1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 9 InterviewSolution

401.	What will be the correct order of vapour pressure of water, acetone and ether at `30^(@)C`. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point ?A. Water `lt` ether`lt ` acetoneB. Water `lt` acetone `lt` etherC. Ether `lt` acetone `lt` waterD. Acetone `lt` ether `lt` water
Answer» Correct Answer - B Greater the boiling point, less is the vapour pressure. Hence, the correct order of vapour pressures will be : water`lt ` acetone `lt` ether.

Discussion

402.	What will be the correct order of vapour pressure of water, acetone and ether at `30.^(@)C`. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point ?A. Water `lt` Ether `lt` AcetoneB. Water `lt` Acetone `lt` EtherC. Ether `lt` Acetone `lt` WaterD. Acetone `lt` Ether `lt` Water
Answer» Correct Answer - B Higher the boiling point lower will be the vapour pressure. Thus, the correct order of vapour pressure is `underset("(max.b.pt.)")("Water")lt"Acetone"ltunderset("(min.b.pt.)")("Ether")`

Discussion

403.	One mole of `H_(2)`, two moles of `I_(2)` and three moles of HI are injected in a litre flask. What will be the concentration of `H_(2) ,I_(2)` and HI at equilibrium at `490^(@)C` ? The equiibrium constant for the reaction at `490^(@)` is 45.9
Answer» `{:(,H_(2),+,I_(2)(g),hArr,2 HI(g)),("Intial conc.",1,2,,,3"moles"L^(1)), (" Concs.at eqm",(1-x),,(2-x),,(3+2x) "moe"L^(1)):}` `K = ((3 +2x)^(2))/((1-x)(2-x))=(9+ 4x^(2) + 12x)/(2+ x^(2) - 3 x ) = 459 ` ` :. 9 + 4 x^(2) + 12 x = 91 8 + 459x^(2) - 137 7 x ` or ` 419 x^(2) - 140 7 x + 82* 8 = 0 ` ` x = (149* 7 pm sqrt(1497)^(2)- 4 xx 419 xx 82 * 8 )/(2 xx 41 * 9) = (1497 pm sqrt( 2241009-13877 28))/(838)` ` (1497 pm 924)/(838)= 2 89 and 068 ` But `x = 2 89` is impossible . Hence , ` x= 0* 684` ` :. " Concentrations at equilibrium will be " ` ` [H_(2)] = 1- x=1 - 0684= 0 316 "mol" L^(-1)` ` [I_(2) ] = 2 - x = 2 - 0* 684 = 1* 316 "mol" L^(-1)` ` [HI] = 3 +2 x = 3 + 2 xx 0684 = 4 368 "mol"L^(-1)`

Discussion

404.	Fill in the blanks in the given table with the appropriate choice. A. `{:(p,q,r,s,t),(H_(2)CO_(3),SO_(4)^(2-),NH_(4)^(+),NH_(2)^(-),H_(3)O^(+)):}`B. `{:(p,q,r,s,t),(HCO_(3)^(-),H_(2)SO_(3),NH_(2)^(-),NH_(4)^(+),H_(3)O^(+)):}`C. `{:(p,q,r,s,t),(H_(2)CO_(3),H_(2)SO_(3),NH_(2)^(-),NH_(4)^(+),H_(3)O^(+)):}`D. `{:(p,q,r,s,t),(HCO_(3)^(-),H_(2)SO_(4),NH_(2)^(+),NH_(2)^(-),OH^(-)):}`
Answer» Correct Answer - A `{:("Species","Conjugate acid","Conjugate base",),(HCO_(3)^(-),H_(2)CO_(3),CO_(3)^(2-),),(HSO_(4)^(-),H_(2)SO_(4),SO_(4)^(2-),),(NH_(3),NH_(4)^(+),NH_(2)^(-),),(H_(2)O,H_(3)O^(+),OH^(-),):}`

Discussion

405.	Nucleophiles are __________ while electrophiles are __________ .A. Lewis bases, Lewis acidsB. Lewis acids, Lewis basesC. Bronsted acids, Bronsted basesD. Lewis acids Bronsted bases
Answer» Correct Answer - A Nucleophiles are Lewis bases while electrophiles are Lewis acids.

Discussion

406.	In which of the following equilibrium equation, `K_pgtK_c`?A. `2SO_3(g)hArr2SO_2(g)+O_2(g)`B. `PCl_3(g)+Cl_2(g)hArrPCl_5(g)`C. `H_2(g)+I_2(g)hArr2HI(g)`D. `N_2(g)+3H_2(g)hArr2NH_3(g)`
Answer» Correct Answer - A `K_p=K_c(RT)^(Deltan_g)` where `Deltan_g=(n_(gas, pro d))-(n_(gas, react))` `K_pgtK_c` when `Deltan_g` is positive. This is true for reaction (1) only. `Deltan_g=(2+1)-(2)` `=1` Note that `Deltan_g` refers to the number of moles of gaseous substances in the balanced equation not in the reaction vessel. For reactions (2) and (4), `Deltan_g` is `-ve`. Thus, `K_p lt K_c`. For reaction (3), `Deltan_g` is zero. Thus, `K_p=K_c`.

