1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 20 InterviewSolution

951.	Consider the following reactions in which all the reactants and the products are in gasous state ` 2 PQ hArr P_(2) + Q_(2) , K_(1) = 2.5 xx 10^(5)` ` PQ + 1/2 R_(2) hArr PQR, K_(2) = 5 xx 10^(-3)` The value of `K_(3)` for the equilibrium ` 1//2 P_(2) + 1//2 Q_(2) + 1//2 R_(2) hArr PQR, ` isA. `2.5 xx 10^(-3)`B. `2.5 xx 10^(3)`C. `1.0 xx 10^(-5)`D. ` 5 xx 10^(3)`
Answer» Correct Answer - C For `2 PQ hArr P_(2) + Q_(2) , K_(1) = ([P_(2)] [Q_(2)])/([PQ]^(2))` For ` PQ + 1/2 R_(2) hArr PQR, ` `K_(2) = ([PQR])/([PQ] [R_(2)]^(1//2))` For the required equilibrium ` 1/2 P_(2) + 1/2 Q_(2) + 1/2 R_(2) hArr PQR` ` K = ([PQR])/([P_(2)]^(1//2) [Q_(2)]^(1//2) [R_(2)]^(1//2))=(K_(2))/(sqrt(K_(1)))` ` = ( 5xx 10^ (3))/(sqrt(2*5 xx 10^(5) ))=(5xx 10^(-3))/(5xx 10^(2))= 10^(-5)`

Discussion

952.	Mercurous chloride , ` Hg_(2)Cl_(2)`, in a saturated solution has the equilibrium called solubility equilibrium . The equilibrium constant for this solubility equilibrium will beA. `[Hg^(+)] [Cl^(-)]`B. `[Hg^(+)]^(2) [ Cl^(-)]^(2)`C. ` [ Hg_(2)^(2).^(+)] [ Cl^(-)]^(2)`D. `2 [Hg^(+)] xx2 [Cl^(-)]`
Answer» Correct Answer - C The equilibrium reaction ` Hg _(2)Cl_(2) (s) hArr Hg_(2)^(2+) (aq) + 2 Cl^(-) (aq) ` Hence, `K= [ Hg_(2)^(2+) ] [Cl^(-)]^(2`

Discussion

953.	What will be the concentration of iodine in solution?A. `1.1M`B. `0.011M`C. `0.0011M`D. `0.11M`
Answer» Correct Answer - C Molarity `(M)=(n_(I_2))/(V_L)` `n_(I_2)=(Mass_(I_2))/(Molar mass_(I_2))=(0.070g)/(254g mol^(-1))` `V_L=250 mLxx(IL)/(1000mL)` `:. M=(0.070g)/(254 g mol^(-1)xx250mL)xx((1000mL)/(1L))` `=0.0011 mol L^(-1)`

Discussion

954.	Which of these is least likely to act as Lewis base?A. `F^(-)`B. `BF_(3)`C. `PF_(3)`D. `CO`
Answer» Correct Answer - B `BF_(3)` cannot act as a Lewis base because its central atom, boron, is electron defient. It is a typical I Lewis acid. If possible, it can donate the lone pair only through its F atoms.

Discussion

955.	In 1 L saturated solution of AgCl `[K_(sp)(AgCl)=1.6xx10^(-10)], 0.1 " mol of " CuCl [ K_(sp)(CuCl)=1.0xx10^(-6)]` is added. The resultant concentration of `Ag^(+)` in the solution is `1.6xx10^(-x)`. The value of` "x"` is
Answer» Correct Answer - 7 As `K_(sp)` of CuCl is much greater than that of AgCl, `Cl^(-)` ions present in the mixture solution are mainly from CuCl and equal to `sqrt(10^(-6))`, i.e., `10^(-3) M`. If `[Ag^(+)]= y "mol" L^(-1)` , then `K_(sp) (AgCl) = [Ag^(+)][Cl^(-)]` i.e., `1.6xx10^(-10) = y xx 10^(-3) or y = 1.6 xx 10^(-7)` . Hence , x = 7 .

Discussion

956.	The solubility product constants of `Ag_(2)CrO_(4)` and AgBr are `1.1xx10^(-12) and 5.0xx10^(-13)` respectively. Calculate the ratio of molarities of their saturated solutions.
Answer» Correct Answer - `(S(Ag_(2)CrO_(4)) )/(S(AgBr))=(6.5xx10^(-5)M)/(7.07xx10^(-7))=91.9` Calculate their solubilities separately and then calculate the ratio.

Discussion

957.	Which of the following is not a Lewis base ?A. `CH_(4)`B. `CN^(-)`C. `ROH`D. `NH_(3)`
Answer» Correct Answer - A Lewis base is a species that has at least one nonbonding valence shell electron pair that can be donated to form a coordinate covalent bond. `CH_(4)` is not a Lewis base because no atom in`CH_(4)` can donate a pair of electrons.

Discussion

958.	The `K_(sp)` of `Ag_(2)CrO_(4),AgCl,AgBr` and AgI are respectively, `1.1xx10^(-12),1.8xx10^(-10),5.0xx10^(-13),8.3xx10^(-17)`. Which one of the following salts will precipitate last if `AgNO_(3)` solution is added to the solution containing equal moles of NaCl,NaBr,NaI and `Na_(2)CrO_(4)` ?A. `AgBr`B. `Ag_(2)CrO_(4)`C. `Agl`D. `AgCI.`
Answer» Correct Answer - B the salt with maxium `K_(sp)` value will be precipitated last. Therefore `AgCI (K_(sp) = 1.8 xx 10^(-10))` will be precipitated last .

Discussion

959.	Which of the following is a Lewis acid ?A. `BF_(3)`B. `SnCI_(4)`C. `CO_(2)`D. All of these
Answer» Correct Answer - D Molecules having central atoms with incomplete octets are Lewis acids. Examples are `BF_(3), BeCI_(2)`, and `AICI_(3)`. Moecules having central atoms capable of octet expansion are Lewis acids, e.g., `SnCI_(4)`, `PF_(5)`, and `PtCI_(4)`. Molecules containing `pi` bonds betweeen atoms having signifacant electronagativity difference can act as Lewis acids, e.g., `CO_(2)`.

Discussion

960.	The `K_(sp)` of `Ag_(2)CrO_(4)`, AgCl, A gBr and AgI are respectively `1.1xx10^(-12), 1.8xx10^(-6), 5.0xx10^(-13) and 8.3xx10^(-17)`. Which one of the following salts will precipitate last if `AgNO_(3)`solution is added to the solution containing equal moles of NaCl, NaBr, NaI and `Na_(2)CrO_(4)`?A. AgBrB. `Ag_(2)CrO_(4)`C. AgID. AgCl
Answer» Correct Answer - B `Ag_(2)CrO_(4)hArr2Ag+CrO_(4)^(2-), K_(sp)=(2s)^(2)xxs=4s^(3), s=((K_(sp))/(4))^(1//3)=((1.1xx10^(-12))/(4))^(1//3)` `=0.65xx10^(-4)` `AgCl hArr Ag^(+)+Cl^(-),K_(sp)=sxxs=s^(2)`, `s=sqrt(K_(sp))=sqrt(1.8xx10^(-10))=1.34xx10^(-5)` `AgBr hArr Ag^(+)+Br^(-), K_(sp)=sxxs=s^(2)`, `s=sqrt(K_(sp))=sqrt(5.0xx10^(-13))=0.71xx10^(-6)` `AgI hArr Ag^(+)+I, K_(sp)=sxxs=s^(2)`, `s=sqrt(K_(sp))=sqrt(8.3xx10^(-17))=0.9xx10^(-8)` As solubility of `Ag_(2)CrO_(4)` is highest, it will be precipitated last of all.

