1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 19 InterviewSolution

901.	The pH of 0.05 M aqueous solution of diethylamine is 12.0 . Calculate its `K_(b)`.
Answer» `(C_(2)H_(5))_(2)NH+H_(2)O hArr(C_(2)H_(5))_(2)NH_(2)^(+)+OH^(-)` As `pH = 12, :. [H^(+)]=10^(-12)M or [OH^(-)]=10^(-2)M, [(C_(2)H_(5))_(2)NH]=0.05-0.01 = 0.04 M ` `K_(b) = ([(C_(2)H_(5))_(2)NH_(2)^(+)][OH^(-)])/([(C_(2)H_(5))_(2)NH]) = (10^(-2)xx10^(-2))/(0.04)=2.5xx10^(-3)" " {[C_(2)H_(5))_(2)NH_(2)^(+)]=[OH^(-)]}`

Discussion

902.	An aqueous solution of aniline of concentration 0.24 M is prepared. What concentraton of sodium hydroxide is needed in this solution so that anilinium ion concentration remains at `1xx10^(-8)` M? `(K_(a) "for" C_(6)H_(5)NH_(3)^(+)=2.4xx10^(-6)M)`
Answer» In aqueous solution, aniline is hydrolysed as `C_(6)H_(5)NH_(2)+ H_(2)O hArr C_(6)H_(5)overset(+)N H_(3)+OH^(-)` When NaOH is added, hydrolysis will be supressed so that in the final solution, `[C_(6)H_(5) overset(+)NH_(3)]=10^(-8)M` (Given) In conc. of NaOH added is x mol `L^(-1)`, then at equilibrium `[C_(6)H_(5)NH_(2)]=0.24-10^(-8) ~= 0.24` `[C_(6)H_(5)overset(+)NH_(3)]=10^(-8)M` (Given) `[OH^(-)]=10^(-8)+x ~= x M` Hydrolysis constant, `K_(h) = ([C_(6)H_(5)overset(+)H_(3)][OH^(-)])/([C_(6)H_(5)NH_(2)])` ...(i) Further, we are given `C_(6)H_(5)overset(N)H_(3) hArr C_(6)H_(5)NH_(2)+H^(+)` `K_(a) = ([C_(6)H_(5)NH_(2)][H^(+)])/([C_(6)H_(5)overset(N)H_(3)])` ...(ii) Also `[H^(+)][OH^(-)]=K_(w)` ...(iii) Combining eqns. (i), (ii) and (iii) `k_(h)=(K_(w))/(K_(a))=(10^(-14))/(2.4xx10^(-6))=(10^(-8))/(2.4)` Substituting the values in eqn. (i), we get, `(10^(-8))/(2.4)=(10^(-8)xx x)/(0.24)` which gives `x=10^(-2)M`

Discussion

903.	Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:A. increase in mass of `BaO_(2)`B. increase in mass of `BaO`C. increase in temperature on equilibriumD. increase in mass of `BaO_(2)` and `BaO` both.
Answer» Correct Answer - C Is the correct answer

Discussion

904.	For the reaction `A(g) + B(s) hArr C (g) + D (g), K_(c) = 49 mol dm^(-3)` at `127^(@)C`. Calculate `K_(p)`.
Answer» Correct Answer - `1.61 xx 10^(3)` atm `Delta n = n_(p) - n_(r) = 2-1 = 1, K_(p) = K_(c) (RT)^(Delta n) = (49 mol dm^(-3)) (0.0821 dm^(3) "atm" K^(-1) mol^(-1) xx 400 K)^(1) = 1.61 xx 10^(3)` atm.

Discussion

905.	A system at equilibrium is described by the equation `SO_(2)Cl_(2) hArr SO_(2)+Cl_(2),Delta H =+ve`. When `Cl_(2)` is added to the equilibrium mixture at constant volume, the temperture of the system
Answer» Temperature of the system will increase because on adding `Cl_(2).` equlibrium will shift in the backward direction producing more heat.

Discussion

906.	For a reversible reactionn at 298 K the equilibrium constant K is 200. What is value of `DeltaG^(@)` at 298 K ?A. `-13.13kcal`B. `-0.13kcal`C. `-3.158kcal`D. `-0.413kcal`
Answer» Correct Answer - C Applying `DeltaG^(@)=-2.303RTxxlogK` `" "=-2.303xx2xx298xxlog200` `" "=-3158.4" cal "=-3.158" kcal"`

Discussion

907.	The reaction , `SO_(2) + Cl_(2) hArr SO_(2)Cl_(2)` is exothermic and reversible . A mixture of `SO_(2)(g) , Cl_(2) hArr SO_(2) Cl_(2)(g) ` is at equilibrium in a closed container . Now a certain quantity of extra `SO_(2)` is introduced into the container , the volume remaining the same. Which of the following is / are/ true ?A. The pressure inside the container will not changeB. The temperature will not changeC. The temperature will increasesD. The temperature will decrease.
Answer» Correct Answer - C `SO_(2) +Cl_(2) hArr SO_(2)Cl_(2) + "Heat"`. On adding `SO_(2)` , the equilibrium will shift forward , i.e., more heat will be evolved . So temperature will increase .

Discussion

908.	Which of the following is not true about a reversible reaction ?A. The reaction does not proceed to completion.B. It cannot be influenced by catalystC. Number of moles of reactants and products is always euqal.D. It can be attained only in a closed container.
Answer» Correct Answer - C

Discussion

909.	The value of ` K_(p) " for the reaction ", CO_(2) (g) + C (s) hArr 2 CO (g),` is 3.0 at 1000 K. Initially `, p_(co_(2)) = 0.48" bar "and P_(CO)= 0` bar and opure graphite is present , calculate the equilibrium partial pressures of `CO and CO_(2)`
Answer» `{:(,CO_(2)(g) + C(s),hArr,2 CO (g)),("Intial",0.48 " bar ",,0 "bar"),("AT.eqm.",(0.48-p)"bar",,2 " p bar"):}` ` K_(p) = (2 p)^(2)/(0.48-p )=3.0 " (Given) ".` ` :. 4 p^(2) = 1.44 - 3 p or 4 p^(2) + 3 p - 1.44=0` ` :. p = ( -b pm sqrt(b^(2) - 4 ac))/(2a)=(-3 pm sqrt ( 9-4 xx 4 ( - 1. 44)))/8` ` = ( -3 pm sqrt(32*04))/8= (-3 pm 5.66)/8 = 2.66/8 " "` (neglecting - ve value) = 0.33 atm `:. p_(CO_(2))= 0.48 - 0.33 = 0.15 " bar" , p_(CO_(2)) = 2 xx 0.33 = 0. 66 " bar " `

Discussion

910.	Which of the following are reversible reactions ?A. ` AgNO_(3) (aq) + NaCl (aq) to` `AgCl (s) + NaNO_(3) (aq)`B. `KNO_(3) (aq) + NaCl (aq) to ` ` KCl (aq) + NaNO_(3)`C. ` BaCl_(2) (aq) + Na_(2) SO_(4) to` `BaSO_(4) (s) + 2 NaCl (aq) `D. `AgCl (s) + Water to Ag^(+) (aq) + Cl^(-) (aq)`
Answer» Correct Answer - B::D In (b), ions are in equilibrium . In (d) , ions are in equilibrium with the undissolved salt .

