94420 + Interview Questions in Chemistry in Class 12

1.	For the reaction, A+Brarr C+D. The variation of the concentration of the products with time is given by the curve:
Answer» `X` `Y` `Z` `W` SOLUTION :The curve `Y` SHOWS a gradual increase in the concentration of products with TIME.

Discussion

2.	For the reaction A + B to products, what will be the order of reaction with respect to A and B? {:("Exp",[A](mol L^(-1)),[B](mol L^(-1)),"Initial rate (mol L^(-1) s^(-1))),(1,2.5 xx 10^(-4), 3 xx 10^(-5), 5 xx 10^(-4)),(2,5 xx 10^(-4),6 xx 10^(-5), 4 xx 10^(-3)),(3, 1xx 10^(-3), 6 xx 10^(-5), 1.6 xx 10^(-2)):}
Answer» 1 with respect to A and 2 with respect to B 2 with respect to A and 1 with respect to B 1 with respect to A and 1 with respect to B 2 with respect to A and 2 with respect to B Solution :Rate `= k [A]^(x) [B]^(y)` From EXP.(1), `5 xx 10^(-4) = k (2.5 xx 10^(-4))^(x (3 xx 10^(-5))^(y)` Frox exp. (2), `4 xx 10^(-3) = k (5 xx 10^(-4))^(x) (6 xx 10^(-5))^(y)`….(ii) DIVIDING (ii) by (i), `(4 xx 10^(-3))/(5 xx 10^(-4))` From exp. (3) `1.6 xx 10^(-2) = k (1 xx 10^(-3))^(x) (6 xx 10^(-5)^(-6))^(y)`...(iii) Dividing (iii) by (ii), `(1.6 xx 10^(-2))/(4 xx 10^(-3)) = 2^(x) = 4` or x = 2,y = 1 Hence order with respect to A is 2 and with respect to B is 1.

Discussion

3.	For the reaction, A+BhArr C+D, the initial concentration of A and B are equal, but the equilibrium concentration of C istwice that of equlibrium concentration of A. The equlibrium constant is
Answer» Solution :`{:(A,+,B,hArr,C,+,D),(a,,a,,0,,0),((a-x),,(a-x),,x,,x):}` GIVEN `x=2(a-x)or x=(2a)/(3)` `K_(c)=(x^(2))/((a-x)^(2))=((2a//3)^(2))/((a-2a//3)^(2))=4`

Discussion

4.	For the reaction A+BhArr3C at 25^(@)C, a 3 litre volume reaction vessel contains 1,2 and 4 moles of A,B and C respectively at equilibrium, calculate the equilibrium constant K_(c) of the reaction at 25^(@)C.
Answer» ANSWER :`10.66" MOL DM"^(-3)`

Discussion

5.	For the reaction A + B to products , it is observed that : (i) On doubling the initial concentration of A only , the rate of reaction is also doubled and (ii) On doubling the intial concentration of both A and B , there is a change by a factor of 8 in the rate of the reaction The rate of this reaction is given by
Answer» Rate = `k [A]^(2) [B]` Rate = ` k [A] [B]^(2)` Rate = k `[A]^(2) [B]^(2)` Rate = `k [A] [B]` Solution :When concentration A is doubled , rate is doubled . Hence order with respect to A is one. When concentration of both A and B are doubled , rate INCREASED by 8 TIMES hence TOTAL order is 3 . `THEREFORE`Rate = k `[A]^(1) [B]^(2)` Order = 1 + 2 = 3 .

Discussion

6.	For the reaction , A+B toProducts, it is found that the order of A is 2 and of B is 3 in the rate expression. When concentration of both is doubled the rate will increased by
Answer» `10` `6` `32` `16` Answer :C

Discussion

7.	For the reaction A+B to Products, the rate law is : Rate =k[A][B]^(3//2). Can the reaction be elementary reaction? Explain.
Answer» SOLUTION :Had this been an ELEMENTARY reaction, the order with respect to B would have been 1. But it is given as `(3)/(2)` in the PROBLEM. HENCE it is not an elementary reaction.

