52671 + Interview Questions in Chemistry in Class 11 Page 1025 InterviewSolution

51201.	3.0 g sample of KOCl and CaOCl_(2) is dissolved in water to prepare 100 mL solution, which requried 100 " mL of " 0.15 M acidified K_(2)C_(2)O_(4). For the point. The clear solution is now treated with excess of AgNO_(3) solution which precipitates 2.87 g of AgCl. Calculate the mass percentage of KOCl and CaOCl_(2) in the mixture.
Answer» Solution :(a). `2e^(-)+undersetunderset(x=+1)(x-2=-1)(overset(+1)(ClO^(ɵ))toundersetx=-1)(Cl^(ɵ))(n=2)` (b). `2e^(-)+undersetnderset(2x=0)(2x-2=-2)(Cl_(2)O^(-2))tounderset(2x=-2)(Cl^(ɵ))(n=2)` (c). `C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)(n=2)` Let a and b millimoles of KOCl and `CaCOCl_(2)` are present in the mixture. `m" Eq of "KOCl+m" Eq of "CaOCl_(2)=m" Eq of "K_(2)C_(2)O_(4)` `2a+2b=100xx0.15xx2` (n-factor) `therefore2a+2b=30`.(i) ALSO millomoles of `Cl^(ɵ)` from `KOCl+` millimoles of `Cl^(ɵ)` from `CaOCl_(2)-=` millimoles of AgCl` `a(lCl^(ɵ)` ions) `+2b(2Cl^(ɵ)` ions)`=(2.87)/(143.5)xx10^(3)` `[MW of AgCl=143.5]` `thereforea+2b=20`(ii) From EQUATION (i) and (ii) `a=10,b=5` `% of KOCl=(10xx10^(-3)xx90.5xx100)/(3)` `[Mw of KOCl=90.5]` `=30.1%` `% of CaOCl_(2)=(5xx10^(-3)xx127)/(3)xx100``[Mw of CaOCl_(2)=127]` `=21.1%`

Discussion

51202.	3.0 g of pyrolusite ore were treated with 20 g of pure ferrous ammonium sulphate (Mol.mass =392 " g mol"^(-1)) and dilute H_(2)SO_(4). After the reaction, the solution was diluted to 500 mL. 50 mL of diluted solution required 10 mL of 0.1 N K_(2)Cr_(2)O_(7) solution. Calculate the % of pure MnO_(2) in pyrolusite.
Answer» Solution :`MnO_(2)` present in pyrolusite oxidises ferrous ammonium SULPHATE into ferric ammonium sulphate, i.e., `Fe^(2+) to Fe^(3+)`. Unreacted ferrous ammonium sulphate is estimated by potassium dichromate solution. `MnO_(2)+2Fe^(2+)+4H^(+) to Mn^(2+) +2H_(2)O+2Fe^(3+)` `Cr_(2)O_(7)^(2-)+6Fe^(2+) +14H^(+) to 2Cr^(3+)+7H_(2)O+6Fe^(3+)` 50 mL DILUTED ferrous ammonium sulphate solution `=10 mL " of" 0.1 N K_(2)Cr_(2)O_(7)` 500 mL diluted ferrous ammonium sulphate solution `=10xx10 mL " of" 0.1 N K_(2)Cr_(2)O_(7)` =100 mL of 0.1 N FEAS `=(0.1xx392)/(1000)xx100=3.92 g` Used FeAS=(20-3.92)=16.08 g `MnO_(2)` present in pyrolusite `=(87)/(392xx2)xx16.08=1.784 g` Percentage of pure `MnO_(2)=(1.784)/(3.0)xx100=59.4 %`

