94420 + Interview Questions in Chemistry in Class 12

1.	For the equilibrium system.N_(2)(g)+O_(2)(g)+Heat hArr 2NO(g)Which of the following factors would cause the value of equilibrium constant to decrease ?
Answer» Adding a catalyst. Decreasing the TEMPERATURE Adding `N_(2)` gas Adding NO(G). ANSWER :B

Discussion

2.	For the equilibrium system :2HCl(g)hArr H_(2)(g)+Cl_(2)(g)the equilibrium constant is 1.0xx10^(-5).What is the concentration of HCl if the equilibrium concentration of H_(2) and Cl_(2) are 1.2xx10^(-3) M and 1.2xx10^(-4) M respectively ?
Answer» `12XX10^(-4)M` `12xx10^(-3)M` `12xx10^(-2)M` `12xx10^(-1)M` Solution :`K=([H_(2)][Cl_(2)])/([HCL]^(2))` `1.0xx10^(-5)=((1.2xx10^(-3))(1.2xx10^(-4)))/([HCl]^(2))` or `[HCl]^(2)=((1.2xx10^(-3))(1.2xx10^(-4)))/(1.0xx10^(-5))` `=1.44xx10^(-2)` `[HCl]=sqrt(1.44xx10^(-2))=12xx10^(-2)`

Discussion

3.	For the equilibrium N_(2)+3H_(2)hArr2NH_(3),K_(c) at 1000K is 2.73xx10^(-3) if at equlibrium [N_(2)]=2M,[H_(2)]=3M, the concentraion of NH_(3) is
Answer» `0.00358` M `0.0358` M `0.358` M `3.58` M Solution :`K_(c)=([NH_(3)]^(3))/([N_(2)][H_(2)]^(3))` `2.37xx10^(-3)=(X^(2))/([2][3]^(3))=x^(2)=0.12798impliesx=0.358M.`

Discussion

4.	For the equilibrium :MgCO_(3)(s)hArr MgO(s)+CO_(2)(g)which of the following expressions is correct ?
Answer» `K_(p)=(P_(MgO)xxP_(CO_(2)))/(P_(MgCO_(3)))` `K_(p)=([MgO][CO_(2)])/([MgCO_(3)])` `K_(p)=pMgO+pCO_(2)` `K_(p)=pCO_(2)` ANSWER :D

Discussion

5.	For the equilibrium MgCO_3(g)oversetDeltahArrMgO(s)CO_2(s)which of the following expressions is correct ?
Answer» `K_p=pco_2` `K_p=([MgO][CO_2])/([MgCO_3])` `K_p=(p_(Mgo).p_(CO_2))/(p_(MgCO_3))` `K_p=(p_(Mgo)+p_(CO_2))/(p_(MgCO_3))` SOLUTION :In hetrogeneous system. `K_(c)and K_(p)` are not depend upon the concentration of pressure of solid substence Hence. At equilibrium their concentratiaon or pressure are ASSUMED as one. `MgCO_3(s)hArrMgO(s)+CO_2(G)` `THEREFORE K_p=Pco_(2)`

Discussion

6.	For the equilibrium: LiCl.3NH_(3(s))hArrLiCl.NH_(3(s))+2NH_(3), K_(p)=9 atm^(2) at 40^(@)C. A 5 litre vessel contains 0.1 mole of LiCl.NH_(3). How many mole of NH_(3) should be added to the flask at this temperture to derive the backward reaction for completion?
Answer» ANSWER :0.79

Discussion

7.	For the equilibrium in a closed vessel PCI_5 (g) hArr PCI_3 (g) +CI_2(g) K_p is found to be double of K_c this is attained when
Answer» T=2k T=12.18K T=24.36 K T=27.3K Answer :C

Discussion

8.	For the equilibrium : HCO_(3)^(-)hArr H^(+)+CO_(3)^(2-)K=4.8xx10^(-11), [CO_(3)^(2-)]=1.1xx10^(-3)M, [HCO_(3)^(-)]=9.8xx10^(-3)MThe pH of the solution is :
Answer» `8.37` `9.37` `6.0` `8.0` SOLUTION :`K=([H^(+)][CO_(3)^(2-)])/([HCO_(3)^(-)])` or `[H^(+)]=(K[HCO_(3)^(-)])/([CO_(3)^(2-)])` `=(4.8xx10^(-11)xx9.8xx10^(-2))/(1.1xx10^(-3))` `=4.28xx10^(-9)` `pH=-LOG[H^(+)]=-log (4.28xx10^(-9))` `=8.37`

Discussion

9.	For the equilibriumH_(2)O(l)subH_(2)O(g) at 1 atm and 298 K
Answer» STANDARD free energy change is equal to zero `(DeltaG^(@)=0)` Free energy change is less than zero `(DeltaG lt 0)` Standard free energy change is less than zero `(DeltaG^(@)lt0)` Standard free energy change is GREATER than zero `(DeltaG^(@)gt0)` Solution :For reaction `H_(2)O(l)rarrH_(2)O(G)` `Deltan=1` MEANS POSITIVE.