Discussion

961.	The solubility product `(K_(sp))` of the following compound s are given at `25^(@)C` `{:("Compound " ,K_(sp)),(AgCl,1.1xx10^(-10)),(Agl,1.0xx10^(-16)),(PbCrO_(4),4.0xx10^(-10)),(Ag_(2)CO_(3),8.0xx10^(-12)),(,):}` The most soluble and least soluble compound are respectivelyA. `AgCl and PbCrO_(4)`B. `AgI and Ag_(2)CO_(3)`C. `AgCl and Ag_(2)CO_(3)`D. `Ag_(2)CO_(3) and AgI`
Answer» Correct Answer - D The 1 st three salts are of the type AB for which `S=sqrt(K_(sp))`, i.e., `10^(-5) M (AgCl), 10^(-8) M (AgI), 2xx10^(-7)M (PbCrO_(4))`. Last salt is of the type `A_(2)B` for which `S=(K_(sp)//4)^(1//3)=(2xx10^(-12))^(1//3)~=10^(-4)` M. Thus, most soluble is `Ag_(2)CO_(3)` and least soluble is AgI.

Discussion

962.	`015` mole of CO taken in `25` l flask is maintained at 750 K along with catalyst so that the following reaction can take place : `CO (g) + 2 H_(2) (g) hArr CH_(3)OH (g)` Hydrogen is introduced until the total pressure of the system is `85` atmosphere at equilibrium and `008` mole of methanol is formed . Calculate (i) `K_(p) and K_(c)` and (ii) the final pressure if the same amount of CO and `H_(2)` as before are used but with no catalyst so that the reaction does not take place.
Answer» ` {:((i),CO,+,2H_(2),hArr,CH_(3)OH) ,("Intialy :",015 "mole",,,,) ,("At eqm:",015 - 008 "mole",,,,), (,=0017 "mole",,,,):}` Total volume, `V= 25 L` Total pressure ` P+ 85 "atm", T= 750 "K" ` ` " Applying " PV= nRT,` we get ` 85 xx 25 = n xx 0* 0821 * 750 or n = 0* 345 "mole" ` ` :. "No. of moles of " H_(2) " at equilibrium " = 0* 345 - ( 0* 017 + 008 ) = 0 248 " mol "` ` P_(CO) = (0* 017 )/(0345 ) xx 85 "atm" = 0* 42 "atm" ` ` p_(H_(2)) = ( 0248 )/(0 345) xx 85 "atm " = 6 11 "atm "` ` P _(CH_(3)OH) = (08)/(0345) xx 85 "atm" = 1 97 "atm" ` ` K_(p) = (P_(CH_(3)OH))/(P_(CO) xx P_(H_(2)))= (197 )/(042 xx (6* 11)^(2) )= 0* 1256 ` ` K_(c) = ([CH_(3) OH])/([CO][H_(2)]^(2))= (0* 08 //25)/((0 017 //25 )(0248 //25)^(2)) = 478 2 ` n(ii) No. of moles of `H_(2) " taken intially " = 0* 248 + 2 xx 0* 08 = 0* 308 ` No. of moles of CO taken intially `= 015` Applying PV = nRt , ` P xx 25 = 0* 458 xx 0* 0821 xx 750 or P = 11* 28 "atm " ` .

Discussion

963.	Give the conjugate bases of (i) `H_(2)O (ii) HNO_(3) (iii) NH_(4)^(+) (Iv) HCO_(3)^(2-).`
Answer» `(i) OH^(-)" "(ii) NO_(3)^(-)" "(iii)NH_(3)" "(iv)CO_(3)^(-2)`

Discussion

964.	Which concept can explain the acitic character of `CO_(2)` ?
Answer» Correct Answer - Lewis concept.

Discussion

965.	According to the law of mass action, the rate of an elementary reaction is directly proportional to the________of the reactants.A. mole fractionsB. molalitiesC. normalitiesD. molarities
Answer» Correct Answer - D According to law of mass action, the rate of an elementary reaction is directly proportional to the active mass of the reactants. For components of ideal solutions, the activity of each component is taken to be the ratio of its molar concentration to a standard concentration of `1M`.

Discussion

966.	`K_(rho)` for the following reaction will be equal to `3Fe(s)+4H_(2)O(g)hArrFe_(3)O_(4)(s)+4H_(2)(g)`A. `(P_(H_(2)))^(4)(P_(Fe_(3)O_(4)))`B. `(P_(H_(2)))/(P_(H_(2O)))`C. `((P_(H_(2)))^(4))/((P_(H_(2O)))^(4))D. `((P_(H_(2)))xxP_(Fe_(3)O_(4)))/(P_(Fe))`
Answer» Correct Answer - C

Discussion

967.	Calculate the pH of the following resultant mixtures : (a) 10 mL of 0.2 M `Ca(OH)_(2)` + 25 mL of 0.1 M HCl (b) 10 mL of 0.01 M `H_(2)SO_(4)` + 10 mL of 0.01 M `Ca(OH)_(2)` (c) 10 mL of 0.1 M `H_(2)SO_(4)` + 10mL of 0.1 M KOH.
Answer» (a) mL of 0.2 M `Ca(OH)_(2)=10xx0.2` millimoles = 2 millimoles of `Ca(OH)_(2)` 25mL of 0.1 M HCl `= 25xx0.1` millimoles = 2.5 millimoles of HCl `Ca(OH)_(2)+2HCl rarr CaCl_(2)+2 H_(2)O` 1 millimole of `Ca(OH)_(2) + 2 HCl rarr CaCl_(2) + 2H_(2)O` 1 millimole of `Ca(OH)_(2)` reacts with 2 millimoles of HCl `:.` 2.5 millimoles of HCl will react with 1.25 millimoles of `Ca(OH)_(2)` `:. Ca(OH)_(2) ` left `= 2-1.25 = 0.75` millimoles (HCl is the limiting reactant) Total volume of the solution = 10 + 25 mL = 35 mL `:.` Molarity of `Ca(OH)_(2)` in the mixture solution `=(0.75)/(35) M = 0.0214M` `:. [OH^(-)]=2xx0.0214 M = 0.0428 M = 4.28 xx 10^(-2)` `pOH = - log (4.28xx10^(-2))=2-0.6314=1.39=686~=1.37` `:. pH = 14 - 1.37 = 12.63` (b) 10 mL of 0.01 M `H_(2)SO_(4) = 10xx0.01` millimole = 0.1 millimole 10 mL of 0.01 M `Ca(OH)_(2) = 10 xx 0.01` millimole = 0.1 millimole `Ca(OH)_(2) + H_(2)SO_(4) rarr CaSO_(4) + 2H_(2)O` 1 mole of `Ca(OH)_(2) ` reacts with 1 mole of `H_(2)SO_(4)` ltbvrgt `:. ` 0.1 millimole of `Ca(OH)_(2)` will react completely with 0.1 millimole of `H_(2)SO_(4)`. Hence, solution will be neutral with pH = 7.0 (c) 10 mL of 0.1 M `H_(2)SO_(4) = 1 ` millimole 10 mL of 0.1 M KOH = 1 millimole `2KOH + H_(2)SO_(4) rarr K_(2)SO_(4) + 2 H_(2)O` 1 millimole of KOH will react with 0.5 millimole of `H_(2)SO_(4)` `:. H_(2)SO_(4) ` left = 1 - 0.5 = 0.5 millimole Volume of reaction mixture = 10 + 10 = 20 mL `:. ` Molarity of `H_(2)SO_(4)` in the mixture solution `=(0.5)/(20) = 2.5xx106(-12) M` `[H^(+)]=2xx(2.5xx10^(-2))=5xx10^(-2)` `pH = - log (5xx10^(-2))=2-0.699=1.3`

