52671 + Interview Questions in Chemistry in Class 11

1.	Eq. wt ("Mol wt")/x What is x for acetaldehyde converted into acetic acid.
Answer» Solution :`CH_(3)CHOrarrCH_(3)COOH` `underset("Acetaldehyde")overset((-1))(C_(2)H_(4)O)rarrunderset("acetic acid")overset((0))(C_(2)H_(4)O_(2))` EQ. wt = `("Mol wt")/(2)`

Discussion

2.	Equivalent weight of X is 9. The vapour density of chloride salt is 66.75 state vulency of X.
Answer» Solution :MOLECULAR MASS `= 2 xx"VAPOURS DENSITY"` `=2xx 66.75 = 133.50` `:. "Valency"=("Molecular mass")/("Equivalent weight")` `=(133.50)/(9+35.5) = (133.50)/(44.5)=3`

Discussion

3.	Equivalent weight of the reaction is same as its formula weight or molecular weight in case of the conversion__________
Answer» `Na_(2)S_(2)O_(3)` into `Na_(2)S_(4)O_(6)`<BR>`FeSO_(4)` into `Fe_(2)(SO_(4))_(3)` `KBr` into `Br_(2)` `Br_(2)` into `KBr` Solution :E = M if N - f = 1 `Na_(2)SO_(3)rarrNa_(2)S_(4)O_(6),n-f=0.5xx2=1` `FeSO_(4)rarrFe_(2)(SO_(4))_(3),n-f=1xx1=1` `KBrrarrBr_(2),n-f=1xx1=1` `Br_(2)rarrKBr,n-f=1xx2=2`

Discussion

4.	Equivalent weight of Pyrophosphoric acid is (H_(4)P_(2)O_(7))
Answer» `M.W//1` `M.W//2` `M.W//4` `M.W//3` Solution :`H_(4)P_(2)O_(7), N`-factor =4

Discussion

5.	Equivalent weights of K_(2)Cr_(2)O_(7) in acidic medium is
Answer» 24.5 49 147 296 Answer :B

Discussion

6.	Equivalent weight of K_(2)Cr_(2)O_(7) as oxidant in acidic medium is
Answer» 24.5 49 147 296 Answer :B

Discussion

7.	Equivalent weight offerrous ion when it acts as salt is X and as reductant is Y. What is the ratio of X and Y?
Answer» Solution :X is 55.8/2. When ferrous ION acts as reductant , ITIS oxidised to FERRIC. Y is 55.8/1. THUS X:Y is 1:2

Discussion

8.	Equivalent weight of element X is 3. If vapour density of volatile chloride of X is 77, Find out the moleculer formula of chloride.
Answer» SOLUTION :Equivalent weight of chloride of X= Equivalent weight of Cl +equivalent weight of X = 35.5+3=38.5 Moleculer weight of chloride = vapour density `xx2=77xx2=154` `"Valency of the element" X=(154)/(38.5)=4` Chlorine is taken as monovalent, SINCE it is chloride MOLECULAR formula of chloride of X is `XCl_(4)`

Discussion

9.	Equivalent weight of Ba(MNO_4)_2 in acidic medium ( M = molar mass)
Answer» M M/3 M/5 M/10 Solution :For one PERMANGANATE ION, n-factor is 5. But BA `(MnO_4)_2`has TWO permanganate ions. So its n-factor is 10.

Discussion

10.	Equivalent weight of As_(2)O_3 in the following equation As_(2)O_3 +2I_(2) + 2H_(2)O rarr As_(2)O_(5) +4HI [arsenci at wt. = 75]
Answer» 49.5 94.9 99 156.6 Answer :A

Discussion

11.	Equivalent weight of a metal chloride is 75.5. How many moles of NaOH is required to completely precipitate one mole of metal hydroxide. Atomic weight of the metal is 120.
Answer» Solution :EQ of metal CHLORIDE = EQ wt of metal + EQ wt of `Cl^(-)` 75.5 = E + 35.5 E = 40 VALENCY = `(120)/(40)=3` `THEREFORE` Formula of metal chloride = `MCl_(3)` `MCl_(3)+3NaOHrarrM(OH)_(3)darr+3NaCl`