Discussion

10.	For the equilibrium, H_2O(l) iff H_2O_((v)) , which of the following is correct?
Answer» `Delta G = 0, Delta lt 0 , Delta S lt 0` `Delta G lt 0 , Delta H gt 0 , Delta S gt 0` `Delta S gt 0, Delta H = 0 , Delta S gt 0 ` `Delta G = 0, Delta H gt 0, Delta S gt 0` Solution :`H_2O(l) iff H_2O (v)` As the process is in EQUILIBRIUM, `Delta G = 0` ` Delta H gt 0`(as the process is ENDOTHERMIC) ` Delta S gt 0`(as entropy increases from LIQUID to gas)

Discussion

11.	For the equilibrium: CaCO_(3(s)) iff CaO_((s)) + CO_(2(g)) , K_p = 1.64 atm at 1000 K, 50 g of CaCO_3 in a 10litre closed vessel is heated to 1000 K. Percentage of ĆaCO_3 that remains unreacted at equilibrium is (Given R=0.082 L atm K^(-1) "mol"^(-1)).
Answer» 40 50 60 20 Solution :`CaCO_3 iff CaO + CO_2` `Kp = pCO_2` No. of MOLES =n `1.64 xx 10 = 0.082 xx 1000 xx n` `n=(1.64 xx 10)/(0.082 xx 1000)=0.2` `:.` No. of moles `CO_2 = 0.2` 50g of `CaCO_3 = 0.5` moles of `CaCO_3` gives 0.2 moles of `CO_2` `IMPLIES` percentage of `CaCO_3` unreacted = 0.3 MOLE = 60%

Discussion

12.	For the equilibrium at 298 K: N_2O_(4(g)) hArr 2NO_(2(g)), G_(N_2O_4)^@ = 100 "kJ mol"^(-1) and G_(NO_2)^@ = 50 "kJ mol"^(-1) . If 5 moles of N_2O_4 and 2 moles of NO_2 are taken initially in one litre container then which statements are correct?
Answer» Reaction proceeds in forward direction `K_c=1` `DELTAG`= - 0.55 kJ , `DeltaG^@=0` At equilibrium `[N_2O_4]` = 4.894 M and `[NO_2]`= 2.212 M Solution :`DeltaG=DeltaG^@ + 2.303 RT log Q` `DeltaG^@ = 2xxG_(NO_2)^@ - G_(N_2O_4)^@ = 2 xx 50 -100 =0` `therefore DeltaG=0+ 2.303 xx 8.314 xx 10^(-3) xx 298 "log" 2^2/5` = 0-0.55 kJ `therefore DeltaG`=- 0.55 kJ, i.e., reaction proceeds in forward direction . Also, `DeltaG^@=0=-2.303 RT log K_c therefore K_c=1` Now, `{:(N_2O_4,hArr , 2NO_2),(5,,2),((5-x),,(2+2X)):}` `therefore 1=(2+2x)^2/((5-x))` or x=0.106 `[NO_2]`= 2+ 2x = 2+ 2 x 0.106 = 2.212 M `[N_2O_4]` = (5-x) =5- 0.106 = 4.894 M

Discussion

13.	For the equilibrium, CaCO_3(s) hArr CaO(s) + CO_2(g) which of the following expressionis correct :
Answer» <P>`K_p=[CAO] [CO_2]/[CaCO_3]` `K_p=p_((cao)+p(co_2))/p_(caco_2)` `K_p=p(co_3)` `K_p=(p_(cao)+p_(co_2))/p_(caco_3) ANSWER :C

Discussion

14.	For the equilibrium 2SO_(3(g))hArr2SO_(2(g))+O_(2(g)), the value of equilibrium constant is 4.8xx10^(-3) at 700^(@)C. At equilibrium, if the concentration of SO_(3) and SO_(2) are 0.60M and 0.15M respectively. Calculate the concentration of O_(2) in the equillibrium mixture.
Answer» ANSWER :0.0768 M

Discussion

15.	For the equilibrium 2SO_(3(g))hArrSO_(2(g))+O_(2(g)), the value of equilibrium constant is 4.8xx10^(-3) at 700^(@)C. At equilibrium, if the concentration of SO_(3) and SO_(2) are 0.60M and 0.15M respectively. Calculate the concentration of O_(2) in the equillibrium mixture.
Answer» ANSWER :0.0768 M

