1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 17 InterviewSolution

801.	Why solution of common salt in water conduct electricity while that of sugar solution does not?
Answer» Common salt (NaCI) gives Na and Cl ions in its aqueous solution. So, it conducts electricity. While sugar (C₁₂H₂₂O₁₁) is a non-electrolyte and does not give ions in aqueous solution so it does not conduct electricity.

Discussion

802.	Why is ammonia termed as a base though it does not contain OH− ions?
Answer» When ammonia (NH₃) dissolves in water it accepts proton (H⁺) from the water to form NH₄⁺ ion due to presence of lone pair of electrons. So, it acts as base. Such type of base is called Bronsted base.

Discussion

803.	Which of the following is correct if reaction quotient `(Q)=1`?A. `DeltaG=0`B. `DeltaG^(@)=0`C. `DeltaGg tDeltaG^(@)`D. `DeltaG=DeltaG^(@)`
Answer» Correct Answer - D

Discussion

804.	Choose the correct order arranged in decreasing order of basicityA. `CH-=C^(-) gt CH_(3)O^(-) gt OH^(-)`B. `OH^(-) gt CH_(3)O^(-) CH - = C^(-)`C. `CH_(3)O ^(-) gt OH^(-) gt CH -= C^(-)`D. `CH_(3)O^(-) gt CH-= C^(-) gt OH^(-)`
Answer» Correct Answer - A The strengths of conjugate acids of the given bases are in the order : `H_(2)OgtCH_(3)OHgtHC-=CH` Stronger the acid, weaker is the conjugate base. Hence, decreasing order of basicity is `HC-=C^(-)CH_(3)O^(-)gtOH^(-)`

Discussion

805.	Why solid NaCl starts separating out from a saturated solution of NaCl if HCl gas is passed through it ?
Answer» HCl in the solution provides `Cl^(-)` ions. This increases the ionic product of NaCl and so the solid NaCl Starts separating out.

Discussion

806.	Why solid NaCl starts separating out from a saturated solution of NaCl if HCl gas is passed through it?
Answer» HCl in the solution provides Cl ions. This increases the ionic product in NaCl and so the solid NaCl starts separating out.

Discussion

807.	What happens HCl gas passed through saturated NaCl solution?
Answer» NaCl will be precipitated out.

Discussion

808.	The equilibrium for the synthesis of ammonia isN2 + 3H2 \(\rightleftharpoons\) NH3 + xkJWhat is the effect of pressure, temperature and concentration on this equilibrium ?
Answer» Effect of pressure : High pressure favours the formation of ammonia. An increase in pressure will favour the reaction that decreases the number of gas molecules. The equilibrium will shift to the right and the yield of ammoina will be increased. Effect of temperature: Low temperature favours the formation of ammonia, a decrease in temperature favours the exothermic reaction because it releases energy, it will favour the forward reaction and the yield of ammonia will be increased. Effect of concentration: High concentration of reactants favours the formation of ammonia, According to Le-Chatelier’s principle, if the concentration of N₂ or H₂ is increased then equilibrium will shift to forward direction and the yield of ammonia will be increased.

808.

The equilibrium for the synthesis of ammonia isN2 + 3H2 \(\rightleftharpoons\) NH3 + xkJWhat is the effect of pressure, temperature and concentration on this equilibrium ?

Answer»

Effect of pressure :
High pressure favours the formation of ammonia. An increase in pressure will favour the reaction that decreases the number of gas molecules. The equilibrium will shift to the right and the yield of ammoina will be increased.

Effect of temperature:
Low temperature favours the formation of ammonia, a decrease in temperature favours the exothermic reaction because it releases energy, it will favour the forward reaction and the yield of ammonia will be increased.

Effect of concentration:
High concentration of reactants favours the formation of ammonia, According to Le-Chatelier’s principle, if the concentration of N₂ or H₂ is increased then equilibrium will shift to forward direction and the yield of ammonia will be increased.