Discussion

911.	The values of ` K_(p_(1)) and K_(p_(2))` for the reactions `X hArr Y+Z" " ` …(1) and `A hArr 2 B " " ` …(2) are in the ratio of 9:1 .If the drgree of dissociation kof X and A be equal , then total pressure at equilibrium (1) and (2) are in the ratioA. `3:1`B. `1:9`C. `36:1`D. `1:1`
Answer» Correct Answer - C 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+alpa)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) 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)))/((1-alpha)/(1+alpha)P_(2)) = (4alpha^(2))/(1-alpha^(2))P_(2)= 4 alpha^(2)P_(2)` ` K_(p_(1))/K_(p_(2)) = (alpha_(2)P_(1))/(4 alpha^(2)P_(2))=P_(1)/(4 P_(2))=9/1 ` (Given ) or ` (P_(1))/(P_(2))=36/1 = 36*1`

Discussion

912.	`H_(2)S` gas when passed through a solution of cations containing `HCl` precipitates the cations of second group in qualitative analysis but not those belonging to the fourth group. It is becauseA. presence of HCI decreases the sulphide ion sulphide ion concentrationB. sulphides of group IV are unstable in HCIC. solubility product of group II sulphides is more than that of group IV sulphidesD. pressence of HCI increases the sulphide ion concentration
Answer» Correct Answer - A `H_(2)S(aq.)hArr 2H^(+)(aq.)+S^(2-)(aq.)` `H_(2)S` is a weak acid. The presence of HCI suppresses the ionization of `H_(2)O` due to common ion effect. As a result, the concentration of sulphide ions decreases. Since `K_(sp)` values of group II sulphides are lower than those of group IV sulphides, only the cations of group II precipitate.

Discussion

913.	The reaction `CH_(3)COOH (l) + C_(2) H_(5) OH (l) hArr CH_(3) COOC_(2)H_(5) (l) + H_(2) O (l)` was carried out at `27^(@)C` by taking one mole of each of the reactants . The reaction reached equilibrium when 2/3 rd of the reactants were consumed. Calculate the free energy change for the reaction `(R= 8*314 " JK "^(-1) " mol "^(-1)).`
Answer» Correct Answer - `-34579 " j K"^(-1) " mol"^(-1)` `K_(c) = (2//3 xx 2//3)/(1//3 xx 1//3)=4. "Then " Delta G=-2303 " RT log " K_(c).`

Discussion

914.	The equilibrium constant for the reaction `CH_(3) COOH + C_(2)H_(5)OH hArrCH_(3)COOC_(2)H_(5) + H_(2)O` is `4*0 " at " 25^(@)C. ` . Calculate the weight of ethyl acetate that will be obtained when 120 g of acetic acid are reacted with 92 g of alchol.
Answer» Correct Answer - `11704 g` Initially, `CH_(3) COOH = 120/60 "mol" = 2 "mol"` `C_(2) H_(5) OH= 92/46 "mol" = 2 "mol"` At equilibrium, `[CH_(3)COOH] = (2-x)//V " mol "L^(-1)` `[C_(2)H_(5)OH]= (2-x)//V " mol "L^(-1),` `[CH_(3)COOC_(2)H_(5)]=[H_(2)O]= x//V " mol " L^(-1)` ` K = (x xx x )/(2-x)^(2) = 4 " (Given)"`. `" This gives "x== 133 " mol"` ` :. " Mass of ethyl acetate " = 133 xx 88 = 11704 g " "("Molar mass of " CH_(3)COOC_(2)H_(5)= 88" g mol"^(-1))`

Discussion

915.	The equilibrium constant for the reaction : `CH_(3) COOH + C_(2)H_(5)OH hArr CH_(3) COOC_(2)H_(5) +H_(2)O` is 4.0 at `25^(@)C`. Calculate the weight of ethyl acetate that will be obtained when 120 g of acetic acid are reacted with 92 g of ethyl alcohol.
Answer» The given reaction is : `CH_(3)COOH + C_(2)H_(5)OH hArr CH_(3) COOC_(2) H_(5) +H_(2) O` Applying the Law of Chemical equilibrium, `K_(c) = [[CH_(3)COOC_(2)H_(5)][H_(2)O]]/[[CH_(3)COOH][C_(2)H_(5)OH]]=4.0` Initial moles of `CH_(3)COOH = (120g)/(60 g mol^(-1)) = 2` mol (Molar mass of `CH_(3) COOH = 60 g mol^(-1))` Initial moles of `C_(2) H_(5) OH = (92g)/(46 mol^(-1)) =2` mol (Molar mass of `C_(2)H_(5)OH= 46 g mol^(-1))` Let the moles of `CH_(3) COOH` and `C_(2)H_(5)OH` reacted be x. Therefore , according to the equation x,moles of `CH_(3)COOC_(2)H_(5)" and "H_(2)O` will be formed .

Discussion

916.	A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?2HI ⇌ H2(g) + I2(g).
Answer» pHI = 0.04 atm pH₂ = 0.08 atm pI₂ = 0.08 atm K_p = \(\frac{pH_2 \times pI_2}{p^2HI}\) = \(\frac{(0.08\, atm)\times (0.08\, atm)}{(0.04\,atm)\times (0.04\,atm)}\) = 4

916.

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?2HI ⇌ H2(g) + I2(g).