Discussion

16.	For the equilibrium: 2NO(g) + O_2(g) + 2NO_2(g), K_c=6.45 xx 10^5 . (i) At what O_2concentration is the NO_2concentration equal to the NOconcentration?(ii) At what O_2concentration is the NO_2concentration 100 times the NO concentration?
Answer» SOLUTION :`1.55 XX 10^(-6) , 1.55 xx 10^(-2)`

Discussion

17.	For the equilibrium 2NOCl_((g))hArr2NO_((g))+Cl_(2(g)) the value of the equilibrium constant K_(c) is 3.75xx10^(-6) at 790^(@)C. Calculate K_(p) for this equilibrium at the same temperature.
Answer» <P> SOLUTION :`K_(p)=K_(C)(RT)^(Deltang)`

Discussion

18.	For the equilibrium2NO_(2(g))hArrN_(2)O_(4(g))+14.6 kcal the increase in temperaturee would
Answer» Favour the FORMATION of `N_(2)O_(4)` Favour the decomposition of `N_(2)O_(4)` Not alter the EQULIBRIUM Stop the REACTION Solution :The REACTIN is endothemic in reverse directin and HENCE increase in temperature will favour reverse reaction.

Discussion

19.	For the equilibrium, 2H_(2)(g)+O_(2)(g)hArr2H_(2)O(l) at 25^(@)C,DeltaG^(@) is -474.78kJ mol^(-1). Calculate log K for it (R=8.314 JK^(-1)mol^(-1))
Answer» SOLUTION :`-DELTAG^(@)=2.303RT" LOG K",""474780=2.303xx8.314xx298" log K or log K"=83.2`

Discussion

20.	For the equilibrium 2NO_2(g) ⇌N_2O_4(g) +14.6 kcalAn increase of temperature will:
Answer» FAVOUR the FORMATION of `N_2O_4` Favour the decomposotion of `N_2O_4` Not AFFECT the equilibrium Stop the reaction Answer :B

Discussion

21.	For the endothermic reactionA_(2) rarr 2A, which of the following will increase yield of monomer?
Answer» INCREASE in both TEMPERATURE and CONCENTRATION of reactant Increase in temperature and DECREASE in concentration of reactant. Decrease in temperature and increase in concentration of reactant. Decrease in both temperature and concentration of reactant. Answer :A

Discussion

22.	For the elementary reaction M to N , the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M . The order of the reaction with respect to M is
Answer» 4 2 2 1 Solution :`M to N` `R = K[M]^(x)` as [M] is doubled RATE increases by a factor of 8 . i.e. 8r = `K [2M]^(x) implies 8 (2)^(x) implies x = 3`.

Discussion

23.	For the emperical formula Pt(NH_(3))_(2)Cl_(2), the no. of possible coordination isomers would be (ON of pt is +2 in all all isomeric forms)
Answer» 1 2 3 4 Solution :`[Pt(NH_(3))_(2)Cl_(2)]` can have two GEOMETRICAL isomers.

Discussion

24.	For the elementary step (CH_3)_3 CBr_((aq)) to (CH_3)_3C_((aq))^++Br_((aq))^(-) the molecularity is
Answer» zero 1 2 connot ASCERTAINED Solution :Molecularity presents the NUMBER of molecules of reactants taking PART in an elementary STEP.

Discussion

25.	For the elementary step (CH_3)_3CBr(aq)rarr(CH)_3)_3C^+(aq)+Br^(-)(aq) the molecularity is :
Answer» Zero 1 2 Cannot ascertained Answer :B

Discussion

26.	For the elementary step(CH_3)_3CBr(aq)rarr(CH)_3)_3C^+(aq)+Br^(-)(aq) the molecularity is :
Answer» Zero 1 2 Cannot ascertained Answer :B

Discussion

27.	For the elementary reaction M to N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of reaction with respect to M is
Answer» 4 3 2 1 Solution :Initially, rate `r = K[M]^(alpha) = ka^(alpha) ""…(i)` On DOUBLING the CONCENTRATION of M Dividing eqn. (ii) by eqn. (i), `8 =2^(alpha) or 2^(alpha) = 2^(3) or alpha = 3`

