InterviewSolution
Saved Bookmarks
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.
| 201. |
In a `20` litre vessel initially each have `1-1` mole. `CO,H_(2)OCO_(2)` is present, then for the equilibrium of `CO+H_(2)OhArrCO_(2)+H_(2)` following is true:A. `H_(2)`, more then `1` moleB. `CO,H_(2)O,H_(2)` less then `1` moleC. `CO_(2) "&" H_(2)O` both more then `1` moleD. All of these |
|
Answer» Correct Answer - B `Co+H_(2)OhArrCO_(2)+H_(2)` `{:(t=0,1,1,1,0),(t=teq,1-x,1-x,1+x,x):}` at equilibrium, only `CO_(2)` has `(1+x)` moles. |
|
| 202. |
`4.5` moles each of hydrogen and iodine heated in a sealed 10 litrevesel. At equilibrium, 3 moles of HI was foun. The equilibrium constant for `H_(2)(g) + I_(2) (g)hArr2HI(g)` isA. `1`B. `10`C. `5`D. `0.33` |
|
Answer» Correct Answer - A `H_(2)(g)+l_(2)(g)hArr2HI(g)` `{:(t=0,4.5,4.5,0),(t=teq.,4.5-x,4.5-x,2x),(put,x=1.5,,),(,4.5-1.5,4.5-1.5,2xx1.5=3),(,darr,darr,darr),(,3,3,3):}` `K_(c)=((3)^(2))/(3xx3)=1` |
|
| 203. |
For the reaction `PCl_(5)hArrPCl_(3)+Cl_(2)`, Supposing at constant temperature, if the volume is increased `16` times the initial volume, the degree of dissociation for this reaction will becomes:A. `4` timesB. `(1)/(4)` timesC. `2` timesD. `(1)/(2)` times |
| Answer» Correct Answer - A::B::D | |
| 204. |
A vessel of `10L` was filled with `6` mole of `Sb_(2)S_(3)` and `6` mole of `H_(2)` to attain the equilibrium at `440^(@)C` as: `Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)` After equilibrium the `H_(2)S` formed was analysed by dissolving it in water and treating with excess of `Pb^(2+)` to give `708 g "of" PbS` as precipitate. What is value of `K_(c)` of the reaction at `440^(@)C`?(At. weight of `Pb=206)`.A. `0.08`B. `0.8`C. `0.4`D. `0.04` |
|
Answer» Correct Answer - A::B::D Mole of `PbS=708//236="mole of" H_(2)S` `Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)` `{:("Initial",6,6,0,0),("at eq.",5,3,2,3):}` `K_(c)=((3//10)^(3)xx(2//10)^(2))/((5//10)xx(3//10)^(3))=(4)/(50)=0.08` |
|
| 205. |
Form the given data of equilibrium constants of the following reactions: `CuO(s)+H_(2)(g)hArrCu(s)+H_(2)O(g),K=67` `CuO(s)+CO(g)hArrCu(s)+CO_(2)(g),K=490` Calculate the equilibrium constant of the reaction, `CO_(2)(g)+H_(2)(g)hArrCu(s)+CO_(2)(g)+H_(2)O(g)` |
|
Answer» `Cu(s)+H_(2)(g)(g)hArrCuO(s)+H_(2)O(g),K_(1)=67` Now reversing the second reaction, `Cu(s)+CO_(2)(g)hArrCuO(s)+CO(g),` `K_(2)=1/490` Adding the two reactions, we get, `Co_(2)(g)+H_(2)hArrCO(g)+H_(2)O(g)` for which `K=K_(1).K_(2)=67xx1/490=0.137` |
|
| 206. |
Sulphide ions in alkaline solution react with solid sulphur to form polyvalent sulphide ions. The equilibrium constant for the formation of `S_(2)^(2-)` and `S_(3)^(2-)` from `S` and `S^(2-)` ions is `1.7` and `5.3` respectively. Calculate equilibrium constant for the formation of `S_(3)^(2-)` from `S_(2)^(2-)` and `S`.A. `1.33`B. `3.11`C. `4.21`D. `1.63` |
|