Discussion

51203.	3.0 g of pure acetic acid and 4.1 g of anhydrous sodium acetate are dissolved together in water and the solution is made up to 500 ml. Calculate the pH of the solution. Given K_(a) of acetic acid is 1.75xx10^(-5).
Answer» Solution :Calculation of the concentration of acetic acid : MASS/`DM^(3)=Nxxg` eq. mass `"mass"//500cm^(3)=(Nxxg" eq. mass")/2rArrN_(CH_(3)COOH)=("Mass"//500cm^(3)xx2)/("g.eq.mass")` Molecular wt. of `CH_(3)COOH=12xx2+1xx4+16xx2=24+04+32=60` `[CH_(3)COOH]=(3.0xx2)/60=6.0/60=0.1N` Calculation of concentration of sodium ACETATE. `"mass"//500cm^(3)=(Nxxg" eq.mass")/2rArrN_(CH_(3)COONa)=("Mass"//500cm^(3)xx2)/("g.eq.mass")` Molecular wt. of `[CH_(3)COONa]=12xx2+1xx3+16xx2+1xx23=82` `[CH_(3)COONa]=(41xx2)/82=8.2/82=0.1N` `pH=pK_(a)+"log"(["SALT"])/(["Acid"])rArrpH=-log_(10)K_(a)+"log"([CH_(3)COONa])/([CH_(3)COOH])` `pH=-log_(10)(1.75xx10^(-5))+"log"([0.1])/([0.1])=-log1.75-log10^(-5)+log1` `pH=-0.2430+5+0=4.7570`

Discussion

51204.	30 g of marble stone on heating produced 11g of CO_2. The percentage of CaCO_3 in marble is
Answer» 0.75 0.8 0.833 0.866 Answer :C

Discussion

51205.	3.0 g of a sample of impure ammonium chloride were boiled with excess of caustic soda solution. Ammonia gas so evolved was passed into 120 mL of N/2 H_(2)SO_(4)mL of N/2 NaOH were required to neutralise excess of the acid. Calculate the percent purity of the given sample of ammonium chloride.
Answer» ANSWER :`82 %`

Discussion

51206.	3.0 g of a sample of impure ammonium chloride were boiled with excess of caustic soda solution. Ammonia gas so evolved was passed into 120 mL of N//2 H_(2)SO_(4). 28 mL of N//2 NaOH were required to neutralise residual acid. Calculate the percentage of purity of the given sample of ammonium chloride.
Answer» SOLUTION :`(120-28) mL N//2 H_(2)SO_(4)-=92 mL N//2 NH_(4)CL`

Discussion

51207.	30 cc of (M)/(3) HCl , 20 cc of (M)/(2) HNO_(3) and40 cc of (M)/(4) NaOH solutions are mixed and the resulting solution is
Answer» 2 1 3 8 Solution :Total millimolar of `H^(+)=(30xx(1)/(3))+(20xx(1)/(2))` `=10+10=20` Total MILLIMOLES of `OH^(-)=40xx(1)/(4)=10` `:. H^(+)` IONS left after neutralization = 10 MILLIMOLE Volume of solution `=1 dm^(3) = 1000cc` Hence, molarityof `H^(+)` ions `=(10)/(1000)M= 10^(-2)M` `:. PH =- log [H^(+)]=-log 10^(-2)=2`

Discussion

51208.	3-Phenylpropene reacts with HBr in the presence of peroxide, the major product formed is
Answer» 2-bromo-1-phenylpropane 1,2-dibromo-3-phenylpropane 3-(o-bromophenyl)propane 1-bromo-3-phenylpropane Solution :`C_6H_5CH_2CH=CH_2+HBr UNDERSET"(Anti-Markovnikov ADDITION)"OVERSET"Peroxide effect"to underset"1-Bromo-3-phenylpropane"(C_6H_5CH_2CH_2CH_2Br)`

Discussion

51209.	3-Phenylpropene on reaction with HBr gives (as a major product )
Answer» `C_6H_5CH_2CH(BR)CH_3` `C_6H_5CH(Br)CH_2CH_3` `C_6H_5CH_2CH_2CH_2Br` `C_6H_5CH(Br)CH=CH_2` SOLUTION :

Discussion

51210.	3 moles of water is boiled at 373 K and is changed to vapour state having the same temperature. What will be the change in entropy of the system ? [The molecules heat of vaporizations of water is 40.668 KJ/mol]
Answer» SOLUTION :`DeltaS_("VAP") = 0.327` KJ/Kelvin

Discussion

51211.	3 moles of NH_3 are allowed to dissociate in a 5 litre vessel and the equilibrium concentration of N_2 is 0.2 mole/lit . Then the total number of moles at equilibrium is
Answer» 2.5 5 1.5 7.5 Answer :B