Discussion

28.	for the electrorode reaction, M^(n+)(aq)+n e^(-)rarrM(s) Nernst equation is:E=E^o +frac(RT)(nF)ln frac(1)([M^(n+)],E=E^o +RT ln[M^(n+)],E=E^o +frac(RT)(nF)l n[M^(n+)],E/E^o=frac(RT)(nF)l n[M^(n+)]..
Answer» `E=E^o +frac(RT)(NF)LOG frac(1)([M^(n+)]` `E=E^o +RT In[M^(n+)]` `E=E^o +frac(RT)(nF)I n[M^(n+)]` `E/E^o=frac(RT)(nF)I n[M^(n+)].` ANSWER :C

Discussion

29.	For the electrolytic production of NaClO_(4) from NaClO_(3) as per the following equation: NaClO_(3) + H_(2)O rightarrow NaClO_(4) + H_(2) How many faradays of electricity will be required to produce 0.5 "mole" of NaClO_(4) assuming 60% efficiency?
Answer» 0.835 F 1.67 F 3.34 F 1.6 F Answer :B

Discussion

30.	For the electrolytic production of NaClO_(4) from NaClO_(3) as per reactions : ClO_(3)^(-) + H_(2)O rarr ClO_(4)^(-) + 2H^(+) + 2e^(-) (i) How many faradays of electricity would be required to produce 1 mole of NaClO_(4) ? (ii) What volume of H_(2) at STP would be liberated at the cathode in the time that it takes to form 12.25 g of NaClO_(4) ?
Answer» SOLUTION :2 F, 2.24 LITRES

Discussion

31.	For the electrolytic production of NaClO_4from NaClO_3according to the equation NaClO_3 +H_2O to NaClO_4 +H_2,the number of Faradays of electricity required to produce 0.5 mole of NaClO_4is
Answer» 1 2 3 1.5 Answer :A

Discussion

32.	For the electrode process, H^(+)+e=1/2H_(2),E_(H_(H)^(+),H_(2)) = x volt then for 2H^(+)+2e=H_(2),E_(2H^(+),H_(2)) is equal to
Answer» X VOLT 2x volt `x/2` volt 0 volt Answer :A

Discussion

33.	For the electrochemicalcell,M[M^+]X^-\|X,E^@_(M+//M)=0.44 V and E_(X//X^-)^@ =0.33 V. From this data, one can duduce that :
Answer» `M+ X rarr M^+ +X` is the SPONTANEOUS reaction `M^+X rarr M+X` is the spontaneous reaction `E_cell=0.77V` `E_(CELL)=-0.77V` ANSWER :B

Discussion

34.	For the electrochemical cell, Mg(s)\|Mg^(2+)(aq,1M)\|\|Cu^(2+)(aq,1M)\|Cu(s) standard emf of the cell is 2.70 V at 300K. When the concentration of Mg^(2+) is changed to x M, the cell potential changes to 2.67V at 300K. The value of x is_____ (Given, (F)/(R)=11500KV^(-1), where F is the Faraday constant and R is the gas constant, ln(10)=2.30)
Answer» Solution :`underset(E_(cell)^(o)=2.70)((10)MG(s)+)Cu^(2+)underset(E_(cell)=2.67)((aq)to)Mg^(2+)undersetunderset(Cu^(2+)=1M)(Mg^(2+)=xM)((aq)+Cu(s))` `E_(cell)=E_(cell)^(o)-(RT)/(nF)LN" "x` `2.67=2.70-(RT)/(2F)ln" "x` `-0.03=-(Rxx300)/(2F)xxln" "x` ln `x=(0.03xx2)/(300)xx(F)/(R)=(0.03xx2xx11500)/(300xx1)` ln `x=2.30=ln(10)impliesx=10`

Discussion

35.	For the electrochemical cell, M\|M^(+)\|\|X^(-)\|X,E^(@)(M^(+)//M)=0.44V and E^(@)(X//X^(-))=0.33V. From this data one can deduce that
Answer» `M+XtoM^(+)+X^(-)` is the spontaneous reaction `M^(+)+X^(-)TOM+X` is the spontaneous reaction `E_(cell)=0.77V` `E_(cell)=-0.77V` Solution :For the given cell `M\|M^(+)\|\|X^(-)\|X`, the cell reaction is derived as follows: R.H.S. reduction `X+e^(-)TOX^(-)` . . (i) L.H.S. oxidation `MtoM^(+)+e^(-)`. . . . (ii) Add (i) and (ii) M+X`toM^(+)+X^(-)` the cell POTENTIAL `=-0.11V` Since, `E_(cell)=-ve`, the cell reaction derived above is not spontaneous. In fact, the reverse reaction will occur SPONTANEOUSLY.