Answer» `{:("Given",S_((s)),+,S^(2-),hArr,S_(2)^(2-),, ,K_(c_(1))=1.7),(,2S_((s)),+,S^(2-),hArr,S_(3)^(2-),, ,K_(c_(2))=5.3):}` `:. K_(c_(1))=([S_(2)^(2-)])/([S^(2-)])=1.7` ….(`1`) `K_(c_(2))=([S_(3)^(2-)])/([S^(2-)])=5.3` ….(`2`) Nor for `S_(2)^(2-)+S_((s))hArrS_(3)^(2-)` `K_(c)=([S_(3)^(2-)])/([S_(2)^(2-)])` `:.` By eqs. (`2`) and (`1`) , `(K_(c_(2)))/(K_(c_(1)))=([S_(3)^(2-)])/([S_(2)^(2-)])=K_(c)` `:. K_(c)=3.11` |
|
| 207. |
When sulphur in the form of `S_(8)` is heated at `900 K`, the initial pressure of `1` atm falls by `10%` at equilibrium. This is because of conversion of some `S_(8)` to `S_(2)`. Find the value of equilibrium constant for this reaction. |
|
Answer» `S_(8)(g)hArr4S_(2)(g)` `{:(Initila "mole",,1,,0),(At equilibrium,,1-x,,4x),(,,(1-0.10),,(4xx0.10)),(,,=0.9,,=0.40):}` Applying the law of mass action, `K_(p)=([p_(S_(2))]^(4))/([p_(S_(8))])=((0.4)^(4))/0.9=0.0256/0.9~~0.03a` |
|
| 208. |
When sulphur is heated at 800 K, the initial pressure of 1 atm decreases by `20%` at equilibrium `K_(p)` of the reaction `S_(8)(g)rArr4S_(2)(g)` is :-A. 0.2B. 0.512C. 1.51D. 2 |
| Answer» Correct Answer - 2 | |
| 209. |
`ArArr2B,K_(p),CrArrD+E,K_(p)` . If degrees of dissociation of A and C are same and `K_(p)=2K_(p)`, then the ratio of total presure `p//p=`?A. `(1)/(2)`B. `(1)/(3)`C. `(1)/(4)`D. 2 |
| Answer» Correct Answer - 1 | |
| 210. |
`2NO_2(g)+7H_2(g)hArr 2NH_3(g)+4H_2O(l)` What is the correct equilibrium expression for this reaction ?A. `K_C=([NH_3]^(2))/([NO_2]^2[H_2]^7)`B. `K_C=([NO_2]^2[H_2]^7)/([NO_3]^(2))`C. `K_C=([NH_3]^2[H_2O]^4)/([NO_2]^2[H_2]^7)`D. `K_C=([NH_3]^2[H_2O]^4)/([NO_2]^2)` |
|
Answer» Correct Answer - A |
|
| 211. |
The degree of dissociation of `PCI_(5) (alpha)` obeying the equilibrium, `PCI_(5)hArrPCI_(3)+CI_(2)`, is approximately related to the presure at equilibrium by (given `alphaltlt1`):A. `alphapropP`B. `alphaprop(1)/sqrt(P)`C. `alphaprop(1)/P^(2)`D. `alphaprop(1)/(P^(4)` |
| Answer» Correct Answer - B | |
| 212. |
Formation of `SO_(3)` take place according to the reaction `2SO_(2) + O_(2)hArr2SO_(3)`, `Delta H = - 45.2` kcal Which of the following factors favours the formation of `SO_(3)` ?A. Increasing in temperatureB. Increase in pressureC. Removal of oxygenD. Increase in volume |
|
Answer» Correct Answer - B According to Le-Charelier principle when we increase pressure, reaction proceeds in that direction where vollume is decreseing. |
|
| 213. |
In which reaction will an increase in the volume of the container favor the formation of products?A. `C(s)+H_(2)O(g) hArr CO(g)+H_(2)(g)`B. `H_(2)O(g)+I_(2)(g)hArr 2HI(g)`C. `4NH_(3)(g)+5O_(2)(g)hArr 4NO(g)+6H_(2)O(l)`D. `3O_(2)(g) hArr 2O_(3)(g)` |
|
Answer» Correct Answer - 1 `(V propn)i.e.. Deltangt0` `Vgt0` in option `(1)` |
|
| 214. |
Which of the following is not favourble for `SO_(3)` formation ? `2SO_(2)(g) + O_(2)(g)hArr2SO_(3) (g)` , `delta H = - 45.0` kcalA. High pressureB. High temperatureC. Decrease in `SO_(3)` concentrationD. Increase in reactant concentration |