Discussion

12.	Equivalent masses ofNH_3in the reactions are (i) 4NH_(3) +5O_(3)rarr4NO + 6H_(2)O
Answer» `5:6` `6:5` `5:3` `3:5` SOLUTION :`OVERSET(-3)(NH_3)RARROVERSET(+2)(NO)` `E_1=M/5,overset(-3)(2NH_3)rarroverset(0)N_2` `E_2=M/3,E_1/E_2=3:5`

Discussion

13.	Equivalent mass of oxidising agent in the reaction SO_(2)+2H_(2)Srarr3S+2H_(2)O is
Answer» 321 64 16 8 Solution :`OVERSET(+4)SO_(2)^(-2)+overset(+1)2H_(2)overset(-2)Srarr3overset(0)S+overset(-2)2H_(2)O` Inthis reaction , `SO_(2)` acts as oxidising agent `overset(+4)SO_(2)rarroverset(0)S` `SO_(2)+4e^(-)rarrS` therefore Eq mass of `SO_(2)`=`("MOLECULAR mass")/(4)` `=(32+32)/(4) =16`

Discussion

14.	Equivalent weight and formula weight of reactants are same in the conversions
Answer» `AgNO_(3)rarrAg` metal `CuOrarrCu_(2)O` `H_(2)O_(2)rarrH_(2)O` `Na_(2)S_(2)O_(3)rarrNa_(2)S_(4)O_(6)` Solution :`E=Mimplies n-f=1` `AgNO_(3),n-f=1` `CUO,n-f=1` `H_(2)O_(2),n-f=2` `Na_(2)S_(2)O_(3),n-f=1`

Discussion

15.	Equivalent mass of N, in the change N_(2)- NH_3is
Answer» 28/6 28 28/2 28/3 Solution :`OVERSET(0)(N_2)rarr2overset(+3)NH_3,E=(28)/6`

Discussion

16.	Equivalent mass of KMnO_(4) when it is converted to MnSO4 is equal to molar mass divide by……….
Answer» 6 4 5 2 Solution :`Koverset(+7)(Mn)O_(4) to Mnoverset(+2)(S)O_(4)` `KMnO_(4) + 5e^(-) to MnSO_(4)` So, EQUIVALENT mass of `KMnO_(4)=("molar mass of" KMnO_(4))/5`

Discussion

17.	Equivalent mass of KMnO_(4), when it is converted to MnSO_(4), is :
Answer» `M//5` `M//3` `M//6` `M//2` SOLUTION :N//A

Discussion

18.	Equivalent mass of KMnO_(4) in acidic medium, concentrated alkaline medium and dilute basic medium respectively are M, M, M. Reduced products can be
Answer» `MnO_(2), MnO_(4)^(-) , Mn^(2+)` `MnO_(2), Mn^(2+), MnO_(4)^(2-)` `Mn^(2+), MnO_(2), MnO_(4)^(2-)` `Mn^(2+), MnO_(4)^(2-), MnO_(2)` SOLUTION :`MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O`(acidic medium) `MnO_(4)^(-) + 4H^(+) + 3e^(-) to MnO_(2) + 2H_(2)O`(CONCENTRATED basic medium) `MnO_(4)^(-) + e^(-) to MnO_(4)^(2-)`(DILUTE basic medium)

Discussion

19.	Equivalent mass of H_(3)PO_(2) when it disproportionates into PH_(3) " and" H_(3)PO_(3) is (Molecular mass=M) :
Answer» M `(M)/(2)` `(M)/(4)` `(3M)/(4)` SOLUTION :`overset(+1)(H_(3)PO_(2))to overset(+3)(H_(3)PO_(3)), " Eq. wt. "=(M)/(2)` (Change in OXIDATION number=2) `overset(+1)(H_(3)PO_(2) to overset(-3)(PH_(3)), " Eq. wt." = (M)/(4)` (Change in oxidation number=4) The EQUIVALENT mass of `H_(3)PO_(2)` in the PROCESS of disproportionation`=(M)/(2)+(M)/(4)=(3M)/(4)`

Discussion

20.	Equivalent mass of C_(2)O_(4)^(2-) ion in the reaction, C_(2)O_(4)^(2-)to2CO_(2)+2e^(-), is
Answer» 11 22 44 88 Answer :C