Discussion

36.	For the electrochemical cell, Mabs(M^(+))abs(X^(-)) X, E^(o) (M^(+)//M) = 0.44 V and E^(o) (X//X^(-))=0.33V. From this data, we can deduce that:
Answer» `M+ XrarrM^(+)+ X^(-)`is a SPONTANEOUS reaction `M^(+)+ X^(-) RARR M+ X`is a spontaneous reaction `E_(cell) = 0.77` V None of these Answer :B

Discussion

37.	For the electrochemical cellM// M^(+) // // X^(-) // X , E^(0) M^(+)// M = 0.44 V and E^(0) (x//x^(-)) = 0.33 V. From the one can deduce that
Answer» `M+X rarr M^(+) +X^(-)` is the SPONTANEOUS REACTION `M^(+)+X^(-) rarr M +X` is the spontaneous reaction `E_(CELL)=0.11V` `E_(cell)= - 0.77V`

Discussion

38.	For the dissolution of an ionic solid in water
Answer» HYDRATION ENERGY should be more than LATTICE energy lattice energy should be more than hydration energy lattice energy should be EQUAL to hydration energy none of these Solution :A LEVEL information

Discussion

39.	for the dissociation reaction , H_(2)(g)Leftrightarrow2H(g),DeltaH= 162 kcal heat of atomisation of H is
Answer» 81 KCAL 162 kcal 208 kcal 218 kcal Solution :`DeltaH=DeltaH_("PRODUCT")DeltaH_(("REACTANT"))` ` 162 =2 xx DeltaH_(H)-DeltaH_(H_(2))` `DeltaH_(H)=162/2=81 kcal` `(DeltaH_(H_(H_2))=0)`

Discussion

40.	For the dissociation equilibrium, N_(2)O_(4(g))iff2NO_(2(g)), the variation of free energy with the fraction of N_(2)O_(4) dissociated under standard conditions is shown in the figure : Which of the following statements is/are correct?
Answer» The free ENERGY CHANGE for the forward reaction is NEGATIVE The free energy change for the backward reaction is negative The NET free energy change for the complete reaction is positive Forward reaction is more SPONTANEOUS than backward reaction Solution :`DeltaG^(@)` for conversion of 1 mole of `N_(2)O_(4)` into equilibrium mixture (forward reaction) `=-0.84kJ,i.e.,-ve` `DeltaG^(@)` for conversion of 1 mole of `N_(2)O_(4)` completely into 2 moles of `NO_(2)=+5.40kJ`. `DeltaG^(@)` for complete conversion is positive therefore, complete conversion is not possible. As `DeltaG^(@)` for backward reaction is more negative than for forward reaction, i.e., formation of `N_(2)O_(4)` is more spontaneous.

Discussion

41.	For the discharge of equal masses of the following ions, the number of electrons required is maximum in the case of
Answer» `H^(+)` `Cu^(2+)` `AG^(+)` `AL^(3+)`

Discussion

42.	For the detection of sulphur by Lassaigne's test, the addition of sodium nitroprusside to the sodium extract gives purple colouration. This is due to the formation of :
Answer» `Na_(3)[FE(CN)_(6)]` `Na_(4)[Fe(CN)_(5)NOS]` `Fe(SCN)_(3)` `Na_(4)[Fe(CN)_(6)]` ANSWER :B

Discussion

43.	For the decomposition reaction NH_2COONH_4 (s) hArr 2NH_3(g) +CO_2(g) The K_p=2.9 xx 10^(-5) atm^3. The total pressure of gases at equilibrium when 1 mole of NH_2COONH_4 (s) was taken to start with would be
Answer» 0.0194 atm 0.0388 atm 0.0582 atm 0.0766 atm Solution :`{:(NH_2COOHN_4(s), HARR 2NH_3(G), +CO_2(g)),(1,2,1):}` `K_p=2.9xx10^(-5)atm^3` If P is the total pressure at equilibrium `K_p=((2p)/3)^2(p/3)` `THEREFORE P_3=(27xx2.9xx10^(10^(-5)))/4=1.9575` `P=3sqrt(1.9575)=0.0582`

Discussion

44.	For the decomposition of N_(2)O_(5)(g) it is given that 2NOI_(2)O_(5(g))to4NO_(2(g))+O_(2(g)).Activation energy E_(a), N_(2)O_(5(g))to2NO_(2(g))+1/2O_(2(g)) Activating energy E_(a)
Answer» `E_(a)=E_(a)^(')` `E_(a)gtE_(a)^(')` `E_(a)ltE_(a)^(')` `E_(a)=2E_(a)^(')` Solution :`Ea_(1)=Ea_(2)`