|
Answer» Correct Answer - B Because reaction is exothermic. |
|
| 215. |
For the reaction `H_(2)(g)+I_(2)(g) hArr 2HI(g)`, the equilibrium can be shifted in favour of product byA. Increasing the `[H_(2)]`B. Increasing the pressureC. Increasing the `[I_(2)]`D. By using the catalyst |
|
Answer» Correct Answer - A::B Increasing the concentration of reactants favours the product formation. |
|
| 216. |
Which of the following factor is shifted the reaction `PC1_(3) + C1_(2)hArr` `PC1_(5)` at left side ?A. Adding `PC1_(3)`B. Increase pressureC. Constant temperatureD. Catalyst |
|
Answer» Correct Answer - A `K_(p) = (P_(CO)^(2))/(P_(CO_(2))) = (8^(2))/(4) = 16` |
|
| 217. |
The formation of `SO_(3)` takes place according to the following reaction, `2SO_(2) + O_(2)hArr2SO_(3)` , `Delta H = 45.2` kcal The formation of `SO_(3)` is favoured byA. Increasing in temperatureB. Removal of oxygenC. Increasing of volumeD. Increase of pressure |
|
Answer» Correct Answer - D In this reaction `Delta H` is negative so reaction moves forward by decreases in temperature while value of `Delta n = 2 - 3 = - 1`, i.e., negative so the reaction moves forward by increase in pressure. |
|
| 218. |
Which of the following equilibrium is not shifted by increase in the pressure ?A. `H_(2(g)) + I_(2(g))hArr2HI_((g))`B. `N_(2(g)) + 3H_(2(g))hArr2NH_(3(g))`C. `2CO_((g)) + O_(2(g))hArr2CO_(2(g))`D. `2C_((s)) + O_(2(g))hArr2CO_((g))` |
|
Answer» Correct Answer - A `H_(2(g)) + I_(2)hArr2HI` `Delta n =0` , `.: K_(c)=K_(p)` |
|
| 219. |
In a reaction `PCI_(5)hArrPCI_(3) + CI_(2)` degree of dissociation is `30%` . If initial moles of `PCI_(3)` is one then total moles at equilibrium isA. `1.3`B. `0.7`C. `1.6`D. `1.0` |
|
Answer» Correct Answer - A `PCI_(5)(g) rarr PCI_(3)(g) + CI_(2)(g)` t = 0 1 mol `t = oo` `1 - 0.3` `0.3` `0.3` Total moles at equilibrium `= 1.3` |
|
| 220. |
For the given reaction at constant pressure, `{:(,nA(g)hArr,A_n(g)),("Initial moles",1,0),("Final moles",1-alpha,alpha/n):}` Then the correct relation between initial density `(d_i)` and final density `(d_f)` of the system is :A. `[(n-1)/n][(d_f-d_i)/d_f]=alpha`B. `[n/(n-1)][(d_f-d_i)/d_f]=alpha`C. `[(n-1)/n][(d_i-d_f)/d_i]=alpha`D. `1/((n-1))[(d_i-d_f)/d_i]=alpha` |
|
Answer» Correct Answer - B |
|
| 221. |
In the following reaction, `3A (g)+B(g) hArr 2C(g) +D(g)`, Initial moles of `B` is double at A . At equilibrium, moles of A and C are equal. Hence `%` dissociation is `:`A. `10%`B. `20%`C. `40%`D. `5%` |
| Answer» Correct Answer - 1 | |
| 222. |
For the reaction `PCI_(3)(g) + CI_(2)hArrPCI_(5)(g)` the position of equilibrium can be shifted to the right byA. Increasing the temperatureB. Doubling the volumeC. Addition of `CI_(2)` at constant volumeD. Addition of equimolar quantities of `PCI_(3)` and `PCI_(5)` |
|
