94420 + Interview Questions in Chemistry in Class 12 Page 493 InterviewSolution

24601.	The rate of reactino of which of the following is not affected by pressure
Answer» `PCl_(3)+Cl_(2)hArrPCl_(5)` `N_(2)+3H_(2)hArr2NH_(3)` `N_(2)+O_(2)hArr2NO` `2SO_(2)+O_(2)hArr2SO_(2)` Solution :`HN_(2)+O_(2)hArr2NO,Deltan=0`

Discussion

24602.	The rate of radioactive decomposition corresponding to 3.7 xx 10^(10) disintegration per second is called a curie. What weight of ""^(225)Ra, whose t_((1)/(2))=1620 yr, will be required to yield 1 millicurie of radiation?
Answer» SOLUTION :`1 xx 10^(-3)G`

Discussion

24603.	The rate of physisorption
Answer» DECREASES with increase of pressure is independent at high pressure is maximum at ONE atmospheric pressure INCREASES with increases of pressure Answer :B

Discussion

24604.	The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.
Answer» SOLUTION :`logK_2/K_1=E_a/(2.303R)[1/T_1-1/T_2]` `E_a=(2.303xx8.314xx298xx308xx0.3010)/1000` `E_a`=52.89 KJ/mol

Discussion

24605.	The rate of most reactions becomes double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.
Answer» Solution :`T_(1) = 298 K, T_(2) = 308 K`. `R = 8.314 J mol^(-1) K^(-1)` Activation energy `K_(2) = 2K_(1)` `log.(K_(2))/(K_(1)) = (E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1) XX T_(2))]` `log 2 = (E_(a))/(2.303 xx 8.314)[(308 - 298)/(308 xx 298)]` `0.3010 = (E_(a))/(19.147)[(10)/(91784)]` `0.3010 = (E_(a))/(19.147) xx 1.08 xx 10^(-4)` `E_(a) = (0.3010 xx 19.147)/(1.089 xx 10^(-4))` `= (5.763)/(1.089 xx 10^(-4))` `= 52922.37 J mol^(-1)`

Discussion

24606.	The rate of forward reaction is two times that of the reverse reaction at a given temperature and identical concentration. K_("equilibrium") is
Answer» `0.5` `1.5` `2.5` `2.0` Solution :`K_(eqm)=(k_(f))/(k_(B))=(2)/(1)=2`

Discussion

24607.	The rate of Hoffmann's bromamide degradation with following amide will follow the order:
Answer» `III gtIgtIVgtII` `IIgtIVgtI LT III` `IgtIIltIIIltIV` `IgtIIIgtIVgtII` Solution :GROUP having -I EFFECT slow down the REACTION.

Discussion

24608.	The rate of forward reaction is two times that of reverse reaction at a given temperaturee and identical concentration. K_("equilibriuim") is
Answer» `2.5` `2.0` `0.5` `1.5` Solution :The RATE of forward reaction is two TIMES that of reverse reaction at a given temperature and identical concentration `K_("equlibrium")` is 2 because the reaction is REVERSIBLE. So `K=(K_(1))/(K_(2))=2/1=2.`

Discussion

24609.	The ratio of the forward reaction is two times that of the revers reaction at a given temperature and identical concentration.K_("equilibrium") is
Answer» 2.5 `2.0` 0.5 1.5 Solution :The rate of a forward reaction is two times that of reverse reaction at a given temperature. `(d[A])/(DT)=2(d[B])/(dt)` `K_("EQUILIBRIUM")=(K_(1))/(K_(2))=(d[A]//dt)/(d[B]//dt)=2(d[B]//dt)/(d[B]//dt)=2`. `thereforeK_("equilibrium")=2`.

