutionB. Increases with dilutionC. Increases with decrease in `K_(b)` of the basesD. Decreases with decrease in temperature |
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Answer» Correct Answer - B::C::D Let BA be this salt `BA to B^(+)+A^(-)` `A^(-)` does not undergo hydrolysis because HA is strong acid. `B^(+)` undergoes hydrolysis `B^(+) +H_(2)O=BOH+H^(+)` `h=sqrt((K_(h)V)/(c))-(1)h=` degreee of hydrolysis where `K_(h)= " Hydrolysis constant "= (K_(w))/(K_(b))` `h propsqrt(K_(h))-(3)` greater the `K_(b)` lesser the h `h propsqrtV-(4)V=` volume of salt solution hence h increases with dilution. `K_(h)=(K_(w))/(K_(b)). Both K_(w)and B_(b)` change with temperature, so `K_(h)` change with temperature. Hence, statement (b) is correct, h increases if `K_(b)` decreases, statement (c) correct. It is found that as temperatue, increases, `k_(w) and K+_(b)` increase but increase in `K_(w)` is greater than increase in `K_(b).` Hence, h increase with increase in temp. or h decreases with decrease in temp., hence statement (d) is correct, (a) is not correct from explanation of (b). |
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| 21. |
1 mole of `H_(2) and 2` moles of `I_(2)` are taken initially in 2 litre vessel. The number of moles of `H_(2)` at equlibrium is `0.2.` Then the number of moles of `I_(2) and HI` at equliberium areA. `1.2,1.6`B. `1.8,2.4`C. `0.4,2.4`D. `0.8,2.0` |
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Answer» Correct Answer - A At equlibrium number of moles of `H_(2)=0.2` moles 0.8 moles of `H_(2)` reacts with `(2-0.8)` moles of `I_(2)` to from `(2xx0.8)` moles of HI `H_(2)+I_(2)hArr2HI` `{:("Inital no.",1,2,0),("of mole",,,),("No of moles"0.,2,,(-20.8)2xx0.8),("at equlibrium",,=1.2,=1.6):}` |
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| 22. |
When 3 moles of A and 1 mole of B are mixed in 1 litre vessel the following reaction takes place `A_((b))+B_((g))hArr2C_((g)).1.5` moles of C are formed. The equilibrium constant for the reaction isA. `0.12`B. `0.25`C. `0.50`D. `4.0` |
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Answer» Correct Answer - D `underset((3-0.75))(A)+underset((1-0.75))(B)tounderset(1.5)(2C)` `K=([C]^(2))/([A][B])=((1.5)^(2))/(2.25xx0.25)=(2.25)/(2.25xx0.25)=4.0.` |
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| 23. |
For the reaction, `CO_((g))+Cl_(2(g))hArrCOCl_(2(g))the K_(p)//K_(c)` is equal toA. `sqrt(RT)`B. RTC. `1//RT`D. `1.0` |
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Answer» Correct Answer - C `CO_((g))+Cl_(2(g))hArrCOCl_(2(g))` `Deltan=1-2=-1` `K_(p)=K_(c)[RT]^(Deltan), therefore(K_(p))/(K_(c))=[RT]^(-1)=(1)/(RT)` |
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| 24. |
For the system at equlibrium, which of the following are correctA. `logK=(1)/(2.303R)(DeltaS-(DeltaH)/(T))`B. On increasing the temperature of an endothemic reaction, the equlibrium shifts in forwared direction because Q decreasesC. On increasing the temperture of an endothermic reaction, the equilibrium shifts in forward direction because K increaseD. On increasing the temperature of an endothemic reaction, the conncentration in moles per litre of the reactants increase |
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Answer» Correct Answer - A::C `DeltaG=DeltaH-TDeltaS=-2.303RTlogK` `TDeltaS-DeltaH=2.303RTlogKimplies` `(TDeltaS-DeltaH)/(2.303RT)=logKimplies(1)/(1.303R)[DeltaS-(DeltaH)/(T)]=logK` Hence choice (a) is correct. On changing temperatue K changes and not Q. Also `log""(K_(2))/(K_(1))=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` For endothemic reactions on increasing T, K increase hte concentration of products. Thus choice (c) is correct while (b) and (d) are not correct. |
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| 25. |
`A0.1M` sodium acetate solution was prepared. The `K_(h)=5.6xx10^(-10)`A. The degree of hydrolysis is `7.48xx10^(-5)`B. The `[OH^(-)]` concentration is `7.48xx10^(-3)M`C. The `[OH^(-)]` concentration is `7.48xx10^(-6)M`D. The pH is approximately `8.88` |