Answer»

pHI = 0.04 atm

pH₂ = 0.08 atm

pI₂ = 0.08 atm

K_p = \(\frac{pH_2 \times pI_2}{p^2HI}\)

= \(\frac{(0.08\, atm)\times (0.08\, atm)}{(0.04\,atm)\times (0.04\,atm)}\) = 4

Discussion

917.	A sample of `HI(g)` is placed in flask at a pressure of `0.2 atm`. At equilibrium. The partial pressure of `HI(g)` is `0.04 atm`. What is `K_(p)` for the given equilibrium? `2HI(g) hArr H_(2)(g)+I_(2)(g)`
Answer» `pHI =0.04 atm pH_(2) = 0.08 atm , pI_(2) =0.08 atm` `K_(p) = (pH_(2)xx pI_(2))/(P_(HI)^(2)) =((0.08 atm)xx(0.08 atm))/((0.04 atm)xx(0.04 atm))=4.0`

Discussion

918.	The `pH` of `0.004M` hydrazine `(NH_(2).NH_(2))` solution is `9.7`. Calculate its ionisation constant `K_(b)` and `pK_(b)`.
Answer» Strategy: From the `pH`, calculate the hydrogen ion concentration. Knowing the `H^(+)` ion concentration and the ionic product of water, calculate the concentration of hydroxyl ions. Using the concentration of `OH^(-)` ions, calculate `K_b` and `pK_b`. Solution: `pH=-log a_(H^+)` `9.7=-log a_(H^+)` Taking the antilog of both sides, we get `C_(H^+)=1.67xx10^(-10)` The ionic product of water is given as `K_w=C_(H^+)+C_(OH^(-))` `:. C_(OH^(-))=(K_w)/(C_(H^+))=(1xx10^(-14))/(1.67xx10^(-10))` `=5.98xx10^(-5)` Write the reaction summary: `{:(NH_(2)NH_(2)(aq.)+H_(2)O(l)hArrNH_(2)overset(+)NH_(3)(aq.)+OH^(-)),("Initial (M) 0.004 0.00 0.000"),("Change (M)"-5.98xx10^(-5) +5.98xx10^(-5)+5.98xx10^(-5)),(bar("Equilibrium (M)" (0.004-598xx10^(-5))5.98xx10^(-5)5.98xx10^(-5))):}` Note that the concentration of the hydrazinium ion is also the same as that of the hydroxide ion. Since the concentration of both these ions is very small, the concentration of the unionized base can be taken (approximately) equal to `0.004M`. Writing the basically constant `K_b=([NH_2overset(+)NH_3][OH^(-)])/([NH_2NH_2])` `=(5.98xx10^(-5))(5.98xx10^(-5))/(0.004)` `=8.94xx10^(-7)` `pK_b=-logK_b=-log(8.96xx10^(-7))` `=6.1`

Discussion

919.	The thermal dissociation equilibrium of `CaCO_(3)(s)` is strudied under different conditions `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)` For this equilibrium, the correct statements are (i) K is dependent on the pressure of `CO_(2)` at a given T. (ii) `Delta H` is dependent on T. (iii) `Delta H` is independent of the catalyst, if any. (iv) K is independent of the inintial amount of `CaCO_(3)`.A. (i), (ii), (iii), (iv)B. (i), (ii), (iii)C. (ii), (iii), (iv)D. (i), (ii), (iv)
Answer» Correct Answer - C `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)` Since `CaCO_(3)` and `CaO` are pure solids, we have `K_(eq)=[CO_(2)]` Thus, the equilibrium constant (K) is dependent on the pressure of `CO_(2)` at a given T but is independent of the intial amount of K. `Delta H` does not depend on catalyst but depends on temperature.

Discussion

920.	Calculate the concentration of all species present in `0.010M H_2SO_4` solution. `(K_(a_2)=1.3xx10^(-2))`
Answer» Strategy: Sulphuric acid `(H_2SO_4)` is an example of a diprotic acid because each unit of the acid yields two `H^(+)` ions, in two separate steps: `H_2SO_4(aq.)rarrH^(+)(aq.)+HSO_4^(-)(aq.)` `HSO_4^(-)(aq.)hArr H^(+)(aq.)+SO_4^(2-)(aq.)` `H_2SO_4` is a strong electrolyte (or strong acid) as the first step of ionization is complete, but `HSO_4^(-)` is weak acid, we need a double arrow to represent its incomplete ionization. Because the first ionization step of `H_2SO_4` is complete, we read the concentration for the first step from the balanced equation. The second ionization step is not complete. Thus, we write the ionization equation, the `K_(a_2)` expression, and the algebratic representations of equilibrium concentrations. Then we substitute into `K_(a_2)` for `H_2SO_4`. Solution: Summarize the changes in the first stage of ionization which is complete: `{:(,H_(2)SO_(4)(aq.)overset(100%)rarrH^(+)(aq.)+HSO_(4)^(-)(aq.)),("Initial (M)"," 0.010 0.00 0.00"),("Change (M)"," -0.010 +0.010 +0.010"),("Final (M)", bar(" 0.00 0.010 0.010 ")):}` For the second stage of ionization (which is not complete), we proceed as for a weak monoprotic acid. Step 1: Let `x=C_(HSO_4^(-))` that ionizes (or the concentration in `mol L^(-1)` of `H^(+)` and `SO_4^(2-)` produced by the ionization of `HSO_4^(-)`). The total concentration of `H^(+)` ions at equilibrium must be the sum of the `H^(+)` ion concentration produced in the first and second steps. So we represent the equilibrium concentrations as `{:(,HSO_(4)^(-)(aq.)hArrH^(+)(aq.)+SO_(4)^(2-)(aq.)),("Initial (M)"," 0.10 0.010 0.00"),("Change (M)"," -x +x +x"),("Equilibrium (M)", bar(" (0.010-x) (0.010+x)" " x ")):}` Step 2: Substitution into the ionization constant expression for `K_(a_2)` gives `K_(a_2)=(C_(H^+)C_(SO_4^(2-)))/(C_(HSO_4^(-))` `1.3xx10^(-2)=((0.010+x)(x))/((0.010-x))` Since the `K_(a_2)` of `H_2SO_4` is quite large, x cannot be neglected. We must solve the quadratic equation, which simplies to `x^2+0.023x-1.3xx10^(-4)=0` Applying the quadratic formula `x=(-b+-sqrt(b^2-4ac))/(2a)` we have `x=(-0.023+-sqrt((0.023)^2-4(1)(-1.3xx10^(-4))))/(2(1))` `=4.7xx10^(-3)M` `:. (C_(H^+))_(2nd)=C_(SO_4^(2-))=4.7xx10^(-3)M` The concentrations of different species in `0.010M H_2SO_4` are `C_(H_2SO_4)~~0M`, `C_(HSO_4^(-))=(0.010-x)` `=(0.010-4.7xx10^(-3))` `=5.3xx10^(-3)` `C_(H^(+))=(0.010+x)` `=(0.010+4.7xx10^(-3))` `=0.015` `C_(SO_4^(2-))=4.7xx10^(-3)` `C_(OH^(-))=(K_w)/(C_(H^+))=(1.0xx10^(-14))/(0.015)=6.7xx10^(-13)`

Discussion

921.	Give one example of everyday life in which there is gas y..’ >'. solution equilibrium.
Answer» Soda-water bottle.

Discussion

922.	Which measurable property becomes constant in water, water vapour equilibrium at constant temperature?
Answer» Vapour pressure.