Discussion

21.	Equivalent mass of FeC_(2)O_4in the change, FeC_(2)O_(4) rarrFe^(3+) + CO_(2)is
Answer» M/3 M/6 M/2 M/1 Solution :`FeC_2O_4 ""{:(FE^(2+),overset((1))rarr,Fe^(3+)),(overset(+3)(C_(2))O_(4)^(2-),overset((2))rarr,2overset(+4)(CO_2)):}}3e^(-)`

Discussion

22.	Equivalent mass of an organic base can be determined by:
Answer» SILVER SALT method depression in FREEZING point elevation in BOILING point platinichloride method Answer :D

Discussion

23.	Equivalent mass of an organic acid can be determined by :
Answer» SILVER SALT method cryoscopic method ebullioscopic method platinichloride method Answer :A

Discussion

24.	Equivalent mass of a substance may be calculated as, Equivalent mass=("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor") n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. Equivalentmass of H_(3)PO_(2) when it undergoes disporportionation to PH_(3) " and " H_(3)PO_(3) will be :
Answer» `M.w.//2` `M.w. //4` `M. w. //24` `3M.w. //4` Solution :`H_(3)PO_(2)+4H^(+) +4e^(-) to PH_(3)+2H_(2)O "" N_(1)=4` `H_(3)PO_(2)+H_(2)O to H_(3)PO_(3)+2H^(+)+2e^(-) "" n_(2)=2` N-factor`=(n_(1)xxn_(2))/(n_(1)+n_(2))=(2xx4)/(2+4)=(8)/(6)=(4)/(3)` E.w. `M.w.//(4)/(3)=(3M.w.)/(4)`

Discussion

25.	Equivalent mass of a substance may be calculated as, Equivalent mass=("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor") n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. BrO_(3)^(-) ion reacts with Br^(-) to form Br_(2), in acid medium. The equivalent mass of Br_(2) in this reaction is :
Answer» `(4 M.w.)/(6)` `(3M.w.)/(5)` `(5M.w.)/(3)` `(5M.w.)/(8)` Solution :`2BrO_(3)^(-)+12H^(+)+10e^(-) to Br_(2)+6H_(2)O "" n_(1)=10` `2Br^(-) to Br_(2)+6e^(-) "" n_(2)=2` N-factor`=(n_(1)xxn_(2))/(n_(1)+n_(2))=(10xx2)/(10+2)=(20)/(12)=(5)/(3)` `E.w.=(M.w.)/(5//3)=(3M.w.)/(5)`

Discussion

26.	Equivalent mass of a substance may be calculated as, Equivalent mass=("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor") n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. When KMnO_(4) is titrated against ferrous ammonium sulphate in acid medium then equivalent mass of KMnO_(4) will be :
Answer» `("Molecular mass")/(10)` `("molecular mass")/(5)` `("Molecular mass")/(3)` `("molecular mass")/(2)` Solution :`MnO_(4)^(-)+8H^(+)+5E^(-) to Mn^(2+) +4H_(2)O`n-factor=5

Discussion

27.	Equimotal aquous solutions ofNaCl and KCl are prepared If the freezing point ofNaClis -2^(@) Cthe freezing point of KCI solution is expectedtobe
Answer» `-2^(@) C ` ` -4^(@) C ` ` -1 ^(@) C ` ` 0^(@) C ` ANSWER :A

Discussion

28.	Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH ?
Answer» `SrCl_2` `BaCl_2` `MgCl_2` `CaCl_2` Solution :The highest pH will be recorded by the most basic solution. The basic NATURE of hydroxides of alkaline earth METALS INCREASE as we move from Mg to BA and THUS the solution of `BaCl_2` in water will be most basic and so it will have highest pH.

Discussion

29.	Equimolar solutions of the following substances were prepared separately. Which one of these has highest pH value?
Answer» `BaCI_(2)` `AICI_(2)` `LiCI` `BeCI_(2)` Solution :The bases and acids which constitute these salts are: `BaCI_(2) : BA(OH)_(2) +2HCI` `AICI_(3) , AI(OH)_(3) + 3HCI` `LiCI : Li(OH) + HCI` `BeCI_(2) : Be(OH)_(2) + 2HCI` Since `Ba(OH)_(2)` is the strongest base , `BaCI_(2)` solution has the maximum pH VALUE.