Discussion

24610.	The rate of formation of SO_(3) in the reaction 2SO_(2)+O_(2)to2SO_(3) is 100 g "min"^(-1) Hence rate of disappearance of O_(2) is
Answer» `50g"min"^(-1)` `100g "min"^(-1)` `20G"min"^(-1)` `40G"min"^(-1)` Answer :C

Discussion

24611.	The rate of formation of NO(g) in the reaction 2NOBr(h) to 2NO(g)+Br_(2)(g) was reported as 1.6xx10^(-4)mol^(-1)sec^(-1). What is the rate of reaction and rate of consumption of NOBr?
Answer» SOLUTION :`8XX10^(-5)`, `1.6xx10^(-4)MSEC^(-1)`

Discussion

24612.	The rate of formation of ammonia by the reaction: N_2 +3H_2 to 2NH_3 expressed as (d[NH_3])/(dt)= 2.5xx 10^(-4)molL^(-1)s^(-1) The rate of consumption expressed in terms of H_2 as (-d[H_2])/(dt) will be
Answer» double THREE times Same one and a HALF TIME of that EXPRESSED in terms of `NH_3` SOLUTION :`1/2(dNH_3])/(dt)=-1/3(d[H_2])/(dt) or -(d[H_2])/(dt)=3/2xx(d[NH_3])/(dt)`

Discussion

24613.	The rate of formation of a dimer in a second order reaction 7.5xx10^(-3) "mol L"^(-1)s^(-1)at 0.05"mol L "^(--1)monomer concentration. Calculate the rate constant.
Answer» Solution :LET us consider the DIMERISATION of a MONOMER M . `2M rarr(M)_2` Rate `=k[M]^n` Given that n = 2 [ M] `=0.05 "mol L"^(-1)` Rate `=7.5xx10^(-3) "mol L"^(-1)s^(-1)` `k=("Rate")/([M]^n)impliesk=(7.5xx10^(-3))/((0.05)^2)=3 "mol"^(-1)Ls^(-1)`

Discussion

24614.	The rate of formation of a dimer in a second order reaction is 7.5xx10^(-3)"mol L"^(-1)s^(-1) at 0.05"mol L"^(-1) monomer concentration. Calculate the rate constant.
Answer» SOLUTION :Let US consider the dimerisation of a monomer M `2Mrarr(M)_(2)` `"Rate "=k[M]^(n)` `"Given that n = 2 and [M]"="0.05 MOL L"^(-1)` `"Rate "=7.5xx10^(-3)" mol L"^(-1)s^(-1)` `k=("Rate")/([M]^(n))RARR k=(7.5xx10^(-3))/((0.05)^(2))="3 mol"^(-1)L s"^(-1)`

Discussion

24615.	The rate of formation of a dimer in a second order dimerisation reaction is 95xx10^(-5)molL^(-1)s^(-1)" at "001molL^(-1) monomer concentration. Calculate the rate constant.
Answer» Solution :If the monomer is represented by A, then the reaction is : `2Ato(A)_(2)` As the reaction is of SECOND ORDER, the rate of reaction will be given by : Rate `=k[A]^(2)` Rate of reaction `=-(1)/(2)(d[A])/(dt)=+(d[A_(2)])/(dt)=95xx10^(-5)molL^(-1)s^(-1)` `:.95xx10^(-5)"MOL "L^(-1)s^(-1)=k(001"mol "L^(-1))^(2)=kxx10^(-4)("mol "L^(-1))^(2)" or "k=095L" mol"^(-1)s^(-1)`

Discussion

24616.	The rate of first order reaction increases:
Answer» If the temperature is increased If the CONCENTRATION of the REACTANTS is decreased If the concentration of the reactants is increased With time Solution :rate `prop` temperature , rate `prop` CONC. Of reactants

Discussion

24617.	The rate of fchemical reaction depends on the nature of chemical reactions, because
Answer» the threshold energy LEVEL differs from one reaction to another some of the reactant are solid at room temperature some of the REACTANTS are coloured all Solution :Difference in thershold energybarrier given difference values of energy of activation. Also `K = Ae^(-E//RT)` and rate `= k ["reactant"]^n`

Discussion

24618.	The rate of first order reaction A rarr Products, is7.5xx10^-4 mole litre^-1sec^-1.If the concentration of A is 0.5 mole litre-1the rate constant is:
Answer» `3.75xx10^-4sec^-1` `2.5xx10^-5sec^-1` `1.5xx10^-3sec^-1` `8.0xx10^-4sec^-1` ANSWER :C