Answer» Correct Answer - B The reaction is not reversible. |
|
| 223. |
For the given reaction at constant pressure, `{:(,nA(g)hArr,A_n(g)),("Initial moles",1,0),("Final moles",1-alpha,alpha/n):}` Then the correct relation between initial density `(d_i)` and final density `(d_f)` of the system is :A. `[(n-1)/(n)][(d_(f)-d_(i))/(d_(f))]=alpha`B. `(n)/(n-1)([d_(f)-d_(i)])/(d_(f))=alpha`C. `[(n-1)/(n)][(d_(i)-d_(f))/(d_(i))]=alpha`D. `(1)/((n-1))[(d_(i)-d_(f))/(d_(i))]=alpha` |
|
Answer» Correct Answer - B Total moles at equilibrium `=1-alpha+alpha//n=1+[(1)/(n)-1]alpha` So using `(d_(i))/(d_(f))=1+[(1)/(n)-1]alpha` |
|
| 224. |
An equilibrium mixture of the reaction `2H_(2)S(g)("number of moles of" O_(2))/("volume" ("in litre")) = (96)/(32)xx(1)/(2) = 1.5 mol//L2H_(2)(g) + S_(2)(g)` had `0.5` mole `H_(2)S, 0.10` mole `H_(2)` and `0.4` mole `S_(2)` in one litre vessel. The value of equilibrium constant (K) in mole `"litre"^(-1)` isA. `0.004`B. `0.008`C. `0.016`D. `0.160` |
|
Answer» Correct Answer - C `K = [H_(2]^(2)[S_(2)])/([H_(2)S]^(2)) = ([0.10]^(2)[0.4])/([0.5]^(2)) = 0.016` |
|
| 225. |
At equilibrium `X+Y hArr 3Z, 1` mol of `X, 2` mol of Y and `4` mol of Z are contained in a `3-L` vessel. Among the given values of reaction coefficient Q, given at three different instants, which value refers to system at equilibrium?A. `10`B. `15`C. `10.67`D. N/A |
|
Answer» Correct Answer - C Volume `=3 L` `{:(,X,+,Y,hArr,3Z),("At eq.",1/3,,2/3,,4/3):}` `K=([Z]^(2))/([X][Y])=((4/3)^(3))/([2/3][1/3])=10.67` a. `Q=10, :. Q lt K` The reaction proceed forward direction to attain the equilibrium value of K b. `Q=15, :. Q lt K` The reaction proceeds backward direction to attain the equilibrium value of K `(10.67)`. c. `Q=10.67, :. Q=K` The system is at equilibrium |
|
| 226. |
Ina `0.25` litre dissociation of 4 moles of NO takes place. If its degree of dissociation is `10%`. The value of `K_(p)` for reaction `2NO hArrN_(2) + O_(2)` is:A. `(1)/((18)^(2))`B. `(1)/((8)^(2))`C. `(1)/(16)`D. `(1)/(32)` |
|
Answer» Correct Answer - A `{:(,2NO,hArr,N_(2),+,O_(2),alpha=10%),(t=0,4-.4,,.2,,.2,),(,3.6,,0.2,,0.2,):}` `Deltan=0`, `thereforeK_(p)=K_(c), K_(c)=((.2//V)^(2))/((3.6//V)^(2))=(4)/(36xx36)` |
|
| 227. |
Molar concentration of 96 g of `O_(2)` contained in a 2 L vessel is:A. (A) `16mol//litre`B. (B) `1.5mol//litre`C. (C) `4 mol//litre`D. (D) `24 mol//litre` |
|
Answer» Correct Answer - B Molar conc. `=("no. of moles of" O_(2))/("Volume(in litre)")=(96)/(32)xx(1)/(2)=1.5M` |
|
| 228. |
`0.2 "mole"` of `NH_(4)Cl` are introduced into an empty container of `10 litre` and heated to `327^(@)C` to attain equilibrium as : `NH_(4)Cl_((s))hArrNH_(3(g))+HCl_((g))`, `(K_(p)=0.36 atm^(2))`. The quantity of solid `NH_(4)Cl` left is :A. `0.078 "mole"`B. `0.02 "mole"`C. `0.095 "mole"`D. `0.035 "mole"` |
|