Discussion

809.	Which states of a substance are at equilibrium at its melting point?
Answer» Solid \(\rightleftharpoons\) Liquid

Discussion

810.	When hydrogen molecules decompose into its atoms, which conditions give the maximum yield of hydrogen atoms ?A. High temperatue and low pressureB. Low temperatuer and high pressureC. High temperatuer and high pressureD. Low temperature and low pressure
Answer» Correct Answer - A

Discussion

811.	What are the conditions for getting maximum yield of NHs by Haber’s process?
Answer» High concentrations of N₂ and H₂, low temperature, high pressure.

Discussion

812.	What are the conditions for getting maximum yield of NH3 by Haber’s process?
Answer» High conc. Of N₂ &H₂, Low Temp. & High Pressure.

Discussion

813.	What are the conditions for more yield of ammonia by Haber’s process?
Answer» Haber’s Process: N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) Low temperature favours the exothermic reection because it releases energy. So the forward reaction is forward and the yield of ammonia will increase. High pressure favours the reaction that decreaces the number of gas molecules so the forward reaction is favoured. The equilibrium will shift to the right and the yield of ammonia will increase.

813.

What are the conditions for more yield of ammonia by Haber’s process?

Answer»

Haber’s Process:

N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)

Low temperature favours the exothermic reection because it releases energy. So the forward reaction is forward and the yield of ammonia will increase.

High pressure favours the reaction that decreaces the number of gas molecules so the forward reaction is favoured. The equilibrium will shift to the right and the yield of ammonia will increase.

Discussion

814.	Which of the following will occur if a 0.1 M solution of a weak acid is diluted to 0.01 M at constant temperature ?A. `[H^(+)]` will decrease to 0.01 MB. pH will decreaseC. percentage ionization will increaseD. `K_(a)` will increase.
Answer» Correct Answer - C

Discussion

815.	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 falseAssertion (A). If to the equilibrium PCl5(g) ⇋PCl3(g) + Cl2(g), in a closed vessel, an inert gas is added, total pressure will increase and hence, equilibrium will shift backward.Reason (R). Addition of an inert gas to an equilibrium mixture at constant volume shifts the equilibrium in the backward direction.
Answer» (d) If both assertion and reason are false.

Discussion

816.	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 falseAssertion (A). The active mass of pure solid and pure liquid is taken unity.Reason (R). The active mass of pure solids and liquids depends on density and molecular mass. The density and molecular mass of pure liquids and solids are constant.
Answer» (a) If both assertion and reason are true, and reason is the true explanation of the assertion.

Discussion

817.	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 falseAssertion (A). If standard free energy change of a reaction is zero, this implies that equilibrium constant of the reaction is unity.Reason (R). For a reaction in equilibrium, equilibrium constant is always unity.
Answer» (c) If assertion is true, but reason is false.

Discussion

818.	Define equilibrium constant.
Answer» It is the ratio of the product of molar concentrations of the product to the product of molar concentrations of the reactants raised to the respective stoichiometric coefficients as mentioned in the balanced chemical equation.

Discussion

819.	State law of equilibrium.
Answer» Law of equilibrium states that “at a given temperature ,the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant valve”.

Discussion

820.

(a) Define the equilibrium constant.(b) For the general reaction:aA(g) + bB(g) ⇌ cC(g) + dD(g)Derive the relationship between Kp and Kc.

Answer»

(a) Equilibrium constant: It is the ratio of the rate constants of forward reaction to that of backward reaction.

(b) For the general reaction, aA(g) + bB(g) ⇌ cC(g) + dD(g) Equilibrium constant

K_c= \(\frac{[C]^c[D]^d}{[A]^a[B]^b}\)

K_p = \(\frac{[P_C]^c[P_D]^d}{[P_A]^a[P_B]^b}\)

= \(\frac{[C]^c[D]^d\,(RT)^{c+d}}{[A]^a[B]^b\,(RT)^{(a+b)}}\)

= \(\frac{[C]^c[D]^d}{[A]^a[B]^b}\)(RT)^(c+d)-(a+b)

⁼\(\frac{[C]^c[D]^d}{[A]^a[B]^b}\) (RT)^Δn

K_p = K_c.(RT)^Δn

Discussion

821.	Predict the nature of solution when NH4NO3 undergo hydrolysis.
Answer» NH₄NO₃ gives strong acid (HNO₃) and weak base (NH₄OH) on hydrolysis. So, the solution is acidic in nature.