Discussion

407.	The value of ` K_(p) " is " 1 xx10^(-3) " atm"^(-1) " at " 25^(@) C. " for the reaction " : 2 NO + Cl_(2) hArr2 NOCl`. A flask contains NO at `002 " atm at " 25^(@)(C).` Calculate the mol of `Cl_(2)` that must be added if 1 % of the NO is to be converted to NOCl at equilibrium. The volume of the flask is such that ` 02` mol of the gas produce 1 atm pressure at ` 25^(@)C`. ( Ignore the probable association of NO to ` N_(2)O_(2)`)
Answer» Suppose intial pressure of `Cl_(2)` added is p atm . `{:("Then ",2NO,+,Cl_(2),hArr,2 NOCl),("Intial",002 "atm",,p " atm",,),("A eqm. " ,002 - (002)/100,,(p-(001)/100),,(002)/100),(,=2 xx 10^(-2)-2 xx10^(-4),,=p-10^(-4),,-2 xx 10^(-4)"atm"),(,=2 xx 10^(-4)(100-1),,,,),(,=198 xx10^(-4) "atm",,,,):}` ` K_(p) = (p_(NOCl)^(2))/(p_(NO)^(2)xx p_(Cl_(2)) )` ` 10^(-3) = ( 2xx 10^(-4))^(2)/ ((198 xx 10^(-4))^(2)xx(p-10^(-4))) ` or ` (p- 10^(-4)) = 4/(198)^(2) xx 1/ (10^(-3))= 0 102 or p= 0* 102 + 0* 0001 = 0* 1021 "atm "` Volume of the vessel can be calculated as follows : ` PV = nRT or V = (nRT)/P = (0* 2 xx 0* 082 xx 273 )/1 L = 4* 887 L ` To calculate the number of moles of `Cl_(2)` , again apply ` PV = nRT or n = (PV)/ (RT) = (01021 xx 4 887 )/(0082 xx 298) = 0 204 "mol"`

Discussion

408.	At `77^(@)C` and one atmospheric pressure , `N_(2)O_(4)` is 70% dissociated into `NO_(2)` What will be the volume occupied by the mixture under these conditions if we start with 10 g of `N_(2)O_(4)` ?
Answer» Molar mass of ` N_(2)O_(4) = 28 + 64 = 92 g "mol"^(-1)` ` {:(,N_(2)O_(4),hArr,2NO_(2)),("Intial moles",10/12,,0),("After dissociation",10/92-70/100xx10/92,,2xx0076),(,=0109 - 0076=0033,,= 0152):}` ` :. " Total moles after dissociation " = 0033 + 0* 152 = 0185, T = 77^(@)C = 77 + 273 K= 350 K` ` Pv = nRT or V= (nRT)/P = ( 0185 "mole" xx 0* 0821 L "atm " K^(-1) "mol"^(-1) xx350 K )/(1atm) = 5* 32 L.`

Discussion

409.	Which of the following will produce a buffer solution when mixed in equal volumes?(i) 0.1 mol dm–3 NH4OH and 0.1 mol dm–3 HCl(ii) 0.05 mol dm–3 NH4OH and 0.1 mol dm–3 HCl(iii) 0.1 mol dm–3 NH4OH and 0.05 mol dm–3 HCl(iv) 0.1 mol dm–3 CH4COONa and 0.1 mol dm–3 NaOH
Answer» (iii) 0.1 mol dm^–3 NH₄OH and 0.05 mol dm^–3 HCl

Discussion

410.	In which of the following solvents is silver chloride most soluble?(i) 0.1 mol dm–3 AgNO3 solution(ii) 0.1 mol dm–3 HCl solution(iii) H2O(iv) Aqueous ammonia
Answer» (iv) Aqueous ammonia

Discussion

411.	Which of the following salts will give basic solution on hydrolysis ?A. `NH_(4)Cl`B. `KCl`C. `K_(2)CO_(3)`D. `(NH_(4))_(2)CO_(3)`
Answer» Correct Answer - C `K_(2)CO_(3)+2H_(2)OhArrunderset("Weak acid")(H_(2)CO_(3))+underset("Strong base")(2KOH)` `KOHhArrK^(+)+OH^(-)` (Basic solution)

Discussion

412.	Which of the following will produce a buffer solution when mixed in equal volumes ?A. 0.1 mol `dm^(-3) NH_(4)OH` and 0.1 mol `dm^(-3)` HClB. 0.05 mol `dm^(-3) NH_(4)OH` and 0.1 mol `dm^(-3)` HClC. 0.1 mol `dm^(-3) NH_(4)OH` and 0.05 mol `dm^(-3)` HClD. 0.1 mol `dm^(-3) CH_(3)CO ON a` and 0.1 mol `dm^(-3)` NaOH
Answer» Correct Answer - C In (c), all HCl will be neutralized and `NH_(4)Cl` will be formed. Also some `NH_(4)OH` will be left unneutralized. Thus, the final solution will contain `NH_(4)OH and NH_(4)Cl` and hence will form a buffer.

Discussion

413.	A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures of acids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is `pH=pK_(a) + log. (["Salt"])/(["Acid"])` a similar equation is used for basic buffers. The pH of aqueous solution of single salts is calculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is `pH = 7-(1)/(2) pK_(b)-(1)/(2) log c` For weak acids and bases used by a chemist, data are given below: `K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5)` Also logarithmic values of some numbers are given below : log 1.8 = 0.2553, log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH value in each of the following cases. 100 mL of 0.05 M `NH_(4)OH` mixed with 100 mL of 0.10 M HCl solutionA. 1.6B. 2.6C. 3.6D. 4.6
Answer» Correct Answer - A 100 mL of 0.05 M `NH_(4)OH = 5` mmol of `NH_(4)OH` 100 mL of 0.10 M HCl = 10 mmol of HCl 5 mmol of `NH_(4)OH ` will react with 5 mmol of HCl to form 5 mmol of `NH_(4)Cl` HCl left = 5 mmol . Volume of solution = 200 mL `:. [HCl]=(5)/(200) = 0.025M`, `[H^(+)] =0.025 M`, `pH = - log (0.025) = 1.602`