Discussion

923.	How does a catalyst influence the equilibrium constant of a reversible reaction?
Answer» A catalyst does not influence the equilibrium constant of reversible reaction.

Discussion

924.	For the reaction, `SO_(2)(g)+1/2O_(2)(g)hArrSO_(3)(g)` if `K_(p)=K_(C)(RT)^(x)` where, the symbols have usual meaning, then the value of `x` is (assuming ideality)A. `=-1`B. `-1//2`C. `1//2`D. `1`
Answer» Correct Answer - B `X(Delta^(ng))=2-(3//2) =- 1//2.`

Discussion

925.	Explain with an example that strong acids have very weak conjugate bases and a strong base have a very weak conjugate acid.
Answer» HCl + HOH --------> Cl^- + H₃O⁺ CH₃COOH + HOH --------> CH₃COO^- + H₃O⁺ HCl is strong acid but chloride ion is a weak base similarly CH₃COOH is weak acid but CH3COO- ion is a strong base, these example shows that every strong acid has a weak conjugate base and weak acid has a strong conjugate acid.

925.

Explain with an example that strong acids have very weak conjugate bases and a strong base have a very weak conjugate acid.

Answer»

HCl + HOH --------> Cl^- + H₃O⁺

CH₃COOH + HOH --------> CH₃COO^- + H₃O⁺

HCl is strong acid but chloride ion is a weak base similarly CH₃COOH is weak acid but CH3COO- ion is a strong base, these example shows that every strong acid has a weak conjugate base and weak acid has a strong conjugate acid.

Discussion

926.	Which of the following salts does show its correct nature mentioned against it ?A. KBr solution - NeutralB. NaCN solution - AcidicC. `NH_(4)NO_(3)` solution - AcisdicD. KF solution - Basic
Answer» Correct Answer - B NaCN solution is basic in nature since HCN formed is a weak acid and does not hydrolyse. `NaCN+H_(2)OhArrNaOH+HCN` `NaOHhArrNa^(+)+OH^(-)` (Basic solution)

Discussion

927.	When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temerature increase of `0.7^(@)`C was measured for the beaker and its contents (Expt. 1 ) . Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ `"mol"^(-1)`), the experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid `(K_(a)=2.0xx10^(-5))` was mixed with 100 mL of 1.0 M NaOH . (under identical conditions of Expt.1) where hte temperature rise of `5.6^(@)C` was measured. (Consider heat capacity of all solutions as `4.2 J g^(-1) K^(-1)` and density of all solutions as 1.0 m `mL^(-1)`) The pH of the solution after Expt. 2 isA. 2.8B. 4.7C. `5.0`D. `7.0`
Answer» Correct Answer - B `CH_(3)CO OH ` present in 100 mL of 2.0 M the solution = 0.2 mole NaOH present in 100 mL of 1.0 M solution = 0.1 mole 0.1 mole of NaOH neutralizes 0.1 mole of NaOH to form 0.1 mole of `CH_(3)CO ON a` `:. ` In the final solution, `CH_(3)CO OH` = 0.1 mole, `CH_(3)CO ONa`= 0.1 mole, i.e., it is a buffer solution `pH = pK_(a) + log. (["Salt"])/(["Acid"])=-log(2xx10^(-5))+log.(0.1)/(0.1)` `=-log(2xx10^(-5))` `=5-0.301=4.699~= 4.7`

Discussion

928.	Which of the following salts is the most basic in aqueous solution ?A. `Al(CN)_(3)`B. `CH_(3)CO OK`C. `FeCl_(3)`D. `Pb(CH_(3)CO O)_(2)`
Answer» Correct Answer - B `Al(CN)_(3)`-salt of weak acid and weak base `CH_(3)CO OK` - salt of weak acid and strong base. `FeCl_(3)`-salt of weak base and strong acid `Pb(CH_(3)CO O)_(2)`-salt of weak acid and weak base Thus, solution of `CH_(3)CO OK` is most basic.

Discussion

929.	Which of the following salts with a concentration .1 M will give a basic solution ?A. Ammonium acetateB. Ammonium chlorideC. Ammonium sulphateD. Sodium acetate
Answer» Correct Answer - D `CH_(3)COONa+H_(2)OhArrCH_(3)COOH+NaOH` `NaOHhArrNa^(+)+OH^(-)` (Basic solution)

Discussion

930.	Which of the following relations is correct during the hydrolysis of salts of weak acid and strong bases ?A. `K_(h)=(K_(w))/(K_(a))`B. `K_(h)=(K_(w))/(K_(a)K_(b))`C. `K_(h)=(K_(w))/(K_(b))`D. `K_(h)=(K_(a))/(K_(w))`
Answer» Correct Answer - A `A^(-)+H_(2)OhArrHA+OH^(-)` `K_(b)=([HA][OH^(-)])/([A^(-)])=([HA][OH^(-)])/([A^(-)])xx([H^(+)])/([H^(+)])` `= ([HA])/([H^(+)][A^(-)])xx[H^(+)][OH^(-)]` `=(1)/(K_(a))xxK_(w)` `= (K_(w))/(K_(a))` Expression `(2)` is valid for the hydrolysis of salts of weak bases and weak acids. Expression `(3)` is valid for the hydrolysis of salts of weak bases and strong acids

Discussion

931.	In which of the following solvent is silver chloride most soluble ?A. 0.1 mol `dm^(-3) AgNO_(3)` solutionB. 0.1 mol `dm^(-3)` HCl solutionC. `H_(2)O`D. Aqueous ammonia
Answer» Correct Answer - D AgCl will be most soluble in aqueous ammonia because it forms a complex, `[Ag(NH_(3))_(2)]^(+)Cl^(-)`.