Discussion

24619.	The rate of elementary reaction, A to B, increases by 400 times when the concentration of A is increased twenty folds. The order of the reaction with respect to A is
Answer» 1 2 10 100 Solution :`r = K [A]^m`…(i) ALSO, `400r = k [20A]^m` …(ii) From (ii)`-:`(i) `400 = (20)^m` `(20)^2 = (20)^m` ltbRgt `:.m = 2`

Discussion

24620.	The rate of elementary chemical reaction is directly proportional to:
Answer» ACTIVE masses of REACTANTS EQUILIBRIUM constant Active masses of PRODUCTS PRESSURE of reaction Answer :A

Discussion

24621.	The rate of effusion of a particular gas whas measured to be 40ml/min. Under same condition the rate of effusion of pure methane was 20ml/min. Then find molar mass of gas.
Answer» 4amu 2amu 4GM 2gm Solution :`(r_(CH_(4)))/(r_(GAS))=sqrt((M_(gas))/(16))=(20)/(40)=(1)/(2)` `M_(gas)=4gm`

Discussion

24622.	The rate of disapperance of SO_(2) in the reaction 2SO_(2) + O_(2) to 2SO_(3) is 1.28 xx 10^(-3)g/secthen the rate of formation of SO_(3) is
Answer» `0.64xx 10^(-3)` G /sec `0.80xx 10^(-3)` g /sec `1.28xx 10^(-3)` g /sec `1.60xx 10^(-3)` g /sec SOLUTION :The RATE of formation of `SO_(3)` is `1.28 XX 10^(-3)` g/sec.

Discussion

24623.	The rate of disappearance of CO in the reaction, 2 CO +O_2 to 2CO_2 is 1.28 xx 10^(-3)g//sec. Then the rate of formation of CO_2 is
Answer» `0.64 xx 10^(-3) g// sec` `0.80 xx 10^(-3) g //sec` `1.28 xx 10^(-1) g//sec` `1.28 xx 10^(-3) g//sec` Solution :`Rate = -1/2(DT[CO])/(dt)=-(d[O_2])/(dt)=+1/2(d[CO_2])/(dt)` Rate of FORMATION of CO = rate of disappearance of `CO_2` `=1.28xx10^(-3)g//sec`

Discussion

24624.	The rate of diffusion of two gases x and y is in the ratio of 1 : 5 and that of y and z in the ratio of 1 : 6 the ratio of the rate of diffusion of z with respect to x is :
Answer» `5:6` `1:30` `6:5` `30:1` Solution :`(r_(x))/(r_(y))=(1)/(5)""....(i)""(r_(g))/(r_(z))=(1)/(6)""....(ii)` `(i)xx(ii) : (r_(x))/(r_(z))=(1)/(30)IMPLIES (r_(z))/(r_(x))=(30)/(1)`

Discussion

24625.	The rate of diffusion of SO_2 and O_2 at the same P and T are in the ratio (S = 32, O = 16):
Answer» 1:32 1:4 1:2 `1:SQRT2` ANSWER :D

Discussion

24626.	The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is
Answer» 64 32 4 8 Answer :A

Discussion

24627.	The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is :
Answer» 64 32 4 8 Answer :A

Discussion

24628.	The rate of diffusion of a sample of ozonised oxygen is 0.98 times more than that of pure oxygen. Find the percentage (by volume) of ozone in the ozonised sample.
Answer» ANSWER :`8.25%`

Discussion

24629.	The rate of diffusion of gas is ……… proportional to both ....... of molecular mass.
Answer» SOLUTION :INTERCEPT =RT

Discussion

24630.	The rate of decomposition of N_(2)O_(5) in C Cl_(4) solution has been studied at 318 K and the following results have been obtained : {:(t//min,,,0,,,135,,,342,,,683,,,1693),(c//M,,,2.08,,,1.91,,,1.67,,,1.35,,,0.57):} Find the order of the reaction and calculate its rate constant. What is the half-life period ?
Answer» Solution :As RATE does not remain CONSTANT with time, it is not a ZERO ORDER reaction. HENCE, we try integrated first order equation, i.e., `k=(2.303)/(t)log""(c_(0))/( c )`

Discussion

24631.	The rate of dehydration of alcohols to form alkenes depends on ........
Answer» Both the CONCENTRATION of dehydrating AGENT and alcohol Only on the concentration of alcohol The concentration of protonated alcohol Only on the concentration of dehydrating agent. Solution :The formation of carbocation from the protonated alcohol is the slow step of the reaction and THUS it is the RATE DETERMINING step.