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Answer» Correct Answer - A::C::D `underset(0.1(1-h))(CH_(3)COO^(-))+H_(2)OhArrunderset(0.1h)(CH_(3)COOH)+underset(0.1h)(OH^(+))` `K_(h)=((0.1h)(0.1h))/(0.1(1-h))=0.1h^(2)` `implies5.6xx10^(-10)=0.1h^(2)impliesh=7.48xx10^(-5)` `[thereforeltltltlt1]` `[OH^(-)]=ch=7.48xx10^(-5)xx10^(-1)=7.48xx10^(-6)implies` `[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(7.48xx10^(-6))=1.33xx10^(-9)`implies pH=8.8` approx Hence choices (a), (c) and (d) are correct while (b) is wrong. |
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| 26. |
Match the entries listed in Column I with appropriate entries listed in Column II. |
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Answer» Correct Answer - A::B::C::D (A) Increae in pressure shifts the equlibrium in te direaction where lesser no. of gaseous moles are present. `(Deltang=n_(p)-n_(r)=2-4=-2)` (B) Increase in temperature shifts the equlibrium in the direction of endothermic reaction. `(Deltang=n_(p)-n_(r)=3-3=0,` hence pressure is unchanged) (C) Decrease in pressure shifts the equlibrium in the direaction where larger no. of gaseous moles are present. `(Deltang=n_(p)-n_(r)=2-1=1)` (D) No gaseous moles present hence no effect of pressure. |
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| 27. |
The rate of forward reaction is two times that of reverse reaction at a given temperaturee and identical concentration. `K_("equilibriuim")` isA. `2.5`B. `2.0`C. `0.5`D. `1.5` |
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Answer» Correct Answer - B The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration `K_("equlibrium")` is 2 because the reaction is reversible. So `K=(K_(1))/(K_(2))=2/1=2.` |
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| 28. |
In `H_(3)PO_(4),` which of the following is trueA. `K_(a)=K_(a_(1))xxK_(a_(2))xxK_(a_(3))`B. `K_(a_(1))ltK_(a_(2))ltK_(a_(3))`C. `K_(a_(1))gtK_(a_(2))ltK_(a_(3))`D. `K_(a_(1))=K_(a_(2))=K_(a_(3))` |
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Answer» Correct Answer - A::C `H_(3)PO_(4)hArrH^(+)+H_(2)PO_(4)^(-)Ka_(1)=([H^(+)][H_(2)PO_(4)^(-)])/([H_(3)PO_(4)])" "...(i)` `h_(2)PO_(4)^(-)hArrH^(+)+HPO_(4)^(2-)K_(a_(2))=([H^(+)][HPO_(4)^(-2)])/([H_(2)PO_(4)^(-)])" "...(ii)` `HPO_(4)^(2-)hArrH^(+)+PO_(4)^(3-)K_(a_(3))=([H^(+)][HPO_(4)^(2-)])/([HPO_(4)^(2-)])" "...(iii)` `H_(3)PO_(4)hArr3H^(+)+P_(4)^(3-)K_(a)=([H^(+)]^(3)[PO_(4)^(3-)])/([H_(3)PO_(4)])" "...(iv)` Multiplying (i),(ii), (ii) and equate with (iv) `K_(a)=K_(aq)xxK_(a_(2))xxK_(a_(3)),therefore(a)` is correct `1^(st)` acid dissociation constant is highest because dissociation is maximum in first step, less in `2^(nd)` step and minimum in `3^(rd)` step because it is difficult for a negatively changed ion tolose `H_(+)` ion. `K_(a_(1))gtK_(a_(2))gtK_(a_(3)), therefore(c)` is correct, (b) and (d) are ruled out. |
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| 29. |
Which of the following will fovour the reverse reaction in a cheical equilibriumA. Increasing the concentration of the reactantsB. Removal of at least one of the products at regular intervalsC. Increasing the concentration of one or more of the productsD. None of these |
| Answer» Correct Answer - C | |
| 30. |
For a reaction `C(s)+CO_(2)(g)hArr2CO(g)`A. The equlibrium shifts in backward direction if more of CO is addedB. The equilibrium shifts in backward direction if more of CO is addedC. The equlibrium shifts in forward direction if more of corbon is addedD. On adding carbon, the equlibrium is not disturbed |
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Answer» Correct Answer - A::B::D `Q_(p)=(P_(CO)^(2))/(P_(CO_(2))),Q_(C)=([CO]^(2))/([CO_(2)])` When more of `CO_(2)` is added `Q_(p)` decreases, hence in order to reach equlibrium, system moves in forward direction, hence choice (a) is correct. When more of CO is added, `Q_(p)` increases and therefore to reach equlibrium, system moves in backward directin, so choice (b) is correct. Since active mass of colids is 1, hence changing the concentrations of carbon does not change Q. Hence adding carbon does not shift the equlibrium, (c) is incorrect. Whereas (d) is correct. |
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| 31. |
