966 + Interview Questions in CHEMICAL-EQUILIBRIUM IN GENERAL AWARENESS Page 4 InterviewSolution

151.	The equilibrium composition for the reaction is `{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.20,,0.10,,0.40 mol L^(-1)):}` What will be the equilibrium concentration of `PCl_(5)` on adding `0.10 mol` of `Cl_(2)` at the same temperature?
Answer» Correct Answer - A::D `{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.2,,0.1,,0.4):}` `K_(c ) 0.4/(0.2xx0.1)xx20` After adding `0.1` mole of `Cl_(2)`, New initial concentration of `Cl_(2)=0.1+0.1 =0.2 "mole"` New initial concentration of `PCl_(3)` and `PCl_(5)` remains the same. Now, suppose `x` mol of `PCl_(3)` reacts, the new equilibrium concentration. will be `[PCl_(3)]=0.2-x, [Cl_(2)]=0.2-x, [PCl_(5)]=0.4+x` `K_(c )=([PCl_(5)])/([PCl_(3)][Cl_(2)])=((0.4+x))/((0.2-x)(0.2-x))` `:. 20=((0.4+x))/((0.2-x)(0.2-x))` solve for `x` `x=0.45 "mol" L^(-1)`

Discussion

152.	Write the expression for equilibrium constant for the following reactions. If the concentrations are expressed in `mol L^(-1)`, give the units in each case. a.
Answer» a. `K_(c)=([NO_(2)]^(2))/([N_(2)O_(4)])`, units, `"mol" L` b. `K_(c )=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])`, units, `"mol" L^(-1)` c. `K_(c)=([NO]^(4)[H_(2)O]^(6))/([NH_(3)]^(4)[O_(2)]^(5))`, units, `"mol" L^(-1)` d. `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`, untis, `"mol"^(-2) L^(2)` e. `K_(c)=([H_(2)][I_(2)])/([HI]^(2))`, no units f. `K_(c)=[CO_(2)]`, units, `"mol" L^(-1)` g. `K_(c)=([H_(2)]^(4))/([H_(2)O]^(4))`, no units h. `K_(c)=([NO_(2)]^(4)[O_(2)])/([N_(2)O_(4)]^(2))`, units `"mol"^(3) L^(-3)`

Discussion

153.	Equilibrium constant can also be expressed in terms of `K_(x)` , when concentrations of the species are taken in mole fraction `F_(2)(g)hArr2F(g),K_(x) = (X_(F)^(2))/(X_(F_(2)))` For the above equilibrium mixture, aberage molar mass at 1000 K was `36.74` g `mol^(-1)` . Thus, `K_(x)` isA. `14.08`B. `2.124xx10^(2)`C. `7.1xx10^(-2)`D. `4.708xx10^(-3)`
Answer» Correct Answer - D `Let , f_(2)(g)=x("mole friction")`, `"molar mass"=38.09 "mol"^(-1)` `F(g)=(1-x),"molar mass"=19.0g "mol"^(-1)` `therefore "molar mass"=(M_(1)x_(1)+M_(2)x_(2))/(x_(1)+x_(2))` `36.74=38x+19(1-x)` `therefore x=0.337("molar fraction of" F_(2))` `(1-x)=0.0633("mole fraction of "F)` `therefore K_(x)=x_(F)^(2)/x_(f_(2))=(0.0663)^(2)/(0.9337)=4.708xx10^(-3)`

Discussion

154.	The equilibrium constant expression for a gas reaction is : `K_(c) = ([NH_(3)]^4[O_(2)]^5)/([NO]^(4)[H_(2)O]^(6))` Write the balanced chemical equation corresponding to this expression.
Answer» `4NO(g)+6H_(2)O(g)+5O_(2)(g)`

Discussion

155.	The equilibrium `SO_(2) CI_(2) (g) hArr SO_(2)(g) +CI_(2)(g)` is attained at `25^(@)C` in a closed container and inert gas helium is introduced. Which of the following statement (s) is`//`are correct ? (1).concentrations of `SO_(2), CI_(2) " and " SO_(2) CI_(2)` change (2). More chlorine is formed (3).Concentration of `SO_(2)` is reduced (4).More `SO_(2) CI_(2)` is formedA. I,II,IIIB. II,III,IVC. III,IVD. None
Answer» Correct Answer - D

Discussion

156.	If `0.5 `mole `H_(2)` is reacted with 0.5 mole `I_(2)` in a ten `-` litre container at `444^(@)C` and at same temperature value of equilibrium constant `K_(C)` is 49, the ratio of `[Hl]` and `[l_(2)]` will be `:`A. 7B. `(1)/(7)`C. `sqrt((1)/(7))`D. 49
Answer» Correct Answer - 1