Discussion

822.	The species: H2O can act both as Bronsted-acid and base. For each case give the corresponding conjugate acid and conjugate base.
Answer» H₂O can act both as Bronsted -acid and base as it can accept and donate proton. It undergoes auto-ionisation to form corresponding conjugate acid (H₃O⁺) and conjugate base (OH^-). H₂O₍₁₎ Acid + H₂O ₍₁₎Acid ⇋ H₃O+ (aq) Conjugate acid + OH^-(aq) Conjugate base

822.

The species: H2O can act both as Bronsted-acid and base. For each case give the corresponding conjugate acid and conjugate base.

Answer»

H₂O can act both as Bronsted -acid and base as it can accept and donate proton. It undergoes auto-ionisation to form corresponding conjugate acid (H₃O⁺) and conjugate base (OH^-).

H₂O₍₁₎ Acid + H₂O ₍₁₎Acid ⇋ H₃O+ (aq) Conjugate acid + OH^-(aq) Conjugate base

Discussion

823.	If K1 and K2 are the respective equilibrium constants for the two reactionsXeF6(g) + H2O (g) ⇋ XeOF4(g) + 2HF(g)XeO4 (g) + XeF6 (g) ⇋ XeOF4 (g) + XeO3F2(g), the equilibrium constant of the reactionXeO4 (g) + 2HF (g) ⇋ XeO3F2 (g) will be –(a) K1/ K22 (b) K1.K2 (c) K1/K2 (d) K2/K1
Answer» The correct answer is (d) K₂/K₁

823.

If K1 and K2 are the respective equilibrium constants for the two reactionsXeF6(g) + H2O (g) ⇋ XeOF4(g) + 2HF(g)XeO4 (g) + XeF6 (g) ⇋ XeOF4 (g) + XeO3F2(g), the equilibrium constant of the reactionXeO4 (g) + 2HF (g) ⇋ XeO3F2 (g) will be –(a) K1/ K22 (b) K1.K2 (c) K1/K2 (d) K2/K1

Answer»

The correct answer is

(d) K₂/K₁

Discussion

824.	Give the generalizations concerning the composition of equilibrium mixtures.
Answer» (i) If Kc > 10³, products predominates over reactants i.e; if Kc is very large, the reaction proceeds nearly to completion. (ii) If Kc < 10^-3, reactants predominates over products i.e; if Kc is very small, the reaction proceeds rarely. (iii) If Kc is in the range of 10^-3 to 10³, appreciable concentration of both reactants and products are present.

824.

Give the generalizations concerning the composition of equilibrium mixtures.

Answer»

(i) If Kc > 10³, products predominates over reactants i.e; if Kc is very large, the reaction proceeds nearly to completion.

(ii) If Kc < 10^-3, reactants predominates over products i.e; if Kc is very small, the reaction proceeds rarely.

(iii) If Kc is in the range of 10^-3 to 10³, appreciable concentration of both reactants and products are present.

Discussion

825.	The relation `K_(p) " and " K_(c) `for the reaction: `2NO (g) + CI_(2) (g) hArr 2NOCI(g) " is":`A. `K_(p) =K_(c)`B. `K_(p) =K_(c) (RT)`C. `K_(p) =K_(c)//RT`D. `K_(p) =K_(c)//RT^(2)`
Answer» Correct Answer - C `2NO(g) +CI_(2) (g) hArr 2NOCI(g)` `Deltan =2 -3 =- 1, K_(p) =K_(c)(RT)^(-1)` `K_(p) =K_(c)//RT`