Discussion

414.	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)`) Enthalpy of dissociation (in kJ `"mol"^(-1)`) of acetic acid obtained from Expt. 2 isA. `1.0`B. `10.0`C. `24.5`D. `51.4`
Answer» Correct Answer - A Moles of HCl or NaOH neutralized (n) `=100xx1` mmol = 0.1 mole Heat evolved = 0.1 mole `xx` 57.0 kJ `"mol"^(-1)` = 5.7 kJ = 5700 J Heat used to increase temperature of the solution (200 mL ) by `5.7^(@)` `=200 xx 4.2 xx 5.7` (sp. heat capacity of solution `= 4.2 JK^(-1)g^(-1)`) = 4788 J `:.` Heat used to increase the temperature of the calorimeter `=5700-4788 J = 912 J` ms `Delta t = 912` or ms `(5.7) = 912` or ms (calorimeter constant ) `= (912)/(5.7) = 160 J//^(@)C` In Expt. 2, heat evolved by neutralization of `CH_(3)CO OH` with NaOH `{:(=,200xx4.2xx5.6,+,160xx5.6),(,("Absorbed by solution"),,("Absorbed by calorimeter")):}` `= 4704+896 J = 5600 J` `CH_(3)CO OH` present in 100 mL of 2.0 M solution `=100xx2 ` mmol = 0.2 mol NaOH present in 100 mL of 1.0 M NaOH `=100xx1` mmol = 0.1 mol `:. CH_(3)CO OH` neutralized by NaOH = 0.1 mol Thus, heat evolved by neutralisation of 0.1 mol of `CH_(3)CO OH = 5600 J` `:.` Heat used in the dissociation of 0.1 mol of `CH_(3)CO OH =5700-5600J = 100 J ` `:.` Enthalpy of dissociation of acetic acid per mole = 1000 J = 1 kJ

Discussion

415.	Which of the following salts undergoes anionic hydrolysis ?A. `AICI_(3)`B. `CuSO_(4)`C. `Na_(2)CO_(3)`D. `NH_(4)CI`
Answer» Correct Answer - C `Na_(2)CO_(3)` is a salt of strong base `(NaOH)` and weak acid `(H_(2)CO_(3))`. Such a salt undergoes anionic hydrolysis to give basicv solution `CO_(3)^(2-)+H_(2)OhArrHCO_(3)^(-)OH^(-)` Rest of the salts undergo cationic hydrolysis to yield acidic solutions.

Discussion

416.	Which of the following salts is basic in nature? (a) NH4NO3 (b) Na2CO3 (c) Na2SO4 (d) NaCl
Answer» The correct answer is (b) Na₂CO₃

416.

Which of the following salts is basic in nature? (a) NH4NO3 (b) Na2CO3 (c) Na2SO4 (d) NaCl

Answer»

The correct answer is

(b) Na₂CO₃

Discussion

417.	What is the difference between a conjugate acid and a conjugate base?
Answer» A conjugate acid and base differ by a proton.

Discussion

418.	What is the `pH` of `0.001 M` aniline solution? The ionization constant of aniline `4.27xx10^(-10)`. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer» Correct Answer - `alpha = 6.53 xx 10^(-4), K_(a) = 2.35 xx 10^(-5)` `K_(b) = 4.27 xx 10^(-10)` `c = 0.001 M` `pH = ?` `alpha = ?` `k_(b) = calpha^(2)` `4.27 xx 10^(-10) = 0.001 xx alpha^(2)` `4270 xx 10^(-10) = alpha^(2)` `65.34 xx 10^(-5) = alpha = 6.53 xx 10^(-4)` Then, `["anion"] = calpha = .001 xx 65.34 xx 10^(-5)` `= .065 xx 10^(-5)` `pOH = - "log" (.065 xx 10^(-5))` `= 6.187` `pH = 7.813` Now, `K_(a) xx K_(b) = K_(w)` `:. 4.27 xx 10^(-10) xx K_(a) = K_(w)` `:. 4.27 xx 10^(-10) xx K_(a) = K_(w)` `K_(a) = (10^(-14))/(4.27 xx 10^(-10))` `= 2.34 xx 10^(-5)` Thus, the ionization constant of the conjugate acid of aniline is `2.34 xx 10^(-5)`

Discussion

419.	Which conjugate base is stronger CN- or F- ?
Answer» F – < CN – basic character.

Discussion

420.	Conjugate acid of a weak base is always stronger. What will be the decrinsing order of basic strength of the following conjugate bases? `OH^(-)`,`RO^(-)`,`CH_(3)`,`COO^(-)`,`CI^(-)`
Answer» The corresponding congugate acids are , `HOH, ROH, CH_(3)COOH,HCI`. The decrising order of acidic strength , `HCI,CH_(3)COOH,HOH,ROH` The decrising order of basic strength : `RO^(-),OH^(-),CH_(3)COO^(-),CI^(-)` Please remember that acidic and basic strengths of the conjugate acids and bases are inversely proportional to each other.

Discussion

421.	Calculate the pH of 0.125M H2SO4.
Answer» H₂SO₄→ 2H⁺ +SO₄^2- [H⁺] = 2[H₂SO₄] = 2 × 0.125 = 0.25 pH = -log₁₀[H⁺] = -log₁₀[0.25] = -log₁₀ [2.5 × 10^-1] pH = -log2.5 - log10^-1 = -0.3979 + 1 = 0.6021.

421.

Calculate the pH of 0.125M H2SO4.

Answer»

H₂SO₄→ 2H⁺ +SO₄^2-

[H⁺] = 2[H₂SO₄]

= 2 × 0.125

= 0.25

pH = -log₁₀[H⁺]

= -log₁₀[0.25]

= -log₁₀ [2.5 × 10^-1]

pH = -log2.5 - log10^-1

= -0.3979 + 1

= 0.6021.

Discussion

422.	Which of the following will produce a buffer sollution when mixed in equal volumes ?A. `0.1" mol dm"^(-3) NH_(4)OH " and " 0.1 " mol dm"^(-3)HCI`B. `0.05 " mol dm"^(-3) NH_(4)OH " and " 0.1" mol dm"^(-3)HCI`C. `0.1 " mol dm"^(-3) NH_(4)OH" and "0.05" mol dm"^(-3)HCI`D. `0.1 " mol dm"^(-3)CH_(3)COONA" and "0.1" mol dm"^(-3) CH_(3)COOH`.
Answer» Correct Answer - D is the correct answer.