Discussion

45.	For the decomposition of H_2O_2(aq) it was found that V_(O_2) (t=15 min) was 100 mL (at 0^@C and 1 atm) while V_(O_2) (maximum) was 200 mL (at 0^@C and 2 atm). If the same reaction had been followed by the titration method of if V_(KMnO_(4))^((cM))(t=0) had been 40 mL, what would V_(KMnO_(4))^((cM))(t=15 min) have been ?
Answer» 30 ML 25 mL 20 mL 15 mL Solution :`1/4th` reaction has completed UPTO 15 min.Hence `V_(KMnO_4)` will be `3/4xx40=30` mL

Discussion

46.	For the decomposition of N_2O_5(g), it is given that :2N_2O_5(g)rarr4NO_2(g)+O_2(g),Activation energy E_aN_2O_5(g)rarr2NO_2(g)+(1/2)O_2(g),Activation energyE_athen:
Answer» `E_a=E_a` `E_agtE_a` `E_altE_a` `E_a=2E_a` ANSWER :A

Discussion

47.	For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data is obtained. Calculate the rate constant.
Answer» Solution :The decomposition reaction is of gaseous nature and expression of the rate equation for the reaction is: `k =2.303/t log (p_(i))/(2p_(i)-p_(t))` Rate CONSTANT after 360 s i.e. `k_(360) = 2.303/(360 s) log(35 atm)/(70-54)atm = 2.303/(360S) log 35/16 = 2.303/(360 s) (log 2.1875)` `=(2.303 xx 0.33995)/(360 s) = 2.17 xx 10^(-3) s^(-1)` Rate constant after 720 s i.e. `k_(720) = (2.303)/(720 s) log (35 atm)/(70-63) atm` `=(2.303)/(720 s) log 5 = (2.303 xx 6900)/(720s) log 5 = 2.24 xx 10^(-3)s^(-1)`

Discussion

48.	For the decomposition of azoisopropane to hexane and nitrogen at 543 K, following data are obtained : {:("t (sec)",,,,"P (mm of Hg)"),(0,,,,""35.0),(360,,,,""54.0),(720,,,,""63.0):} Calculate the rate constant.
Answer» Solution :`("CH"_(3))_(2)"CH N"="N CH"("CH"_(3))_(2)(g)to"N"_(2)(g)+"C"_(6)"H"_(16)(g)` `{:("Initial pressure",,," ""P"_(0),,,,0,,,0),("After time t",,,"P"_(0)-"p",,,,"p",,,"p"):}` Total pressure after time `t(P_(t))=(P_(0)-p)+p+p=P_(0)+p" or "p=P_(t)-P_(0)` `apropP_(0)" and "(a-x)propP_(0)-p` or substituting the value of p, `a-xprop P_(0)-(P_(t)-P_(0)),i.e.,(a-x)prop P_(0)-P_(t)` As DECOMPOSITION of AZOISOPROPANE is a first order REACTION, `K=(2.303)/(t)log""(a)/(a-x)=(2.303)/(t)log""(P_(0))/(2P_(0)-P_(t))` When `t=360" sec,"k=(2.303)/(360" s")log""(350)/(2xx35.0xx54.0)=(2.303)/(360" s")log""(35)/(16)=(2.303)/(360" s")(0.3400)=2.175xx10^(-3)s^(-1)` When `t=720" sec,"k=(2.303)/(720" s")log""(35.0)/(2xx35.0-63.0)=(2.303)/(720" s")log5=(2.303)/(720)(0.6990)=2.235xx10^(-3)s^(-1)` `:." Average value of "k=(2.175+2.235)/(2)xx10^(-3)s^(-3)=2.20xx10^(-3)s^(-1).`

Discussion

49.	For the decomposition of a compound AB at 600 K, the following data were obtained. The order of the decomposition of AB is
Answer» 0 1 2 1.5 Solution :Let rate equation be rate `=k[AB]^(n)` CASE (i) `2.75xx10^(-8)=k[0.2]^(n)` Case (II) `11xx10^(8)=k[0.40]^(n)` Case (iii) `24.75xx10^(-8)=k[0.6]^(n)` DIVIDE ii by I `=(11xx10^(-8))/(2.75xx10^(-8))=([0.40]^(n))/([0.2]^(n))` `implies2^(n)=4impliesn=2:.` ORDER of reaction is 2.

Discussion

50.	For the decomposition of a compound AB at 600 K the following data were obtained : The order for the decomposition of AB is
Answer» 0 1 2 1.5 Solution :(C ) LET Rate = k `[AB]^(N)` `2.75xx10^(-8) = k (0.2)^(n)` `11xx10^(-8) = k(0.40)^(n)` `24.75xx10^(-8) = k (0.6)^(n)` Dividing (II ) by (i) `(11xx10^(-8) )/(2.75xx10^(-8)) = ((0.4)^(n))/((0.2)^(n))` 4= `2^(n)` n=2

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For the equilibrium system.N_(2)(g)+O_(2)(g)+Heat hArr 2NO(g)Which of the following factors would cause the value of equilibrium constant to decrease ?