Discussion

8.	For the reaction, A+B to C, the following dara were obtained. In the first experiment, when the initial concentrations of both A and B are 0.1M the observed initial rate of formation of C is 1xx10^(-4)mol "lt"^(-1)minute^(-1). In second experiment when the initial concentrations of (A) and (B) are 0.1M and 0.3M, the initial rate is 9.0xx10^(-4) mol litre^(-1)minute^(-1). In the third experiment, when the initial concentrations of both A and B are 0.3M, the initial rate is 2.7xx10^(-3)mol litre^(-1) minute^(-1). (a) Write rate law for this reaction. (ii) Calculate the value of specific rate constant for this reaction.
Answer» Solution :Let Rate `-k[A]^(x)[B]^(y)` `r_(1)=1xx10^(-4)=k(0.1)^(x)(0.1)^(y)` …………`(1)` `r_(2)=9xx10^(-4)=k(0.1)^(x)(0.3)^(y)`………..`(2)` `r_(3)=2.7xx10^(-3)=k(0.3)^(x)(0.3)^(y)`…………`(3)` By Eqs. `(1)` and `(2)`, `(r_(1))/(r_(2))=(1xx10^(-4))/(9xx10^(-4))=((1)/(3))^(y) :. y=1` By Eqs. `(2)` and `(3)` `(r_(2))/(r_(3))=(9xx10^(-4))/(27xx10^(-4))=((1)/(3))^(x) :. x=1` `:.` Rate `=k[A]^(1)[B]^(2)` ALSO, `1xx10^(-4)=k(0.1)^(1)(0.1)^(2)` `:. k=10^(-1)=0.1Lt^(2)MOL^(-1)min^(-1)`

Discussion

9.	For the reaction A  B, the rate law expression is-(d[A])/(dt) =k [A]^(1//2). If initial concentration of [A] is [A]_(0), then
Answer» The integerated rate expression is `K=(2)/(t) (A_(0)^(1//2)-A^(1//2))` The graph of `SQRT(A)` Vs t will be The half life period `t_(1//2)=(K)/(2[A]_(0)^(1//2))` The TIME taken for 75% completion of reaction `t_(3//4) =(sqrt[A]_(0))/(k)` Answer :A::B::D

Discussion

10.	For the reaction A + B to 2C + D which of the following statements is/are correct?
Answer» Rate of disappearance of B = `1//2` X rate of APPEARANCE of C Rate of disappearance of B =`1//2` x rate of appearance of C Rate of disappearance of A = rate of appearance of B Rate of disapperance of A = rate of disapperance of B. Solution :(a,c,d) are correct options.

Discussion

11.	For the reaction A+B to C+D, doubling the concentration of both the reactants increases the reaction rate by 8 times and doubling the concentration of only B simply doubles the reaction rate. The rate law is given as
Answer» `r=k[A]^(1//2)[B]^(1//2)` `r=k[A][B]^(2)` `r=k[A]^(2)[B]` `r=k[A][B]` Solution :`(i) r = k a^(alpha) b^(BETA)` (ii) `8 r = k ( 2a)^(alpha) ( 2B)^(beta)` `(iii) 2R = ka^(alpha)(2b)^(beta)` Eqn. (ii) Eqn. (iii) GIVES `(8)/(2) = 2^(alpha)` or `2^(alpha) = 4 = 2^(2) or alpha = 2.` , Eqn. (iii) Eqn. (i) gives `2 = 2^(beta) or beta = 1`, Hence rate ` = k[A]^(2) [B].`

Discussion

12.	For the reaction A+ B rightarrowproduct , the folowing initial rates were obtained at various given initial concentrations: Determine the half - life period.
Answer» SOLUTION : Rate = `k[A]^(x)[0.1]^y` `0.10=k[0.2]^(x)[0.1]^y` `0.05= K[0.1]6(x)[0.2]^(y)` Dividing equation (II) by (i) WEE get `(0.10)/(0.05)=2^(x)` or `2=2^(2x)` or x=1 Dividing equation (ii) by (i) , we get `(0.10)/(90.05) = 2^(x)` ir `2=2^(x)` or x=1 Dividing equation (iii) by (i) , we get `(0.05)/(0.05)= 2^(y)` or `1=2^(y)` or y=0 `THEREFORE` Rate =` K [A]^(1)[B]^(0)` , It is a forst order reaction . HALF - life period Substituting the values in equation (i) we get " 0.05=k[0.1]or `k=(0.05)/(0.1)=0.5s^(-1)` `t_(1//2)= (0.693)/(k) = (0.693)/(0.5)` or `t_(1//2)=1.386s`