Discussion

423.	Which of the following will produce a buffer sollution when mixed in equal volumes ?A. `0.1" mol dm"^(-3)NH_(4)OHand0.1" mol dm"^(-3)HCl`B. `0.05" mol dm"^(-3)NH_(4)OHand0.1" mol dm"^(-3)HCl`C. `0.1"mol dm"^(-3)NH_(4)OHand0.05" mol dm"^(-3)HCl`D. `0.1"mol dm"^(-3)CH_(3)COONaand0.1" mol dm"^(-3)NaOH`
Answer» Correct Answer - C In 0.1 mol `dm^(-3)NH_(4)OH and 0.05` mol `dm^(-3)` HCl, total amount of HCl reacts with `NH_(4)OH` to form `NH_(4)Cl` and some `NH_(4)OH` will be left unreacted. Thus, the resultant solution contains `NH_(4)Cl and NH_(4)OH` which will produce a buffer solution.

Discussion

424.	The standard Gibbs energy change at 300 K for the reaction `2 A hArr B+ C " is" 2494 2 J` At a given time, the composition of the reaction mixture is `[A] =1/2 , [B] =2 and [C] =1/2 .` The reaction proceeds in the : `[R=8314 J//K//mol, e=2*718]`A. Forward direction because `Q gt K_(c)`B. Reverse direction because `Q gt K_(c)`C. Forward direction because `Q lt K_(c)`D. Reverse direction because `Q lt K_(c)`
Answer» Correct Answer - B `Delta G^(@) = -230" RT "log K_(c)` ` 2494.2 = - 2303 (8.314) (300) log K_(c)` or log `K_(c) =-04342` or `K_(c)= "antilog" (bar 15658) = 0*3679` Now , when `[A] =1/2, [B] = 2 and [C] = 1/2` `Q_(c)= ([B][C])/([A]^(2)) = (2xx1/2)/((1/2)^(2))=4` As `Q_(c) gt K_(c),` reaction will proceed in the reverse direction.

Discussion

425.	Fizz is observed when soda water bottle is opened. Why?
Answer» An unopened soda water bottle is virtually bubble-free because the pressure inside the bottle keeps the carbon dioxide dissolved in the liquid. When it is opened, pressure is released and the gas bubbles are allowed to wiggle free from the liquid and rise to the surface in the form of fizz.

Discussion

426.	In which of the following acid-base titration, the pH is greater than 8 at the equivalence point ?A. Acetic acid versus ammoniaB. Acetic acidversus sodium hydroxideC. Hydrohloric acid versus ammoniaD. Hydrochloric acid versus sodium hydroxide.
Answer» `CH_(3)COOH+NH_(3)rarrCH_(3)COO^(-)N^(+)H_(4)` `CH_(3)COOH+NaOHrarr Na^(+)CH_(3)COO^(-)` `HCl+NH_(3)rarr NH_(4)Cl` `HCl+NaOH rarr NaCl` pH greater than 8 at the equivalence point implies that salt formed during the titration is undergoing anionic hydrolysis. Only the salt of strong base (NaOH) and weak acid `(CH_(3)COOH)` undergoes anionic hydrolysis to give basic solution.

Discussion

427.	The least soluble compound (salt) of the following isA. `CsCI(K_(sp)=10^(-12))`B. `HgS(K_(sp)=1xx10^(-52))`C. `PbCI_(2)(K_(sp)=1.7xx10^(-5))`D. `ZnS(K_(sp)=1.2xx10^(-23))`
Answer» Correct Answer - B Let the solubility of the salt be S mol `L^(-1)` `CsCl(s)hArrCs^(+)(aq.)+Cl^(-)(aq.)` `K_(sp)=S^(2)` or `S= sqrt(K_(sp))=sqrt(10^(-12))=10^(-6)mol L^(-1)` `HgS(s)hArr Hg^(2+)(aq.)+S^(2-)(aq.)` `K_(sp)=S^(2)` or `S=sqrt(K_(sp))=sqrt(10^(-52))=10^(-26)mol L^(-1)` `PbCl_(2)hArr Pb^(2+)(aq.)+2Cl^(-)(aq.)` `K_(sp)=(S)(2S)^(2)= 4S^(3)` or `S= ((K_(sp))/(4))^(1//3) = ((1.7xx10^(-5))/(4))^(1//3)` `ZnS(s) hArr Zn^(2+)(aq.)+S^(2-)(aq.)` `K_(sp)=S^(2)` or `S=sqrt(K_(sp))= sqrt(1.2xx10^(-23))` `=sqrt(12xx10^(-24))` `=3.5xx10^(-12)mol L^(-1)` Thus, the least soluble compound is HgS.

Discussion

428.	Which measurable property becomes constant in water `hArr` water vapour equilibrium at constant temperature ?
Answer» Correct Answer - Vapour pressure.

Discussion

429.	Which measurable property become constant in Water ---> <---- Water vapors equilibrium at constant temperature.
Answer» Vapour pressure.

Discussion

430.	In the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as –(a) If both assertion and reason are true, and reason is the true explanation of the assertion.(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.(c) If assertion is true, but reason is false.(d) If both assertion and reason are false Assertion (A). A catalyst does not influence the values of equilibrium constant.Reason (R). Catalysts influence the rate of both forward and backward reactions equally.
Answer» (a) If both assertion and reason are true, and reason is the true explanation of the assertion.

Discussion

431.	For the reaction:N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat; the equilibrium constant K = 0.50 at 673K(a) Write the expression for Kc and Kp.(b) What will be the equilibrium constant value for the reverse reaction at 673K?(c) What is the effect if increasing temperature on the yield of NH3?(d) What is the effect adding N2(g) and H2(g) on the yield of NH3?(e) What is the effect of increasing pressure on the yield of NH3?
Answer» (a) K_c = \(\frac{[NH_3]^2}{[N_2][H_2]^3}\) And K_p = \(\frac{[P_{NH_3}]^2}{P_{N_2}\times(P_{H_2})^3}\) (b) K_eq. (For reverse reaction) = \(\frac{1}{K}\) = \(\frac{1}{0.50}\) = 2 (c) As this is an exothermic reaction, on increasing the temperature, the reaction will proceed in reverse direction. This results in decrease in concentration of NH₃. (d) On increasing the concentration of N₂(g) and H₂(g), the yield of NH₃ will increase. (e) On increasing the pressure, the reaction will proceed in forward reaction in which number of moles of gas or pressure decreases.