Discussion

968.	Calculate the pH of the resultant mixture: a. `10 mL` of `0.2M Ca(OH)_(2)+25 mL` of `0.1 M HCl` b. `10 mL` of `0.01 M H_(2)SO_(4) + 10 mL` of `0.01 M Ca(OH)_(2)`. c. `10 mL` of `0.1 M H_(2)SO_(4)+ 10 mL` of `0.1 M KOH`.
Answer» Correct Answer - a) `12.6` , b) `7.00` , c) `1.3` (a) Moles of `H_(3)O^(+)=(25xx0.1)/(1000)=.0025` mol Moles of `OH^(-)=(10xx0.2xx2)/(1000)=.0040` mol Thus, excess of `OH^(-)` = .0015 mol `[OH^(-)]=(.0015)/(35xx10^(-3))"mol"//"L"` = .0428 pOH = `-log[OH]` = 1.36 pH = 14 - 1.36 = 12.63 (not matched) (b) Moles of `H_(3)O^(+)=(2xx10xx0.1)/(1000)=.0002` mol Moles of `OH^(-)=(2xx10xx0.1)/(1000)=.0002` mol Since there is neither an excess of `H_(3)O^(+)orOH^(-)`, the solution is neutral. Hence, pH = 7 (c) Moles of `H_(3)O^(+)=(2xx10xx0.1)/(1000)=.002` Moles of `OH^(-)=(10xx0.1)/(1000)=0.001` mol Excess of `H_(3)O^(+)` = .001 mol Thus, `[H_(3)O^(+)]=(.001)/(20xx10^(-3))=(10^(-3))/(20xx10^(-3))=.05` `thereforepH =-log(0.05)` = 1.30

Discussion

969.	20 ml of 0.5 N HCl and 35 ml of 0.1 N NaOH are mixed . The resulting solution willA. be neutralB. be basicC. turn phenolphthalein solution pinkD. turn methyl orange red
Answer» Correct Answer - D 20 ml of 0.5 N HCl contain `20xx0.5` milli eq. = 10 milli eq. of HCl . 15 ml of 0.1 N NaOH contains `35xx0.1` milli eq = 3.5 milli eq of NaOH . Thus, 3.5 milli eq of NaOH will nutralize 3.5 milli eq of HCl. (10-35)= 6.5 milli eq of HCl will be left. Hence, solution will be acidic and it will turn methyl orange red.

Discussion

970.	Determine the pH of the solution that results from the addition of 20.00 mL of 0.01 M Ca `(OH)_(2) ` to 30.00 mL of 0.01 M HClA. `11.30`B. `10.53`C. `2.70`D. `8.35`
Answer» Correct Answer - A `20.0 mL "of" 0.01 M Ca (OH)_(2) "has" Ca(OH)_(2)` `=20xx0.01` millimoles = 0.2 millimole `:. OH^(-)` ions present `=2xx0.2` millimole `=0.4` millimole 30.0 mL of 0.01 M HCl has HCl `=30xx0.01` millimoles = 0.3 millimole `:. H^(+)` ions present = 0.3 millimole 0.3 millimole `H^(+)` will neutralize 0.3 millimoe of `OH^(-)`. Hence, `OH^(-)` ions present after mixing = 0.1 millimole Volume of solution = 20 +30 = 50 mL Hence, molarity of `OH^(-)` ions `=(0.1)/(50)=2xx10^(-3)M` `pOH = - log (2xx10^(-3))=3-0.30=2.70` `:. pH = 14 - 2.70 = 11.30`

Discussion

971.	Which of the following is not a general characteristic of equilibrium involving physical processes ?A. Equilibrium is possible only in a closed system at a given temperature.B. All measurable properties of system remain constant.C. All the physical processes stop at equilibrium.D. The opposing processes occur at the same rate and there is dynamic but stable condition.
Answer» Correct Answer - C All the pgysical processes do not stop at equilibrium.

Discussion

972.	Which of the following is not a general characteristic of equilibrium involving physical processes ?A. Equilibrium is possible only in a closed system at a given temperature.B. The equilibrium is dynamic in nature.C. Measurable properties of the system keep changing.D. Equilibrium can be attained from both sides of the reaction.
Answer» Correct Answer - C Measurable properties of the system become constant at equilibrium.

Discussion

973.	Mention the general characteristics of equilibria involving physical processes.
Answer» (a) For solid ⇌ liquid equilibrium, there is only one temperature at 1 atm at which two phases can co-exist. If there is no exchange of heat with the surroundings, the mass of the two phases remain constant. (b) For liquid ⇌ vapors equilibrium, the vapors pressure is constant at a given temperature. (c) For dissolution of solids in liquids, the solubility is constant at a given temperature. (d) For dissolution of gases in liquids, the concentration of a gas in liquid is proportional to pressure of the gas over the liquid.

973.

Mention the general characteristics of equilibria involving physical processes.

Answer»

(a) For solid ⇌ liquid equilibrium, there is only one temperature at 1 atm at which two phases can co-exist. If there is no exchange of heat with the surroundings, the mass of the two phases remain constant.

(b) For liquid ⇌ vapors equilibrium, the vapors pressure is constant at a given temperature.

(c) For dissolution of solids in liquids, the solubility is constant at a given temperature.

(d) For dissolution of gases in liquids, the concentration of a gas in liquid is proportional to pressure of the gas over the liquid.

Discussion

974.	Which of the following is not a general characteristic of equilibria involving physical processes ?A. Equilibrium is possible only in a closed system at a given temperature.B. All measurable properties of the system remain constant.C. All the physical processes stop at equilibrium.D. The opposing processes occur at the same rate and there is dynamic but stable condition.
Answer» Correct Answer - C All the physical processes like melting of ice and freezing of water etc. do not stop at equilibrium.

Discussion

975.	Which of the following statements is incorrect? (i) In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time. (ii) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate. (iii) On addition of catalyst the equilibrium constant value is not affected. (iv) Equilibrium constant for a reaction with negative ΔH value decreases as the temperature increases.
Answer» (ii) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.