Discussion

826.	`K_(p) " and " K_(c)` are related to each other as :A. `K_(p) =K_(c) (RT)^(-Deltan)`B. `K_(c) =K_(p) (RT)^(-Deltan)`C. `K_(p) =(RT)^(Deltan)//K_(c)`D. `K_(p) -K_(c) =(RT)Delta^(n)`
Answer» Correct Answer - B `K_(p) =K_(c) (RT)^(Deltan) " or " K_(c) =K_(p)(RT)^(-Deltan)`

Discussion

827.	With the help of Lowry-Bronsted concept show that `HCIO_(4)` is a stronger acid than `HCIO_(3)` when dissolved is water.
Answer» The dissociation of both the acids in aqueous solution may be shown as : `{:(HCIO_(4),+,H_(2)O ,hArr,H_(3)O^(+) ,+,CIO_(4)^(-)),("Acid"-I,,"Base-II",,"Acid-II",,"Base-II"):}` `{:(HCIO_(3),+,H_(2)O ,hArr,H_(3)O^(+) ,+,CIO_(3)^(-)),("Acid"-I,,"Base-II",,"Acid-II",,"Base-II"):}` Perchlorate `(CIO_(4)^(-))` is a weak base than chlorate `(CIO_(3)^(-))`ion because the negative charge on it is dispersed on four oxygen atoms whereas in chlorate `(CIO_(3)^(-))` ion , three oxygen atoms are available to accommodate negative charge. According to the theory a weak base has a strong conjugate acid. Therefore `HCIO_(4)` is a stronger acid than `HCIO_(3).`

Discussion

828.	The `pH` of `0.1 M` solution of the following salts increases in the orderA. `HCI lt NaCI lt NaCN lt NH_(4)CI`B. `NaCN lt NH_(4)CI lt NaCI lt HCl`C. `HCI lt NH_(4)CI lt NaCI lt NaCN`D. `NaCI lt NH_(4)CI lt NaCN lt HCI`
Answer» Correct Answer - C The aqueous solution of NaCI is neutral `(pH=7)` as it does not undergo hydrolysis. The aqueous solution of NaCN is basic `(pHgt7)` on account of anionic hydrolysis: `CN^(-)+H_(2)OhArrHCN+OH^(-)` The aqueous solution of `NH_(4)Cl` is slightly acidic due to cationic hydrolysis: `NH_(4)^(+)(aq.)+H_(2)O(l)hArr NH_(4)OH+H^(+)(aq.)` The aqueous solution of HCl (a strong acid) has the lowest pH value.

Discussion

829.	When a volatile liquid is introduced into an evacuated closed vessel at a particular temperature, both evaporation and condensation take place simultaneously. The system reaches equilibrium state whenA. the liquid is completely transformed into the corresponding vaporB. equal amounts of liquid and vapor are present in the systemC. the rate of evaporation becomes equal to the rate of condensationD. liquid cannot be converted into vapor and vice versa.
Answer» Correct Answer - C At equilibrium, the number of molecules leaving the liquid equals the number of molecules returning to the liquid from the vapor.

Discussion

830.	What is the value of `k_(c)` for the reaction at 1473 K `I_(2)(g) hArr 2I(g)` when one mode of iodic gas is introduced into an evacuated one litre flask so that only 5% of it gets dissociated ?
Answer» `{:(,I_(2)(g) ,hArr, 2I(g)),("Initial no. of moles"// "litre", 1 ,, 0),("No. of moles"// "litre at equilibrium",1-0.05,, 0.05 xx 2),(,=0.95,,=0.1):}` Applying Law of chemical equilibrium `K_(c) =[[(g)]^(2)]/[[I_(2)(g)]] =((0.1 "mol "L^(-1))(0.1 "mol " L^(-1)))/((0.95 "mol "L^(-1)))` `=0.0105 =1.05 xx 10^(-2) "mol "L`

Discussion

831.	Which of the following will decrease the pH of a 50 ml solution of `0.01 M HCI` ?A. Addition of MgB. Addition of 50 ml `0.002 M HCI`C. Addition of 50 ml `0.01M HCI`D. Addition of 5 ml 1 M HCI
Answer» Correct Answer - D To decreases the pH, we must add more `H^(+)` ions. This can be done by adding more concentrated acid. The addition of Mg (option, I) will increase the pH as it reacts with HCI to decrease the `H^(+)` concentration. `Mg(s)+2H^(+)(aq.)rarr H_(2)(g)+Mg^(2+)(aq.)` The addition of `0.002` M HCI (option 2), a less concentrated solution, will increase the pH by decreasing `H^(+)` ion concentration. The addition of `0.01` M HCI (option 3), the same concentrated solultion, does not change the `H^(+)` ion concentration and, hence, the pH.