Discussion

157.	At `700 K`, the equilibrium constant `K_(p)` for the reaction `2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)` is `1.80xx10^(-3) kPa`. What is the numerical value of `K_(c )` in moles per litre for this reaction at the same temperature?
Answer» For the reaction `2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)` Here `n_(p)=3 "mol", n_(r)=2 "mol"` `:.Deltan=n_(p)-n_(r)=3-2=1 "mole"` `K_(p)=1.80xx10^(-3) kPa` `R=8.314 J K^(-1) "mol"^(-1)` `T=700 K` Using the relation, `K_(p)=K_(c )(RT)^(Deltan)` `K_(c )=K_(p)/(RT)=(1.80xx10^(-3))/(8.314xx700)=3.09xx10^(-7) "mol" L^(-1)`

Discussion

158.	For the reaction `CO(g)+Cl_(2)(g)hArrCOCl_(2)(g)`the value of `(K_(c)/(K_(P)))` is equal to :A. `sqrt(RT)`B. RTC. `(1)/(RT)`D. `1.0`
Answer» Correct Answer - b

Discussion

159.	The equilibrium of formation of phosgene is represented as : `CO(g)+Cl_(2)(g)hArrCOCl_(2)(g)` The reaction is carried out in a `500 mL` flask. At equilibrium, `0.3` mol of phosgene, `0.1 mol` of `CO`, and `0.1` mol of `Cl_(2)` are present. The equilibrium constant of the reaction isA. `30`B. `15`C. `5`D. `25`
Answer» Correct Answer - B For the reaction `CO(g)+Cl_(2)(g)hArrCOCl_(2)(g)` The equation constant, `K_(c )=([COCl_(2)])/([CO][Cl_(2)])…(i)` The concentration of `[COCl_(2)]=("mol")/(V(L))=0.3/0.5 "mol" L^(-1)` `[CO]=0.1/0.5 "mol" L^(-1)` `[Cl_(2)]=0.1/0.5 "mol" L^(-1)` Therefore, on substituting all the value in expression (1), we get `(0.3/0.5)/((0.1/0.5)(0.1/0.5))=0.3/0.5xx0.5/0.1xx0.5/0.1` `=0.15/0.01=15`

Discussion

160.	4.4 grams of `CO_(2)` are introduced into a 0.82 L flask containing excess solid carbon at `627^(circ)C`, so that the equilibrium: The density of equilibrium gaseous mixture corresponds to an average molecular weight of 36. `K_(p)=P_(CO)^(2)/P_(CO_2)and K_(C)([CO]^(2))/([CO_(2)])` `[R=0.082"Lt-atm/mol-K",C=12,0=16]` If in the problem,where actually 1.2g of solid carbon is present initially, how many total moles of `CO_(2)` would have to be inroduced initiallly so that at equilibrium only a trace of carbon remained?A. `0.25`B. `0.7`C. `0.6`D. `0.4`
Answer» Correct Answer - B

Discussion

161.	4.4 grams of `CO_(2)` are introduced into a 0.82 L flask containing excess solid carbon at `627^(circ)C`, so that the equilibrium: The density of equilibrium gaseous mixture corresponds to an average molecular weight of 36. `K_(p)=P_(CO)^(2)/P_(CO_2)and K_(C)([CO]^(2))/([CO_(2)])` `[R=0.082"Lt-atm/mol-K",C=12,0=16]` `K_(p)` of the reaction `C(s) +CO_(2)(g)hArr2CO(g)` is :A. 6atmB. 12atmC. 24atmD. 15atm
Answer» Correct Answer - A

Discussion

162.	For the reaction : `A(g)hArr2B(g),K_(p)=4.0` bar. The equilibrium pressure pf A(g) is 1.0 bar. Now the equilibrium mixture is compressed reversibly and isothermally such that the final total pressure of system becomes 8 bar. The partial pressure of A(g) (in bar)at this new equilibrium is :
Answer» Correct Answer - 4

Discussion

163.	The figure shows the change in concentration of species A and B as a function of time. The equilibrium constant `K_(c)`for the reaction` A(g)hArr2B(g)` is : A. `K_(c)gt1`B. `Klt1`C. `K=1`D. data insufficient
Answer» Correct Answer - a

Discussion

164.	Consider the following two equilibrium established together in a closed container `A(s)hArr2B(g)+3C(g) , K_(P_(1)` `A(s)hArr3D(g) , K_(P_(2)` Starting with only `A(s)`, molar ratio of `B(g)` & `D(g)` at equilibrium is found to be in a ratio `1:6` determine `(K_(P_(2)))/(8K_(P_(1)))`
Answer» Correct Answer - 8 `P_(B):P_(D)=1:6` Let the partial at pressure of `B` eq. be `P_(0)` `K_(p_(1))=(P_(0))^(2)((3P_(0))/(2))^(3)` `K_(p_(2))=(6P_(0))^(3)` `(K_(p_(2)))/(K_(p_(1)))=(6^(3))/(((3)/(2))^(3))=64` `(K_(p_(2)))/(8K_(p_(1)))=(64)/(8)=8`

Discussion

165.	For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^(@)C` and `527^(@)C` are, respectively, `6.1xx10^(-12)` and `1.0xx10^(-7)`. What is the nature of the reaction?
Answer» We have `"log" K_(p2)-log K_(p1)=(DeltaH)/(2.303 R)xx(T_(2)-T_(1))/(T_(2)T_(1))` As we know from the above equation that if on increasing temperature, `K_(p)` increases, `DeltaH` becomes positive, i.e., the reaction is endothermic. Thus, from the given data, we see that the reaction is endothermic.