Discussion

13.	For the reaction : A+B rarr Cthe initial concentration of A and B are 2 and 1 moles per littre. At equilibrium, the concentration of B has been found to be 0.5 mol/litre. The K for the reaction is :
Answer» `0.5` `2.0` `1.0` `1.5` ANSWER :C

Discussion

14.	For the reaction A + B rarr C the ratio constant for the second-order forward reaciton is k_(2) = 10.00 exp. ( - ( 90500)/( RT)) dm^(3) mol^(-1) s^(-1). The pre-exponential factor and activation energy are, respectivley
Answer» `10^(10) dm^(3) mol^(-1) s^(-1)` and `- 90.50 kJ mol^(-1)` `LOG 10 dm^(3) mol^(-1) s^(-1)` and `- 45.25 kJ mol^(-1)` `10 dm^(3) mol^(-1) s^(-1) ` and `90.50 kJ mol^(-1)` `10 dm^(3) mol^(-1) s^(-1) `and `- 90.25 kJ mol^(-1)` ANSWER :C

Discussion

15.	For the reaction A + B rarrProduct, it is found that the order of A is 2 and of B is 3 in the rate expression. When concentration of both is doubled, the rate will increase by:
Answer» 10 6 32 16 Answer :C

Discussion

16.	For the reaction A+B+Q hArrC+D, if the temperatue is increased, then concentration of the products will
Answer» Increase Decrease Remain same BECOME Zero Solution :`A+UNDERSET(2)(B)+QhArrC underset(2)(+)D` The reaction is ENDOTHERMIC so on increase temperature CONCENTRATION of product will increase.

Discussion

17.	For the reaction A+BhArrC+D, the forward reaction is exothermic. The activation energy of formation of A+B is ……….that for the formation of C+D
Answer» EQUAL to less than greater than double Answer :C

Discussion

18.	For the reaction A+B products, it is observed that (a) On doubling the initial concentration of A only, the rate of reaction is also doubled (b) On doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is given by
Answer» `"rate"=K[A][B]` `"rate"=k[A]^(2)[B]` `"rate"=k[A][B]^(2)` `"rate"=k[A]^(2)[B]^(2)` ANSWER :C

Discussion

19.	For the reaction A+B+CtoD. The following observation were made (i) When the concentrations of A was doubled, the rate of formation of D was doubled (ii) When the concentration of B was halved, the rate of formation of d becomes one fourth (iii) Doublingthe concentration of C and no effect on rate Select the correct statement (s).
Answer» RATE EQN, `r=k[A][B]^(1//2)` REACTANT C must INVOLVE after rate determining step Only A and B participate in the rate determining step Order of C is 1. Solution :`r=K.[A]^(1)[B]^(2)[C]^(0)`

Discussion

20.	For the reaction, A+2BrarrC, the relative rate of reaction can be represented by
Answer» `(d[A])/DT=+1/2(d[B])/dt=-(d[C])/dt` `-(d[A])/dt=-2(d[B])/dt=+(d[C])/dt` `-(d[A])/dt=-1/2(d[B])/dt=+(d[C])/dt` `-(d[A])/dt=+1/2(d[B])/dt=-(d[C])/dt` ANSWER :C

Discussion

21.	For the reaction A +2B to C, the reaction rate is doubled if the concentration of A is doubled. The rate becomes four times when the concentration of both A and B are made four times. The order of reaction is:
Answer» 3 0 1 2 Solution :`A + 2B to C` SINCE the reaction rate becomes double when the CONCENTRATION of A is made twice order w.r.t A=1 Since the reaction rate becomes four TIMES when concentration of both A and B are made four times. The reaction rate is independent of the concentration of B. Order w.r.t. B=0 Overall order of reaction =1