Discussion

976.	Which of the following is not a general characteristic of equilibria involving physical processes?(i) Equilibrium is possible only in a closed system at a given temperature.(ii) All measurable properties of the system remain constant.(iii) All the physical processes stop at equilibrium.(iv) The opposing processes occur at the same rate and there is dynamic but stable condition.
Answer» (iii) All the physical processes stop at equilibrium.

Discussion

977.	PCl5, PCl3 and Cl2 are at equilibrium at 500K in a closed container and their concentrations are 0.8 × 10–3 mol L–1, 1.2 × 10–3 mol L–1 and 1.2 × 10–3 mol L–1 respectively. The value of Kc for the reaction PCl5 (g) ⇄ PCl3 (g) + Cl2 (g) will be (i) 1.8 × 103 mol L–1 (ii) 1.8 × 10–3 (iii) 1.8 × 10–3 L mol–1 (iv) 0.55 × 104
Answer» (ii) 1.8 × 10^–3

Discussion

978.	Which of the following reactions involve homogeneous equilibrium or heterogeneous equilibrium?(i) 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)(ii) C (s) + CO2 (g) ⇌ 2CO (g)(iii) CH3COOC2H5 (aq) + H2O(1)CH3COOH (aq) + C2H5OH (aq)(iv) BaCO3 (s) ⇌ Bao (s) + CO2 (g)
Answer» (i) Homogeneous equilibrium (ii) Heterogeneous equilibrium (iii) Homogeneous equilibrium (iv) Heterogeneous equilibrium

978.

Which of the following reactions involve homogeneous equilibrium or heterogeneous equilibrium?(i) 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)(ii) C (s) + CO2 (g) ⇌ 2CO (g)(iii) CH3COOC2H5 (aq) + H2O(1)CH3COOH (aq) + C2H5OH (aq)(iv) BaCO3 (s) ⇌ Bao (s) + CO2 (g)

Answer»

(i) Homogeneous equilibrium

(ii) Heterogeneous equilibrium

(iii) Homogeneous equilibrium

(iv) Heterogeneous equilibrium

Discussion

979.	Does the number of moles of reaction products increase, decrease or remain same when PCl5(g) ⇌ PCl3(g) + Cl2(g)equilibria is subjected to a decrease in pressure by increasing the volume?
Answer» PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) Here, the forward reaction occurs with increase in number of moles. If the pressure on the system is decreased, the equilibrium will shift in the forward direction which is accompanied by increase in total number of moles. In other words, decrease favours the dissociation of PCl₅ into PCl₃ and Cl₂, increases the dissociation of PCl₅.

979.

Does the number of moles of reaction products increase, decrease or remain same when PCl5(g) ⇌ PCl3(g) + Cl2(g)equilibria is subjected to a decrease in pressure by increasing the volume?

Answer»

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Here, the forward reaction occurs with increase in number of moles. If the pressure on the system is decreased, the equilibrium will shift in the forward direction which is accompanied by increase in total number of moles. In other words, decrease favours the dissociation of PCl₅ into PCl₃ and Cl₂, increases the dissociation of PCl₅.

Discussion

980.	STATEMENT-1: Equlibrium constant does not depend upon concentration of various reactants, presence of catalyst, direction from which equilibrium is reached. STATEMENT-2 : Equlibrium constant is only dependent upon the temperature.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-2B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - B

Discussion

981.	PCl5, PCl3 and Cl2 are at equilibrium at 500K and having concentration 1.59 M PCl3, 1.59 M Cl2 and 1.41 M PCl5. Calculate KC for the reaction, PCl5 ⇌ PCl3 + Cl2.
Answer» For the reaction, PCl₅ ⇌ PCl₃ + Cl₂ Equilibrium constant = K_c = \(\frac{[PCl_3][Cl_2]}{[PCl_5]}\) = \(\frac{(1.59)\times(1.59)}{(1.41)}\) = 1.79

981.

PCl5, PCl3 and Cl2 are at equilibrium at 500K and having concentration 1.59 M PCl3, 1.59 M Cl2 and 1.41 M PCl5. Calculate KC for the reaction, PCl5 ⇌ PCl3 + Cl2.

Answer»

For the reaction, PCl₅ ⇌ PCl₃ + Cl₂

Equilibrium constant = K_c = \(\frac{[PCl_3][Cl_2]}{[PCl_5]}\)

= \(\frac{(1.59)\times(1.59)}{(1.41)}\)

= 1.79

Discussion

982.	`NH_(4)COONH_(4)(s) hArr 2NH_(3)(g)+CO_(2)(g)`. If equilibrium pressure is `3` atm for the above reaction, `K_(p)` will beA. 27B. 4C. 3D. 9
Answer» Correct Answer - B `NH_(4)COONH_(2(g))hArrunderset(2p)(2NH_(3(g)))+underset(p)(CO_(2(g)))` When volume and temperature are constant the number of moles of a gas is proportional to its partial pressure. So, `2p+p=3` `3p=3," ":." "p=1" atm"` `K_(p)=(2p)^(2)xxp=4p^(3)=4xx(1)^(3)=4" atm"^(3)`

Discussion

983.	For which of the following reactions does the equilibrium constant depend upon the units of concentration? `(a) CO(g) + H_(2)O(g) hArr CO_(2)(g) + H_(2)(g)` `(b) COCl_(2) (g) hArr CO(g) + Cl_(2)(g)` `(C) NO(g) hArr 1//2 N_(2) (g) + 1//2O_(2)(g)`
Answer» For reaction (b), value of equilibrium constant depends upon units of concentration because the reactant and product moles is not the same.

Discussion

984.	`PCl_(3), PCl_(3)` and `Cl_(2)` are at eqilibrium at 500 K and above have concentration 1.59 M for `PCl_(3), 1.59 M` for `Cl_(2)` and 1.41 M for `PCl_(5).` Calculate `K_(c)` for the reaction : `PCl_(5) hArr PCl_(3) + Cl_(2)`
Answer» `K_(c) = [[PCl_(3)][Cl_(2)]]/[[PCl_(5)]] = ((1.59 M)xx(1.59M))/((1.41M)) = 1.79`

Discussion

985.	For the reaction : `CO(g) +H_(2)O(g) hArr CO_(2)(g) +H_(2)(g)` the value of `K_(c) = 4.24 at 600 K.` Calculate the equilibrium concentration of `CO_(2), H_(2), CO` and `H_(2)O` at 800 K, if only CO and `H_(2)` are present initially at a concentration of 0.10 M each.
Answer» `{:(,CO(g),+,H_(2)O(g),hArr,CO_(2)(g),+,H_(2)(g)),("Initial molar conc": ,0.1M,,0.1M,,0,,0),("Eqm. molar conc" : ,(0.1-x),, (0.1-x),,xM,,xM):}` `K_(c) =[[CO_(2)][H_(2)]]/[[CO][H_(2)O]]=(x xx x)/((0.1-x)(0.1-x)) = (x^(2))/((0.1-x)^(2))` `(x^(2))/((0.1-x)^(2))=4.24 or (x)/((0.1-x))=(4.24)^(1//2)=2.06` , `x= 2.06(0.1-x) or 3.06x = 0.206 or x = (0.206)/(3.06) = 0.067` Eqm. conc of CO =0.1 -0.067 = 0.033 M Eqm. conc. of `H_(2)O` = 0.1 - 0.067 = 0.033 M Eqm. conc. of `CO_(2)` = 0.067 M Eqm. conc. of `H_(2)O` = 0.067 M`