431.

For the reaction:N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat; the equilibrium constant K = 0.50 at 673K(a) Write the expression for Kc and Kp.(b) What will be the equilibrium constant value for the reverse reaction at 673K?(c) What is the effect if increasing temperature on the yield of NH3?(d) What is the effect adding N2(g) and H2(g) on the yield of NH3?(e) What is the effect of increasing pressure on the yield of NH3?

Answer»

(a) K_c = \(\frac{[NH_3]^2}{[N_2][H_2]^3}\)

And K_p = \(\frac{[P_{NH_3}]^2}{P_{N_2}\times(P_{H_2})^3}\)

(b) K_eq. (For reverse reaction) = \(\frac{1}{K}\) = \(\frac{1}{0.50}\) = 2

(c) As this is an exothermic reaction, on increasing the temperature, the reaction will proceed in reverse direction. This results in decrease in concentration of NH₃.

(d) On increasing the concentration of N₂(g) and H₂(g), the yield of NH₃ will increase.

(e) On increasing the pressure, the reaction will proceed in forward reaction in which number of moles of gas or pressure decreases.

Discussion

432.	What is the effect of temperature on the reactions?N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat
Answer» The increase in temperature will favour backward reaction because the reaction is exothermic.

Discussion

433.	The solubility product of `AgCl` is `1.56xx10^(-10)`find solubility in g/ltrA. `143.5`B. 108C. `1.57xx10^(-8)`D. `1.79xx10^(-3)`
Answer» Correct Answer - D `AgClhArrunderset(s)(Ag)^(+)+underset(s)(Cl^(-))` `s^(2)=1.5625xx10^(-10)` `s=1.25xx10^(-5)"mol L"^(-1)` Solubility in g `L^(-1)` = Molar mass `xx` s `=143.5xx1.25xx10^(-5)=1.79xx10^(-3)"g L"^(-1)`

Discussion

434.	The pH of a `10^(-8)M` solution of HCl is water isA. 8B. -8C. between 7 and 8D. between 6 and 7.
Answer» Correct Answer - D

Discussion

435.	pH is the negative logarithm of `H^(+)`` pH=log[H^(+)]` `HCllt H^(+)+Cl^(-)` `H_(2)OhArrH^(+)+OH^(-)` ` K_(W)=[H^(+)][OH^(-)]` `K_(W)` depend on the temperatue. With rise in temperature `K_(W)` increases. At 298 K, pH of pure water = 7 The exact concentration of `H^(+)` in `10^(-6)` M HCl given byA. `10^(-6)+10^(-8)`B. `10^(-6)+10^(-7)`C. `10^(-6)`D. `10^(-6)-10^(-7)`
Answer» Correct Answer - B

Discussion

436.	pH is the negative logarithm of `H^(+)`` pH=log[H^(+)]` `HCl lt H^(+)+Cl^(-)` `H_(2)OhArrH^(+)+OH^(-)` ` K_(W)=[H^(+)][OH^(-)]` `K_(W)` depend on the temperatue. With rise in temperature `K_(W)` increases. At 298 K, pH of pure water = 7 The pH at first equivalance of `H_(3)PO_(4)` vs NaOH will beA. 7B. `gt7`C. `lt7`D. Depend on the concentration of titrant
Answer» Correct Answer - C

Discussion

437.	pH is the negative logarithm of `H^(+)`` pH=log[H^(+)]` `HCllt H^(+)+Cl^(-)` `H_(2)OhArrH^(+)+OH^(-)` ` K_(W)=[H^(+)][OH^(-)]` `K_(W)` depend on the temperatue. With rise in temperature `K_(W)` increases. At 298 K, pH of pure water = 7 At 373 K, pH of the pure water isA. 7B. `gt7`C. `lt7`D. Cannot be stated
Answer» Correct Answer - C

Discussion

438.	The solubility of `CaF_(2) " in water at " 298 K is 1.7 xx 10^(-3)` grams per 100 mL of the solution. Calculate solubility product of `CaF_(2).`
Answer» Solubility of `CaF_(2) = 1.7xx10^(-3) g//100mL =17 xx 10^(-3) gL^(-1)` Molar mass of `CaF_(2) =(40 +2 xx 19)=78 g mol^(-1)` Solubility of `CaF_(2)=((17xx10^(-3) gL^(-1)))/((78.0 g mol^(-1))) =2.18 xx 10^(-4) mol L^(-1)` The solubility equibrium for the saturated solution of `CaF_(2)` is : `CaF_(2)(s)overset(aq)(hArr) Ca^(2+) (aq) +2F^(-) (aq)` `[Ca^(2+)(aq)] =2.18 xx10^(-4) mol L^(-1)` `[F^(-)(aq)] =2 xx2.18 xx 10^(-4) mol L^(-1)` Applying Law of chemical equilibrium : `K_(sp) =[Ca^(2+) (aq)] [F^(-) (aq)]^(2)` `=(2.18 xx 10^(-4) mol L^(-1)) xx (2.18 xx 10^(-4) mol L^(-1))^(2) = 4.14 xx 10^(-11) mol^(3).`

Discussion

439.	The following concentrations were obtained for the formation of `NH_(3)` from `N_(2)` and `H_(2)` at equilibrium at `500 K`. `[N_(2)]=1.5xx10^(-2) M, [H_(2)]=3.0xx10^(-2)M,` and `[NH_(3)]=1.2xx10^(-2)M`. Calculate the equilibrium constant.
Answer» `N_(2)(g) + 3H_(2) hArr 2NH_(3)(g)` `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=([1.2xx10^(-2)M])/([1.5xx10^(-2)M][3.0xx10^(-2)M])=3.55 xx 10^(2)M`