Discussion

22.	For the reaction A + 2Btoproduct, the reaction rate was halved as the concentration of A was doubled. What is the order of reaction with respect to A?
Answer» ANSWER :`-1`

Discussion

23.	For the reaction A+2BhArrC, the expression for equlibrium constant is
Answer» `([A][B]^(2))/([C])` `([A][B])/([C])` `([C])/([C][B]^(2))` `([C])/(2[B][A])` ANSWER :C

Discussion

24.	For the reaction A + 2Bto C, 5 molesof A andB moles of B willproduce
Answer» 5MOLES of C 4 MOLES of C 8 moles of C 13 molesof C Answer :B

Discussion

25.	For the reaction A + 2B to 2C + D, the concentration of A is kept constant and that of B is tripled, the rate of reaction will become
Answer» three times SIX times eight times nine times Solution :INITIAL `"RATE"_1= k [A]^x[B]^(2Y)` New `"rate"_2=k[A]^x[3B]^(2y)` `:. ("rate"_2)/("rate"_1)= 3^2=9:."rate"_2 = 9 XX "rate"_1`

Discussion

26.	For the reaction A + 2B rarr C, the reaction rate is doubled if the concentration of A is doubled. The rate is increased by four times when concentrations of both A and B are increased by four times. The order of the reaction is
Answer» 3 0 1 2 Solution :`A+2BrarrC` REACTION rate is doubled when the concentration of A is doubled. Again when both A and B are made FOUR times, reaction rate also becomes four times. It MEANS reaction rate is independent of the concentration of B. With respect to A, the order is definitely ONE. `therefore` w.r.t.A,order=1 w.r.t.B,order=0 Overall order = 1

Discussion

27.	For the reaction, A+2B +C to D +2E, the rate of formation of D is found to be i)doubled when [A] is doubled keeping [B] and [C] constant ii)doubled when [C] is doubled keeping [A] and [B] constant iii) the same when [B] is doubled keeping [A] adn [C] constant. Which one is the rate equation for the reaction
Answer» rate `=k[A] [B] [C]` rate `=k [A]^0[B] [C]` rate `=k[A] [B]^0[C]` rate `=[A][B][C]^0` Solution :The given reaction is Ist order in A because on doubling the concentration of A because on doubling the concentration of A, rate of formation of 'D' is DOUBLED. Similary, if is FIRST order in C. But, the reaction is zero order in B because the doubling of its concentration does not affect the reaction rate. So, the rate eqaution for the reaction BECOMES, `(dx)/(DT)=k[A][B]^0[C]`

Discussion

28.	For the reaction A +2B rarr C , 5 mole of A and 8 mole of B will produce :
Answer» 5mole of C 4 MOLE of C 8 moleof C 13 mole of C Answer :B

Discussion

29.	For the reaction A + 2B hArr2C + D, initial concentration of A is a and that of B is 1.5 times that of A. Concentration of A and D are same at equillibrium. What should be the concentration of B at equillibrium :-
Answer» `a/4` `a/2` `(3A)/(4)` All of the above Answer :B

Discussion

30.	For the reaction A+2" B"to" C", the reaction rate is doubled if the concentration of A is doubled. The rate is increased by four times when condcentrations of both A and B are increased by four times. The order of reaction is
Answer» 3 0 1 2 Solution :Rate, `r = K [A]^(a)[B]^(b)""…(i)` `2r = [2 A]^(a) [B]^(b)""…(ii)` `4R = [ 4 A ]^(a) [ 4 B]^(b) ""…(iii)` Divding eqn. (ii) by eqn. (i), `2 = 2^(a) therefore a = 1` Divding eqn. (ii) by eqn. (i) `4 = 4^(a) 4^(b) = 4 xx 4^(b) = 4^(b+1)` `therefore b + 1 = 1 or b = 0` Hence, overall ORDER= 1.