Discussion

986.	At `817^(@)C, K_(p)` for the reaction between `CO_(2) (g)` and excess hot graphite (s) is 10 atm (a) What are the equilibrium concentrations of the gases at ` 817^(@)C` and a total pressure of 5 atm ? (b) At what total pressure, the gas contains 5 % `CO_(2)` by voume ?
Answer» (a) ` CO_(2) + C(g) hArr 2 CO (g) ` Suppose at equilibrium , pressure of `CO_(p_CO_(2)) = p "atm "` Then pressure of ` CO_(2) (p_(CO_(2)))= 5 - p " atm " ` ` K_(p) (p_(CO)^(2))/(p_(CO_(2))) = (p^(2))/((5 - p))" " = 10 or p^(2) + 10 p - 50 = 0 ` or ` p = ( -b pm sqrt (b^(2) - 4ac))/(2a) = (-10 pm sqrt ( 100- (-200)))/2 = 3* 66" atm " ` Thus , at eqm. ` p _(CO) = 3 * 6 "atm " ` ` p_(CO_(2)) = 5 - 3* 66 = 1* 34 "atm "` Applying ` PV = nRT or n/V = P /(RT) , i.e ., "molar conc." = P/(RT) ` Molar conc. of `CO = (366)/(0 0821 xx ( 817 + 273)) = 0041 " mol"L ^(-1)` Molar conc. of ` CO_(2) = (134)/(00821 xx 1090 ) = 0 015 " mol" L^(-1)` (b) When the gas contains 5 % `CO_(2)` by volume , this means that pressure exerted by `CO_(2) ` is also 5 % of the total pressure . Thus, if P is the total pressure , then at equilibrium , ` p_(CO_(2)) = 005 P and p_(CO) = 0 95 P ` ` K_(P) = (p_(CO)^(2))/p_(CO^(2))= (095 P)^(2)/((0 05 P)) = 10 or 18* 05 P= 10 or P = 0* 554 "atm "`

Discussion

987.	A vessel at 1000 K contains carbon dioxide with a pressure of 0.5 atm. Some of the carbon dioxide is converted to carbon monoxide on addition of graphite. Calculate the value of K it the total pressure at equilibrium is `0*8` atm.
Answer» `CO_(2) (g) + C (s) hArr 2 CO (g)` Suppose decrease in pressure of `CO_(2)` after reaction = p atm Then increase in pressure due to CO= 2 p Pressure of `CO_(2)` at equilibrium `=(0.5 - p)` atm `:." Final total pressure " (0.5- p) + 2p = 05 + p = 0.8 " atm"` (Given) `:. p= 03 " atm. Hence we have "` `p_(CO_2) = 05 - 03 = 02 " atm "` `and " "p_(CO) = 2 xx 03 = 06 " atm" ` `:. K = (P_(cO)^(2))/P_(CO_(2)) = (0.6)^(2)/(02) = 1*8`

Discussion

988.	For the reaction `AB (g) hArr A(g) + B(g), AB ` is 33 % dissociated at a totalpressure of P . Therefore, P is related to `K_(p)`by one of the following optionsA. `P=K_(p)`B. ` P=3 K_(p)`C. ` P=4K_(p)`D. ` P=8 K_(p)`
Answer» Correct Answer - D `{:(,AB (g),hArr,A(g),+,B(g)),("Intial moles:",1,,0,,0),("Moles at eqm:",1-0.33,,0.33,,0.33):}` `0.67` Total moles at eqm. =1.33 `p_(A) =(0.33/1.33)P, p_(B)=(0.33/1.33) P,p_(AB)=(0.67/1.33)P` `K_(p) = (p_(A) xxp_(B))/p_(AB) = (0.33 xx 0.33 xx P)/(0.67 xx 1.33)` `P/K_(p)=8 or P =8 K_(p).`

Discussion

989.	A vessel at 1000 K contains `CO_(2)` with a pressure of 0.5 atm . Some of the `CO_(2)` is converted into CO on the addition of graphite . If the total pressure at equilibrium is `08` atm , the value of K isA. 3 atmB. `03 `atmC. `018 atm`D. `18 ` atm
Answer» Correct Answer - D `{:(,CO_(2)(g),+,C(s),hArr,2CO(g)),(" Intial ",0.5 atm,,,,),("At eqm.",0.5 -p,,,,2 p):}` `:. (0.5 -p) +2 p = 0.8" atm " or p=0.3 " atm "` `K=(p^(2) CO)/p_(CO_(2)) = (0.6)^(2)/((0.5 -0.3)) = (0.6 xx0.6)/0.2 = 1.8 " atm", `

Discussion

990.	` 138 g " of "N_(2)O_(4)` was placed in 1 L reaction vessel in 1 L reaction vessel at 400 K and allowed to attain equilibrium : `N_(2)O_(4) (g) hArr 2 NO(g).` The total pressure at equilibrium was found to be` 9 15` bar. Calculate `K_(c) , K_(p)` and partial pressures at equilibrium.
Answer» `138 "g" N_(2)O_(4) = (138)/92" mol"=015" mol" "` `138 N_(2)O_(4) = (138)/92" mol"=015" mol " "(Molar mass of "N_(2)O_(4) = 92 g " mol"^(-1))` PV= nRT ` :. P xx 1 L = 0* 15 mol xx 0* 083 " bar"L mol^(-1) K^(-1) xx 400 K or P = 498 " bar"` ` {:(,N_(2)O_(4)(g),hArr,2NO_(2)),("Initial pressures",498 "bar",,0),(" At equilibrium",(498-p),,2p):}` ` :. 498 - p+ 2 p = 915 " bar" or p= 417 " bar"` `:. (p_(N_(2)O_(4)))eq = 4* 98 - 417 = 081 "bar", (p_(NO_(2)))eq = 2 xx 417 = 834 "bar"` `K_(p) = p_(NO_(2))^(2)//_(P_(N_(2)O_(4)))=(834)^(2)//)81=8587,` ` K_(p) = K_(c) (RT)^(Delta n) :. 8587 = K_(c) (0083 xx400)^(1) or K_(c)= 2586 = 2*6`

Discussion

991.

13.8 g of N2O4 was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium- N2O4(g) ⇌ 2NO2(g)The total pressure at equilibrium was found to be 9.15 bar. Calculate Kc . Kp and partial pressure at equilibrium.