For the equilibrium system :2HCl(g)hArr H_(2)(g)+Cl_(2)(g)the equilibrium constant is 1.0xx10^(-5).What is the concentration of HCl if the equilibrium concentration of H_(2) and Cl_(2) are 1.2xx10^(-3) M and 1.2xx10^(-4) M respectively ?

For the equilibrium N_(2)+3H_(2)hArr2NH_(3),K_(c) at 1000K is 2.73xx10^(-3) if at equlibrium [N_(2)]=2M,[H_(2)]=3M, the concentraion of NH_(3) is

For the equilibrium :MgCO_(3)(s)hArr MgO(s)+CO_(2)(g)which of the following expressions is correct ?

For the equilibrium MgCO_3(g)oversetDeltahArrMgO(s)CO_2(s)which of the following expressions is correct ?

For the equilibrium: LiCl.3NH_(3(s))hArrLiCl.NH_(3(s))+2NH_(3), K_(p)=9 atm^(2) at 40^(@)C. A 5 litre vessel contains 0.1 mole of LiCl.NH_(3). How many mole of NH_(3) should be added to the flask at this temperture to derive the backward reaction for completion?

For the equilibrium in a closed vessel PCI_5 (g) hArr PCI_3 (g) +CI_2(g) K_p is found to be double of K_c this is attained when

For the equilibrium : HCO_(3)^(-)hArr H^(+)+CO_(3)^(2-)K=4.8xx10^(-11), [CO_(3)^(2-)]=1.1xx10^(-3)M, [HCO_(3)^(-)]=9.8xx10^(-3)MThe pH of the solution is :

For the equilibriumH_(2)O(l)subH_(2)O(g) at 1 atm and 298 K

For the equilibrium, H_2O(l) iff H_2O_((v)) , which of the following is correct?

For the equilibrium: CaCO_(3(s)) iff CaO_((s)) + CO_(2(g)) , K_p = 1.64 atm at 1000 K, 50 g of CaCO_3 in a 10litre closed vessel is heated to 1000 K. Percentage of ĆaCO_3 that remains unreacted at equilibrium is (Given R=0.082 L atm K^(-1) "mol"^(-1)).

For the equilibrium at 298 K: N_2O_(4(g)) hArr 2NO_(2(g)), G_(N_2O_4)^@ = 100 "kJ mol"^(-1) and G_(NO_2)^@ = 50 "kJ mol"^(-1) . If 5 moles of N_2O_4 and 2 moles of NO_2 are taken initially in one litre container then which statements are correct?

For the equilibrium, CaCO_3(s) hArr CaO(s) + CO_2(g) which of the following expressionis correct :

For the equilibrium 2SO_(3(g))hArr2SO_(2(g))+O_(2(g)), the value of equilibrium constant is 4.8xx10^(-3) at 700^(@)C. At equilibrium, if the concentration of SO_(3) and SO_(2) are 0.60M and 0.15M respectively. Calculate the concentration of O_(2) in the equillibrium mixture.

For the equilibrium 2SO_(3(g))hArrSO_(2(g))+O_(2(g)), the value of equilibrium constant is 4.8xx10^(-3) at 700^(@)C. At equilibrium, if the concentration of SO_(3) and SO_(2) are 0.60M and 0.15M respectively. Calculate the concentration of O_(2) in the equillibrium mixture.

For the equilibrium: 2NO(g) + O_2(g) + 2NO_2(g), K_c=6.45 xx 10^5 . (i) At what O_2concentration is the NO_2concentration equal to the NOconcentration?(ii) At what O_2concentration is the NO_2concentration 100 times the NO concentration?

For the equilibrium 2NOCl_((g))hArr2NO_((g))+Cl_(2(g)) the value of the equilibrium constant K_(c) is 3.75xx10^(-6) at 790^(@)C. Calculate K_(p) for this equilibrium at the same temperature.

For the equilibrium2NO_(2(g))hArrN_(2)O_(4(g))+14.6 kcal the increase in temperaturee would

For the equilibrium, 2H_(2)(g)+O_(2)(g)hArr2H_(2)O(l) at 25^(@)C,DeltaG^(@) is -474.78kJ mol^(-1). Calculate log K for it (R=8.314 JK^(-1)mol^(-1))

For the equilibrium 2NO_2(g) ⇌N_2O_4(g) +14.6 kcalAn increase of temperature will:

For the endothermic reactionA_(2) rarr 2A, which of the following will increase yield of monomer?