Discussion

832.	An acid having pH = 6 is diluted 1000 times. What will be the pH of the final solution ?
Answer» pH = 6 means `[H_(3)O^(+)=10^(-6)M`. After diluting 1000 times, `[H_(3)O^(+)]=10^(-6)//1000M = 10^(-9)M`. `:. [H_(3)O^(+)]` from `H_(2)O` cannot be neglected. Total `[H_(3)O^(+)]=10^(-9)+10^(-7)(10^(-2)+1)=10^(-7)(1.01)` `:. pH = - log (1.01xx10^(-7))=7-0.0043=6.9957`.

Discussion

833.	What will be the volume of `pH` of `0.01` mold`m^(-3)` `CH_(3)COOH(K_(a) = 1.74 xx 10^(-5))`A. `3.4`B. `3.6`C. `3.9`D. `3.0`
Answer» Correct Answer - A `H^(+)= sqrt(K_(a)xxC)` `=sqrt(174xx10^(-7))=sqrt(174.4xx10^(-8))` `=4.2xx10^(-4)` `ph=-log[4.2xx10^(-4)]` `=(4-log4.2)` `=4-0.6232=3.3768` `approx 3.4`

Discussion

834.	What will be the conjugate bases for the following Bronsted acids ? `HF, H_(2)SO_(4) and HCO_(3)^(-)`
Answer» Conjugate acid `hArr` Conjugate base + `H^(+)` or Conjugate base = Conjugate acid - `H^(+)` `:.` Conjugate bases of the given acids will be `F^(-), HSO_(4)^(-), CO_(3)^(2-)`.

Discussion

835.	Strongest conjugate base isA. `Cl^(-)`B. `Br^(-)`C. `F^(-)`D. `I^(-)`
Answer» Correct Answer - C Their conjugate acids are HCI, HBr, HF and HI Out of these HF is the weakest due to maximum bond strength (and hydrogen bonding). Therefore, its conjugate base,` F^(-)` is the strongest

Discussion

836.	`K_(p) " for the reaction " N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at "400^(@)C " is " 1*64 xx10^(-4) atm^(-2)." Find " K_(c). " Also calculate " DeltaG^(@) " using " K_(p) and K_(c) " values and interpret the difference ".`
Answer» `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)` `Delta n= 2-4 = -2` `K_(p) = K_(c) (RT)^(Delta n)` `164 xx 10^(-4) " atm"^(-2) = K_(c) (00821) L " atm "K^(-1) mol^(-1) xx 673 K ) ^ (-2) " "(T=400 + 273 K = 673 K)` or `K_(c) = (164 xx 10^(-4) "atm"^(-2))/((0821 xx 673 " atm"^(-1) mol^(-1))^(2))=05372 " mol"^(2)L^(-2)` Now, `Delta G^(@) = - 2303" RT " log K` If `K=K_(p)` ` Delta G^(@) = - 2303 xx (8314 " JK"^(-1) mol^(-1)) (673 K) xx log (164 xx 10^(-4))` `=-2303 xx 8314 xx 673 xx (-3 7852) " J "mol^(-1) = +4878 " kj "mol^(-1)` If `K=K_(c)` , `Delta G^(@)= - 2303 xx (8314 xx 673 xx (-027) " J "mol^(-1) = + 3479 " J "mol^(-1)` Interpreting the difference .`DeltaG^(@)` is the free energy change when all the reactants and products are in their standard state. In case of `K_(p)`, standard state pressure is used which is 1 atm (or now 1 bar) wheras in case of `K_(c)` , standard concentration is used which is 1 mol `L^(-1)` (as already explained on page 7/22).