For a system at equilibriumA. `((dellnK_(p))/(delP))=-(del)/(delP)((DeltaG)/(RT))_(T)`B. `DeltaG=0`C. `G_((P))=G_((R))`D. The free energy of the ststem is minimum |
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Answer» Correct Answer - A::B::C::D `DeltaG=-RTlnK_(p)implies-(DeltaG)/(RT)=lnK_(p)` `implies((dellnK_(p))/(delP))_(T)=-(del)/(delP)((DeltaG)/(RT))_(T)` Hence statement (a) is true At equlibrium `G_(p)=G_(R)` i.e., free energy of products and reactants are equal. `impliesDeltaG=G_(p)-G_(R)=0implies` Statement (b) and(c) are correct. At equlibrium, the free energy of system is minimum. Statement (d) is correct. |
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| 32. |
Which of the following will act as bufferA. `NaCl+NaOH`B. Borax+ Boric acidC. `NaH_(2)PO_(4)+Na_(2)HPO_(4)`D. `NH_(4)Cl+NH_(4)OH` |
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Answer» Correct Answer - B::D (b) and (d) are correct because they have common ion and one of them is weak electrolyte. (a) is not correct because NaCl and NaOH both are strong electrolytes and will not have common ion effect. © isnot correct because both are salt and strong electrolytes., cannot from buffer solution. |
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| 33. |
Choose the correct statementA. pH of acidic buffer solution decrease if more salt is addedB. pH of acidic buffer solution increeses if more salt is addedC. pH of basic buffer decreases if more salt is addedD. pH of basic buffer increases if more salt is added |
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Answer» Correct Answer - B::C (b) is correct because `pH=pH_(a)+log""(["Salt"])/(["Acid"])` for acidic buffer. If [Salt] increase, pOH will increase, pH will decrease as `pH=14-pOH` Therefore © is correct but (d) is wrong. (a) is correct explained below `pH=pK_(a)+log""(["Salt"])/(["Acis"])` If [salt] increase, pH will increase. |
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| 34. |
Value of `K_(p)` in the reaction `MgCO_(3(s))hArrMgO_((s))+CO_(2(g))` isA. `K_(p)=P_(CO_(2))`B. `K_(p)=P_(CO_(2))xx(P_(CO_(2))xxP_(MgO))/(P_(MgCO_(3)))`C. `K_(p)=(P_(CO_(2))xxP_(MgO))/(P_(MgCO_(3)))`D. `K_(p)=(P_(MgCO_(3)))/(P_(CO_(2))xxP_(MgO))` |
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Answer» Correct Answer - A In this reaction gaseous molecule count `MgCO_(3)to MgO_((s))+CO_(2(g))K_(p)=P_(CO_(2))` |
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| 35. |
The equlibrium constant for the reaction between `CH_(4)(g)and H_(2)S(g)` to from `CS_(2)(g)and H_(2)(g),` at 1173 K is 3.6. for the following composition of th reaciton mixture, decide which of the following option is correct `[CH_(4)]=1.07M,[H_(2)S]=1.20M,` `[CS_(2)]=0.09 M, [H_(2)]=1.78M`A. Reaction is in equlibriumB. Reaction will shift to from more of `CS_(2)`C. Reaction will shift to from more of `H_(2)S`D. No reaction takes place |
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Answer» Correct Answer - C `CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` `Q_(c)=((0.90)(1.78)^(4))/((1.04)(1.20))=5.86. As Q_(c)gtK_(c),` equlibrium will go in to backward direction. |
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| 36. |
The chemical reaction : `BaO_(2(s))hArrBaO_((s))+1/2O_(2(g)),DeltaH=+ve.` In equilibrium condition, pressure of `O_(2)` depends uponA. Increase mass of BaOB. Increase mass of `BaO_(2)`C. Increase in temperatureeD. Increase mass of `BaO_(2)` and BaO both |
| Answer» Correct Answer - C | |
| 37. |
In which of the following equlibrium `K_(c)and K_(p)` are not equalA. `2C_((s))+O_(2(g))hArr2CO_(2(g))`B. `2NO_((g))hArrN_(2(g))+O_(2(g))`C. `SO_(2(g))+NO_(2(g))hArrSO_(3(g))+NO_((g))`D. `H_(2(g))+I_(2(g))hArr2HI_((g))` |
| Answer» Correct Answer - A | |
| 38. |
For `N_(2)+3H_(2)hArr2NH_(3)+` heatA. `K_(p)=K_(c)(RT)`B. `K_(p)=K_(c)(RT)`C. `K_(p)=K_(c)(RT)^(-2)`D. `K_(p)=K_(c)(RT)^(-1)` |
| Answer» Correct Answer - C | |
| 39. |
In the reaction `H_(2(g))+3H_(2)hArr2NH_(3(g)),` the value of the equlibrium constatn depends onA. Volume of the reaction vesselB. Total pressure of the systemC. The initial concentration of nitrogen and hydrogenD. The tempereatue |