For the elementary reaction M to N , the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M . The order of the reaction with respect to M is

For the emperical formula Pt(NH_(3))_(2)Cl_(2), the no. of possible coordination isomers would be (ON of pt is +2 in all all isomeric forms)

For the elementary step (CH_3)_3 CBr_((aq)) to (CH_3)_3C_((aq))^++Br_((aq))^(-) the molecularity is

For the elementary step (CH_3)_3CBr(aq)rarr(CH)_3)_3C^+(aq)+Br^(-)(aq) the molecularity is :

For the elementary step(CH_3)_3CBr(aq)rarr(CH)_3)_3C^+(aq)+Br^(-)(aq) the molecularity is :

For the elementary reaction M to N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of reaction with respect to M is

for the electrorode reaction, M^(n+)(aq)+n e^(-)rarrM(s) Nernst equation is:E=E^o +frac(RT)(nF)ln frac(1)([M^(n+)],E=E^o +RT ln[M^(n+)],E=E^o +frac(RT)(nF)l n[M^(n+)],E/E^o=frac(RT)(nF)l n[M^(n+)]..

For the electrolytic production of NaClO_(4) from NaClO_(3) as per the following equation: NaClO_(3) + H_(2)O rightarrow NaClO_(4) + H_(2) How many faradays of electricity will be required to produce 0.5 "mole" of NaClO_(4) assuming 60% efficiency?

For the electrolytic production of NaClO_4from NaClO_3according to the equation NaClO_3 +H_2O to NaClO_4 +H_2,the number of Faradays of electricity required to produce 0.5 mole of NaClO_4is

For the electrode process, H^(+)+e=1/2H_(2),E_(H_(H)^(+),H_(2)) = x volt then for 2H^(+)+2e=H_(2),E_(2H^(+),H_(2)) is equal to

For the electrochemicalcell,M[M^+]X^-|X,E^@_(M+//M)=0.44 V and E_(X//X^-)^@ =0.33 V. From this data, one can duduce that :

For the electrochemical cell, M|M^(+)||X^(-)|X,E^(@)(M^(+)//M)=0.44V and E^(@)(X//X^(-))=0.33V. From this data one can deduce that

For the electrochemical cell, Mabs(M^(+))abs(X^(-)) X, E^(o) (M^(+)//M) = 0.44 V and E^(o) (X//X^(-))=0.33V. From this data, we can deduce that:

For the electrochemical cellM// M^(+) // // X^(-) // X , E^(0) M^(+)// M = 0.44 V and E^(0) (x//x^(-)) = 0.33 V. From the one can deduce that

For the dissolution of an ionic solid in water

for the dissociation reaction , H_(2)(g)Leftrightarrow2H(g),DeltaH= 162 kcal heat of atomisation of H is

For the dissociation equilibrium, N_(2)O_(4(g))iff2NO_(2(g)), the variation of free energy with the fraction of N_(2)O_(4) dissociated under standard conditions is shown in the figure : Which of the following statements is/are correct?

For the discharge of equal masses of the following ions, the number of electrons required is maximum in the case of

For the detection of sulphur by Lassaigne's test, the addition of sodium nitroprusside to the sodium extract gives purple colouration. This is due to the formation of :

For the decomposition reaction NH_2COONH_4 (s) hArr 2NH_3(g) +CO_2(g) The K_p=2.9 xx 10^(-5) atm^3. The total pressure of gases at equilibrium when 1 mole of NH_2COONH_4 (s) was taken to start with would be

For the decomposition of N_(2)O_(5)(g) it is given that 2NOI_(2)O_(5(g))to4NO_(2(g))+O_(2(g)).Activation energy E_(a), N_(2)O_(5(g))to2NO_(2(g))+1/2O_(2(g)) Activating energy E_(a)

For the decomposition of N_2O_5(g), it is given that :2N_2O_5(g)rarr4NO_2(g)+O_2(g),Activation energy E_aN_2O_5(g)rarr2NO_2(g)+(1/2)O_2(g),Activation energyE_athen:

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data is obtained. Calculate the rate constant.

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, following data are obtained : {:("t (sec)",,,,"P (mm of Hg)"),(0,,,,""35.0),(360,,,,""54.0),(720,,,,""63.0):} Calculate the rate constant.

For the decomposition of a compound AB at 600 K, the following data were obtained. The order of the decomposition of AB is

For the decomposition of a compound AB at 600 K the following data were obtained : The order for the decomposition of AB is