Discussion

31.	For the reaction 4Al(s)+2O_2(g)+6H_2O+4OH^-
Answer» ANSWER :A::C

Discussion

32.	For the reaction:
Answer» `CH_3-CH=CH-CH_3 ` predominates `CH_2=CH-CH_2-CH_3` predominates Both are FORMED in equal amounts The product RATIO is DEPENDENT on the HALOGEN `X` Answer :A

Discussion

33.	For the reaction 4NH_(3)+5O_(2)to4NO+6H_(2)O the rate of reaction with respection with respect to NH_(3) is 2xx10^(-3)Ms^(-1). Then the rate of the reaction with respect to oxygen is _______Ms^(-1)
Answer» `2XX10^(-3)` `1.5xx10^(-3)` `2.5xx10^(-3)` `3XX10^(-3)` ANSWER :A

Discussion

34.	For the reaction, 4A + Brarr2C+2D,The statement not correct is:
Answer» The rate of disappearance of B is one FOURTH the rate of disappearance of A The rate of appearance of C is HALF the rate of disappearance of B The rate of formation of D is half the rate of consumption of A The rates of formation of C and D are equal Answer :B

Discussion

35.	For the reaction 3BrO^(-)rarr BrO_(3)^(-) +2Br^(-) in an alkaline aquesous solution, the value of the second order (in BrO^(–)) rate constant at 80^(@)C in the rate law for -(Delta [BrO^(-)])/(Delta t) was found to be 0.056 L "mol"^(-1)s^(-1). What is the rate of constant when the rate law is written for (a) (Delta [BrO_(3)^(-)])/(Delta t)," "(b) (Delta [Br^(-)])/(Delta t) ?
Answer» SOLUTION :(a) `0.019" MOL L"^(-1)s^(-1)`, (B) `0.037" mol L"^(-1) s^(-1)`

Discussion

36.	For the reaction, 3O_2(g) hArr 2O_3(g)KC at 25^@C for this equilibrium is 2 x× 10^(–50). If equilibrium concentration of O_2 at 25^@C is 1.6 x× 10^(–2), what is the equilibrium concentration of O_3 ?
Answer» `8.19xx10^(-56)` `2.86 xx 10^(-28)` `2.86 xx 10^(-14)` `8.19xx10^(-44)` Answer :B

Discussion

37.	For the reaction, 3O_2rarr2O_3, triangleH=+ve. We can say that:
Answer» Ozone is more stable than oxygen Ozone is less stable than oxygen and ozone DECOMPOSES FORMING oxygen readily Oxygen is less stable than ozone and oxygen readily FORMS ozone. None Answer :B

Discussion

38.	For the reaction 3Br_(2)+6OH^(-)-5Br^(-)+BrO_(3)+3H_(2)O Equivalent weight of Br_(2) (mol. Wt. M) is
Answer» `M/2`<BR>`M/10` `(M/2+M/10)` `M/6` Solution :As `Br_(2)` disproportionates (simultaneous oxidation and REDUCTION) its EQUIVALENT weight is the sum of equivalent weights of the two HALF reactions.

Discussion

39.	For the reaction 2x+yrarrL find the rate law from the following data. {:([x](min),[y](min),Rate(Ms^(-1))),(0.2,0.02,0.15),(0.4,0.02,0.30),(0.4,0.08,1.20):}
Answer» SOLUTION :Rate `=k[X]^m[y]^m` `0.15=k[0.2]^N[0.02]^m""...(1)` `0.30=k[0.4]^n[0.02]^m""...(2)` `1.20=k[0.4]^n[0.08]^m""...(3)` `{:("Dividing Eq(3) by Eq(2) we get","Dividing Eq(2) by Eq(1) we get"),(1.2/0.3=(k[0.4]^n[0.08]^m)/(k[0.4]^n[0.02]^m),0.30/0.15=(k[0.4]^n[0.02]^m)/(k[0.2]^n[0.02]^m)),(4=(([0.08])/([0.02]))^m,2=(([0.4])/([0.2]))^n),(thereforem=1,thereforen=1):}` Rate `=k[x]^1[y]^1` `0.15=k[0.2]^1[0.02]^1` `(0.15)/([0.2]^1[0.02]^1)=k` `k=37.5"MOL ^(-1) Ls^(-1)`