Answer» `{:(,NH_(4)Cl,hArr,NH_(3),+,HCl),("Moles at t=0",0.20,,0,,0),("Mole at eqm.",(0.20-a),,a,,a):}` Also, `K_(p)=P_(NH_(3))xxP_(HCl)=P^(2)` `P=sqrt(K_(p))=sqrt(0.36)=0.6 atm` Now, `NH_(3)` formed `n=(PV)/(RT)=(0.6xx10)/(0.0821xx600)` `=0.122 "mole" ="moles of" NH_(4)Cl` decomposed `:. NH_(4)Cl` left `=0.2-0.122=0.078 "mole"` |
|
| 229. |
At the equilibrium of the reaction , `N_(2)O_(4)(g)rArr2NO_(2)(g)`, the observed molar mass of `N_(2)O_(4)` is 77.70 g . The percentage dissociation of `N_(2)O_(4)` is :-A. 28 .4B. 46 .7C. 22.4D. 18 .4 |
| Answer» Correct Answer - 4 | |
| 230. |
Molar concentration of 96 g of `O_(2)` contained in a 2 L vessel is:A. (A) `16mol//litre`B. (B) `1.5mol//litre`C. (C) `4 mol//litre`D. (D) `24 mol//litre` |
|
Answer» Correct Answer - B Molar conc. `=("no. of moles of" O_(2))/("Volume(in litre)")=(96)/(32)xx(1)/(2)=1.5M` |
|
| 231. |
Number of moles when divided by the total volume in litre gives …………. of the respective species. |
| Answer» Correct Answer - Active mass | |
| 232. |
One mole of `SO_(3)` was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel `2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)` The equilibrium mixture reacts with `0.2` mole `KMnO_(4)` in acidic medium. Hence, `K_(c)` is :A. `0.50`B. `0.25`C. `0.125`D. None of these |
|
Answer» Correct Answer - C |
|
| 233. |
Active mass of a `6%` solution of a compound is `2` then calculate molar mass of compound.A. (A) `15`B. (B) `30`C. (C) `60`D. (D) `120` |
| Answer» Correct Answer - B | |
| 234. |
48 g of `SO_(3)` , 12.8 g of `SO_(2)` and 9.6 g of `O_(2)` are present in one litre . The respective active masses will beA. 1.0 , 05 and 0.3B. 0.6 , 0.2 and 0.3C. 0.6 , 0.4 and 0.2D. 1.0 , 0.5 and 1.5 |
| Answer» Correct Answer - 2 | |
| 235. |
For the reaction`H_(2)(g)+CO(g)hArrCO(g)+H_(2)O(g),` if the initial concentration of`[H_(2)]=[CO_(2)]`and x moles /litres of hydrogen is consummed at equilibrium , the correct expression of `K_(p)` is :A. `(x^(2))/((1-x)^(2))`B. `((1-x)^(2))/((1-x)^(2))`C. `(x^(2))/((2+x)^(2))`D. `(x^(2))/((1-x)^(2))` |
|
Answer» Correct Answer - a |
|
| 236. |
Reaction stoichiometry, kinetics and thermodynamics ltbgt Nitrosyl chloride `(NOCl)`, is a yellow gas that is most commonly encountered as a decomposition product of aqua regia. It is toxic and irritating to the lungs. On heating `NOCl` decomposes as `2NOClrarr2NO+Cl_(2)`. The enthalpy change `(DeltaH)` for the formation of `1` mole of `Cl_(2)` by the decomposition of `NOCl` is `75.3KJ` between `100.K "to" 600K`. The standard entropies `(S^(@)_(298K))` of different species are as given below: `{:("Substance",NOCl,NO,Cl_(2)),(S_(298K)^(@),264,211,223):}` Calculate the temperature at which `K_(p)` will be double the value at `298K`. |
|
Answer» Correct Answer - `"log"(KP_(2))/(KP_(1))=(DeltaH)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `therefore "log" 2=(75300)/(2.303xx8.314)[(1)/(298)-(1)/(T_(2))] thereforeT_(2)=305K`. |
|
| 237. |