Discussion

24632.	The rate of decompostion of ozone drops sharply in ………… .
Answer» ACIDIC MEDIUM ALKALINE medium neutral medium Ether medium Solution :alkaline medium

Discussion

24633.	The rate of decomposition of ammonia is found to depend upon the concentration of NH_(3) according to the equation -(d[NH_(3)])/(dt)=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]) What will be the order of reaction when (i) concentration of NH_(3) is very high ? (ii) concentration of NH_(3) is very low ?
Answer» Solution :The given rate law equation can be written as `-(d[NH_(3)])/(DT)=(k_(1))/(1//[NH_(3)]+k_(2))` (i) If `[NH_(3)]` is very high, `1//[NH_(3)]` becomes negligible `:.-(d[NH_(3)])/(dt)=(k_(1))/(k_(2))=k` i.e., rate becomes independent of concentration. Hence,it is of zero ORDER. (ii) If `[NH_(3)]` is very SMALL, `1//[NH_(3)]` will be very large `(gtgtk_(2))`, so that `k_(2)` can be neglected in comparison to `1//[NH_(3)]`. Hence, `-(d[NH_(3)])/(dt)=(k_(1))/(1//[NH_(3)])=k_(1)[NH_(3)]` Thus, REACTION is of 1st order.

Discussion

24634.	The rate of decomposition of hydrogen peroxide decreases in presence of
Answer» Platinum Iron `MnO_2` Glycerine Answer :D

Discussion

24635.	The rate of decomposition of ammonia is found to depend upon the concentration of ammonia as : - (d[NH_(3)])/(dt) = (k_(1) [NH_(3)])/(1 + k_(2) [NH_(3)]) which of the following statement is correct ?
Answer» The reaction is zero ORDER at very lowas WELL as very high `NH_(3)` CONCENTRATION The reaction is first order at very low as well as very high `NH_(3)` concentration . The reaction is zero order at very low concentration of `NH_(3)` and is of first order at very high concentration of `NH_(3)` . The reaction is first order at very low `NH_(3)` concentration and is of zero order at very high `NH_(3)` concentration . Answer :D

Discussion

24636.	The rate of decomposition of a gas at a certain temperature is 5.14 and 7.25 in some units for 20% and 5% decomposition respectively. Calculate the order of the reaction.
Answer» ANSWER :2

Discussion

24637.	The rate of decompositionof a substance increases by a factor 2.25 for 1.5 times increases in concentration of substance at same temperature. Find order of the reaction
Answer» ANSWER :2

Discussion

24638.	The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k_(1) and k_(2) respectively. The energy of activation for the two reactions are 152.30 kJ mol^(-1) and 157.7 kg mol^(-1) as well as frequency factors are 10^(13) and 10^(14) respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be same for the two reactions.
Answer» 256 K 354 K 282 K 674 K Answer :C

Discussion

24639.	The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant ("in" sec^(-1)) k_(1) "and"k_(2) respectively. The energy of activationsfor the two reactions are 152.30 kJ "mol"^(-1) "and" 157.7 kJ "mol"^(-1) as well as frequency factors are 10^(13)"and" 10^(14) respectively for the decomposition of methyl and ethyl nitrite. Calculate the approx. temperature at which rate constant will be same for the two reactions.
Answer» ANSWER :282

Discussion

24640.	The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant ("in" sec^(-1)) K_(1) and K_(2) respectively. The energy of activation for the two reaction are 152.30 kJ mol^(-1) and 157.7 kJ mol^(-1) as well as frequency factor are 10^(13) and 10^(14) respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be same for the two reactions.
Answer» SOLUTION :N//A