Discussion

837.	For the reaction `C(s) + CO_(2) (g) to 2 CO (g) K_(p) = 63 " atm at " 100 K .` If at equilibrium `p_(CO) = 10_(p_(CO_(2))` , then the toal pressure of the gases at equilibrium isA. `63` atmB. `693` atmC. `063 `atmD. `0693` atm
Answer» Correct Answer - B `C(s) +CO_(2) (g) hArr 2 CO (g)` `K_(p)=(p^(2) CO)/p_(CO_(2)) or 63 = ((10p_(CO_(2))))/(p_(CO_(2)))` or `p_(CO_(2)) = 63/100 = 0.63 " atm"` ` p_(CO)= 10p_(CO_(2))= 10 xx 0.63 " atm " =6.3 " atm"` ` :. p_("total")= p_(CO_(2)) + p_(CO) = 0.63 + 6.3 = 6.93 " atm"`

Discussion

838.	What is the difference between conjugate acid and the conjugate base.
Answer» These differ from each other by proton. Acid ----------> Conjugate base + proton Base ------------> Conjugate acid+ proton

Discussion

839.	Determine the concentration of `CO_(2) " which will be in equilibrium with " 2* 5 xx 10^(-2) mol L^(-1) " of CO at "100^(@)C " for the reaction, " FeO(s) + CO (g) hArrFe (s) + CO_(2) (g) , K_(c) = 5*0`
Answer» `K_(c) = ([CO_(2)])/([CO]) , i.e.,5 = ([CO_(2)])/(25 xx 10^(-2)) or [CO_(2)] = 5 xx 25 xx 10^(-2) = 12*5 xx 10^(-2) mol L^(-1)`

Discussion

840.	Which of the following on reaction with `H_(2)S` does not produce metallic sulphide ?A. `CdCI_(2)`B. `ZnCI_(2)`C. `COCI_(2)`D. `CuCI_(2)`
Answer» Correct Answer - C Carbonyl chloride `(COCI_(2))`. Carbonyl sulphide (COS) is not a metallic sulphide.

Discussion

841.	The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500K. [N2] = 1.5 x 10-2 M, [H2] = 3.0 x 10-2 M and [NH3] = 1.2 x 10-2 M. Calculate equilibrium constant.
Answer» For the reaction, N₂ (g) + 3H₂ (g) ⇌ 2NH₃(g) Equilibrium constant K_c = \(\frac{[NH_3]^2}{[N_2][H_2]^3}\) = \(\frac{(1.2\times10^{-2})^2}{(1.5\times10^{-12})(3\times10^{-2})^3}\) = 0.106 x 10⁴ = 1.06 x 10³

841.

The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500K. [N2] = 1.5 x 10-2 M, [H2] = 3.0 x 10-2 M and [NH3] = 1.2 x 10-2 M. Calculate equilibrium constant.

Answer»

For the reaction,

N₂ (g) + 3H₂ (g) ⇌ 2NH₃(g)

Equilibrium constant

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

= \(\frac{(1.2\times10^{-2})^2}{(1.5\times10^{-12})(3\times10^{-2})^3}\)

= 0.106 x 10⁴

= 1.06 x 10³

Discussion

842.	0.05 mole of NaOH is added to 5 liters of water What will be the pH of the solution ?A. 12B. 7C. 2D. 10
Answer» Correct Answer - A Conc. of `NaOH=(0.05)/(5)=0.01or10^(-2)M` `[OH^(-)]=10^(-2)M` `[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(10^(-2))=10^(-12)` `pH=-log[H^(+)]=-log10^(-12)=12`