Discussion

166.	The equilibrium constants for the reaction, `A_2hArr2A` A at `500K and 700K` are `1xx10^(-10)` and `1xx10^(-5)`. The given reaction isA. exothermicB. slowC. endothermicD. fast
Answer» Correct Answer - B For the reaction. `A-2hArr2A` `K=([A])^2/([A_2])` The value of equilibrium constant is very less and hence, the product concentration is also very less. So, the reaction is slow.

Discussion

167.	The gas `A_(2)` in the left flask allowed to react with gas `B_(2)` present in right flask as `A_(2)(g)+B_(2)(g)hArr2AB(g),K_(c)=4` at `27^(@)C.` What is the concentration of AB when equilibrium is established ?A. 1.33 MB. 0.66 MC. 0.33 MD. 2.66 M
Answer» Correct Answer - D

Discussion

168.	In order to prepare 25.92 gm of HBr in 20 litre container by following reactions what minimum mass of equimolar mixture of `H_2` and `Br_2` should be taken ? Given : `H_2(g)+Br_2(g)hArr 2HBr(g),K_(eq)=64` [H=1,Br=80]A. 64 gB. 32.4 gC. 80 gD. 80.4 g
Answer» Correct Answer - B

Discussion

169.	Consider the pertial decomposition of `A` as `2A(g)hArr2B(g)+C(g)` At equilibrium `700`mL gaseous mixture contains `100`mL of gas C at `10` atm and `300K` what is the value of `K_(p)` for the reaction ?A. `(40)/(7)`B. `(1)/(28)`C. `(10)/(28)`D. `(28)/(10)`
Answer» Correct Answer - C

Discussion

170.	An equilibrium mixture at 700 K of 0.05M `N_(2)(g) and 0.2 M NH_(3)(g)` is present in a container .Now if this equilibrium is disturbed by adding `N_(@)` (g) so that its concentration becomes 0.15M just after addition then which of the following graph represents the above situation more appropriately:A. B. C. D.
Answer» Correct Answer - a

Discussion

171.	The equilibrium constants for the reaction `Br_(2)hArr 2Br` at 500 K and 700 K are `1xx10^(-10)` and `1xx10^(-5)` respectively. The reaction is:A. EndothermicB. ExothermicC. FastD. Slow
Answer» Correct Answer - A

Discussion

172.	Assertion (A) : The equilibrium constant is fixed and characteristic for any given chemical reaction at a specified temperature. Reason (R) : The composition of the final equilibrium mixture at a particular temperature depends upon the starting amount of reactants.A. If both (A) and (R) are correct, and (R) is the correct explanation for (A)B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A)C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct.
Answer» Correct Answer - A Equilibrium constant depends only upon temperature.

Discussion

173.	At `700 K`, hydrogen and bromine react to form hydrogen bromine. The value of equilibrium constant for this reaction is `5xx10^(8)`. Calculate the amount of the `H_(2), Br_(2)` and `HBr` at equilibrium if a mixture of `0.6 mol` of `H_(2)` and `0.2 mol` of `Br_(2)` is heated to `700K`.
Answer» Correct Answer - A::B::D `H_(2)+Br_(2) hArr 2HBr` `{:("Initial",0.6,0.2,0),("At equilibrium",(0.6-x)/V,(0.2-x)/V,(2x)/V "mole"),("mol. conc.",,,):}` `K=(((2x)/v)^(2))/(((0.6-x)/v)((0.2-x)/v))` `(4x^(2))/((0.6-x)(0.2-x))=5xx10^(8)` `(x^(2)-0.8x-0.12)xx5xx10^(8)=4x^(2)` Neglecting `4x^(2)` in comparison to `5xx10^(8)x^(2)`, we get `x^(2)-0.8-0.12=0` `x=(0.8 +- sqrt((0.8)^(2)-4xx0.12))/2` `=(0.8 +- 0.693)/(2)` `=0.7465` and `0.0535` `x=0.7465` is impossible hence, `x=0.0535` `[H_(2)]_(eq)=0.6-0.0535=0.5465` mol `[Br_(2)]_(eq)=0.2-0.0535=0.1465` mol `[HBr]_(eq)=2xx0.0535=0.1070` mol

Discussion

174.	Assertion (A) : `K_(p)` is always greater than `K_(c)`. Reason (R) : The effect of pressure is greater on the rate of reaction than the effect of concentration.A. If both (A) and (R) are correct, and (R) is the correct explanation for (A)B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A)C. If (A) is correct, but (R) is incorrectD. If both (A) and (R) are incorrect.
Answer» Correct Answer - D `K_(p)` can be greater, less or equal to `K_(c )`

Discussion

175.	Assertion (A) : A catalyst does not influences the values of equilibrium constant Reason (R) : Catalyst influences the rate of both forward and backward reactions equally.A. If both (A) and (R) are correct, and (R) is the correct explanation for (A)B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A)C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct.
Answer» Correct Answer - A Catalyst only increases the rate of reaction.