Discussion

932.	In which of the following solvents is silver chloride most soluble ?A. `0.1"mol dm"^(-3)AgNO_(3)" solution"`B. `0.1" mol dm"^(-3)HCl" solution"`C. `H_(2)O`D. Aqueous ammonia
Answer» Correct Answer - D `AgCl+2NH_(3)tounderset("(Soluble)")([Ag(NH_(3))_(2)])Cl` Thus, AgCl is most soluble in aqueous ammonia.

Discussion

933.	A solution has been prepared by dissolving 0.063 g of `HNO_(3)` in 1000 mL of It . Calculate the `[H^(+)]` and `[OH^(-)]` of the solution.
Answer» Molar concetration of `HNO_(3) = ("Mass of" HNO_(3) // "litre of solution")/("Molar mass of"HNO_(3)) = ((0.063g))/((63g" mol "^(-1))` `=1.0xx 10^(-3)M` `HNO_(3)` is a strong acid and is ompletely ionised in water. `{:(HNO_(3),overset(aq)(to),Na^(+) (aq),+, OH^(-)(aq)),(,,1.0xx10^(-3)M,,1.0xx10^(-3)M):}` `:." "[H_(3)O^(+)] = [HNO_(3)] = 1.0 xx 10^(-3)M` `"["OH^(-)"]"=(K_(w))/[[H_(3)O^(+)]]=((1.0xx10^(-14)M^(2)))/((1.0xx10^(-3)M)) = 10^(-11)M`

Discussion

934.	What is the `pH` of `0.01 M` glycine solution? For glycine, `K_(a_(1))=4.5xx10^(-3)` and `K_(a_(2))=1.7xx10^(-10)` at `298 K`A. `3.0`B. `10.0`C. `6.1`D. `7.2`
Answer» Correct Answer - C `pK_(a_(1))= -log(4.5xx10^(-3))` `=3-log4.5` `pK_(a_(2))= -log(1.7xx10^(-10))` `=10-log 1.7` We have `pH=(1)/(2)(pK_(a_(1))+pK_(a_(2)))` `=(1)/(2)[3-log(4.5)+10-log(1.7)]` `=(1)/(2)[3-log(4.5xx1.7)]` `=(1)/(2)[13-log(7.65)]` `=6.1` Alternatively, overall ionization constant (K) is given as `K= K_(a_(1))xxK_(a_(2))` `=(4.5xx10^(-3))(1.7xx10^(-10))` we have `C_(H^(+))=sqrt(KC)=sqrt((7.65xx10^(-13))(0.01))` `=sqrt(0.765xx10^(-14))` `=0.875xx10^(-7)M` `pH= -log(0.875xx10^(-7))=7-0.9=6.1`

Discussion

935.	Each question given below contains STATEMENT -1 (Assertion) and STATEMENT -2 (Reason). It has four choice (a), (b), ( c ) and (d) out of which ONLY ONE is correct . Choose the correct option as under : Statement -1. Reaction quotiet of a reaction at any time decides the direction in which the reaction will proceed. Statement -2. The value of reaction quotient cannot be greater than the equilibrium constant .
Answer» Correct Answer - c Correct statement -2. The value of reaction quotient `(Q_(p) or Q_(c))` can be zero or greater than or less than 1.

Discussion

936.	Equilibirum constants `K_(1)` and `K_(2)` for the following equilibria `NO(g) + (1)/(2)O_(2)hArrNO_(2)(g)` and `2NO_(2)(g)hArr` `2NO(g) + O_(2)(g)` are related asA. `K_(2)=1//K_(1)`B. `K_(2)=K_(1)//2`C. `K_(2)=1//K_(1)^(2)`D. `K_(2)=K_(1)^(2)`
Answer» Correct Answer - C Second equilibrium equation can be genrated from the first one in two stages: Stage 1: Reverse the equilibrium equation `NO_(2)(g)hArr NO(g)+(1)/(2)O_(2)(g)` `:. K_(eq)= (1)/(K_(1))` Stage 2: Multiply the reversed equilibrium equation by 2. `2NO_(2)(g)hArr 2NO(g)+O_(2)(g)` `:. K_(eq)= ((1)/(K_(1)))^(2)` This implies `K_(2)= (1)/(K_(1)^(2))`

Discussion

937.	The `K_(p) " for the reaction," N_(2) O_(4) hArr 2NO_(2) ` is 640 mm at 775 K. Calculate the percentage dissociation of `N_(2)O_(4)` at equilibrium pressure of 160 mm. At what pressure the dissociation will be 50% ?
Answer» Suppose intially `N_(2)O_(4)` taken = 1 mole and its degree of dissociation = `alpha` ` {:(,N_(2)O_(4),hArr,2NO_(2),),("Intial ",1 "mole",,,),("At eqm.",1-alpha,2alpha,"Total " =1-alpha + 2 alpha = 1+alpha,):}` If P is the total pressure at equilibrium , then ` p_(N_(2)O_(4)) = (1-alpha)/(1+alpha) xxP and p_(NO_(2)) = (2 alpha)/(1+alpha) xxP ` Now, ` K_(p) = (p_(NO_(2))^(2))/(p_(N_(2)O_(4)))= ((2 alpha)/(1+alpha).P)^(2)/((1-alpha)/(1+alpha) P)= (4 a^(2))/((1+alpha)(1-alpha)) = (4a^(2))/(1-alpha^(2)) xx P` Putting ` K_(p) = 640 " mm (Given ) and equilibrium pressure , P+ 160 mm, we get "` ` 640 = (4a^(2))/(1-alpha)^(2) xx 160 or (alpha^(2))/(1-alpha)^(2) = or alpha^(2) = 1 - alpha^(2) or 2 alpha^(2) = 1 or alpha^(2) = 05 or alpha = 0* 707 = 70* 7 % ` For dissociation to be ` 50 % , alpha = 050 , K_(p) = 640` mm ( constant ) ` :. 640 = (4 ( 05)^(2))/(1-(0*5)^(2)) xx P or 640 = 1/(1-1/4) P= 4/3 P or P =480 "mm"` .