Discussion

440.	In each of the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer as Assertion . For the reaction , ` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) , " units of " K_(c) = L^(2) mol^(-2)` Reason . For the reaction , ` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g),` Equilibrium constant , `K_(c) - ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`.A. If both assertion and reason are true, and reason is the true explanation of the assertion .B. If both assertion and reason are true but reason is the true explanation of the assertion .C. If assertion is true, but reason is false.D. If both assertion and reason are false .
Answer» Correct Answer - A ` K_(c) = [NH_(3)]^(2)/([N_(2)][H_(2)]^(3))=( " mol" L^(-1))^(2)/(("mol"L^(-1))("mol"L^(-1))^(3))` ` = (mol L^(-1))^(-2)=L^(2) mol^(-2)` Hence , R is the correct explanation of A.

Discussion

441.	In a chemical reaction under equilibrium , there is no change in moler conertration of products and reactants. Does the reaction stop?
Answer» No, it does not stop. The equilibrium is of dynamic nature.

Discussion

442.	`0.1` mole of `CH_(3)NH_(2) (K_(b)=5xx10^(-4))` is mixed with `0.08` mole of `HCl` and diluted to one litre. The `[H^(+)]` in solution isA. `8xx 10^(-2)`B. `2xx10^(11)`C. `1.23 xx 10^(-4)`D. `8xx 10^(-11)`
Answer» Correct Answer - D `{:(CH_(3)NH_(2),+,HCIto ,CH_(3)NH, +CI^(-)),(0.1"mol",,0.08"mol",,0),(0.02"mol",,0,,0.08):}` `pOh =pK_(b) + log .("salt")/("base")` `=- log 5 xx 10^(-4) + log .(0.08)/(0.02)` `=(4 -log 5) +log 4` `=4- 0.6989 + 0.6020 =3.903` `pOH =3.903, pH =10.0967` `[H^(+)] = 8xx 10^(-11)`

Discussion

443.	For the reaction `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)` the forward reaction at constant temperature is favoured byA. Introducing an inert gas at constatn volumeB. Introducing `PCl_(3)(g)` gas at constnt volumeC. Introducing `PCl_(5)(g)` gas at constant volumeD. Introducing `Cl_(2)(g)` gas at constant volume
Answer» Correct Answer - C

Discussion

444.	The solubility of `Mg(OH)_(2) " is " 8.352 xx 10^(-3) g L^(-1)` at 298 K. Calculate the `K_(sp)` of `Mg(OH)_(2)` at this temerature.
Answer» Correct Answer - `1.194 xx 10^(-11) "moll"^(3) L^(-3)` The solubility of Mg `(OH)_(2) =8,352 xx 10^(-3) gL^(-1)` , `=((8.352 xx 10^(-3) gL^(-1)))/((58 g " mol ")) = 1.44 xx 10^(-4) " mol"//L^(-1)` The solubility equilibrium in saturated solution is : `Mg(OH)_(2) (s) hArr Mg^(2+) (aq) +2OH^(-) (aq)` `[Mg^(2+)(aq)] =1.44 xx 10^(-4) " mol " L^(-1)` `[OH^(-)(aq)] =2xx 1.44 xx 10^(-4) =2.88 xx 10^(-4) " mol " L^(-1)` `K_(sp) " of " Mg (OH)_(2) (s) =[Mg^(2+) (aq)] OH^(-) (aq)]^(2)` `= (1.44 xx10^(-4) " mol " L^(-1)) xx (2.88 xx 10^(-4) " mol " L^(-1))^(2) =1.194 xx 10^(-11) (" mol " L^(-1))^(3)`

Discussion

445.	Species acting as both Bronsted acid and base is:A. `(HSO_(4))^(2-)`B. `Na_(2)CO_(3)`C. `(NH_(2))^(1-)`D. `(OH)^(1-)`
Answer» Correct Answer - A Is the correct answer.

Discussion

446.	If the solubility of an aqueoues solution of `Mg(OH)_(2)` be X mole litre, then `K_(SP)` of `Mg(OH)_(2)` is:A. `4x^(3)`B. `108 x^(5)`C. `27x^(4)`D. `9X.`
Answer» Correct Answer - A `Mg (OH)_(2) hArr Mg^(2+) + 2[OH]^(-)` `K_(sp) =[Mg^(2+)] [OH^(-)]^(2) = (x) xx (2x)^(2) =4x^(3).`

Discussion

447.	The pH of `10^(-10)` M NaOH solution lies between........and ........... .
Answer» Correct Answer - 7 and 8

Discussion

448.	An acid solution of `pH =6` is diluted `100` times. The `pH` of solution becomes
Answer» pH of solution =6 , `[H_(3)O^(+)] =10^(-6)` After diluting the solution to 100 times. `[H_(3)O^(+)] =10^(-6)//100 =10^(-8) M` `[H_(3)O^(+)]` by the dissociation of water `=10^(-7) M` Total `[H_(3)O^(+)] =10^(-8) M+ 10^(-7) (10^(-1) +1) = 1.1 xx 10^(-7) M` pH =- log `(1.1xx10^(-7)) =(7 -log 1.1) =(7-0.0414) = 6.9586`