Discussion

51212.	3 moles of ethanol reacts with one mole of phosphorous tribromide to form 3 moles of bromoethane and one mole of X. Which of the following is X?
Answer» `H_(3)PO_(4)` `H_(3)PO_(2)` `HPO_(3)` `H_(3)PO_(3)` Solution :`3CH_(3)CH_(2)OH+PBr_(3) rarr 3CH_(3)CH_(2)Br+H_(3)PO_(3)(X)`

Discussion

51213.	3 molesof anideal gas(gamma=5//3) is subjected tofollowing chargeof state.Identify thecorrectstatement:satete A(400 K,2 bar) underset("isothermalcooling") overset("Reversible")tostateB(T_(B),T_(B)) underset ("isothermal expansion") overset("Reversible")to State C (300 K, 1 "bar")underset("free expansion") overset("Asiabatic")to State D (T_(D), 0.5 "bar")
Answer» `T_(B)` and `T_(D)` will BESAME and equal to 300 K `DeltaH_(AD)=750 R` `DeltaS_(AC)=0` `DeltaE_(BD)=0` Answer :a,d

Discussion

51214.	3 moles of a mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) required 100 mL of 2 M KMnO_(4) solution in acidic medium. Hence, mole fraction of FeSO_(4) in the mixture is
Answer» `(1)/(3)` `(2)/(3)` `(2)/(5)` `(3)/(5)` Solution :LET moles of `FeSO_(4)=x` EQ. `FeSO_(4)` = eq. `KMnO_(4)` `x xx1=0.1xx2xx5impliesx=1` `therefore` mole FRACTION =1/3

Discussion

51215.	3 moles of a gas are present in a vessel at a temperature of 27^@C. What will be the value of R, the gas constant, in terms of the kinetic energy of the molecules of the gas ?
Answer» Solution :Kinetic energy of n moles of a gas is GIVEN by `K.E. = n 3/2 RT` `:. " The kinetic energy of 3 moles of a gas at " 27^@C (300 K)` will be given by `K.E. = 3xx 3/2 XX R xx 300` `:. "" R =(2 xx K.E.)/(3xx3xx300)= 7.41 xx 10&(-4) xx K.E`. HENCE, the value of R is `7.41 xx 10^(-4) " K.E. " K^(-1)`

Discussion

51216.	3 moles of A and 4 moles of Bare mixed together and allowed to come into equilibrium according to the following reaction A(g) + 4 B(g) hArr 2 C (g) + 3 D (g) When equilirium is reached , there is 1 mole of C. The equilibriumextent of the reaction is
Answer» `1//4` `1//3` `1//2` 1 Solution :`{:(,A,+,4 B,hArr,2C,+,3D),("INTIAL moles:" ,3,,4,,0,,0):}` 4 moles of B react with 1 mole of A. Hence , B is the limiting reactant . From 4 moles of B, if reaction were COMPLETE , 2 moles of C would be formed . Actual C formed = 1 mole . Hence extent of reaction completed `=1/2` . ALSO , when 1 mole of C is formed , out of 4 moles of B taken, 2 moles of B must have reacted . This also shows that the equilibrium extent is 1/2 .

Discussion

51217.	3 mole of reactant A and one mole of reactant B are mixed in a vessel of volume 1 litre . The reaction taking place is A + B hArr 2 C . If 1.5 mol of C is formed at equilibrium , the value of K_(c) is
Answer» `0.12` `0.50` `4.00` `0.25` ANSWER :C

Discussion

51218.	3-Methylpentane on monochlorination gives four possible products. The reaction follows free radical mechanism.The relative reactivities for replacement of -H are 3^(ul@):2^(ul@):1^(ul@) =6:4:1. Relative amounts of A,B,C and D formed are
Answer» 6/31,16/31,6/31,3/31 16/31,6/31,6/31,3/31 6/31,16/31,3/31,6/31 6/31,3/31,6/31,16/31 Answer :C

Discussion

51219.	3-methyl-pent-2-ene on reaction with HBr in the presence of peroxide forms an addition product. The number of possible stereoisomers for the product is
Answer» TWO FOUR six zero Answer :B