Discussion

176.	At some temperature and under a pressure of `4` atm, `PCl_(5)` is `10%` dissociated. Calculated the pressure at which `PCl_(5)` will be `20%` dissociated temperature remaining same.
Answer» Correct Answer - D When `PCl_(5)` is `10%` dissociated `alpha=0.1` `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",1,,0,,0),("At equilibrium",-0.1,,0.1,,0.1),(,=0.9 "mol",,,,):}` Total number of moles at equilibrium `=0.9+0.1+0.1` `=1.1` "mol" `p_(PCl_(5))=0.9/1.1xx4 "atm"=(0.9xx4)/1.1` `p_(PCl_(3))=0.1/1.1xx4 "atm" =0.4/1.1` `p_(Cl_(2))=0.1/1.1xx4 "atm"=0.4/1.1` `K_(p)=(p_(PCl_(3))xxp_(Cl_(2)))/p_(PCl_(5))=(0.4/1.1xx0.4/1.1)/((0.9xx4)/1.1)=0.0404` Second case: When `PCl_(5)` is `20%` dissociated: Suppose total pressure =P atm, then `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",1 "mol",,0,,0),("At equilibrium",1-0.2,,0.2,,0.2),(,=0.8,,,,):}` Total number of "moles" `=0.8+0.2+0.2=1.2 "mol"` `p_(PCl_(3))=0.8/1.2xxP "atm"` `p_(PCl_(3))=0.2/1.2xxP "atm"` `p_(Cl_(2))=0.2/1.2xx P "atm"` `K_(p)=((0.2P)/1.2xx(0.2P)/1.2)/((0.8 P)/1.2)=0.2/1.2xx0.2/0.8xxP` `(=0.404 "calculate above")` `P=0.97 "atm"`

Discussion

177.	The degree of dissociation of `PCl_(5)` at `1` atm pressure is `0.2`. Calculate the pressure at which `PCl_(5)` is dissociated to `50%`?
Answer» Correct Answer - A::B::D `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("moles before",1,,0,,0),("dissociation",,,,,),("moles after",1-alpha,,alpha,,alpha),("dissociation",,,,,):}` Given `=0.2` at `1` atm temperature `:. K_(p)=(n_(PCl_(3))xxn_(Cl_(2)))/n_(PCl_(5))xx[P/(Sigman)]^(Deltan)` `=(alpha.alpha)/((1-alpha))[P/(1+alpha)]=(Palpha^(2))/(1-alpha^(2))=(1xx(0.2)^(2))/(1-(0.2)^(2))` `K_(p)=0.0416 "atm"` Again when `alpha` is desired at `0.5, K_(p)` remains constant and thus, `K_(SP)=(Palpha^(2))/(1-alpha^(2))` `0.0416=(Pxx(0.5)^(2))/(1-(0.5)^(2)), :. P=0.1248 "atm"`

Discussion

178.	A flask is initially filled with pure `N_2O_3(g)` having pressure 2 bar and following equilibria are established. `N_2O_3(g)hArrNO_2(g)+NO(g)" "K_(P_1)=2.5bar` `2NO_2(g)hArrN_2O_4(g)" "K_(P_2)=?` If at equilibrium partial pressure of NO(g) was found to be 1.5 bar,then:A. Equilibrium parial pressure of `N_2O_3(g)` is 0.5 bar.B. Equilibrium partial pressure of `NO_2(g)` is 0.83 bar.C. Equilibrium parital pressure of `N_2O_4` is 0.33 bar.D. Value of `K_(P_2)` is 0.48 bar
Answer» Correct Answer - A::B::C::D

Discussion

179.	Consider folllowing reaction at 300 K : `N_(2)O_(5)(g)hArrN_(2)O_(5)(g)+O_(2)(g)` `K_(c)=5` `N_(2)O_(5)(g)hArrN_(2)O(g)+O_(2)(g)`. For second reaction, if energy of activation of forward and backward reaction are respectively 85 and `42KJ//"mole"`, then at 400K, `K_(c)` for the second reaction is :A. 10B. `lt10`C. `gt10`D. can be greater or less than 10
Answer» Correct Answer - C

Discussion

180.	Consider folllowing reaction at 300 K : `N_(2)O_(5)(g)hArrN_(2)O_(5)(g)+O_(2)(g)` `K_(c)=5` `N_(2)O_(5)(g)hArrN_(2)O(g)+O_(2)(g)`. For second reaction, if energy of activation of forward and backward reaction are respectively 85 and `42KJ//"mole"`, then at 400K, `K_(c)` for the second reaction is :A. 10B. `lt10`C. `gt10`D. can be greater or less than 10
Answer» Correct Answer - C