| Answer» Correct Answer - C | |
| 40. |
The quantity of K in a rate of expressionA. is independent of concentration of reactantsB. Is called Arrhenius constantC. Is dimesionlessD. Is independent of temperaturee |
| Answer» Correct Answer - A | |
| 41. |
Reaction in which yield of prodict will increase with increase in pressure isA. `H_(2(g))+I_(2(g))hArrHI_((g))`B. `H_(2)O_((g))+CO_((g))hArrCO_(2(g))+H_(2(g))`C. `H_(2)O_((g))+CO_((g))hArrCO_(2(g))+H_(2(g))`D. `CO_((g))+3H_(2(g))hArrCH_(4(g))+H_(2)O_((g))` |
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Answer» Correct Answer - D In reaction `CO+3H_(2)hArrCH_(4)+H_(2)O` Volume is decreasing in forward direction so on increasing pressure the yield of product will increase. |
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| 42. |
`A_((g))+3B_((g))hArr4C_((g)).` Strating concentration of A is equal to B, equlibrium concentration of A and C are same `K_(c)=`A. `0.08`B. `0.8`C. 8D. 80 |
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Answer» Correct Answer - C `{:(A,+,3B,hArr,4C),(a,,b,,0),((a-x),,(b-3x),,4x):}` `K_(c)=([C]^(4))/([A][B]^(3))=(4x.4x.4x.4x)/((a-x)(b-3x))` Given `a=b,a-x=4ximpliesa=5x=b` `K_(c)=(4x.4x.4x.4x.)/((5x-x)(5x-3x))=(4x.4x.4x.4x)/(4x2x.2x2x)=8.` |
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| 43. |
For the reaction `CO_((g))+2H_(2(g))hArrCH_(3)OH_((g)),` true condition isA. `K_(p)=K_(c)`B. `K_(p)gtK_(c)`C. `K_(p)ltK_(c)`D. `K_(c)=0but K_(p)ne0` |
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Answer» Correct Answer - C When `n_(r)gt n_(p)then K_(p)ltK_(c) where n_(r)=` no. of moles of reactant `n_(p)=` no. of moles of product. |
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| 44. |
The free energy for a recersible reaction at equlibrium isA. NegativeB. PositiveC. ZeroD. Either positive or negative but not zero |
| Answer» Correct Answer - B | |
| 45. |
When the pressure is applied over system ice `hArr` wate what will happenA. More water will formB. More ice will formC. There will be no effect over equlibriumD. Water will decompose in `H_(2)and O_(2)` |
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Answer» Correct Answer - A `underset("more volume")(Ice)hArrunderset("less volume")(Water)` On increasing pressure, equlibrium shifts forward. |
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| 46. |
For the synthesis of ammonia by the reaction `N_(2)+3H_(2)hArr hNH_(3)` in the Haber process the attainment of equilibrium is correctly predicted by the curveA. B. C. D. |
| Answer» Correct Answer - C | |
| 47. |
Which of the following reactions proceed at low pressureA. `N_(2)+3H_(2)hArr2NH_(3)`B. `H_(2)+I_(2)hArr2HI`C. `PCl_(5)hArrPCl_(3)+Cl_(2)`D. `N_(2)+O_(2)hArr2NO` |
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Answer» Correct Answer - C At low pressure, reaction proceeds where volume is increasing. This is the favoourable condition for the reaction. `PCl_(5)hArrPCl_(3)+Cl_(2).` |
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| 48. |
All reactions which have chemical disintegrationA. Is reversibleB. Is reversible and endothermicC. Is exothermicD. Is reversible or irreversible and endothermic or exothermic |
| Answer» Correct Answer - D | |
| 49. |
In the given reaction `N_(2)+O_(2)hArr 2NO,` equilbrium means thatA. Concentration of reactants is changing where as concentration of products is constantB. Concentration of all substances is contantC. Concentration of reactants is constant where as concentration of prodcts is changingD. Concentaration of all substances is changing |
| Answer» Correct Answer - B | |
| 50. |
In which of the following reactions, the concentration of the proudct is higher than the concentration of reatsnt at equlibrium (K=equlibrium constant)A. `MhArrB,K=0.001`B. `MhArrN,K=10`C. `XhArrY,K=0.005`D. `RhArrP,K=0.01` |
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Answer» Correct Answer - B For a reaction, `K_(c)=(["Product"])/(["Reactant"]),` Hence, `If K_(c)lt1,"then [Product]"gt"[Reactant"]` |
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