Discussion

51220.	3-Methyl-1-butene on oxymercuration demercuration yields............as the major product
Answer» 3-Methyl-2-butanol 2-Methyl-2-butanol 3-Methyl-1-butanol 2-Methyl-1-butanol Solution :`underset("2-Methyl-1-butene")(CH_(3)-underset(CH_(3))underset(\|)(CH)-CH-CH_(2)) underset((ii) "DEMERCURATION")overset((i) "OXYMERCURATION")RARR underset("3-Methyl-1-butanol")(CH_(3)-underset(CH_(3))underset(\|)(CH)-underset(OH)underset(\|)(CH)-CH_(3))` The addition of WATER is in accordance with MARKOWNIKOFF rule.

Discussion

51221.	3 litres each of Nitrogen and Hydrogen measured at STP are allowed to react together. Find the volumes of the gases after the reaction and also the weight of Ammonia formed in the reaction.
Answer» Solution :Balanced equation is `N_(2(g))+3H_(2(g)) to 2NH_(3(g))` `"1 volt3 vol2 vol"` Ratio of VOLUMES of `N_(2), H_(2) and NH_(3)` is 1:3:2 3 litres of `H_(2)` reacts with 1 litre of `N_(2)` and forms 2 litres of `NH_(3)`. Hence volumes of `N_(2) and H_(2)` after the REACTION are 2 litres (3-1) and .0. litres RESPECTIVELY. WEIGHT of 22.4 litre of `NH_(3)` at STP =17g Weight of 2 lit of `NH_(3)` at STP `=(17 xx 2)/(22.4) =1.517gram`

Discussion

51222.	3 g of an oxide of a metal Is converted to chloride completely and it yielded 5 g of chloride. The equivalent weight of the matel is
Answer» 33.25 3.325 12 20 Solution :a) `("Weight of METAL OXIDE")/("Weight of metal chloride")` `= `("Eq.WT. of metal" + "Eq.wt. of oxide")/("Eq.wt. of metal"+"Eq. wt. of chloride")` `3/5`=`(E+8)/(E+35.5)` Or 5E = 40 = 3E + 106.5 or 2E = 66.5 `therefore` E=33.25

Discussion

51223.	3 g of an oxide of a metal is converted to chloride completely and it yielded 5 g of chloride. Equivalent weight of the metal is :
Answer» `33.25` `3.325` 12 20 Solution :N//A

Discussion

51224.	3 g of activated chasrcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :
Answer» 18 mg 36 mg 42 mg 54 mg SOLUTION :Number of moles of acetic acid adsorbed `= (0.06xx(50)/(1000)-0.042xx(50)/(1000))` `= (0.9)/(1000)` moles. `THEREFORE` Weight of acetic acid adsorbed `= 0.9xx60 mg` = 54 mg Hence, the amount of acetic acid adsorbed per g of charcoal `= (54)/(3)` mg = 18 mg Hence, option (A) is correct.

Discussion

51225.	2xx10^(6) molecules of N_(2) gas enter into the vessel having volume 400 mL at 400 K temperature. Find the pressure of N_(2) gas. [R = 0.082 L atom mol^(-1)K^(-1)] [1 atom = 1.013 bar]
Answer» ANSWER :`27.233xx10^(-16)` ATOM ; `2.76xx10^(-6)` BAR

Discussion

51226.	2xx 10^(8)atoms of carbon arearranged side bysidecaculatethe radiusof carbonatom if thelengthof this arrangementis 2.4cm .
Answer» Solution :Where`2xx 10^(8)` carbonatom = 2.4 CM 1carbonatom= (?)CAM diameter ofcarbon = `(1xx 2.4 cm)/(2xx 10^(8))= 1.2xx 10^(8) cm` radiusof carbon= `(1.2xx 10^(8) cm)/(2 ) ` `6.0 XX 10^(9) cm` `6.0 xx 10^(9) xx 10^(7) NM` `-0.06 nm`

Discussion

51227.	2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)), at 298 K of K_c is 7xx10^25 than calculate for SO_(3(g)) hArr SO_(2(g)) + 1/2O_2
Answer» SOLUTION :`1.1952xx10^(-13)`