Discussion

938.	Equilibrium constants `K_(1)` and `K_(2)` for the following equilibria `NO(g) +1//2O_(2)(g) overset(K_(1))(hArr) NO_(2)(g)` and ` `2NO_(2)(g) overset(K_(2))(hArr)2NO(g)+O_(2)(g)` are related asA. `K_(1) =sqrt(K_(2))`B. `K_(2) =(1)/(K_(1))`C. `K_(1) = 2K_(2)`D. `K_(2) =(1)/(K_(1)^(-2))`
Answer» Correct Answer - D Equation (2) can be obtained by multiplying equation (1) with 2 and reversing it . Therefore `K_(2) " and " K_(1)` are related to each other as : `K_(2) =(1)/(K_(1)^(-2))`

Discussion

939.	Given the equilibrium , `N_(2)O_(4)(g) hArr 2NO_(2) (g) " with " k_(p) =0.15 " atm at " 298 K.` (a) What is `K_(p)` using pressure in torr? (b) What is `K_(c)` using units of moles per litre ?
Answer» `(a) K_(p) = ((760 " torr" ) xx (0.15 atm))/ ((1 atm)) = 1.14 xx 10^(2) " torr "` `(b) K_(p) = K_(c)(RT)^(Deltan)` `K_(C) = (K_(p))/(RT)^(Deltan)=((0.15 atm))/((0.0821 " L atm K"^(-1) " mol"^(-1) xx 298 K)^(2-1))=6.13 xx 10^(-3) "mol"^(-1)`

Discussion

940.	For an equilibrium reaction, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8 xx 10^(-2)` and `1.2 xx 10^(-2) mol//L` respectively. The value of `K_(c)` for the reaction isA. `3xx 10^(-1) " mol " L^(-1)`B. `3xx 10^(-3) " mol "L^(-1)`C. `3xx10^(3) " mol " L^(-1)`D. `3.3 xx 10^(3) " mol " L^(-1)`
Answer» Correct Answer - B `K_(c) =((1.2 xx 10^(-2) "mol "L^(-1))^(2))/((4.8 xx 10^(-2) " mol "L^(-1))) =3.0 xx 10^(-3) "mol "L^(-1)`

Discussion

941.	500 ml vessel contains 1.5 M each of A, B, C and D at equlibrium. If 0.5 M each of C and D are taken out, the value of `K_(c)` for `A+BhArrC+D` will beA. `1.0`B. `1/9`C. `4/9`D. `8/9`
Answer» Correct Answer - A

Discussion

942.	For the reaction `a+bhArrc+d`, initially concentrations of a and b are equal and at equilibrium the concentration of will be twice of that of a. What will be equilibrium constant for the reaction ?A. 2B. 9C. 4D. 3
Answer» Correct Answer - C `underset(x)(a)+underset(x)(b)hArrunderset(2x)(c)+underset(2x)(d)` `K_(c)=(2x.2x)/(x.x)=(4x^(2))/(x^(2))rArrK_(c)=4`

Discussion

943.	The equilibrium constant at `298K` for a reaction, `A+BhArrC+D` is 100. If the initial concentrations of all the four species were 1M each, then equilibirum concentration of `D` (in mol`L^(-1)`) will beA. `0.182`B. `0.818`C. `1.818`D. `1.182`
Answer» Correct Answer - C `{:(,A+B,hArr,C ,+,D),("Initial conc.",1" "1,,1,,1),("Eqm. conc.",(1-X)(1-X),,(1+X),,(1+X)):}` `K_(c) =[[C][D]]/[[A][B]] " or " 100 =((1+X)^(2))/((1-X)^(2))` `10=(1+X)/(1-X) " or " (1+X) =(10 -10X) 11X = 9 " or " X =9/11 = 0.818` Eqm. conc. of D =(1+0.818)= 1.818M

Discussion

944.	The solubility of `Sr(OH)_(2)` at `298 K` is `19.23 g L^(-1)` of solution. Calculate the concentrations cf strontium and hydroxyl ions and the `pH` of the solution.
Answer» Molarity of solution `=("Mass of Sr"(OH_(2))//"Molar Mass")/("Volume of solution in litres")` `=((19.23)//(87.6 + 34g) "mol"^(-1)L)/(1L)` `Sr(OH)_(2) overset((aq))(to) underset(0.1581 M)(Sr^(2) (aq)) + underset(2 xx 1.1581 = 0.3162 M)(2OH^(-) (aq))` `[Sr^(2+)]=0.1581 M, [OH^(-)] = 0.3162 M` `pOH =- log [OH^(-)] =- log (3.162 xx10^(-1))` `=- (log 3.162 - log 10) =(1- log 3.162) = (1-0.5) =0.5` `pH =14 - pOH =14 -0.5 =13.5`

Discussion

945.	Given the reaction between 2 gases represented by `A_(2)` and `B_(2)` to given the compound AB(g). `A_(2)(g) + B_(2)(g)hArr 2AB(g)` At equilibrium, the concentrtation of `A_(2) = 3.0xx10^(-3) M` of `B_(2) = 4.2xx10^(-3) M` of `AB = 2.8xx10^(-3) M` If the reaction takes place in a sealed vessel at `527^(@)C` . then the value of `K_(c)` will beA. `2.0`B. `1.9`C. `0.62`D. `4.5`
Answer» Correct Answer - C `A_(2)(g) +B_(2)(g) hArr 2AB(g)` `K_(c)=[[AB]^(2)]/[[A_(2)][B_(2)]]` `=((2.8 xx 10^(-3) M^(2)))/((3.0 xx 10^(-3)M)xx (4.2 xx 10^(-3) M))` `=(2.8 xx 2.8)/(3.0 xx 4.2) =0.62`

Discussion

946.	The equilibrium constant for the reaction `A_(2)(g) +B_(2) hArr 2AB(g)` at 373 K is 50. If one litre flask containing one mole of `A_(2)` is connected to a two flask containing two moles of `B_(2)`, how many moles of AB will be formed at 373 K?
Answer» Total volume available for reaction mixture =3 L The value of equilibrium constant =50 Let x moles of both the reactants `A_(2)` and `B_(2)` react to form 2x moles of AB according to the equation given above. Therefore, the molar concentration per litre of the different species at equilibrium point is : `{:(,A_(2)(g),+,B_(2)(g),hArr,2AB(g)),("Initial moles//litre" ,1,,2,,0),("Moles //litre at equilibrium point" ,(1-x)/(3),,(2-x)/(3),,(2x)/(3)):}` Applying Law of chemical equilibrium `K_(c) =[[AB]^(2)]/[[A_(2)][B_(2)]]=((2x)/(3))^(2)/(((1-x)/(3))xx((2-x)/(3))^(2))` `50 = (4x^(2))/((1-x)(2-x))` `2x^(2) =25 (2-3x+x^(2)) , 23x^(2) -75x + 50 =0` By solving quadratic equation `.^(**) x= (75 underset(_)(+) sqrt((75^(2)-4 xx 23 xx50)))/(2xx23) =2.326 or 0.934`