Discussion

449.	The number of moles of `Ca(OH)_(2)` required to prepare 250 ml of solution with pH 14 (assuming complete ionization) isA. `0.25`B. `1.0`C. `0.125`D. `10.0`
Answer» `C_(H^(+))=10^(-pH)mol L^(-1)` `=10^(-14)mol L^(-1)` For any aqueous solution at 298K, we have `pH+pOH=14` `:. pOH=14-pH=14-14=0` Since `C_(OH^(-))=10^(-pOH)mol L^(-1)` `=10^(o)mol L^(-1)` `=1 mol L^(-1)` According to dissociation equation, `Ca(OH)_(2)rarr Ca^(2+)(aq.)+2OH^(-)(aq.)` Every 1 mol of `Ca(OH)_(2)` provides 2 mol of `OH^(-)` ions. Thus, `C_(Ca(OH)_(2))=(1)/(2)C_(OH^(-))=(1)/(2)(1 mol L^(-1))` `=0.5 mol L^(-1)` Now, Molarity (M) `= (n_(Ca(OH)_(2)))/(V_(mL))xx(1000mL)/(L)` `0.5mol L^(-1)= (n_(Ca(OH)_(2)))/(250 mL)xx(1000 mL)/(L)` `= 0.125 mol`

Discussion

450.	Calculate the degree of ionization of pure water at `25^@C`.
Answer» Strategy: The degree of ionization of pure water `(alpha)` is the ratio of ionized water to that of unionized water. Solution: The density of pure water is `1g mL^(-1)` or `1000gL^(-1)`. The molar mass of water is `18gmol^(-1)`. Thus, the molarity of pure water can be given as `C_(H_2O)=(d_(H_2O))/("Molar mass"_(H_2O))=(1000gL^(-1))/(18gmol^(-1))` Now, we can calculate the degree of ionization of pure water as `alpha=(C_(H^+))/(C_(H_2O))` or `(C_(OH^-))/(C_(H_2O))=(1.0xx10^-7)/(55.5)` `=1.8xx10^-9`

Discussion

Explore topic-wise InterviewSolutions in .

One mole of `H_(2)`, two moles of `I_(2)` and three moles of HI are injected in a litre flask. What will be the concentration of `H_(2) ,I_(2)` and HI at equilibrium at `490^(@)C` ? The equiibrium constant for the reaction at `490^(@)` is 45.9

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Which of the following salts is basic in nature? (a) NH4NO3 (b) Na2CO3 (c) Na2SO4 (d) NaCl

What is the difference between a conjugate acid and a conjugate base?

What is the `pH` of `0.001 M` aniline solution? The ionization constant of aniline `4.27xx10^(-10)`. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Which conjugate base is stronger CN- or F- ?

Conjugate acid of a weak base is always stronger. What will be the decrinsing order of basic strength of the following conjugate bases? `OH^(-)`,`RO^(-)`,`CH_(3)`,`COO^(-)`,`CI^(-)`

Calculate the pH of 0.125M H2SO4.

Which of the following will produce a buffer sollution when mixed in equal volumes ?A. `0.1" mol dm"^(-3)NH_(4)OHand0.1" mol dm"^(-3)HCl`B. `0.05" mol dm"^(-3)NH_(4)OHand0.1" mol dm"^(-3)HCl`C. `0.1"mol dm"^(-3)NH_(4)OHand0.05" mol dm"^(-3)HCl`D. `0.1"mol dm"^(-3)CH_(3)COONaand0.1" mol dm"^(-3)NaOH`

Fizz is observed when soda water bottle is opened. Why?

In which of the following acid-base titration, the pH is greater than 8 at the equivalence point ?A. Acetic acid versus ammoniaB. Acetic acidversus sodium hydroxideC. Hydrohloric acid versus ammoniaD. Hydrochloric acid versus sodium hydroxide.

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

Which measurable property become constant in Water ---> <---- Water vapors equilibrium at constant temperature.

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The solubility product of `AgCl` is `1.56xx10^(-10)`find solubility in g/ltrA. `143.5`B. 108C. `1.57xx10^(-8)`D. `1.79xx10^(-3)`

The pH of a `10^(-8)M` solution of HCl is water isA. 8B. -8C. between 7 and 8D. between 6 and 7.

The solubility of `CaF_(2) " in water at " 298 K is 1.7 xx 10^(-3)` grams per 100 mL of the solution. Calculate solubility product of `CaF_(2).`

The following concentrations were obtained for the formation of `NH_(3)` from `N_(2)` and `H_(2)` at equilibrium at `500 K`. `[N_(2)]=1.5xx10^(-2) M, [H_(2)]=3.0xx10^(-2)M,` and `[NH_(3)]=1.2xx10^(-2)M`. Calculate the equilibrium constant.

In a chemical reaction under equilibrium , there is no change in moler conertration of products and reactants. Does the reaction stop?

`0.1` mole of `CH_(3)NH_(2) (K_(b)=5xx10^(-4))` is mixed with `0.08` mole of `HCl` and diluted to one litre. The `[H^(+)]` in solution isA. `8xx 10^(-2)`B. `2xx10^(11)`C. `1.23 xx 10^(-4)`D. `8xx 10^(-11)`

The solubility of `Mg(OH)_(2) " is " 8.352 xx 10^(-3) g L^(-1)` at 298 K. Calculate the `K_(sp)` of `Mg(OH)_(2)` at this temerature.

Species acting as both Bronsted acid and base is:A. `(HSO_(4))^(2-)`B. `Na_(2)CO_(3)`C. `(NH_(2))^(1-)`D. `(OH)^(1-)`

If the solubility of an aqueoues solution of `Mg(OH)_(2)` be X mole litre, then `K_(SP)` of `Mg(OH)_(2)` is:A. `4x^(3)`B. `108 x^(5)`C. `27x^(4)`D. `9X.`

The pH of `10^(-10)` M NaOH solution lies between........and ........... .

An acid solution of `pH =6` is diluted `100` times. The `pH` of solution becomes

The number of moles of `Ca(OH)_(2)` required to prepare 250 ml of solution with pH 14 (assuming complete ionization) isA. `0.25`B. `1.0`C. `0.125`D. `10.0`

Calculate the degree of ionization of pure water at `25^@C`.

Explore topic-wise InterviewSolutions in .