Discussion

51228.	2SO_2(g) +O_2(g) leftrightarrow 2SO_3(g) IF the volume of reaction vessel is increased, what happens to Kc. Explain.
Answer» SOLUTION :KC INCREASES

Discussion

51229.	2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)""DeltaHlt0 Which change(s) will increase the quantity of SO_(3)(g) at equilibrium? (P) Increasing the temperature. (Q) Reducing the volume of the container. (R) Adding He to increase the pressure keeping volume
Answer» <P>P only Q only P and Q only Q and R only Answer :B

Discussion

51230.	2RMgCl + SiCl_(4) overset(H_(2)O)rarr B overset("polymerisation")rarr C. C is
Answer» CYCLIC silicone linear silicone cross linked silicone `SiO_(2)` ANSWER :A::C

Discussion

51231.	2R - X + 2Na overset("dry ether") rarr R - R + 2Nax Which of the following alkanes is not obtained by Wurtz reaction ?
Answer» METHANE ETHANE PROPANE BUTANE ANSWER :A

Discussion

51232.	2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-117" kJ". Predict the effect of pressure decrease as a result of increased volume on the equilibrium concentration of NO_(2).
Answer» SOLUTION :`2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-177" KJ"` Decrease in pressure will FAVOUR BACKWARD REACTION, i.e. less `NO_(2)` will be formed.

Discussion

51233.	2O_3 harr 3O_2 Ozone dissociates until the molecular weight of the equilibrium mixture becomes 40. Initial pressure of O_3 is 600 mm Hg. Then
Answer» `K_p = 360` mm degree of DISSOCIATION = 0.4 `P_(eq) = 720` mm `K_p = p_(O)_3 = P_(O)_2` Solution :`(D)/(d)=1+(n-1)alpha, (48)/(40)=1+((3)/(2)-1)alpha, alpha=40%` `2O_(3) harr 3O_(2)` at EQM `UNDERSET(=360)(600(1-alpha))""underset(360)((3alpha)/(2) xx 600)` `K_(P)=((360)^(3))/((360)^(2))=360`

Discussion

51234.	2p, 3p,4p,5p arrangetheseorbitalsin increasingorder of energy ?
Answer» SOLUTION :`2P to3p to4p to5p`energyincreases.

Discussion

51235.	2NO(g)+O_(2)(g)hArr2NO_(2)(g) Which would increase the partial pressure of NO_(3)(g) at equlibrium?
Answer» Decreasing the VOLUME of the system ADDING a mole GAS to INCREASE the pressure of the system Removing some NO(g) from the system Adding an appropriate catalyst Answer :A

Discussion

51236.	2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-117" kJ". Predict the effect of an increase in concentration of NO.
Answer» Solution :`2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-177" KJ"` If we INCREASE the concentration of NO, the rate of forward reaction will increase, i.e. more `NO_(2)` will be FORMED.

Discussion

51237.	2NO_(2)(g)to N_(2)O_(4)(g) DeltaU_(f)^(@)[N_(2)O_(4)(g)]=2kcal//"moleand "DeltaU_(reaction)^(@)=-16kcal//mol, then calculate DeltaH_("formation" )^(@)=-16kcal//mol,then calculate DeltaH_("formation")^(@) of NO_(2)at 727^(@)C:
Answer» 9kcal/mol 4.5kcal/mol 8kcal/mol 10kcal/mol Answer :C

Discussion

51238.	2NO(g)+O_(2)(g)hArr2NO_(2)(g) At a certain temperature the equilibrium concentration for this system are : [NO]=0.25M, [O_(2)]=0.24M,[NO_(2)]=0.18M. What is the value of K_(c) at this temperature?
Answer» 0.063 `0.50` 1.4 `2.0` ANSWER :B

Discussion

51239.	2NO_(2(g)) (brown) hArr N_2O_(4(g)) , DeltaH=57.2 kJ mol^(-1) if the reaction vessel kept in ice and hot water than what is the change in colour ?
Answer» Solution :This REACTION is exothermic. So at lower temperature the forward reaction TAKE PLACE. So in mixture `NO_2` decreases so brown colour. decreases. If the vessel kept in hot water then the reaction take place in REVERSE DIRECTION and concentration of `NO_2` increases so brown colour increases.