Discussion

947.	The equilibrium constant of the reaction, ` A_(2) (g) + B_(2) (g) hArr 2AB (g) " at " 100^(@) C " is " 50.` . If a one litre flask containing one mole of `A_(2) ` is connected to a two litre flask containing two moles of `B_(2)` how many moles of AB will be formed at 373 K ?
Answer» ` {: (,A_(2),+,B_(2),hArr,2AB),(" Intial amounts:",1 "mole",,2 "mole",,0),("Amounts at eqm:",1-x,,2-x,,2x),(" Molar concs at eqm.",(1-x)/3,,(2-x)/3,,(2x)/3 "mol"L^(1)):}` ` K = ([AB]^(2))/([A_(2)][B_(2)]) or 50 = (2x^(2))/(((1-x)/3)((2-x)/3))` On solving , we get `x= 0955 ` mole Hence, no. of moles of AB formed at eqm. `= 2 xx 0 955 = 1*91 ` moles

Discussion

948.	Cl2(g) ⇌ 2Cl(g) Kc = 5 x 10-30Predict this reaction will have appreciable concentration of reactants and products.
Answer» K_c = 5 x 10^-30 which is a small value. Small values of K indicate greater amount of reactant at equilibrium.

Discussion

949.	Predict which of the following reaction will have appreciable concentration of reactants and product : ` (a) Cl_(2) (g) hArr 2 Cl (g), K_(c) = 5 xx 10^(-39)` (b) ` Cl_(2) (g) = 2 NO (g) hArr 2 NOCl (g) , K_(c)= 37 xx 10^(8)` ( c ) ` Cl_(2) (g) + 2 NO_(2) (g) hArr 2 NO_(2)Cl (g) , K_(c) = 18`
Answer» For reaction (c ) , as `k_(c)` is neither high nor very low , reactants and product will be present in comparable amounts.

Discussion

950.	Assertion : If reaction quotient, `Q_(c)` for a particular reaction is greater than `K_(c)` the reaction will proceed in the direction of reactants. Reason : Reaction quotient is defined in the same way as the equilibrium constant `K_(c)` except that the concentrations in `Q_(c)` are not necessarily equilibrium values.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - B If `Q_(c)gtK_(c)` then `Q_(c)` will tend to decrease so as to become `K_(c)`. As a result, the reaction will proceed in the backward direction.

Discussion

Explore topic-wise InterviewSolutions in .

The pH of 0.05 M aqueous solution of diethylamine is 12.0 . Calculate its `K_(b)`.

An aqueous solution of aniline of concentration 0.24 M is prepared. What concentraton of sodium hydroxide is needed in this solution so that anilinium ion concentration remains at `1xx10^(-8)` M? `(K_(a) "for" C_(6)H_(5)NH_(3)^(+)=2.4xx10^(-6)M)`

Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:A. increase in mass of `BaO_(2)`B. increase in mass of `BaO`C. increase in temperature on equilibriumD. increase in mass of `BaO_(2)` and `BaO` both.

For the reaction `A(g) + B(s) hArr C (g) + D (g), K_(c) = 49 mol dm^(-3)` at `127^(@)C`. Calculate `K_(p)`.

A system at equilibrium is described by the equation `SO_(2)Cl_(2) hArr SO_(2)+Cl_(2),Delta H =+ve`. When `Cl_(2)` is added to the equilibrium mixture at constant volume, the temperture of the system

For a reversible reactionn at 298 K the equilibrium constant K is 200. What is value of `DeltaG^(@)` at 298 K ?A. `-13.13kcal`B. `-0.13kcal`C. `-3.158kcal`D. `-0.413kcal`

Which of the following is not true about a reversible reaction ?A. The reaction does not proceed to completion.B. It cannot be influenced by catalystC. Number of moles of reactants and products is always euqal.D. It can be attained only in a closed container.

The value of ` K_(p) " for the reaction ", CO_(2) (g) + C (s) hArr 2 CO (g),` is 3.0 at 1000 K. Initially `, p_(co_(2)) = 0.48" bar "and P_(CO)= 0` bar and opure graphite is present , calculate the equilibrium partial pressures of `CO and CO_(2)`

Which of the following are reversible reactions ?A. ` AgNO_(3) (aq) + NaCl (aq) to` `AgCl (s) + NaNO_(3) (aq)`B. `KNO_(3) (aq) + NaCl (aq) to ` ` KCl (aq) + NaNO_(3)`C. ` BaCl_(2) (aq) + Na_(2) SO_(4) to` `BaSO_(4) (s) + 2 NaCl (aq) `D. `AgCl (s) + Water to Ag^(+) (aq) + Cl^(-) (aq)`

The values of ` K_(p_(1)) and K_(p_(2))` for the reactions `X hArr Y+Z" " ` …(1) and `A hArr 2 B " " ` …(2) are in the ratio of 9:1 .If the drgree of dissociation kof X and A be equal , then total pressure at equilibrium (1) and (2) are in the ratioA. `3:1`B. `1:9`C. `36:1`D. `1:1`

The equilibrium constant for the reaction `CH_(3) COOH + C_(2)H_(5)OH hArrCH_(3)COOC_(2)H_(5) + H_(2)O` is `4*0 " at " 25^(@)C. ` . Calculate the weight of ethyl acetate that will be obtained when 120 g of acetic acid are reacted with 92 g of alchol.

The equilibrium constant for the reaction : `CH_(3) COOH + C_(2)H_(5)OH hArr CH_(3) COOC_(2)H_(5) +H_(2)O` is 4.0 at `25^(@)C`. Calculate the weight of ethyl acetate that will be obtained when 120 g of acetic acid are reacted with 92 g of ethyl alcohol.

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?2HI ⇌ H2(g) + I2(g).

A sample of `HI(g)` is placed in flask at a pressure of `0.2 atm`. At equilibrium. The partial pressure of `HI(g)` is `0.04 atm`. What is `K_(p)` for the given equilibrium? `2HI(g) hArr H_(2)(g)+I_(2)(g)`

The `pH` of `0.004M` hydrazine `(NH_(2).NH_(2))` solution is `9.7`. Calculate its ionisation constant `K_(b)` and `pK_(b)`.

Calculate the concentration of all species present in `0.010M H_2SO_4` solution. `(K_(a_2)=1.3xx10^(-2))`

Give one example of everyday life in which there is gas y..’ &gt;'. solution equilibrium.