Discussion

181.	When `N_(2)O_(5)(g)` is heated it dissociates to give `N_(2)O_(3)` and `O_(2) . K_(c)` for `N_(2)O_(5)rarr N_(2)O_(3)+O_(2)` is `7.75` and `K_(c)` for `N_(2)O_(3)rarr N_(2)O+O_(2)` is `4.0 mol L^(-1)`. (both `K_(c)` are at same temperature) `4` mol `N_(2)O_(5)` in `1.0L` vessel is kept at a certain temperature. the concentration of `O_(2)` was `4.5 mol L^(-1)`. Find the concentration of `N_(2)O_(5), N_(2)O_(3)`, and `N_(2)O` at equilibrium.
Answer» `I[{:(N_(2)O_(5),hArr,N_(2)O_(3),+,O_(2),,K_(c )=7.75),(4-x,,x,,x,):}` `II[{:(N_(2)O_(3),hArr,N_(2)O,+,O_(2),,K_(c )=4),(x,,0,,x,),(x-y,,y,,x+y,):}` `:. x+y =[O_(2)]=4.5` From `I, K_(c )=([N_(2)O_(3)][O_(2)])/([N_(2)O_(5)])` `7.75=([N_(2)O_(3)][O_(2)])/([N_(2)O_(5)])=((x-y)(4.5))/(4-x)` From `II, K_(c )=([N_(2)O][O_(2)])/([N_(2)O_(3)]) 4=(yxx(4.5))/(x-y)` Solve to find `x` and `y [x=3.06, y=1.44]`

Discussion

182.	The endothermic reaction `(M+NhArrP)`is allowed to attain an equilibrium at `25^(@)` . Formation of P can be increased byA. Raising temperatureB. Lowering temperatureC. Keeping temperature constantD. Decreasing the concentration of M and N
Answer» Correct Answer - A In endothermic reaction, rate of forward reaction can be increased by raising temperature.

Discussion

183.	When `N_(2)O_(5)` is heated at certain temperature, it dissociates as `N_(2)O_(5)(g)hArrN_(2)O_(3)(g)+O_(2)(g),K_(c)=2.5` At the same time `N_(2)O_(3)` also decomposes as : `N_(2)O_(3)(g)hArrN_(2)O(g)+O_(2)(g).` "If initially" `4.0` moles of `N_(2)O_(5)` "are taken in" `1.0` litre flask and alowed to dissociate. Concentration of `O_(2)` at equilibrium is `2.5` M. "Equilibrium concentratio of " `N_(2)O_(5)` is :A. `1.0` MB. `1.5`MC. `2.166 M`D. `1.846` M
Answer» Correct Answer - D

Discussion

184.	For `A(g) hArr 2B(g)`, equilibrium constant at total equilibrium pressure `p_(1)` is `K_(p1)& ` for `C(g)hArrD(g)+E(g)`. equilibrium constant at total equilibrium pressure `p_(2)` is `K_(p2)`. If degree of dissciation of `A & C` are same, then the ratio `K_(p1) //K_(p2)` , if `2p_(1)=p_(2)`, is `:`A. 2B. `1//8`C. `1//2`D. 8
Answer» Correct Answer - 1 `K_(p1)=(4alpha.^(2)P_(1))/(1-alpha^(2)),K_(p^(2))=(alpha^(2)P_(2))/(1-alpha^(2))rArr" "(K_(p_(1)))/(K_(p_(2)))=(4P_(1))/(P_(2))=4xx(1)/(2)=2`

Discussion

185.	Two solid compounds X and Y dissociates at a certain temperature as follows `X(s) hArr A(g)+2B(g),K_(p1)=9xx10^(-3)atm^(3)` `Y(s) hArr 2B(g)+C(g),K_(p2)=4.5xx10^(-3)atm^(3)` The total pressure of gases over a mixture of X and Y is `:`A. `4.5`atmB. `0.45` atmC. `0.6` atmD. None of these
Answer» Correct Answer - B

Discussion

186.	Two solid compounds X and Y dissociates at a certain temperature as follows `X(s) hArr A(g)+2B(g),K_(p1)=9xx10^(-3)atm^(3)` `Y(s) hArr 2B(g)+C(g),K_(p2)=4.5xx10^(-3)atm^(3)` The total pressure of gases over a mixture of X and Y is `:`A. `4.5 atm`B. `0.45 atm`C. `0.6 atm`D. none of these
Answer» Correct Answer - B

Discussion

187.	For the reactions `2A(g)+2B(g)=3C(g)` at a certain temperature, K is `2.5xx10^(-2)`.For which conditions will the reaction proceed to the right at the same temperature ?A. `{:([A]","M,[B]","M,[C]","M),(0.10,0.10,0.10):}`B. `{:([A]","M,[B]","M,[C]","M),(1.0,1.0,1.0):}`C. `{:([A]","M,[B]","M,[C]","M),(1.0,0.10,0.10):}`D. `{:([A]","M,[B]","M,[C]","M),(1.0,1.0,0.10):}`
Answer» Correct Answer - D