One mole of `H_(2)`, two moles of `I_(2)` and three moles of HI are injected in a litre flask. What will be the concentration of `H_(2) ,I_(2)` and HI at equilibrium at `490^(@)C` ? The equiibrium constant for the reaction at `490^(@)` is 45.9

Nucleophiles are __________ while electrophiles are __________ .A. Lewis bases, Lewis acidsB. Lewis acids, Lewis basesC. Bronsted acids, Bronsted basesD. Lewis acids Bronsted bases

In which of the following equilibrium equation, `K_pgtK_c`?A. `2SO_3(g)hArr2SO_2(g)+O_2(g)`B. `PCl_3(g)+Cl_2(g)hArrPCl_5(g)`C. `H_2(g)+I_2(g)hArr2HI(g)`D. `N_2(g)+3H_2(g)hArr2NH_3(g)`

At `77^(@)C` and one atmospheric pressure , `N_(2)O_(4)` is 70% dissociated into `NO_(2)` What will be the volume occupied by the mixture under these conditions if we start with 10 g of `N_(2)O_(4)` ?

Which of the following will produce a buffer solution when mixed in equal volumes?(i) 0.1 mol dm–3 NH4OH and 0.1 mol dm–3 HCl(ii) 0.05 mol dm–3 NH4OH and 0.1 mol dm–3 HCl(iii) 0.1 mol dm–3 NH4OH and 0.05 mol dm–3 HCl(iv) 0.1 mol dm–3 CH4COONa and 0.1 mol dm–3 NaOH

In which of the following solvents is silver chloride most soluble?(i) 0.1 mol dm–3 AgNO3 solution(ii) 0.1 mol dm–3 HCl solution(iii) H2O(iv) Aqueous ammonia

Which of the following salts will give basic solution on hydrolysis ?A. `NH_(4)Cl`B. `KCl`C. `K_(2)CO_(3)`D. `(NH_(4))_(2)CO_(3)`

Which of the following salts undergoes anionic hydrolysis ?A. `AICI_(3)`B. `CuSO_(4)`C. `Na_(2)CO_(3)`D. `NH_(4)CI`

Which of the following salts is basic in nature? (a) NH4NO3 (b) Na2CO3 (c) Na2SO4 (d) NaCl

What is the difference between a conjugate acid and a conjugate base?

What is the `pH` of `0.001 M` aniline solution? The ionization constant of aniline `4.27xx10^(-10)`. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Which conjugate base is stronger CN- or F- ?

Conjugate acid of a weak base is always stronger. What will be the decrinsing order of basic strength of the following conjugate bases? `OH^(-)`,`RO^(-)`,`CH_(3)`,`COO^(-)`,`CI^(-)`

Calculate the pH of 0.125M H2SO4.

Fizz is observed when soda water bottle is opened. Why?

In which of the following acid-base titration, the pH is greater than 8 at the equivalence point ?A. Acetic acid versus ammoniaB. Acetic acidversus sodium hydroxideC. Hydrohloric acid versus ammoniaD. Hydrochloric acid versus sodium hydroxide.

The least soluble compound (salt) of the following isA. `CsCI(K_(sp)=10^(-12))`B. `HgS(K_(sp)=1xx10^(-52))`C. `PbCI_(2)(K_(sp)=1.7xx10^(-5))`D. `ZnS(K_(sp)=1.2xx10^(-23))`

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

Which measurable property become constant in Water ---&gt; &lt;---- Water vapors equilibrium at constant temperature.

What is the effect of temperature on the reactions?N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat

The solubility product of `AgCl` is `1.56xx10^(-10)`find solubility in g/ltrA. `143.5`B. 108C. `1.57xx10^(-8)`D. `1.79xx10^(-3)`

The pH of a `10^(-8)M` solution of HCl is water isA. 8B. -8C. between 7 and 8D. between 6 and 7.

The solubility of `CaF_(2) " in water at " 298 K is 1.7 xx 10^(-3)` grams per 100 mL of the solution. Calculate solubility product of `CaF_(2).`

The following concentrations were obtained for the formation of `NH_(3)` from `N_(2)` and `H_(2)` at equilibrium at `500 K`. `[N_(2)]=1.5xx10^(-2) M, [H_(2)]=3.0xx10^(-2)M,` and `[NH_(3)]=1.2xx10^(-2)M`. Calculate the equilibrium constant.

In a chemical reaction under equilibrium , there is no change in moler conertration of products and reactants. Does the reaction stop?

`0.1` mole of `CH_(3)NH_(2) (K_(b)=5xx10^(-4))` is mixed with `0.08` mole of `HCl` and diluted to one litre. The `[H^(+)]` in solution isA. `8xx 10^(-2)`B. `2xx10^(11)`C. `1.23 xx 10^(-4)`D. `8xx 10^(-11)`

The solubility of `Mg(OH)_(2) " is " 8.352 xx 10^(-3) g L^(-1)` at 298 K. Calculate the `K_(sp)` of `Mg(OH)_(2)` at this temerature.

Species acting as both Bronsted acid and base is:A. `(HSO_(4))^(2-)`B. `Na_(2)CO_(3)`C. `(NH_(2))^(1-)`D. `(OH)^(1-)`

If the solubility of an aqueoues solution of `Mg(OH)_(2)` be X mole litre, then `K_(SP)` of `Mg(OH)_(2)` is:A. `4x^(3)`B. `108 x^(5)`C. `27x^(4)`D. `9X.`

The pH of `10^(-10)` M NaOH solution lies between........and ........... .

An acid solution of `pH =6` is diluted `100` times. The `pH` of solution becomes

The number of moles of `Ca(OH)_(2)` required to prepare 250 ml of solution with pH 14 (assuming complete ionization) isA. `0.25`B. `1.0`C. `0.125`D. `10.0`

Calculate the degree of ionization of pure water at `25^@C`.

Nucleophiles are while electrophiles are .A. Lewis bases, Lewis acidsB. Lewis acids, Lewis basesC. Bronsted acids, Bronsted basesD. Lewis acids Bronsted bases

Which measurable property become constant in Water ---> <---- Water vapors equilibrium at constant temperature.