Discussion

30.	Equimolarsolution of non-electrolyte in the same solvent have
Answer» same BOILING POINT and same FREEZING point DIFFERENT boiling point and different freezing point same boiling point but different freezing point same freezing point but different boiling point ANSWER :A

Discussion

31.	Equimolar concentration of H_(2) and I_(2) are heated to equilibrium ina 1 liter flask. What percentage of initial concentration of H_(2) has reacted at equilibrium, the rate constant for the forward reaction is 25xx10^(2) and the equilibrum constant is 50. The rate constant for the reverse reaction is,
Answer» `33%` `66%` `(33)^(2)%` `16.5%` ANSWER :A

Discussion

32.	Equimolar concentraions of H_2 and I_2 are heated to equilibriumin a 1 litre flask. What percentage of initial concentration of H_2 has reacted at equilibrium if rate constant for both farward and reverse reactions are equal
Answer» 0.33 0.66 `(33)^2 %` 0.165 Solution :0.33

Discussion

33.	Equimolar amounts of H_(2) and I_(2) were taken in a bulb maintained at 500 ^(@) C. Dark violet colour faded to light violet which does not change further. This shows that the bulb contains ………….. amounts of …………. .
Answer» ANSWER :FIXED, `H_(2), I_(2) and HI`

Discussion

34.	Equimolar aqueous solutions of NaCl and KCl are prepared. If the freezing point of NaCl is -2^(@)C, the freezing point of KCl solution in expected to be …………
Answer» 1) `-2^(@)C` 2) `-4^(@)C` 3) `-1^(@)C` 4) `0^(@)C` Solution :Equimolar aqueous solution of KCl ALSO shows `2^(@)C` depression in FREEZING point.

Discussion

35.	Equilmolar concentrations of H_(2) and I_(2) are heated to equilibrium in a 2 litre flask . At equilibrium , the forward and the backward rate constants are found to be equal . What percentage of intial concentration of H_(2) has reacted at equilibrium ?
Answer» `33 %` `66 %` `50%` `40%` Solution :`{:(,H_(2)(g),+,I_(2) (g),hArr,2 HI(g)), (" Intial",1 "mol",,1 "mol",,0),("At eqm.",1-X,,1-x,,2x),("Molar conc.",(1-x)/2,,(1-x)/2,,(2x)/2"mol"L^(-1)),(,,,,,=x):}` ` K= (x^(2))/((1-x)/2 xx (1-x)/2 )=(4 x^(2))/(1-x)^(2) approx4x^(2)` But `K= (k_(f))/(k_(b))=1` `:. 4"" x^(2)=1 or x^(2) = 1//4 or x = 1//2 , i.e., 50%`

Discussion

36.	Equilibrium state can be achieved if a reversible reaction is carried out in closed or open container.
Answer» ANSWER :FALSE

Discussion

37.	Equilibrium constants are given (in atm) for the following reactions at 0^(@)C : SrCl_2 . 6H_2O(s) hArr SrCl_2 . 2H_2O(s)+4H_2O(g) K_p=5xx10^(-12) Na_2HPO_4 . 7H_2O(s)+5H_2O(g) K_p=2.43xx10^(-13) Na_2SO_4 . 10H_2O(s) hArr Na_2SO_4(s)+10H_2O(g) K_p=1.024xx10^(-27) The vapour pressure of water at 0^@C is 4.56 torr. At what relative humidities will Na_2SO_4 be deliquescent (i.e. absorb moisture) when exposed to the air at 0^@C ?
Answer» above 33.33% below 33.33% above 66.66% below 66.66% ANSWER :A

Discussion

38.	Equilibrium existing in the hydrolysis of an ester is .......
Answer» Gaseous HOMOGENEOUS Heterogeneous Ionic homogeneous Homogeneous Solution :All the COMPONENTS are liquid, so it is homogeneus further. The reaction take place in presence of `H^+`, so it is ionic homogeneous. In HYDROLYSIS of ester, all the components are liquid STATE and in presence of acid catalyst, reaction is as follow. `CH_3COOC_2H_(5(l)) + H_2O_((l)) OVERSET(H^+)hArr CH_3COOH_((l)) + C_2H_5OH_((l))`