Discussion

843.	A mixture of `SO_(3) , SO_(2) and O_(2)` gases is maintained at equilibrium in 10 litre flask at a temperature at which `K_(c) " for the reaction ," 2SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) " is 100 mol"^(-1) ` litre. At equilibrium . (a) if no. of moles of `SO_(3) and SO_(2)` in the flask are same, how many moles of `O_(2)` are present ? (b) if no. of moles of `SO_(3)` in the flask are twice the number of moles of `SO_(2)` in the flask are twice the number of moles of `SO_(2)` , how many moles `O_(2)` are present ?
Answer» ` 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) , " " K_(c) = ([SO^(2)])/([SO_(2)]^(2)[O_(2)] = 100)(Given)` (a) As ` [ SO_(3)] = [ SO_(2)] , :. 100 = 1/([O_(2)]) or [O_(2)] = 1/ 100 molL^(-1)` `:. O_(2) " present in 10 litre " = 1/100 xx 10 = 01 ` mole (b) If ` [ SO_(3) ] = 2 [ SO_(2)] , i.e, ([SO_(3)])/([ SO_(2)] )= 2 " then " 100 = 4/ ([O_(2)] ) or [O_(2)] = 4/100 "mol"L^(-1)` ` :. O_(2) " present in 10 litre " = 4/100 xx 10 = 04` mole

Discussion

844.	Calculate the PH of the resultant mixture: 10 Ml of 0.1M H2SO4 + 10 Ml 0.1M KOH
Answer» Number of moles of H₊ ion = 2 x number of moles of H₂SO₄ =2 x M(V/1000) = 2 x 0.1x10/1000 = 2x 10^-3 Number of moles of OH ion = number of moles of NaOH = M(V/1000) = 0.1x10/1000 =1 x 10^-3 Number of moles of H⁺ ion after neutralization=2 x 10^-3 - 1x 10^-3 = 1x 10^-3 Molarity of H⁺ ion= n x1000/V = 1x 10^-3x1000/20 =0.05 PH = - log(H⁺ ) = -log (0.05) = 1.301

844.

Calculate the PH of the resultant mixture: 10 Ml of 0.1M H2SO4 + 10 Ml 0.1M KOH

Answer»

Number of moles of H₊ ion = 2 x number of moles of H₂SO₄

=2 x M(V/1000)

= 2 x 0.1x10/1000

= 2x 10^-3

Number of moles of OH ion = number of moles of NaOH

= M(V/1000)

= 0.1x10/1000

=1 x 10^-3

Number of moles of H⁺ ion after neutralization=2 x 10^-3 - 1x 10^-3

= 1x 10^-3

Molarity of H⁺ ion= n x1000/V

= 1x 10^-3x1000/20

=0.05

PH = - log(H⁺ )

= -log (0.05)

= 1.301

Discussion

845.	Consider the equilibrium set up : `2H_(2(g))+CO_((g))hArrCH_(3)OH_((g))` What will be the effect of the following on the equilibrium of the reaction ? (i) Addition of `H_(2)` (ii) Addition of `CH_(3)OH` (iii) Removal of CO (iv) Removal of `CH_(3)OH`A. `{:(,(i),(ii),(iii),(iii)),((a),"Forward direction","Backward direction","Backward direction","Forward direction"):}`B. `{:(,(i),(ii),(iii),(iii)),((a),"Backward direction","Backward direction","Forward direction","Forward direction"):}`C. `{:(,(i),(ii),(iii),(iii)),((a),"Forward direction","Forward direction","Backward direction","Backward direction"):}`D. `{:(,(i),(ii),(iii),(iii)),((a),"Backward direction","Forward direction","Forward direction","Backward direction"):}`
Answer» Correct Answer - A (i) Increase in concentration of reactants favours forward reaction. (ii) Incerease in concentration of products favours backward reaction. (iii) Removal of CO will favour backward reaction (decrease in concentration of reactants). (iv) Removal of `CH_(3)OH` will favour forward reaction (increase in concentration of products.

Discussion

846.	Assertion : Weak acids have very strong conjugate bases while strong acids have conjugate bases. Reason : Conjugate acid - base pair differ only by one proton.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 Strong acids such as perchloric acid `(HClO_(4))`, hydrochloric acid (HCl) give conjugate base ions `ClO_(4)^(-),Cl^(-)` which are much weaker bases acid is only partially dissociated in aqueous medium and thus the solution mainly contains undissociated acetic acid which proves that `CH_(3)COO^(-)` is stronger base than `H_(2)O`.