Discussion

188.	A solid A dissociates to give B and C gases as shown, `A(s) hArr 2B(g)+3C(g)` At equilibrium some B(g) is introduced keeping volume constant so that pressure of B of new equilibrium becomes equal to `8/(5sqrt2)` times original total pressure calculate ratio of initial equilibrium pressure of C to that of its final pressureA. `1:2`B. `2:1`C. `8:sqrt2`D. `sqrt2:8`
Answer» Correct Answer - D

Discussion

189.	For reaction,` 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)` which statements are correct? (P) `K_(c)=[SO_(2)][O_(2)]//[SO_(3)]` (Q) Addition of `O_(2)(g)` to the system at contant temperature and valume would decrease the value of `K_(c)`.A. P onlyB. Q onlyC. Both P and QD. Neither P nor Q
Answer» Correct Answer - D

Discussion

190.	In a closed system: `A(s) hArr 2B(g)+3C(g)`, if the partial pressure of C is doubled at equillibrium, then partial pressure of B will be :A. two times the original valueB. one-half of its original valueC. `(1)/(2sqrt2)` times the original valueD. `2sqrt2` times the original value
Answer» Correct Answer - C

Discussion

191.	At a certain temperature , the following reactions have the equilibrium constants as shown below:`S(s)+O_(2)(g)hArrSO_(2)(g),K_(c)=5xx10^(52)` `2S(s)+3O_(2)(g)hArr2SO_(3)(g),K_(c)=5xx10^(29)` what is the equilibrium constant `K_(c)`for the reaction at tahea same temperature? `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)`A. 2`.5xx10^(76)`B. `4xx10^(23)`C. `4xx10^(-72)`D. None of these
Answer» Correct Answer - C

Discussion

192.	The equilibrium constant for the reaction, `SO_3(g)hArr SO_2(g)+1/2O_2(g)` is `K_C=4.9xx10^(-2)` . The value of `K_C` for the reaction `2SO_2(g) + O_2 hArr 2SO_(3) (g)` will be :A. 416B. `2.40xx10^(-3)`C. `9.8xx10^(-2)`D. `4.9xx10^(-2)`
Answer» Correct Answer - A

Discussion

193.	`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.
Answer» Correct Answer - A::C::D Pressure of `NH_(3)` at `27^(@)C=15 "atm"` Pressure of `NH_(3)` at `347^(@)C=P "atm"` Using relation `P_(1)/T_(1)=P_(2)/T_(2)` `P/620=15/300` `P=31 "atm"` Let a mol of ammonia be present. Total pressure at equilibrium`=50` atm. `{:(,2NH_(3)(g),hArr,N_(2)(g),+,3H_(2)(g)),("At equilibrium",(a-2x),,x,,3x),("Total moles" =,a-2x+x+3x=a+2x,,,,):}` `("Initial number of moles")/("moles at equilibrium")=("Initial pressure")/("Equilibrium pressure")` `a/((a+2x))=31/50` `x=16/62 a` Amount of ammonia decomposed `=2x=2xx19/62a=19/31a` `%` of ammonia decomposed `=(19xxa)/(31xxa)xx100=61.3`

Discussion

194.	`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.A. `65%`B. `61.3%`C. `62.5%`D. `64%`
Answer» Correct Answer - b

Discussion

195.	A schematic plot of In `K_(eq)` versus inverse o ftemperature for a reaction is shown below the reaction must be:A. ExothermicB. EndothermicC. One with negligible enthalpy changeD. Highly spontanceous at ordinary temperature
Answer» Correct Answer - a

Discussion

196.	A reversible reaction is one whichA. proceeds in one directionB. proceeds in both directionsC. proceeds spontaneouslyD. all the statements are wrong
Answer» Correct Answer - b

Discussion

197.	In alkaline solution, the following equilibria exist a. `S^(2-)+S rarr S_(2)^(2-)` equilibrium constant `K_(1)` b. `S_(2)^(2-)+S rarr S_(3)^(2-)` equilibrium constant `K_(2)` `K_(1)` and `K_(2)` have values `12` and `11`, respectively. `S_(3)^(2-) rarr S^(2-)+2S`. What is equilibrium constant for the reactionA. `132`B. `7.58xx10^(-3)`C. `1.09`D. `0.918`
Answer» Correct Answer - B i. `S^(2-)+S rarr S_(2)^(2-) …k_(1)` ii. `S_(2)^(2-)+S rarr S_(3)^(2-) …k_(2)` The equation constant for the reaction, iii. `S_(3)^(2-) rarr S^(2-)+2S`, Since eq. (iii) is obtained by reversing and adding equations (i) and (ii).