A mixture of `SO_(3), SO_(2)` and `O_(2)` gases is maintained in a `10 L` flask at a temperature at which the equilibrium constant for the reaction is `100`: `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` a. If the number of moles of `SO_(2)` and `SO_(3)` in the flask are equal. How many moles of `O_(2)` are present? b. If the number of moles of `SO_(3)` in flask is twice the number of moles of `SO_(2)`, how many moles of oxygen are present? |
|
Answer» Correct Answer - (a) `0.1` (b) `0.4` (a) `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g) K_(C)=100` molelit Initial mole `{:(a,b,0),(a-2x,b-x,2x),((a)/(2),b-(a)/(4),(a)/(2)):}` But according to question. No. of mole of `SO_(2)="No. of mole of" SO_(3)` `=a-2x=2x` `a=4x`. `x=(a)/(4)`. Now, `K_(C)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])` But No. of mole of `SO_(3) and SO_(2)` are equal at eq. so. `K_(C)=(1)/([O_(2)])` `[O_(2)]=(1)/(K_(C)) , [O_(2)]= (1)/(100)` But `[O_(2)]=("mole of"O_(2) "at" eq.)/(10)=(1)/(100)` So No. of mole of `O_(2)=(1)/(10)=0.1` (b) `K_(C)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=``(((2n_(So_(2)))/(V))^(2))/((n_(SO_(2))/(v))^(2)xx(n_(O_(2))/(V))` `K_(C)=(4)/((n_(O_2)/(V))) : n_(O_2)=(4xxV)/(K_(c))=(4xx10)/(100)=0.4` |
|
| 238. |
Reaction stoichiometry, kinetics and thermodynamics ltbgt Nitrosyl chloride `(NOCl)`, is a yellow gas that is most commonly encountered as a decomposition product of aqua regia. It is toxic and irritating to the lungs. On heating `NOCl` decomposes as `2NOClrarr2NO+Cl_(2)`. The enthalpy change `(DeltaH)` for the formation of `1` mole of `Cl_(2)` by the decomposition of `NOCl` is `75.3KJ` between `100.K "to" 600K`. The standard entropies `(S^(@)_(298K))` of different species are as given below: `{:("Substance",NOCl,NO,Cl_(2)),(S_(298K)^(@),264,211,223):}` Calculate `G` of the above decomposition reaction at `298K`. |
|
Answer» Correct Answer - `2nocLRARR2no+Cl_(2)` `DeltaH^(@)=75.3KJ,DeltaS^(@)=(2xx211+223)-(2xx264)=117` `thereforeDeltaG^(@)=75300-298xx117=40,434` `thereforeDeltaG^(@)=-2.303RT log K_(p) thereforeK_(p)=8.18xx10^(-8)` atm. |
|
| 239. |
Reaction stoichiometry, kinetics and thermodynamics ltbgt Nitrosyl chloride `(NOCl)`, is a yellow gas that is most commonly encountered as a decomposition product of aqua regia. It is toxic and irritating to the lungs. On heating `NOCl` decomposes as `2NOClrarr2NO+Cl_(2)`. The enthalpy change `(DeltaH)` for the formation of `1` mole of `Cl_(2)` by the decomposition of `NOCl` is `75.3KJ` between `100.K "to" 600K`. The standard entropies `(S^(@)_(298K))` of different species are as given below: `{:("Substance",NOCl,NO,Cl_(2)),(S_(298K)^(@),264,211,223):}` A gaseous mixture of `NO,Cl_(2)` and `NOCl` with partial pressures (in bar) `1.5,0.88` and `0.065` respectively was taken at `475K`. Deduce whether the net reaction will lead to increase in `NOCl` concentration. |