Answer»

Given,

Total volume (V) = 1L

Mass of N₂O₄ = 13.8g

Molar mass of N₂O₄ = 14(2) + 16(4) = 92g

Number of moles N₂O₄(n) = \(\frac{13.8g}{92g}\) = 0.15

Gas constant (R) = 0.083 bar L mol⁻¹ K⁻¹

Temperature (T) = 400 K

According to ideal gas equation,

pV = nRT

p × 1 L = 0.15 mol × 0.083 bar L mol⁻¹ K⁻¹x 400 K

\(\therefore\) p = 4.98 bar

For this reaction,

N₂O₄ (g) ⇌ 2NO₂ (g)

Initial pressure	4.98 bar	0
At equilibrium	(4.98 – x) bar	2x bar

Hence,

P_totalat equilibrium = \(P_{N_2O_4}\) + \(P_{NO_2}\)

9.15 = (4.98 - x) + 2x

9.15 = 4.98 + x

\(\therefore\) x = 9.15 - 4.98 = 4.17 bar

Partial pressures at equilibrium are,

\(P_{N_2O_4}\) = 4.98 - 4.17 = 0.81 bar

\(P_{NO_2}\) = 2x = 2 x 4.17 = 8.34 bar

K_p = \(\frac{(P_{NO_2})^2}{P_{N_2O_4}}\)

= \(\frac{(8.34)^2}{0.81}\) = 85.87

K_p = K_c (0.083 x 4.00)1

\(\therefore\) K_c = \(\frac{85.87}{0.083\times400}\) = 2.586

Discussion

992.	3.00 mol of `PCl_(5)` kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate the composition of the mixture at equilibrium. `K_(c) = 1.80.`
Answer» `{:(,PCl_(5)(g) ,hArr, PCl_(3)(g) ,+,Cl_(2)(g)),("Initial molar conc.",3.00,,0,,0),("Eqm. molar conc." ,(3.0-x),,x,,x):}` `K_(c)=[[PCl_(3)][Cl_(2)]]/[[PCl_(5)]] or 1.80 =(x xx x )/(3-x) = (x^(2))/(3-x) or x^(2) = 5.40 -1.80 x ror x^(2) +1.80 x -5.40 =0` `:. x= (-bunderset(-)(+)sqrt(b^(2)-4ac))/(2a) =(-1.80 underset(-)(+)sqrt((1.80)^(2)-4(-5.40)))/(2) = 1.59 ("neglecting -ve value")` `:. " At . equilibrium " ,[PCl_(5)] = 3 - 1.59 = 1.41 M , [PCl_(3)] =[Cl_(2)] = 1.59 M`

Discussion

993.	13.8 g of `N_(2)O_(4)` was placed in 1 L reaction vessel at 400K and allowed to attain equilibrium :`N_(2)O_(4) (g) hArr 2NO_(2)(g).` the total pressure at equilibrium was found to be 9.15 bar. Calculate `K_(c),K_(p)` and partial pressure at equilibrium .
Answer» `"No. of moles of " N_(2) O_(4) = (13.8)/(92) " mol " = 0.15 " mol " " " (" Molar mass of " N_(2)O_(4) = 92 g " mol "^(-1))` `PV=nRT or P = (nRT)/(V) or P = ((0.15 "mol") xx (0.083 " bar L mol"^(-1) K^(-1)) xx 400 K)/(1 L)= 4.98 " bar "` `{:(,N_(2)O(g),hArr, 2NO_(2)),("Intital pressure",4.98"bar",,0),("Equilbibrium pressure",(4.98-p),,2_(P)):}` `:. (P_(N_(2)O_(4)) eq = 4.98 -4.17 =0.81 "bar" , (P_(NO_(2)) eq =2x 4.17 =8.34 "bar"` `K_(p) = p_(NO_(2))^(2)//P_(N_(2)O_(4) = (8.34)^(2)//0.81=85.86,` `k_(p) =K_(c) (RT)^(Deltan) :. 85.87 = K_(c) (0.083xx400)^(1)` `or " "K_(c) = (85.87)/(0.083xx 400) = 2.586 =2.6`

Discussion

994.	The values of `K_(p) and K_(p_(2))` for the reaction `X hArr Y+Z ...(1) and A hArr 2B...(2)` Are in the ratio 9: 1 . It the degree of dissociation of X and A be equal , calculate the ratio of the total prssures of (1) and (2) at equilibrium.
Answer» Suppose total pressure at equilibrium for reactions (1) and (2) are `P_(1)and P_(2)` respectively . Then ` {:(,X,hArr,Y,+,Z),("Intial.",1"mole",,0,,0),("At eqm.",1-alpha,,alpha,alpha,"Total"1+alpha):}` ` P_(X) = (1-alpha)/(1+alpha) P_(1), p_(Y) = alpha/(1+alpha) p_(1),p_(Z) = alpha/(1+alpha)P_(1)` ` K_(p_(1)) = ((apha)/(1+alpha) P_(1))^(2)/((1-alpha)/(1+alpha)P_(1))=(alpha^(2)P_(1))/(1- alpha^(2) )cong alpha^(2) P_(1) ` `{:(,A,hArr,2B,),("Intial",1"mole",,0,),("At eqm.",1-alpha,,2 alpha,"Total"=1=alpha):}` ` p_(A) = (1-alpha)/(1 +alpha)p^(2),p^(B) = (2alpha)/(1 + alpha) P^(2)` ` K_(p_(2)) = ((2alpha)/(1+alpha)P_(2))^(2)/((1-alpha)/(1+alpha)P^(2))= (4 alpha^(2))/(1-alpha^(2)) P_(2) = 4 alpha ^(2) P_(2) ` ` (K_(p_(1)))/(K_(p_(2)))= (alpha^(2) P_(1))/(4alpha^(2)P_(2)) = (P_(1))/(4 P_(2))= 9/1 ("Given") or (P_(1))/(P_(2))= 36 /1 = 36 :1 `

Discussion

995.	The values of `K_(p)` and `Kp_(2)` fot the reactions `XhArrY+Z` , (a) and `A hArr 2B` , (b) are in the ration of `9:1`. If the degree of dissociation of X and A is equal, then the total pressure at equilibriums (a) and (b) is in the rationA. `3:1`B. `1:9`C. `36:1`D. `1:1`
Answer» Correct Answer - C Let us assume that `P_(1)` and `P_(2)` are the total pressures at equilibrium for reactions (a) and (b), respectively. First consider reaction (a): `{:(,X hArr,Y+,Z),("Initial moles",1,0,0),("Equilibrium moles",1-alpha,alpha,alpha):}` where `alpha` is the degree of dissociation of X. `K_(p_(1))= (P_(y)P_(z))/(P_(x))` Partial pressure = (Mole fraction) x (Total pressure) `:. P_(x)= (1-alpha)/(1+alpha)P_(1)`, `P_(Y)=(alpha)/(1+alpha)P_(1)`, `P_(Z)=(alpha)/(1+alpha)P_(1)` `:. K_(P_(1))=(((alpha)/(1+alpha)P_(1))^(2))/((1-alpha)/(1+alpha)P_(1)) = (alpha^(2))/((1+alpha)(1-alpha))P_(1)` `= (alpha^(2))/(1-alpha^(2))P_(1)` Now consider reaction (b): `{:(,A hArr,2B),("Initial moles",1,0),("Equilibrium moles",1-alpha,2alpha):}` where `alpha` is the degree of dissociation of A. `P_(A) = (1-alpha)/(1+alpha)P_(2)`, `P_(B)= (2alpha)/(1+alpha)P_(2)` `K_(P_(2))=(P_(B)^(2))/(P_(A))=(((2alpha)/(1+alpha)P_(2))^(2))/(((1-alpha)/(1+alpha)P_(2)))` `= (4alpha^(2))/(1-alpha^(2))P_(2)` Let us take the ratio of `K_(P_(1))` and `K_(P_(2))` : `(K_(P_(1)))/(K_(P_(2)))=((alpha^(2))/(1-alpha^(2))P_(1))/((4alpha^(2))/(1-alpha^(2))P_(2))` `= (P_(1))/(4P_(2))` (as `alpha` of X and A are equal) Since `(K_(P_(1)))/(K_(P_(2)))=(9)/(1)`, we have `(P_(1))/(4P_(2))=(9)/(1)` or `(P_(1))/(P_(2))=(36)/(1)=36 : 1`