Discussion

24641.	The rate of decay per second of a radioactive sample
Answer» proportional to the life time lived by the nucleus DECREASES with the life time lived is independent of the life time lived depends upon the TOTAL number of radioactive NUCLEI present in the sample SOLUTION :`r = LAMBDA , N ,r = ((0.683)/(tau_(1//2)))N therefore r = ((1)/(t)) n`

Discussion

24642.	The rate of chemisorption
Answer» decreases with INCREASE of pressure is INDEPENDENT of pressure is MAXIMUM at one atmospheric pressure increase with increase of pressure Answer :B

Discussion

24643.	The rate of constant is numerically the same for 1 order, II order and III order. Select the correct statements, if
Answer» `[A]LT1,` then `r_(1)gtr_(2)gtr_(3)` `[A]lt1`, then `r_(1)ltr_(2)ltr_(3)` `[A]gt1`, then `r_(1)gtr_(2)gtr_(3)` `[A]=1` then `r_(1)=r_(2)=r_(3)` Solution :`r_(1)=K.[A]^(1),r_(2)=K.[A]^(2),r_(3)=K.[A]^(3)` `AtoB^(+)""C` `C_(0)""0""0`

Discussion

24644.	The rate constant of a reaction depends on
Answer» TEMPERATURE Mass Weight Time Solution :k = `AE^(-E_(a)//RT)` by this equation it is CLEAR that rate CONSTANT of a reaction depends on temperature .

Discussion

24645.	The rate of chemisorption :
Answer» is INDEPENDENT of PRESSURE decreases with INCREASE of pressure INCREASES with increase of pressure none of these Solution :increases with increase of pressure

Discussion

24646.	The rate of chemical reaction is doubled for every 10^@C rise in temperature because of
Answer» decrease in the ACTIVATION energy decrease in the threshold energy INCREASE in the number of molecular collisions increase in the number of ACTIVATED molecules Solution :Rate of REACTION increases because number of activated molecules increases.

Discussion

24647.	The rate of chemical reaction is directly proportional to
Answer» active masses of reactants equlilibrium CONSTANT active masses of PRODUCTS PRESSURE SOLUTION :This is law of mass ACTION.

Discussion

24648.	The rate of chemical reaction (except zero order)
Answer» decreases from MOMENT to moment independent constant throughout independent of the ORDER of reaction reamin unchange SOLUTION :Rate is `(dx)/(dt)=-(d[A])/(dt)` Where x STANDS for product concentration and [A] stands for REACTANT concentration. It continously decreases with decrease in concentrationof reactant with time

Discussion

24649.	The rate-of chemical reaction (except zero order):
Answer» Decreases from MOMENT to moment Remains CONSTANT, throughout Independent of the ORDER of REACTION NONE of the above Answer :A

Discussion

24650.	The rate of chemical reaction depends upon
Answer» TIME pressure concentration All of these Answer :D

Discussion

Explore topic-wise InterviewSolutions in .

The rate of reactino of which of the following is not affected by pressure

The rate of radioactive decomposition corresponding to 3.7 xx 10^(10) disintegration per second is called a curie. What weight of ""^(225)Ra, whose t_((1)/(2))=1620 yr, will be required to yield 1 millicurie of radiation?

The rate of physisorption

The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.

The rate of most reactions becomes double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.

The rate of forward reaction is two times that of the reverse reaction at a given temperature and identical concentration. K_("equilibrium") is

The rate of Hoffmann's bromamide degradation with following amide will follow the order:

The rate of forward reaction is two times that of reverse reaction at a given temperaturee and identical concentration. K_("equilibriuim") is

The ratio of the forward reaction is two times that of the revers reaction at a given temperature and identical concentration.K_("equilibrium") is

The rate of formation of SO_(3) in the reaction 2SO_(2)+O_(2)to2SO_(3) is 100 g "min"^(-1) Hence rate of disappearance of O_(2) is

The rate of formation of NO(g) in the reaction 2NOBr(h) to 2NO(g)+Br_(2)(g) was reported as 1.6xx10^(-4)mol^(-1)sec^(-1). What is the rate of reaction and rate of consumption of NOBr?