Discussion

51240.	2NH_(3)(g)hArrN_(2)(g)+3H_(2)(g) in a V lit container total x mol at eq. N_(2)H_(4)(g)hArrN_(2)+2H_(2)(g) in V lit (other) container total y mol at eq. If both are taken in same container (V lit) then at new equation total mols will be
Answer» `x+y` `gtx+y` `ltx+y` No PREDICTION is possible. Solution :N//A

Discussion

51241.	2NH_3(g) hArr N_2(g)+3H_2(g) If degree of dissociation of ammonia at equilibrium is 0.7 then observed molecular weight of reaction mixture at equilibrium :
Answer» 12 10 15 17 Answer :B

Discussion

51242.	2Na_(2)S_(2)O_(3)+I_(2)toNa_(2)S_(4)O_(6)+2NaI How many equivalents ofHypo is oxidised by one mole of Iodine?
Answer» SOLUTION :`2Na_(2)overset(+2)(S_(2))O_(3)+overset(0)I_(2)rarrNa_(2)overset(+2.5)(S_(4))O_(6)+2NaI^(-1)` EQ wt of `Na_(2)S_(2)O_(3)=("MOL wt of"Na_(2)S_(2)O_(3))/(1)` EQ wt of Iodine = `("Mol wt of iodine")/(2)` 1 mole iodine = 2 equivalents iodine = 2 equivalents hypo

Discussion

51243.	2M(s)+H_(2)SO_(4)(aq)rarrM_(2)SO_(4)(aq)+H_(2)(g) Give therepresentation of the cell which involves the above redox reaction
Answer» SOLUTION :`M (s) \|M_(2)SO_(4) (aq)\|\|H^(+)(aq)\|H_(2)(G)` pt

Discussion

51244.	2 moles of pure KClO_(3) is decomposed to an extent of 66.6%. How many does of O_(2) is released?
Answer» SOLUTION :`2KClO_(3)OVERSET(Delta)rarr2KCl+3O_(2)UARR` `(2xx66.6)/(100)-(2xx66.6)/(100)xx(3)/(2)=~~1.998=2`

Discussion

51245.	2Mn_(2)O_7rarr4MnO_(2)+ 3O_2(If M is mol. wt. of Mn_(2)O_7). Find the equivalent, weight of Mn_2O_7in the above change.
Answer» Solution :`Mn_(2)^(7+) +6 e rarr 2Mn^(4+)` 2 `xx` MOLE of `Mn_(2) O_(7) =4xx` mole of `MnO_2` `=4xx3 xx` eq. of `MnO_2 =12 ` eq. of `MnO_2` `:.` Mole of `Mn_(2)O_(7) = 6 ` eq. of `MnO_2` `:. ` Eq. of `Mn_(2) O_(7) ` = Eq. of `MnO_7` `:.` Eq. wt. of `Mn_(2) O_(7) = M//6`

Discussion

51246.	2KMnO_(4)+16HCl to2MnCl_(2)+2KCl+5Cl_(2)+8H_(2)O How many moles of HCl undergo oxidative in the above reaction ?
Answer» 14 10 5 16 Solution : THUS only 10 MOLES of HCl are Oxidised to 5 moles of `Cl_(2)` ____ (B) From 6HCl, `6Cl^(-)` are remaining as `6Cl^(-)` and the `6Cl^(-)` are not oxidised.

Discussion

51247.	In the reaction, 2HgOrarr2Hg+O_(2),Hg^(2+) act as
Answer» OXIDISING AGENT REDUCING agent Oxidised none of the above Answer :A

Discussion

51248.	2H_(2)Orarr4e^(-)+O_(2)+4H^(+). The equivalent weight of molecules oxygen is
Answer» 32 16 8 4 Solution :n-f = no. of MOLES of `E^(-)` gained/mole `implies n-f" of "O_(2)=4impliesE=(32)/(4)=8`