Which measurable property becomes constant in water, water vapour equilibrium at constant temperature?

How does a catalyst influence the equilibrium constant of a reversible reaction?

For the reaction, `SO_(2)(g)+1/2O_(2)(g)hArrSO_(3)(g)` if `K_(p)=K_(C)(RT)^(x)` where, the symbols have usual meaning, then the value of `x` is (assuming ideality)A. `=-1`B. `-1//2`C. `1//2`D. `1`

Explain with an example that strong acids have very weak conjugate bases and a strong base have a very weak conjugate acid.

Which of the following salts does show its correct nature mentioned against it ?A. KBr solution - NeutralB. NaCN solution - AcidicC. `NH_(4)NO_(3)` solution - AcisdicD. KF solution - Basic

Which of the following salts is the most basic in aqueous solution ?A. `Al(CN)_(3)`B. `CH_(3)CO OK`C. `FeCl_(3)`D. `Pb(CH_(3)CO O)_(2)`

Which of the following salts with a concentration .1 M will give a basic solution ?A. Ammonium acetateB. Ammonium chlorideC. Ammonium sulphateD. Sodium acetate

Which of the following relations is correct during the hydrolysis of salts of weak acid and strong bases ?A. `K_(h)=(K_(w))/(K_(a))`B. `K_(h)=(K_(w))/(K_(a)K_(b))`C. `K_(h)=(K_(w))/(K_(b))`D. `K_(h)=(K_(a))/(K_(w))`

In which of the following solvent is silver chloride most soluble ?A. 0.1 mol `dm^(-3) AgNO_(3)` solutionB. 0.1 mol `dm^(-3)` HCl solutionC. `H_(2)O`D. Aqueous ammonia

In which of the following solvents is silver chloride most soluble ?A. `0.1"mol dm"^(-3)AgNO_(3)" solution"`B. `0.1" mol dm"^(-3)HCl" solution"`C. `H_(2)O`D. Aqueous ammonia

A solution has been prepared by dissolving 0.063 g of `HNO_(3)` in 1000 mL of It . Calculate the `[H^(+)]` and `[OH^(-)]` of the solution.

What is the `pH` of `0.01 M` glycine solution? For glycine, `K_(a_(1))=4.5xx10^(-3)` and `K_(a_(2))=1.7xx10^(-10)` at `298 K`A. `3.0`B. `10.0`C. `6.1`D. `7.2`

Equilibirum constants `K_(1)` and `K_(2)` for the following equilibria `NO(g) + (1)/(2)O_(2)hArrNO_(2)(g)` and `2NO_(2)(g)hArr` `2NO(g) + O_(2)(g)` are related asA. `K_(2)=1//K_(1)`B. `K_(2)=K_(1)//2`C. `K_(2)=1//K_(1)^(2)`D. `K_(2)=K_(1)^(2)`

The `K_(p) " for the reaction," N_(2) O_(4) hArr 2NO_(2) ` is 640 mm at 775 K. Calculate the percentage dissociation of `N_(2)O_(4)` at equilibrium pressure of 160 mm. At what pressure the dissociation will be 50% ?

Equilibrium constants `K_(1)` and `K_(2)` for the following equilibria `NO(g) +1//2O_(2)(g) overset(K_(1))(hArr) NO_(2)(g)` and ` `2NO_(2)(g) overset(K_(2))(hArr)2NO(g)+O_(2)(g)` are related asA. `K_(1) =sqrt(K_(2))`B. `K_(2) =(1)/(K_(1))`C. `K_(1) = 2K_(2)`D. `K_(2) =(1)/(K_(1)^(-2))`

Given the equilibrium , `N_(2)O_(4)(g) hArr 2NO_(2) (g) " with " k_(p) =0.15 " atm at " 298 K.` (a) What is `K_(p)` using pressure in torr? (b) What is `K_(c)` using units of moles per litre ?

500 ml vessel contains 1.5 M each of A, B, C and D at equlibrium. If 0.5 M each of C and D are taken out, the value of `K_(c)` for `A+BhArrC+D` will beA. `1.0`B. `1/9`C. `4/9`D. `8/9`

For the reaction `a+bhArrc+d`, initially concentrations of a and b are equal and at equilibrium the concentration of will be twice of that of a. What will be equilibrium constant for the reaction ?A. 2B. 9C. 4D. 3

The equilibrium constant at `298K` for a reaction, `A+BhArrC+D` is 100. If the initial concentrations of all the four species were 1M each, then equilibirum concentration of `D` (in mol`L^(-1)`) will beA. `0.182`B. `0.818`C. `1.818`D. `1.182`

The solubility of `Sr(OH)_(2)` at `298 K` is `19.23 g L^(-1)` of solution. Calculate the concentrations cf strontium and hydroxyl ions and the `pH` of the solution.

The equilibrium constant for the reaction `A_(2)(g) +B_(2) hArr 2AB(g)` at 373 K is 50. If one litre flask containing one mole of `A_(2)` is connected to a two flask containing two moles of `B_(2)`, how many moles of AB will be formed at 373 K?

The equilibrium constant of the reaction, ` A_(2) (g) + B_(2) (g) hArr 2AB (g) " at " 100^(@) C " is " 50.` . If a one litre flask containing one mole of `A_(2) ` is connected to a two litre flask containing two moles of `B_(2)` how many moles of AB will be formed at 373 K ?

Cl2(g) ⇌ 2Cl(g) Kc = 5 x 10-30Predict this reaction will have appreciable concentration of reactants and products.

Predict which of the following reaction will have appreciable concentration of reactants and product : ` (a) Cl_(2) (g) hArr 2 Cl (g), K_(c) = 5 xx 10^(-39)` (b) ` Cl_(2) (g) = 2 NO (g) hArr 2 NOCl (g) , K_(c)= 3*7 xx 10^(8)` ( c ) ` Cl_(2) (g) + 2 NO_(2) (g) hArr 2 NO_(2)Cl (g) , K_(c) = 1*8`

Give one example of everyday life in which there is gas y..’ >'. solution equilibrium.

Predict which of the following reaction will have appreciable concentration of reactants and product : ` (a) Cl_(2) (g) hArr 2 Cl (g), K_(c) = 5 xx 10^(-39)` (b) ` Cl_(2) (g) = 2 NO (g) hArr 2 NOCl (g) , K_(c)= 37 xx 10^(8)` ( c ) ` Cl_(2) (g) + 2 NO_(2) (g) hArr 2 NO_(2)Cl (g) , K_(c) = 18`