Discussion

198.	For an equilibrium reaction, the rate constants for the forward and the backward reaction are `2.38xx10^(-4)` and `8.15xx10^(-5)`, respectively. Calculate the equilibrium constant for the reaction.
Answer» Correct Answer - B Equilibrium constant `K=k_(f)/k_(b)=(2.38xx10^(-4))/(8.15xx10^(-5))=2.92`

Discussion

199.	Given the equilibrium constants `HgCl^(o+)+Cl^(ɵ) rarr HgCl_(2), K_(1)=3xx10^(6)` `HgCl_(2)+Cl^(ɵ) rarr HgCl_(3)^(ɵ), K_(2)=8.9` The equilibrium constant for the disproportionation equilibrium `2HgCl_(2) rarr HgCl^(o+)+HgCl_(3)^(ɵ)` isA. `-3.3xx10^(5)`B. `3xx10^(-5)`C. `3.3xx10^(5)`D. `3xx10^(-6)`
Answer» Correct Answer - D a. `HgCl^(o+)+Cl^(ɵ) rarr HgCl_(2) … K_(1)` b. `HgCl_(2)+Cl^(ɵ) rarr HgCl_(3) …K_(2)` The eq. constant (k) for the reaction, `2HgCl_(2) rarr HgCl^(o+)+HgCl_(3)^(ɵ)` Can be obtained by reversing equation (a) and adding to equation (b). `K=1/K_(1)xxK_(2)=K_(2)/K_(1)=8.9/(3xx10^(6))~~3xx10^(-6)`

Discussion

200.	In a reaction between `H_(2)` and `I_(2)` at a certain temperature, the amounts of `H_(2), I_(2)` and HI at equilibrium were found to be `0.45` mol, `0.39` mol, and `3.0` mol respectively. Calculate the equilibrium constant for the reaction at the given temperature.
Answer» Correct Answer - A::B::C The reaction between `H_(2)` and `I_(2)` may be represented as `H_(2)+I_(2) hArr 2HI` Amounts of `H_(2), I_(2)`, and `HI` at equilibrium are given to be `H_(2)=0.45 "mol", I_(2)=0.39 "mol"` and `HI=3.0 "mol"` Suppose the volume of the vessel (i.e., reaction mixture) `=V L` Then, the molar concentration at equilibrium will be `[H_(2)]=0.45/V, [I_(2)]=0.39/V`, and `[HI]=3.0/V "mol" L^(-1)` Applying the law of chemical equilibrium to the above reaction `K_(c)=([HI]^(2))/([H_(2)][I_(2)])=((3.0//V^(2)))/((0.45//V)(0.39//V))` `=((3.0)^(2))/(0.45xx0.39)=51.28`

Discussion

Explore topic-wise InterviewSolutions in .

The equilibrium composition for the reaction is `{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.20,,0.10,,0.40 mol L^(-1)):}` What will be the equilibrium concentration of `PCl_(5)` on adding `0.10 mol` of `Cl_(2)` at the same temperature?

Write the expression for equilibrium constant for the following reactions. If the concentrations are expressed in `mol L^(-1)`, give the units in each case. a.

The equilibrium constant expression for a gas reaction is : `K_(c) = ([NH_(3)]^4[O_(2)]^5)/([NO]^(4)[H_(2)O]^(6))` Write the balanced chemical equation corresponding to this expression.

If `0.5 `mole `H_(2)` is reacted with 0.5 mole `I_(2)` in a ten `-` litre container at `444^(@)C` and at same temperature value of equilibrium constant `K_(C)` is 49, the ratio of `[Hl]` and `[l_(2)]` will be `:`A. 7B. `(1)/(7)`C. `sqrt((1)/(7))`D. 49

At `700 K`, the equilibrium constant `K_(p)` for the reaction `2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)` is `1.80xx10^(-3) kPa`. What is the numerical value of `K_(c )` in moles per litre for this reaction at the same temperature?

For the reaction `CO(g)+Cl_(2)(g)hArrCOCl_(2)(g)`the value of `(K_(c)/(K_(P)))` is equal to :A. `sqrt(RT)`B. RTC. `(1)/(RT)`D. `1.0`

For the reaction : `A(g)hArr2B(g),K_(p)=4.0` bar. The equilibrium pressure pf A(g) is 1.0 bar. Now the equilibrium mixture is compressed reversibly and isothermally such that the final total pressure of system becomes 8 bar. The partial pressure of A(g) (in bar)at this new equilibrium is :

The figure shows the change in concentration of species A and B as a function of time. The equilibrium constant `K_(c)`for the reaction` A(g)hArr2B(g)` is : A. `K_(c)gt1`B. `Klt1`C. `K=1`D. data insufficient

Consider the following two equilibrium established together in a closed container `A(s)hArr2B(g)+3C(g) , K_(P_(1)` `A(s)hArr3D(g) , K_(P_(2)` Starting with only `A(s)`, molar ratio of `B(g)` & `D(g)` at equilibrium is found to be in a ratio `1:6` determine `(K_(P_(2)))/(8K_(P_(1)))`

For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^(@)C` and `527^(@)C` are, respectively, `6.1xx10^(-12)` and `1.0xx10^(-7)`. What is the nature of the reaction?