Discussion

847.	The relationship between ionisation and change in concentration of any weak electrolyte is expressed as :A. `alpha= (K_(a))/(C)`B. `alpha= sqrt((K_(a))/(C))`C. `alpha= K_(a) C`D. `alpha=sqrt((K_(a))/(C^(2)))`
Answer» Correct Answer - B Is the correct answer.

Discussion

848.	At a certain temperature and a total pressure of `10^(5) Pa`, iodine vapour contains `40%` by volume of `I "atoms"`, Calculate `K_(p)` for the equilibrium. `I_(2(g))hArr2I_((g))`
Answer» According to available data: Total pressure of equilibrium mixture `=10^(5) Pa` Partial pressure of iodine atoms `(I) =(40)/(100) xx (10^(5) Pa) =0.4 xx 10^(5)Pa` Partial pressure iodine molecules `(I_(2)) =(60)/(100)xx(10^(5) Pa) =0.6 xx 10^(5) Pa` `underset((0.6xx10^(5) Pa))(i_(2) (g)) hArr underset((0.4xx10^(5) Pa))(2I(g))` `K_(p) =(pI^(2))/(pI^(2))= ((0.4 xx 10^(5)Pa)^(2))/((0.6xx10^(5) Pa)) =2.67 xx 10^(4) Pa`

Discussion

849.	At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.I2(g) ⇌ 2I(g)Calculate KP for the equilibrium.
Answer» Total pressure of equilibrium mixture = 10⁵ Pa Partial pressure of iodine atoms (I) = \(\frac{40}{100}\) x (10⁵ Pa) = 0.4 x 10⁵ Pa Partial pressure of iodine molecules (I₂) \(\frac{60}{100}\) x (10⁵ Pa) = 0.6 x 10⁵ Pa I₂(g) (0.6 x 10⁵ Pa) ⇌ 2I(g) (0.4 x 10⁵ Pa) K_P = \(\frac{p^2I}{pI_2}\) = \(\frac{(0.4 \times 10^5\, Pa)^2}{(0.6 \times 10^5\, Pa)}\) = 2.67 x 10⁴ Pa

849.

At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.I2(g) ⇌ 2I(g)Calculate KP for the equilibrium.

Answer»

Total pressure of equilibrium mixture = 10⁵ Pa

Partial pressure of iodine atoms (I) = \(\frac{40}{100}\) x (10⁵ Pa) = 0.4 x 10⁵ Pa

Partial pressure of iodine molecules (I₂)

\(\frac{60}{100}\) x (10⁵ Pa) = 0.6 x 10⁵ Pa

I₂(g) (0.6 x 10⁵ Pa) ⇌ 2I(g) (0.4 x 10⁵ Pa)

K_P = \(\frac{p^2I}{pI_2}\) = \(\frac{(0.4 \times 10^5\, Pa)^2}{(0.6 \times 10^5\, Pa)}\) = 2.67 x 10⁴ Pa

Discussion

850.	A solution of `CoCI_(2).6H_(2)O` in isopropyl alcohol and water is purple. The color change to blue when we addA. concentrated HCIB. `AgNO_(3)(aq.)C. both (1) and (2)D. none of these
Answer» Correct Answer - A The purple color is due to the mixture of `[Co(H_(2)O)_(6)]^(2+)` (pink) and `[CoCl_(4)]^(2-)` (blue): `[Co(OH_(2))_(6)]^(2+)+4Cl^(-)hArr[CoCl_(4)]^(2-)+6H_(2)O` When we add concentrated HCI, excess `Cl^(-)` shifts the equilibrium to the right (blue). Adding `AgNO_(3)` (aq.) removes some `Cl^(-)` by precipitation of `AgCl(s)` and favours the reaction to the left (produces more `[Co(OH_(2))_(6)]^(2+)` and the resulting solution is pink.

Discussion

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