Discussion

39.	Equilibrium constants are given (in atm) for the following reactions at 0^(@)C : SrCl_2 . 6H_2O(s) hArr SrCl_2 . 2H_2O(s)+4H_2O(g) K_p=5xx10^(-12) Na_2HPO_4 . 7H_2O(s)+5H_2O(g) K_p=2.43xx10^(-13) Na_2SO_4 . 10H_2O(s) hArr Na_2SO_4(s)+10H_2O(g) K_p=1.024xx10^(-27) The vapour pressure of water at 0^@C is 4.56 torr. At what relative humidity will Na_2SO_4 . 10H_2O be efflorescent when exposed to air at 0^@ C?
Answer» above 33.33% below 33.33% above 66.66% below 66.66% ANSWER :B

Discussion

40.	Equilibrium constants are given (in atm) for the following reactions at 0^(@)C : SrCl_2 . 6H_2O(s) hArr SrCl_2 . 2H_2O(s)+4H_2O(g) K_p=5xx10^(-12) Na_2HPO_4 . 7H_2O(s)+5H_2O(g) K_p=2.43xx10^(-13) Na_2SO_4 . 10H_2O(s) hArr Na_2SO_4(s)+10H_2O(g) K_p=1.024xx10^(-27) The vapour pressure of water at 0^@C is 4.56 torr. Which is the most effective drying agent at 0^@C ?
Answer» `SrCl_2 . 2H_2O` `Na_HPO_4 . 7H_2O` `Na_2SO_4` All equally Answer :A

Discussion

41.	Equilibrium constant of following reaction is 0.5. CO_((g)) + 2H_(2(g)) hArr CH_3OH_((g)) at equilibrium [CO]=0.18 mol L^(-1) and [H_2] = 0.22 mol L^(-1)Calculate the concentration of CH_3OH.
Answer» SOLUTION :`4.356xx10^(-3) ("mol L"^(-1))^(-2)`

Discussion

42.	Equilibrium constant value depends on _________ .
Answer» TEMPERATURE Volume Pressure Catalyst Answer :A

Discussion

43.	Equilibrium constant K_(X), for the reaction H_(2)(g)+I_(2) (g) Leftrightarrow 2HI(g), is 49. What is the value of K_(C) for the reaction 1/2 H_(2)(g)+1/2I_(2) (g) Leftrightarrow HI(g) and 2HI (g) Leftrightarrow H_(2) (g)+I_(2) (g)?
Answer» SOLUTION :`H_(2) (g)+I_(2) Leftrightarrow 2HI(g)` `K=([HI]^(2))/([H_(2)] [I_(2)])=49` `1/2 H_(2) (g)+1/2 I_(2) (g) Leftrightarrow HI (g), K_(1)=?` `K_(1)=K^(1/2)=sqrt(K)=sqrt(49)=7` `2HI (g) Leftrightarrow H_(2) (g)+I_(2) (g), K_(2)=?` `K_(C)` for the REACTION `2HI(g) Leftrightarrow H_(2)(g)+I_(2)(g)` is `K_(2)=1/K =(1)/(49)=2.04 xx 10^(-2)`

Discussion

44.	Equilibrium constant of a reaction does not change with ……… but changes with ……………
Answer» Answer :INITIAL CONCENTRATIONS of reactants, CHANGE in TEMPERATURE

Discussion

45.	Equilibrium constant, K_(c) the reaction N_(2)(g)+3H_(2) (g) Leftrightarrow 2NH_(3) (g)" is "2 xx 10^(-2) mol^(-2) lit^(2). What is the value of K_(c) for the reaction 2NH_(3) (g) Leftrightarrow N_(2) (g)+3H_(2)(g)?
Answer» SOLUTION :`N_(2)(G)+3H_(2) (g) Leftrightarrow 2NH_(3) (g)`, `K_(C)= ([NH_(3)]^(2))/[N_(2)][H_(2)]^(3)=2 xx 10^(-2)" mol"^(2) lit^(2)` `NH_(3) (g) Leftrightarrow N_(2) (g)+3H_(2)(g), K_(2)=?` `K_(2)=([N_(2)] [H_(2)]^(3))/([NH_(3)]^(2))=(1)/(K_(C)0=(1)/(2 xx 10^(-2))=50 mol^(2) lit^(-2)`