Discussion

996.	At `250^@C`, `K_c` for `PCl_5(g)hArr PCl_3(g)+Cl_2(g)` is `0.04`. How many moles of `PCl_5` must be added to a `3-L` flask to obtain `0.15M Cl_2` at equilibrium?A. `2.1 mol`B. `1.7mol`C. `0.9 mol`D. `3.5 mol`
Answer» Correct Answer - A Molarity `(M)=n/V_L` or `n=M*V_L` Thus, the number of moles of `Cl_2` at equilibrium, `n_(Cl_2)=(0.15M)(3L)` `=0.45mol` `{:(,PCl_(5)hArrPCl_(3)+Cl_(2)),("Initial moles"," a 0 0"),("Change"," -0.45 +0.45 +0.45"),("Equilibrium moles", bar(" a-0.45 0.45 0.45 ")),("Equilibrium concentration"," "(a-0.45)/(3)" 0.15 0.15"):}` Equilibrium constant expression `K_c=(C_(PCl_3)C_(Cl_2))/(C_(PCl_5))` Substituting, we get `0.04=((0.15)(0.15))/(((a-0.45)/(3)))` `0.04((a-0.45)/(3))=0.0225` `a=2.1`

Discussion

997.	The values of `K_(p1)` and `K_(p2)` for the two equilibrium reactions `X hArr +Z and A hArr 2B` are in the ratio 9, 1,. If degree of dissociation of X and A be equal , calculate the ratio of the total pressure of the equilibrium mixture in the two cases.
Answer» Let the degree of disscoiation in the two cases be `alpha` (i) Calculattion of `K_(p1)` `{:("For equilibrium", X,hArr, Y, +,Z),("initial moles" : ,1,,0,,0),("moles at eqm. point",(1:alpha),,alpha,,alpha):}` Total no. of ,moles `=1 -alpha +alpha+alpha=(1+alpha)` `K_(p1) = (P_(y)xxP_(z))/(P_(x)) = (((alpha)/(1+alpha)P_(1))xx((alpha)/(1+alpha)P_(1)))/(((1-alpha)/(1+alpha)P_(1)))` `=(P_(1)^(2)alpha^(2))/((1+alpha)^(2))xx((1+alpha)/((1-alpha)P_(1)))=(P_(1)alpha)^(2)/((1+alpha)(1-alpha))` (II) Calculation of `K_(p2)`. `{:("For equilibrium",A, hArr, 2B),("initial moles",1,,0),("Moles at eqm. point",(1-alpha),,2alpha):}` Total no. of moes `=1 alpha +2alpha =(1+alpha)` `K_(p2) = (pB)^(2)/(pA)=((2alpha)/(1+alpha)P_2)^(2)/(((1-alpha)/(1+alpha)P_(2)))` `P_(2)^(2) xx (4alpha^(2))/(1+alpha)^(2) xx ((1+alpha))/((1-alpha)P_(2))= (P_(2)xx4alpha^(2))/((1+alpha)(1-alpha))` Dividing eqn. (i) with eqn. (ii), `(K_p1)/(K_p2) = ((P_(1)alpha)^(2))/((1-alpha^(2)))xx ((1-alpha^(2)))/(P_(2)xx4alpha^(2)) = P_(1)/(4_(p_(2))` `" Now " " "(K_(p1))/(K_(p2))=9/1 ("given") :. P_(1)/(4P_(2))=9/1 or P_(1)/P_(2) = (36)/(1)`

Discussion

998.	(a) Prove that in the aqueous solution of `NH_(4)CI` concentration of `H_(3)O^(+)` ions is `sqrt(K_(h) xx c)`. (b) Find out the ration of `[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]]` in the aqueous solution of carbonic acid whose pH is `7.4. (K_(a) = 4.5 xx 10^(-7))`
Answer» (a) `NH_(4)CI +H_(2)O hArr NH_(4) OH +HCI` The ionisation in aqueous solution is `NH_(4)^(+) + CI^(-) + H_(2)O hArr NH_(4)OH + H^(+) + CI^(-)` It may be wrtten as `{:(,NH_(4)^(+),+, H_(2)O ,hArr, NH_(3) ,+, H_(3)O^(+)),("Initial molar conc :",1mol,,,,0,,0),("Equilibrium molar conc :",c(1-h),,,,c.h,,c.h):}` (Here c = cone. of salt h = degree of hydrolysis) Hygrolysis contant `(K_(h)) =[[NH_(3)][H_(3)O^(+)]^(+)]/((NH_(4)^(+))) =(c.hxx c.h)/(c(1-h))` since h is very small , it can be neglected as compared to 1 `K_(h) =cg^(2) " or " h = sqrt((K_(h))/(c))` From equilibrium reaction `[H_(3)O^(+)] =ch =c sqrt((K_(h))/(c)) = sqrt(K_(h) xx c)` (b) it is acidic buffer `pH =pK_(a) + log .("Conjugate base")/("Acid")` `pH =7.4 K_(a) =4.5 xx 10^(-7) pK_(a) =7 - log 4.5 =6.35` `pH = pK_(a) + log .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]] 7.4 =6.35 + log .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]]` `log .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]] =7.4 -6.35 =1.05 " or " .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]] = " Antilog " 1.05 =11.22`

Discussion

999.	The equilibrium constants `K_(p1) " and " K_(p2)` for the two equilibrium reactions are `X hArr Y + Z " and " A hArr 2B` in the ratio 9 : 1. If the degree of dissociation of X and A be equal then the ratios of the total pressure at these equilibria will be :A. `9:1`B. `36:1`C. `1:1`D. `3:1`
Answer» Correct Answer - B

Discussion

1000.	The expression for equilibrium constant, `K_(c)` for the following reaction is `Fe_((aq))^(3+)+3OH_((aq))^(-)hArrFe(OH)_(3(s))`A. `K_(c)=([Fe(OH)_(3)])/([Fe^(3+)][OH^(-)]^(3))`B. `K_(c)=([Fe(OH)_(3)])/([Fe^(3+)][OH^(-)])`C. `K_(c)=(1)/([Fe^(3+)][OH^(-)]^(3))`D. `K_(c)=[Fe(OH)_(3)]`
Answer» Correct Answer - C For `Fe_((aq))^(3+)+3OH_((aq))^(-)hArrFe(OH)_(3(s))` `K_(c)=(1)/([Fe^(3+)][OH^(-)]^(3))`

Discussion

Explore topic-wise InterviewSolutions in .