| Answer» Correct Answer - `K_(p) "at" 475K=0.67 "and" Q=((1.5)^(2)xx.88)/((0.065)^(2))=468.6` | |
| 240. |
Reaction stoichiometry, kinetics and thermodynamics ltbgt Nitrosyl chloride `(NOCl)`, is a yellow gas that is most commonly encountered as a decomposition product of aqua regia. It is toxic and irritating to the lungs. On heating `NOCl` decomposes as `2NOClrarr2NO+Cl_(2)`. The enthalpy change `(DeltaH)` for the formation of `1` mole of `Cl_(2)` by the decomposition of `NOCl` is `75.3KJ` between `100.K "to" 600K`. The standard entropies `(S^(@)_(298K))` of different species are as given below: `{:("Substance",NOCl,NO,Cl_(2)),(S_(298K)^(@),264,211,223):}` Calculate the temperature above which the reaction will become spontaneous. |
|
Answer» Correct Answer - For spontaneous Reachtion,`DeltaG^(@)lt0` `therefore75300-Txx117lt0 thereforeTgt645.6K` |
|
| 241. |
Write the expressions for equilbrium constant `K_(c)` and `K_(P)` and classify in Homogeneous and Hetereogeneous equilbrium: (i) `N_(2)O_(4)(g)hArr2NO_(2)(g)` (ii) `3Fe(S)+4H_(2)O(g)hArrFe_(3)O_(4)(S)+4H_(2)(g)` (iii) `NH_(4)HS(S)hArrNH_(3)(g)+H_(2)S(g)` (iv) ` CH_(3)COOH(f)+C_(2)H_(5)OH(f)hArrCH_(3)COOC_(2)H_(5)(f)+H_(2)O(f)` (V) `MgCO_(3)(S)hArrMgO(S)+CO_(2))g)` (vi) `2H_(2)S(g)hArr2H_(2)S(g)hArr2H_(2)(g)+S_(2)(g)` (vii) `SO_(2)(g)+NO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)` (viii) ` NH_(4)NO_(2)(S)hArrN_(2)(g)+2H_(2)O(f)` |
|
Answer» Correct Answer - (i) Homogeneous equilibrium `K_(C)=([NO_(2)]^(2))/([N_(2)O_(4)]) K_(P)=((P_NO_(2)^(2)))/((P_(N_2O_4)))` (ii) Hetereogeneous equilibrium `K_(C)=([H_(2)^(4))/([H_(2)O]) K_(P)=(P_(H_2))^(4))/((P_(H_2O)^(4))` (iii) Hetereogeneous equilibrium `K_(C)=[NH_(3)] [H_(2)S] K_(P)=(P_(NH_(3))(P_(H_(2S))` (iv) Homogeneous equilibrium `K_(C)=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH] [C_(2)H_(5)OH])` But K_`(P)` is not define for liquid system (v) Hetereogeneous equilibrium `K_(C)=[CO_(2)] K_(P)=(P_(CO_2)) (vi) Homogeneous equilibrium ``K_(C)=([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2)) K_(P)((P_(H_(2)))(P_(S_(2))))/((P_(H_(2)S)^(2))` (vii) Homogeneous equilibrium `K_(C)=([SO_(3)][NO])/([SO_(2)][NO_(2)]) K_(P)((P_(SO_(3)))(P_(NO)))/((P_(SO_(2)))(P_(NO_(2)))` Hetereogeneous equilibrium `K_(C)=[N_(2)] K_(P)=(P_N_(2))` |
|
| 242. |
What is the `K_(eq)` expression for the reaction, `C(s)+CO_2(g)to2CO(g)` ?A. `K_(eq)=(2[CO])/([CO_2])`B. `K_(eq)=(2[CO][CO])/([CO_2])`C. `K_(eq)=([CO]^2)/([CO_2])`D. `K_(eq)=([C][CO]^2)/([CO_2])` |
|
Answer» Correct Answer - C |
|
| 243. |
`Fe_(2)O_(3)(s)` may be converted to Fe by the reaction `Fe_(2)O_(3)(s)+3H_(2)(g)hArr2Fe(s)+3H_(2)O(g)` for which `K_(c)=8 `at temp . `720^(@)c`. What percentage of the `H_(2)` ramains unreacted after the reaction hascome to equilibrium ?A. `~22%`B. `~34%`C. `~66%`D. `~78%` |
|
Answer» Correct Answer - b |
|
| 244. |
`SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g)` If observed vapour density of mixture at equilibrium is `35` then find out value of `alpha`A. `0.28`B. `0.38`C. `0.48`D. `0.58` |
|
Answer» Correct Answer - A `d_(o)=(D)/(1+(n-1)alpha) n=1+(1)/(2)=1.5` `35=(40)/(1+0.5alpha)` `1+0.5alpha=(40)/(35)` `0.5alpha=1.14-1` `0.5alpha=0.14` `alpha=0.28` |
|
| 245. |
The equilibrium concentraion of the reactants and products for the given equilibrium in a two litre container are shown below: `underset(2M)(PCl_(3(g)))+underset(1M)(Cl_(2(g)))hArrunderset(4M)(PCl_(5(g)))` (`i`) If `2` mole of `Cl_(2)` are added in the container, find the new equilibrium concentration of each. (`ii`) If the equilibrium mixture reported initially is transferred into `4 litre` vessel, what would be the new concentration at equilibrium? |
|
Answer» (`i`) `[PCl_(3)]=1.5 M`, `[Cl_(2)]=1.5 M`, `[PCl_(5)]=4.5 M`, (`ii`) `[PCl_(3)]-1.225 M`, `[Cl_(2)]=0.725 M` `[PCl_(5)]=1.775M` , |
|
| 246. |
In which reaction at equilibrium will the amount kof reactants present increase with an increase in the container volume?A. `C(s)+CO_(2)(g) hArr 2CO(g)`B. `H_(2)(g)+F_(2)(g) hArr 2HF(g)`C. `CO(g)+NO_(2)(g) hArr CO_(2)(g)+NO(g)`D. `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` |
|
Answer» Correct Answer - D |
|
| 247. |
The equilibrium `SO_(2)(g)+(1)/(2)O_(2)(g)hArrSO_(3)(g)` is established in a container of `4L` at a particular temperature. If the number of moles of `SO_(2),O_(2) "and" SO_(3)` at equilibrium are `2,1 "and" 4` respectively then find the value of equilibrium constant. |
|
Answer» Correct Answer - 4 `SO_(2)(g)+(1)/(2)O_(2)(g)hArrSO_(3)(g)` `{:("No. of mole",2,1,4),("conc.,(2)/(4),(1)/(4),(4)/(4)):}` `K_(C)=([SO_(3)])/([SO_(2)][O_(2)])^(1//2)=(1)/((1//2)xx(1//2))=4` |
|
| 248. |
For the reaction `XCO_(3)hArrXO(s)+CO_(2)(g), K_(p)=1.642 atm" "at727^(@)C " If 4 moles of" XCo_(3)(s)` was put into a 50 litre container and heated to `727^(@)C` What mole percent of the `XCO_(3)` remains unreacted at equilibrium ?A. 20B. 25C. 50D. none of these |
|
Answer» Correct Answer - d |
|
| 249. |
Initially the reactions in the container `a "&" b` are at equilibrium when the products & reactants are put together in a container `C` then at the equilibrium the total number of different compounds are-A. (A) `5`B. (B) `7`C. (C) `6`D. (D) `8` |
|
Answer» Correct Answer - D `{:(N_(2),NHD_(2)),(H_(2),NH_(2)D),(D_(2),HD),(NH_(3),),(NO_(3),):}}=8` |
|
| 250. |
Which of the following conditions represaents an equilibrium ?A. Freezing of ice an open vessel, temperature of ice is constantB. Few drops of water is present along with air in a balloon, temperature of balloon is constantC. Water is boiling in an open vessel over stove, temperature of water is constantD. All the statements (a), (b) and (c) are correct for the equilibrium |
|
Answer» Correct Answer - C In a reversible reaction some amount of the reactants remain unconverted into products. |
|