Discussion

51249.	2H_(2)O(g) hArr 2H_(2)(g) + O_(2)(g) K_(C) = 4.1 xx 10^(-48) At 599 K N_(2)(g) +O_(2)(g) hArr hArr 2NO(g) K_(c)= 1 xx 10^(-30) at 1000 K Predict the extent of the above two reactions.
Answer» SOLUTION :In the reactions , decomposition of water at 500 K and oxidation of NITROGEN at 1000 K, the value, of `K_(c)` is very less `K_(c) lt 10^(-3)` . So reverse reaction is FAVOURED `therefore` Products << REACTANTS

Discussion

51250.	2H_(2)Orarr 4e^(-)+O_(2)+4H^(+).The equivalent weight of oxygen is
Answer» 32 16 8 4 Answer :C

Discussion

Explore topic-wise InterviewSolutions in .

3.0 g of pure acetic acid and 4.1 g of anhydrous sodium acetate are dissolved together in water and the solution is made up to 500 ml. Calculate the pH of the solution. Given K_(a) of acetic acid is 1.75xx10^(-5).

30 g of marble stone on heating produced 11g of CO_2. The percentage of CaCO_3 in marble is

3.0 g of a sample of impure ammonium chloride were boiled with excess of caustic soda solution. Ammonia gas so evolved was passed into 120 mL of N/2 H_(2)SO_(4)mL of N/2 NaOH were required to neutralise excess of the acid. Calculate the percent purity of the given sample of ammonium chloride.

30 cc of (M)/(3) HCl , 20 cc of (M)/(2) HNO_(3) and40 cc of (M)/(4) NaOH solutions are mixed and the resulting solution is

3-Phenylpropene reacts with HBr in the presence of peroxide, the major product formed is

3-Phenylpropene on reaction with HBr gives (as a major product )

3 moles of water is boiled at 373 K and is changed to vapour state having the same temperature. What will be the change in entropy of the system ? [The molecules heat of vaporizations of water is 40.668 KJ/mol]

3 moles of NH_3 are allowed to dissociate in a 5 litre vessel and the equilibrium concentration of N_2 is 0.2 mole/lit . Then the total number of moles at equilibrium is

3 moles of ethanol reacts with one mole of phosphorous tribromide to form 3 moles of bromoethane and one mole of X. Which of the following is X?

3 moles of a mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) required 100 mL of 2 M KMnO_(4) solution in acidic medium. Hence, mole fraction of FeSO_(4) in the mixture is

3 moles of a gas are present in a vessel at a temperature of 27^@C. What will be the value of R, the gas constant, in terms of the kinetic energy of the molecules of the gas ?

3 moles of A and 4 moles of Bare mixed together and allowed to come into equilibrium according to the following reaction A(g) + 4 B(g) hArr 2 C (g) + 3 D (g) When equilirium is reached , there is 1 mole of C. The equilibriumextent of the reaction is

3 mole of reactant A and one mole of reactant B are mixed in a vessel of volume 1 litre . The reaction taking place is A + B hArr 2 C . If 1.5 mol of C is formed at equilibrium , the value of K_(c) is

3-Methylpentane on monochlorination gives four possible products. The reaction follows free radical mechanism.The relative reactivities for replacement of -H are 3^(ul@):2^(ul@):1^(ul@) =6:4:1. Relative amounts of A,B,C and D formed are

3-methyl-pent-2-ene on reaction with HBr in the presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

3-Methyl-1-butene on oxymercuration demercuration yields............as the major product

3 litres each of Nitrogen and Hydrogen measured at STP are allowed to react together. Find the volumes of the gases after the reaction and also the weight of Ammonia formed in the reaction.

3 g of an oxide of a metal Is converted to chloride completely and it yielded 5 g of chloride. The equivalent weight of the matel is

3 g of an oxide of a metal is converted to chloride completely and it yielded 5 g of chloride. Equivalent weight of the metal is :

3 g of activated chasrcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

2xx10^(6) molecules of N_(2) gas enter into the vessel having volume 400 mL at 400 K temperature. Find the pressure of N_(2) gas. [R = 0.082 L atom mol^(-1)K^(-1)] [1 atom = 1.013 bar]

2xx 10^(8)atoms of carbon arearranged side bysidecaculatethe radiusof carbonatom if thelengthof this arrangementis 2.4cm .