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Calculate the pH of the resultant mixture: a. `10 mL` of `0.2M Ca(OH)_(2)+25 mL` of `0.1 M HCl` b. `10 mL` of `0.01 M H_(2)SO_(4) + 10 mL` of `0.01 M Ca(OH)_(2)`. c. `10 mL` of `0.1 M H_(2)SO_(4)+ 10 mL` of `0.1 M KOH`.

20 ml of 0.5 N HCl and 35 ml of 0.1 N NaOH are mixed . The resulting solution willA. be neutralB. be basicC. turn phenolphthalein solution pinkD. turn methyl orange red

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Mention the general characteristics of equilibria involving physical processes.

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For which of the following reactions does the equilibrium constant depend upon the units of concentration? `(a) CO(g) + H_(2)O(g) hArr CO_(2)(g) + H_(2)(g)` `(b) COCl_(2) (g) hArr CO(g) + Cl_(2)(g)` `(C) NO(g) hArr 1//2 N_(2) (g) + 1//2O_(2)(g)`

`PCl_(3), PCl_(3)` and `Cl_(2)` are at eqilibrium at 500 K and above have concentration 1.59 M for `PCl_(3), 1.59 M` for `Cl_(2)` and 1.41 M for `PCl_(5).` Calculate `K_(c)` for the reaction : `PCl_(5) hArr PCl_(3) + Cl_(2)`

For the reaction : `CO(g) +H_(2)O(g) hArr CO_(2)(g) +H_(2)(g)` the value of `K_(c) = 4.24 at 600 K.` Calculate the equilibrium concentration of `CO_(2), H_(2), CO` and `H_(2)O` at 800 K, if only CO and `H_(2)` are present initially at a concentration of 0.10 M each.

At `817^(@)C, K_(p)` for the reaction between `CO_(2) (g)` and excess hot graphite (s) is 10 atm (a) What are the equilibrium concentrations of the gases at ` 817^(@)C` and a total pressure of 5 atm ? (b) At what total pressure, the gas contains 5 % `CO_(2)` by voume ?

A vessel at 1000 K contains carbon dioxide with a pressure of 0.5 atm. Some of the carbon dioxide is converted to carbon monoxide on addition of graphite. Calculate the value of K it the total pressure at equilibrium is `0*8` atm.

For the reaction `AB (g) hArr A(g) + B(g), AB ` is 33 % dissociated at a totalpressure of P . Therefore, P is related to `K_(p)`by one of the following optionsA. `P=K_(p)`B. ` P=3 K_(p)`C. ` P=4K_(p)`D. ` P=8 K_(p)`

A vessel at 1000 K contains `CO_(2)` with a pressure of 0.5 atm . Some of the `CO_(2)` is converted into CO on the addition of graphite . If the total pressure at equilibrium is `0*8` atm , the value of K isA. 3 atmB. `0*3 `atmC. `0*18 atm`D. `1*8 ` atm

` 13*8 g " of "N_(2)O_(4)` was placed in 1 L reaction vessel in 1 L reaction vessel at 400 K and allowed to attain equilibrium : `N_(2)O_(4) (g) hArr 2 NO(g).` The total pressure at equilibrium was found to be` 9* 15` bar. Calculate `K_(c) , K_(p)` and partial pressures at equilibrium.

13.8 g of N2O4 was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium- N2O4(g) ⇌ 2NO2(g)The total pressure at equilibrium was found to be 9.15 bar. Calculate Kc . Kp and partial pressure at equilibrium.

3.00 mol of `PCl_(5)` kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate the composition of the mixture at equilibrium. `K_(c) = 1.80.`

13.8 g of `N_(2)O_(4)` was placed in 1 L reaction vessel at 400K and allowed to attain equilibrium :`N_(2)O_(4) (g) hArr 2NO_(2)(g).` the total pressure at equilibrium was found to be 9.15 bar. Calculate `K_(c),K_(p)` and partial pressure at equilibrium .

The values of `K_(p) and K_(p_(2))` for the reaction `X hArr Y+Z ...(1) and A hArr 2B...(2)` Are in the ratio 9: 1 . It the degree of dissociation of X and A be equal , calculate the ratio of the total prssures of (1) and (2) at equilibrium.

The values of `K_(p)` and `Kp_(2)` fot the reactions `XhArrY+Z` , (a) and `A hArr 2B` , (b) are in the ration of `9:1`. If the degree of dissociation of X and A is equal, then the total pressure at equilibriums (a) and (b) is in the rationA. `3:1`B. `1:9`C. `36:1`D. `1:1`

At `250^@C`, `K_c` for `PCl_5(g)hArr PCl_3(g)+Cl_2(g)` is `0.04`. How many moles of `PCl_5` must be added to a `3-L` flask to obtain `0.15M Cl_2` at equilibrium?A. `2.1 mol`B. `1.7mol`C. `0.9 mol`D. `3.5 mol`

The values of `K_(p1)` and `K_(p2)` for the two equilibrium reactions `X hArr +Z and A hArr 2B` are in the ratio 9, 1,. If degree of dissociation of X and A be equal , calculate the ratio of the total pressure of the equilibrium mixture in the two cases.

(a) Prove that in the aqueous solution of `NH_(4)CI` concentration of `H_(3)O^(+)` ions is `sqrt(K_(h) xx c)`. (b) Find out the ration of `[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]]` in the aqueous solution of carbonic acid whose pH is `7.4. (K_(a) = 4.5 xx 10^(-7))`

The equilibrium constants `K_(p1) " and " K_(p2)` for the two equilibrium reactions are `X hArr Y + Z " and " A hArr 2B` in the ratio 9 : 1. If the degree of dissociation of X and A be equal then the ratios of the total pressure at these equilibria will be :A. `9:1`B. `36:1`C. `1:1`D. `3:1`

The expression for equilibrium constant, `K_(c)` for the following reaction is `Fe_((aq))^(3+)+3OH_((aq))^(-)hArrFe(OH)_(3(s))`A. `K_(c)=([Fe(OH)_(3)])/([Fe^(3+)][OH^(-)]^(3))`B. `K_(c)=([Fe(OH)_(3)])/([Fe^(3+)][OH^(-)])`C. `K_(c)=(1)/([Fe^(3+)][OH^(-)]^(3))`D. `K_(c)=[Fe(OH)_(3)]`

A vessel at 1000 K contains `CO_(2)` with a pressure of 0.5 atm . Some of the `CO_(2)` is converted into CO on the addition of graphite . If the total pressure at equilibrium is `08` atm , the value of K isA. 3 atmB. `03 `atmC. `018 atm`D. `18 ` atm

` 138 g " of "N_(2)O_(4)` was placed in 1 L reaction vessel in 1 L reaction vessel at 400 K and allowed to attain equilibrium : `N_(2)O_(4) (g) hArr 2 NO(g).` The total pressure at equilibrium was found to be` 9 15` bar. Calculate `K_(c) , K_(p)` and partial pressures at equilibrium.