Discussion

40.	For the reaction 3Ato 2B .What will be the reaction rate with reference to B?
Answer» `-(3)/(2)(d[A])/(DT)` `-(2)/(3)(d[A])/(dt)` `-(1)/(3)(d[A])/(dt)` `(2d[A])/(dt)` Solution :3A`to` 2B reaction rate=`(-1)/(3)(d[A])/(dt)=(1)/(2)(d[B])/(dt)` `THEREFORE (d[B])/(dt)=-(2)/(3)(d[A])/(dt)`

Discussion

41.	For thereaction2X+ Y to 2Zthe rateof disappearance of Xsi 0.08 M//s(a) Whatis therateof formationof Z (b) What isthe rateof the reaction
Answer» SOLUTION :(a)0.08 `M//s(B)0.04 M//s`

Discussion

42.	For the reaction 2x + y rarr L. Find the rate law from the following data. {:([X],[Y],"rate"),("(min)","(min)",(M s^(-1))),(0.2,0.02,0.15),(0.4,0.02,0.30),(0.4,0.08,1.20):}
Answer» Solution :Rate `k[x]^(n)[y]^(m)` `0.15=k[0.2]^(n)[0.02]^(m)""`…(1) `0.30=k[0.4]^(n)[0.2]^(m) ""` ….(2) `1.20=k[0.4]^(n)[0.08]^(m) ""`….(3) By dividing EQUATION `((3))/((2))` `(1.2)/(0.3)=(k[0.4]^(n)[0.08]^(m))/(k[0.4]^(n)[0.02]^(m))` `4=(([0.08])/([0.02]))^(m)` ` 4 = (4)^(m)` `therefore m = 1` By DIVING equation `((2))/((1))` `rArr (0.30)/(0.15)=(k[0.4]^(n)[0.02]^(m))/(k[0.2]^(n)[0.02]^(m))` `rArr 2=(([0.4])/([0.2]))^(n)` `2=(2)^(n)` `therefore n = 1` Rate `= k [x]^(1)[y]^(1)` `0.15 = k[0.2]^(1)[0.02]^(1)` `(0.15)/([0.2]^(1)[0.02]^(1))=k` `k = 37.5 mol^(-1)L s^(-1)`

Discussion

43.	For the reaction 2SO_(2)(g)+O_(2)(g)to2SO_(3)(g) the entropy-
Answer» increases decreases remains unchanged change cannot be predicted Answer :B

Discussion

44.	For the reaction 2SO_(2(g))+O_(2(g))iff2SO_(3(g)) at 300 K, the value of DeltaG^(@) is -690.9R. The equilibrium constant value for the reaction at that temperature is (R is gas constant)
Answer» `10ATM^(-1)` 10 atm 10 1 Solution :`2SO_(2(g))+O_(2(g))iff2SO_(3(g))` `DeltaG=-RTlnK` `-690.9R=-RTlnK` `(690.9)/(300)=lnK` `2.303=lnK=2.303logK,K=10` `K_(p)=(p_(SO_(3))^(2))/(p_(SO_(2))^(2).p_(O_(2)))xx(atm^(2))/(atm^(3))=atm^(-1)`

Discussion

45.	For the reaction 2SO_2(g) + O_2 (g) iff 2SO_3 (g) at 300K, the value of DeltaG^@ is - 690.9R. The equilibrium constant value for the reaction at that temperature is (R is gas constant)
Answer» `10ATM^(-1)` 10atm 10 1 Solution :`2SO_2(g)+O_2(g) `/_\G=-RTlnk" or "=690.9R=-RTlnk` or `(690.7)/(300)=INK " or "2.303=Ink" or "K=10` `K_p=(p^2sO_3)/(p^2so_2xxpo_2)xx(atm^2)/(atm^3)=atm^(-1)`