Discussion

46.	Equilibrium constant K_(p)for following reaction MgCO_(3(s))hArrMgO_((s))+CO_(2(g))
Answer» `K_(p)=P_(CO_(2))` `K_(p)=(P_(CO_(3))xxP_(CO_(2))xxP_("Mgo"))/(P_("Mg "CO_(3)))` `K_(p)=(P_("Mg")CO_(3))/(P_(CO_(2))P_("MgO"))` `K_(p)=(P_(CO_(3))P_("MgO"))/(P_("Mg")CO_(3))` Solution :Partial PRESSURE of pure SOLID is unity.

Discussion

47.	Equilibrium constant K_(C) for the reaction, N_(2(g))+3H_(2(g))hArr2NH_(3(g)) at 500K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is 3.0" mol L"^(-1) of N_(2),2.0" mol L"^(-10 of H_(2),0.50" mol L"^(-1) of NH_(3).is the reaction at equilibrium ?
Answer» Solution :The given REACTION is `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` According to available data `N_(2)=(3.0)H_(2)=(2.0)NH_(3)=(0.50)` `Q_(C)=((NH_(3(g)))^(2))/([N_(2(g))][H_(2(g))]^(3))` `=([0.50]^(2))/((3.0)(2.0))=(0.25)/(24)=0.0104`

Discussion

48.	Equilibrium constant, K_c for the reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L^(-1) N_2, 2.0 mol L^(-1) H_2,and 0.5 mol L^(-1) NH_3. Is the reaction at equilibrium ? If not in which direction does the reaction tend to proceed to reach equilibrium ?
Answer» SOLUTION :`{:("Equilibrium reaction :",N_(2(G))+ , 3H_(2(g)) HARR , 2NH_(3(g))),("At t time composition :",3.0,2.0,0.5):}` At t time the reaction quotient Q of composition of reaction MIXTURE is, `Q_c=[NH_3]^2/([N_2][H_2])^3=(0.5)^2/((3.0)(2)^3)`=0.0104 `therefore (Q_c=0.0104) lt (K_c=0.061)` Here, `Q_c lt K_c` , the reaction will proceed in the direction of the products (Forward reaction till = `Q_c=K_c`) .

Discussion

49.	Equilibrium constant K_(c) for the reaction , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) at 500 K is 0.061, At particular time ,the analysis shows that the composition of the reaction mixture is 3.0 mol L^(-1) of H_(2), 0.50 mol L^(-1) of NH_(3) .Is the reaction at equilibrium ?
Answer» SOLUTION :The given reaction is `: N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)` According to available data `N_(2) = [3.0] , H_(2) = [2,0] , NH_(3) = [0.50]` `Q_(C) = ([NH_(3) (g)]^(2))/([N_(2) (g)] [H_(2) (g)]^(3)) = ([0.50]^(2))/([3.0][2.0]) = (0.25)/(24) = 0.0104`

Discussion

50.	Equilibrium constant, K_(c) " for the reaction" , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at " 500 K " is " 0061. At a particular time, the analysis shows that composition of the reaction mixture is 30" mol "L^(-1) N_(2), 20 " mol "L^(-1) H_(2)and50 "mol " L^(-1) NH_(3).Is the reaction at equilibrium ? If not , in which direction does the reaction tend to reach equilibrium ?
Answer» Solution :`Q_(C)= ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3) )=(05)^(2)/((30)(20)^(3))=00104` As ` Q_(c)!=K_(c) `, REACTION is not in equilibrium As ` Q_(c) ltK_(c)`, reaction willproceed in the forwarddirection.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Eq. wt ("Mol wt")/x What is x for acetaldehyde converted into acetic acid.

Equivalent weight of X is 9. The vapour density of chloride salt is 66.75 state vulency of X.