The rate of formation of ammonia by the reaction: N_2 +3H_2 to 2NH_3 expressed as (d[NH_3])/(dt)= 2.5xx 10^(-4)molL^(-1)s^(-1) The rate of consumption expressed in terms of H_2 as (-d[H_2])/(dt) will be

The rate of formation of a dimer in a second order reaction 7.5xx10^(-3) "mol L"^(-1)s^(-1)at 0.05"mol L "^(--1)monomer concentration. Calculate the rate constant.

The rate of formation of a dimer in a second order reaction is 7.5xx10^(-3)"mol L"^(-1)s^(-1) at 0.05"mol L"^(-1) monomer concentration. Calculate the rate constant.

The rate of formation of a dimer in a second order dimerisation reaction is 9*5xx10^(-5)molL^(-1)s^(-1)" at "0*01molL^(-1) monomer concentration. Calculate the rate constant.

The rate of first order reaction increases:

The rate of fchemical reaction depends on the nature of chemical reactions, because

The rate of first order reaction A rarr Products, is7.5xx10^-4 mole litre^-1sec^-1.If the concentration of A is 0.5 mole litre-1the rate constant is:

The rate of elementary reaction, A to B, increases by 400 times when the concentration of A is increased twenty folds. The order of the reaction with respect to A is

The rate of elementary chemical reaction is directly proportional to:

The rate of effusion of a particular gas whas measured to be 40ml/min. Under same condition the rate of effusion of pure methane was 20ml/min. Then find molar mass of gas.

The rate of disapperance of SO_(2) in the reaction 2SO_(2) + O_(2) to 2SO_(3) is 1.28 xx 10^(-3)g/secthen the rate of formation of SO_(3) is

The rate of disappearance of CO in the reaction, 2 CO +O_2 to 2CO_2 is 1.28 xx 10^(-3)g//sec. Then the rate of formation of CO_2 is

The rate of diffusion of two gases x and y is in the ratio of 1 : 5 and that of y and z in the ratio of 1 : 6 the ratio of the rate of diffusion of z with respect to x is :

The rate of diffusion of SO_2 and O_2 at the same P and T are in the ratio (S = 32, O = 16):

The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is

The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is :

The rate of diffusion of a sample of ozonised oxygen is 0.98 times more than that of pure oxygen. Find the percentage (by volume) of ozone in the ozonised sample.

The rate of diffusion of gas is ……… proportional to both ....... of molecular mass.

The rate of dehydration of alcohols to form alkenes depends on ........

The rate of decompostion of ozone drops sharply in ………… .

The rate of decomposition of ammonia is found to depend upon the concentration of NH_(3) according to the equation -(d[NH_(3)])/(dt)=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]) What will be the order of reaction when (i) concentration of NH_(3) is very high ? (ii) concentration of NH_(3) is very low ?

The rate of decomposition of hydrogen peroxide decreases in presence of

The rate of decomposition of ammonia is found to depend upon the concentration of ammonia as : - (d[NH_(3)])/(dt) = (k_(1) [NH_(3)])/(1 + k_(2) [NH_(3)]) which of the following statement is correct ?

The rate of decomposition of a gas at a certain temperature is 5.14 and 7.25 in some units for 20% and 5% decomposition respectively. Calculate the order of the reaction.

The rate of decompositionof a substance increases by a factor 2.25 for 1.5 times increases in concentration of substance at same temperature. Find order of the reaction

The rate of decay per second of a radioactive sample

The rate of chemisorption

The rate of constant is numerically the same for 1 order, II order and III order. Select the correct statements, if

The rate constant of a reaction depends on

The rate of chemisorption :

The rate of chemical reaction is doubled for every 10^@C rise in temperature because of

The rate of chemical reaction is directly proportional to

The rate of chemical reaction (except zero order)

The rate-of chemical reaction (except zero order):

The rate of chemical reaction depends upon

The rate of formation of a dimer in a second order dimerisation reaction is 95xx10^(-5)molL^(-1)s^(-1)" at "001molL^(-1) monomer concentration. Calculate the rate constant.