Discussion

46.	SO_(2) reacts with O_(2) as follows 2SO_(2) +O_(2)to2SO_(3), the rate of disappearance of SO_(2) is 2.4xx10^(-4) mole "lit"^(-1)"min"^(-1). Then
Answer» `2XX10^(-4) MOL L^(-1)s^(-1)` `4XX10^(-4) mol L^(-1)s^(-1)` `1xx10^(-1) mol L^(-1)s^(-1)` `6xx10^(-4) mol L^(-1)s^(-1)` ANSWER :C

Discussion

47.	For the reaction , 2SO_(2) +O_(2) hArr 2SO_(3), the rate of disappearance of O_(2) is 2xx10^(-4)"mol L"^(-1)s^(-1). The rate of appearance of SO_(3) is
Answer» `2xx10^(4) "mol L"^(-1)s^(-1)` `4XX10^(-4)"mol L"^(-1)s^(-1)` `1xx10^(-4) "mol L"^(-1)s^(-1)` `6XX10^(-4) "mol L"^(-1)s^(-1)` Solution :`(Delta[O_(2)])/(Deltaf)=2xx10^(-4)` `([SO_(3)])/(Deltat)=2(Delta[O_(2)])/(Deltat)` `=2xx2xx10^(-4)` `=4xx10^(-4) " mol "1^(-1) s^(-1)`

Discussion

48.	For the reaction 2NO(g) + Cl_(2)to2NOCl(g) the following data were collected. All the measurnments were taken at 263 K. a) Write the expression for rate law, b) Calculate the value of rate constant and specify its untis. c) What is the initial rate of disappearance of Cl_(2) in exp. 4?
Answer» Solution :The RATE law equation may be expressed as: Rate=`k[A]^(p)[B]^(q)` Comparing experiments 1 and 2 `(Rate)_(1) = k[0.15]^(p)[0.15]^(q) = 0.60` `(Rate)_(2) = k[0.15]^(p)[0.30]^(q) = 1.20` DIVIDING eq. (ii) by eq. (i), `(Rate_(2))/(Rate_(1)) = (k[0.15]^(p)[0.30]^(q))/(k[0.15]^(p)[0.15]^(q))=1.20/0.60 = 2, [2]^(q) = [2]^(1)` or q=1 Comparing experiments 1 and 3 `(Rate_(1)) = k[0.15]^(p)[0.15]^(q)` `(Rate_(3)) = k[0.30]^(p)[0.15]^(q)` Dividing eq. (III) by eq (i) `(Rate_(3))/(Rate_(1)) = (k[0.30]^(p)[0.15]^(q))/(k[0.15]^(p)[0.15]^(q)) = 2.40/0.60 = 4,[2]^(p)=[2]^(2) or p=2` a) Expression for rate law, Rate = `k[NO]^(2)[Cl_(2)]` b) Rate constant (k) for the reaction may be calculated as: `k[0.15]^(p)[0.15]_(q) = 0.60, k0.150^(2)[0.15]^(1) = k60, k = p(0.60)/(0.003375) = 1.78 xx 10^(2)` since, the reaction is of third order : units of rate constant k =`L^(2)mol^(-2)min^(-1)` c) Initial rate of disappearancing of chlorine in experiment-4 Rate `= k[0.20mol L^(-1)]^(2)[0.25 mol L^(-1)] = (1.78 xx 10^(2)L^(2)min^(-1)) xx (0.25 mol L^(-1))^(3)` `= 2.78 mol L^(-1)min^(-1)`

Discussion

49.	For the reaction : 2O_(3) rarr 3O_(2) the rate of reaction is correctly given by the expression
Answer» `-(2)/(3)(d[O_(3)])/(dt)` `-(1)/(3)(d[O_(3)])/(dt)` `-(1)/(2)(d[O_(3)])/(dt)` `(1)/(2) (d[O_(3)])/(dt)` ANSWER :C

Discussion

50.	For the reaction 2NO(g) hArrN_2(g) + O_2(g) and NO(g) + 1//2Br_2(g) hArr NOBr(g)values of K_C are respectively2.4xx10^30and 1.4. Determine K_C for the reaction 1//2N_2(g) + 1//2O_2(g)+ 1//2 Br_2(g) hArr NOBr(g) .
Answer» SOLUTION :`9.03xx10^(-16)`

Discussion

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