2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)), at 298 K of K_c is 7xx10^25 than calculate for SO_(3(g)) hArr SO_(2(g)) + 1/2O_2

2SO_2(g) +O_2(g) leftrightarrow 2SO_3(g) IF the volume of reaction vessel is increased, what happens to Kc. Explain.

2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)""DeltaHlt0 Which change(s) will increase the quantity of SO_(3)(g) at equilibrium? (P) Increasing the temperature. (Q) Reducing the volume of the container. (R) Adding He to increase the pressure keeping volume

2RMgCl + SiCl_(4) overset(H_(2)O)rarr B overset("polymerisation")rarr C. C is

2R - X + 2Na overset("dry ether") rarr R - R + 2Nax Which of the following alkanes is not obtained by Wurtz reaction ?

2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-117" kJ". Predict the effect of pressure decrease as a result of increased volume on the equilibrium concentration of NO_(2).

2O_3 harr 3O_2 Ozone dissociates until the molecular weight of the equilibrium mixture becomes 40. Initial pressure of O_3 is 600 mm Hg. Then

2p, 3p,4p,5p arrangetheseorbitalsin increasingorder of energy ?

2NO(g)+O_(2)(g)hArr2NO_(2)(g) Which would increase the partial pressure of NO_(3)(g) at equlibrium?

2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-117" kJ". Predict the effect of an increase in concentration of NO.

2NO_(2)(g)to N_(2)O_(4)(g) DeltaU_(f)^(@)[N_(2)O_(4)(g)]=2kcal//"moleand "DeltaU_(reaction)^(@)=-16kcal//mol, then calculate DeltaH_("formation" )^(@)=-16kcal//mol,then calculate DeltaH_("formation")^(@) of NO_(2)at 727^(@)C:

2NO(g)+O_(2)(g)hArr2NO_(2)(g) At a certain temperature the equilibrium concentration for this system are : [NO]=0.25M, [O_(2)]=0.24M,[NO_(2)]=0.18M. What is the value of K_(c) at this temperature?

2NO_(2(g)) (brown) hArr N_2O_(4(g)) , DeltaH=57.2 kJ mol^(-1) if the reaction vessel kept in ice and hot water than what is the change in colour ?

2NH_(3)(g)hArrN_(2)(g)+3H_(2)(g) in a V lit container total x mol at eq. N_(2)H_(4)(g)hArrN_(2)+2H_(2)(g) in V lit (other) container total y mol at eq. If both are taken in same container (V lit) then at new equation total mols will be

2NH_3(g) hArr N_2(g)+3H_2(g) If degree of dissociation of ammonia at equilibrium is 0.7 then observed molecular weight of reaction mixture at equilibrium :

2Na_(2)S_(2)O_(3)+I_(2)toNa_(2)S_(4)O_(6)+2NaI How many equivalents ofHypo is oxidised by one mole of Iodine?

2M(s)+H_(2)SO_(4)(aq)rarrM_(2)SO_(4)(aq)+H_(2)(g) Give therepresentation of the cell which involves the above redox reaction

2 moles of pure KClO_(3) is decomposed to an extent of 66.6%. How many does of O_(2) is released?

2Mn_(2)O_7rarr4MnO_(2)+ 3O_2(If M is mol. wt. of Mn_(2)O_7). Find the equivalent, weight of Mn_2O_7in the above change.

2KMnO_(4)+16HCl to2MnCl_(2)+2KCl+5Cl_(2)+8H_(2)O How many moles of HCl undergo oxidative in the above reaction ?

In the reaction, 2HgOrarr2Hg+O_(2),Hg^(2+) act as

2H_(2)Orarr4e^(-)+O_(2)+4H^(+). The equivalent weight of molecules oxygen is

2H_(2)O(g) hArr 2H_(2)(g) + O_(2)(g) K_(C) = 4.1 xx 10^(-48) At 599 K N_(2)(g) +O_(2)(g) hArr hArr 2NO(g) K_(c)= 1 xx 10^(-30) at 1000 K Predict the extent of the above two reactions.

2H_(2)Orarr 4e^(-)+O_(2)+4H^(+).The equivalent weight of oxygen is