The equilibrium constants for the reaction, `A_2hArr2A` A at `500K and 700K` are `1xx10^(-10)` and `1xx10^(-5)`. The given reaction isA. exothermicB. slowC. endothermicD. fast

The gas `A_(2)` in the left flask allowed to react with gas `B_(2)` present in right flask as `A_(2)(g)+B_(2)(g)hArr2AB(g),K_(c)=4` at `27^(@)C.` What is the concentration of AB when equilibrium is established ?A. 1.33 MB. 0.66 MC. 0.33 MD. 2.66 M

In order to prepare 25.92 gm of HBr in 20 litre container by following reactions what minimum mass of equimolar mixture of `H_2` and `Br_2` should be taken ? Given : `H_2(g)+Br_2(g)hArr 2HBr(g),K_(eq)=64` [H=1,Br=80]A. 64 gB. 32.4 gC. 80 gD. 80.4 g

Consider the pertial decomposition of `A` as `2A(g)hArr2B(g)+C(g)` At equilibrium `700`mL gaseous mixture contains `100`mL of gas C at `10` atm and `300K` what is the value of `K_(p)` for the reaction ?A. `(40)/(7)`B. `(1)/(28)`C. `(10)/(28)`D. `(28)/(10)`

The equilibrium constants for the reaction `Br_(2)hArr 2Br` at 500 K and 700 K are `1xx10^(-10)` and `1xx10^(-5)` respectively. The reaction is:A. EndothermicB. ExothermicC. FastD. Slow

At `700 K`, hydrogen and bromine react to form hydrogen bromine. The value of equilibrium constant for this reaction is `5xx10^(8)`. Calculate the amount of the `H_(2), Br_(2)` and `HBr` at equilibrium if a mixture of `0.6 mol` of `H_(2)` and `0.2 mol` of `Br_(2)` is heated to `700K`.

At some temperature and under a pressure of `4` atm, `PCl_(5)` is `10%` dissociated. Calculated the pressure at which `PCl_(5)` will be `20%` dissociated temperature remaining same.

The degree of dissociation of `PCl_(5)` at `1` atm pressure is `0.2`. Calculate the pressure at which `PCl_(5)` is dissociated to `50%`?

The endothermic reaction `(M+NhArrP)`is allowed to attain an equilibrium at `25^(@)` . Formation of P can be increased byA. Raising temperatureB. Lowering temperatureC. Keeping temperature constantD. Decreasing the concentration of M and N

Two solid compounds X and Y dissociates at a certain temperature as follows `X(s) hArr A(g)+2B(g),K_(p1)=9xx10^(-3)atm^(3)` `Y(s) hArr 2B(g)+C(g),K_(p2)=4.5xx10^(-3)atm^(3)` The total pressure of gases over a mixture of X and Y is `:`A. `4.5`atmB. `0.45` atmC. `0.6` atmD. None of these

Two solid compounds X and Y dissociates at a certain temperature as follows `X(s) hArr A(g)+2B(g),K_(p1)=9xx10^(-3)atm^(3)` `Y(s) hArr 2B(g)+C(g),K_(p2)=4.5xx10^(-3)atm^(3)` The total pressure of gases over a mixture of X and Y is `:`A. `4.5 atm`B. `0.45 atm`C. `0.6 atm`D. none of these

For reaction,` 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)` which statements are correct? (P) `K_(c)=[SO_(2)][O_(2)]//[SO_(3)]` (Q) Addition of `O_(2)(g)` to the system at contant temperature and valume would decrease the value of `K_(c)`.A. P onlyB. Q onlyC. Both P and QD. Neither P nor Q

In a closed system: `A(s) hArr 2B(g)+3C(g)`, if the partial pressure of C is doubled at equillibrium, then partial pressure of B will be :A. two times the original valueB. one-half of its original valueC. `(1)/(2sqrt2)` times the original valueD. `2sqrt2` times the original value

The equilibrium constant for the reaction, `SO_3(g)hArr SO_2(g)+1/2O_2(g)` is `K_C=4.9xx10^(-2)` . The value of `K_C` for the reaction `2SO_2(g) + O_2 hArr 2SO_(3) (g)` will be :A. 416B. `2.40xx10^(-3)`C. `9.8xx10^(-2)`D. `4.9xx10^(-2)`

`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.

`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.A. `65%`B. `61.3%`C. `62.5%`D. `64%`

A schematic plot of In `K_(eq)` versus inverse o ftemperature for a reaction is shown below the reaction must be:A. ExothermicB. EndothermicC. One with negligible enthalpy changeD. Highly spontanceous at ordinary temperature

A reversible reaction is one whichA. proceeds in one directionB. proceeds in both directionsC. proceeds spontaneouslyD. all the statements are wrong

For an equilibrium reaction, the rate constants for the forward and the backward reaction are `2.38xx10^(-4)` and `8.15xx10^(-5)`, respectively. Calculate the equilibrium constant for the reaction.

In a reaction between `H_(2)` and `I_(2)` at a certain temperature, the amounts of `H_(2), I_(2)` and HI at equilibrium were found to be `0.45` mol, `0.39` mol, and `3.0` mol respectively. Calculate the equilibrium constant for the reaction at the given temperature.