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Equivalent weight of Ba(MNO_4)_2 in acidic medium ( M = molar mass)

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Equivalent weight of a metal chloride is 75.5. How many moles of NaOH is required to completely precipitate one mole of metal hydroxide. Atomic weight of the metal is 120.

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Equivalent mass of KMnO_(4) when it is converted to MnSO4 is equal to molar mass divide by……….

Equivalent mass of KMnO_(4), when it is converted to MnSO_(4), is :

Equivalent mass of KMnO_(4) in acidic medium, concentrated alkaline medium and dilute basic medium respectively are M, M, M. Reduced products can be

Equivalent mass of H_(3)PO_(2) when it disproportionates into PH_(3) " and" H_(3)PO_(3) is (Molecular mass=M) :

Equivalent mass of C_(2)O_(4)^(2-) ion in the reaction, C_(2)O_(4)^(2-)to2CO_(2)+2e^(-), is

Equivalent mass of FeC_(2)O_4in the change, FeC_(2)O_(4) rarrFe^(3+) + CO_(2)is

Equivalent mass of an organic base can be determined by:

Equivalent mass of an organic acid can be determined by :

Equimotal aquous solutions ofNaCl and KCl are prepared If the freezing point ofNaClis -2^(@) Cthe freezing point of KCI solution is expectedtobe

Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH ?

Equimolar solutions of the following substances were prepared separately. Which one of these has highest pH value?

Equimolarsolution of non-electrolyte in the same solvent have

Equimolar concentraions of H_2 and I_2 are heated to equilibriumin a 1 litre flask. What percentage of initial concentration of H_2 has reacted at equilibrium if rate constant for both farward and reverse reactions are equal

Equimolar amounts of H_(2) and I_(2) were taken in a bulb maintained at 500 ^(@) C. Dark violet colour faded to light violet which does not change further. This shows that the bulb contains ………….. amounts of …………. .

Equimolar aqueous solutions of NaCl and KCl are prepared. If the freezing point of NaCl is -2^(@)C, the freezing point of KCl solution in expected to be …………

Equilmolar concentrations of H_(2) and I_(2) are heated to equilibrium in a 2 litre flask . At equilibrium , the forward and the backward rate constants are found to be equal . What percentage of intial concentration of H_(2) has reacted at equilibrium ?

Equilibrium state can be achieved if a reversible reaction is carried out in closed or open container.

Equilibrium existing in the hydrolysis of an ester is .......

Equilibrium constant of following reaction is 0.5. CO_((g)) + 2H_(2(g)) hArr CH_3OH_((g)) at equilibrium [CO]=0.18 mol L^(-1) and [H_2] = 0.22 mol L^(-1)Calculate the concentration of CH_3OH.

Equilibrium constant value depends on _________ .

Equilibrium constant K_(X), for the reaction H_(2)(g)+I_(2) (g) Leftrightarrow 2HI(g), is 49. What is the value of K_(C) for the reaction 1/2 H_(2)(g)+1/2I_(2) (g) Leftrightarrow HI(g) and 2HI (g) Leftrightarrow H_(2) (g)+I_(2) (g)?

Equilibrium constant of a reaction does not change with ……… but changes with ……………

Equilibrium constant, K_(c) the reaction N_(2)(g)+3H_(2) (g) Leftrightarrow 2NH_(3) (g)" is "2 xx 10^(-2) mol^(-2) lit^(2). What is the value of K_(c) for the reaction 2NH_(3) (g) Leftrightarrow N_(2) (g)+3H_(2)(g)?

Equilibrium constant K_(p)for following reaction MgCO_(3(s))hArrMgO_((s))+CO_(2(g))

Equilibrium constant K_(C) for the reaction, N_(2(g))+3H_(2(g))hArr2NH_(3(g)) at 500K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is 3.0" mol L"^(-1) of N_(2),2.0" mol L"^(-10 of H_(2),0.50" mol L"^(-1) of NH_(3).is the reaction at equilibrium ?

Equilibrium constant K_(c) for the reaction , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) at 500 K is 0.061, At particular time ,the analysis shows that the composition of the reaction mixture is 3.0 mol L^(-1) of H_(2), 0.50 mol L^(-1) of NH_(3) .Is the reaction at equilibrium ?