94420 + Interview Questions in Chemistry in Class 12 Page 496 InterviewSolution

24751.	The rate law for a reaction is found to be : Rate =k [NO_(2)^(-) ],[ I^(-)][H^(+)]^(2) How would the rate of reaction change when(i) Concentration of H^(+) is doubled (ii) Concentration of I^(-) is halved(iii) Concentration of each of NO_(2)^(-),I^(-) and H^(+) are tripled?
Answer» Solution :Suppose intially the CONCENTRATIONS are : `[NO_(2)^(-)]=a molL^(-1),[I^(-)]=B molL^(-1)" and "[H^(+)]=c molL^(-1):." Rate "="k a b "c^(2)` (i) New `[H^(+)]=2c:." New Rate = k a b"(2" c")^(2)=4kabc^(2)=4` times (II) New `I^(-)=(b)/(2)"New Rate "=ka(b)/(2)c^(2)=(1)/(2)kabc^(2)`, i.e., rate of REACTION is halved. (III) New `[NO_(2)^(-)]=3" a",[I^(-)]=3" b",[H^(+)]=3" cNew Rate"=k(3" a")(3" b")(3" c")^(2)=81" k "abc^(2)=81` times.

Discussion

24752.	The rate law for a reaction is Rate = K [A] [B]^(3//2) Can the reaction be an elementary process ? Explain.
Answer» Solution :No, an elementary process WOULD have a rate LAW with orders equal to its molecularities and THEREFORE must be in integral form.

Discussion

24753.	The rate law for a reaction is : Rate = k[A]^(n) The rate of reaction remains same on doubling the concentration of A. The value of n is :
Answer» 0 1 2 3 Answer :A

Discussion

24754.	The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction. aA+bBtocC+dD Rate law takes the form r=k[A]^(x)[B]^(y) where x and y are number that must be determined experimentely k is the rate constant and [A] and [B] are concentration of A & B respectively. The varitation of concentration of 'A' with time in two experiments starting with two different initial concentration of 'A' is given by the following graph.The reaction is represented by A(aq)toB(aq) what is rate of reaction (M//min) when conc. of A in aqueous solution was 1.8 M:
Answer» 0.072 M `"MIN"^(-1)` 0.1296 M `"min"^(-1)` 0.036 M `"min"^(-1)` 1 M `"min"^(-1)` SOLUTION :`r_1=0.3/5 M min^(-1) IMPLIES r_2=0.2/5` `r_1/r_2=[A_1/A_2]^m implies 0.3/0.2=(1.5/1)^mimplies m=1 : " " 0.3/5=k(1.5),k=1/25` `r=1/25xx1.8=0.072 MMIN^(-1)`

Discussion

24755.	The rate law for a reaction between the substances A and B is given by Rate =k[A]^(n)[B]^(m) On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be
Answer» `m+n` `(n-m)` `2^(n-m)` `(1)/(2^((m+n)))` Solution :Earler RATE `= K a^(n) B^(m)` New rate `= k(2A)^(n) (b)^(m)` `("New rate")/("Earlier rate") = (2^(n) a^(n) b^(m) 2^(-m))/(a^(n)b^(m)) = 2^(n) . 2^(-m) = 2^(n-m)`

Discussion

24756.	The rate law for a reaction between the substances A and B is given by rate =k[A]^(n)[B]^(m). On doubling the concentration of A and having the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be
Answer» `m+n` `n-m` `2^((n-m))` `(1)/(2^((m+n)))` ANSWER :C

Discussion

24757.	The rate law expression for a hypothetic reaction: 2A +3B to 2C is (dx)/(dt)=k[A][B]^(2) The order of the reaction is
Answer» 1 2 3 5 Answer :C

Discussion

24758.	The rate law for reaction between the substances A and B is given by : Rate = k[A]^(n) [B]^(m) on doubling the concentration of A and halving the cocentration of B the ratio of the new rate to the earlier rate of reaction will be :
Answer» m+n n-m `2^(n-m)` `(1)/(2^(m-n))` Solution :( C) Earlier rate rate =k `a^(n) B^(m)` New rate rate = `k(2A)^(n) (b)/(2)` `("rate")/("rate")= (2^(n)a^(n)b^(m)2^(-m))/(a^(n)b^(m))` `=2^(n) .2^(-m) = 2^(n-m)`

Discussion

24759.	The rate law equation for the reaction A to B is found to be -(d[A])/(dt)=k[A]^(1//2) If [A]_(0) were the initial concentration of A, derive expressions for (i) rate constant in the integrated form(ii) half-life period of the reaction
Answer» Solution :(i) `-(d[A])/(DT)=k[A]^(1//2)" or "-(d[A])/([A]^(1//2))=k" dt"` Integrating both sides of the equation, `-int[A]^(1//2)d[A]=int k" dt or "-([A]^(1//2))/(1//2)=kt+I` or `-2[A]^(1//2)=kt+I""...(i)` When `t=0,[A]=[A_(0)]:.-2[A_(0)]^(1//2)=0+I` Substittuting this value of I in EQN. (i), we get `-2[A]^(1//2)=kt-2[A_(0)]^(1//2)" or "kt=2"{"[A_(0)]^(1//2)-[A]^(1//2)"}"" or "k=(2)/(t){[A_(0)]^(1//2)-[A]^(1//2)}` (ii) `t=t_(1//2)" when "[A]=[A_(0)]//2` `:.t_(1//2)=(2)/(k){[A_(0)]^(1//2)-([A_(0)]^(1//2))/(2^(1//2))}=(2)/(k)[A_(0)]^(1//2){1-(1)/(sqrt(2))}=(2)/(k)[A_(0)]^(1//2)(sqrt(2)-1)/(sqrt(2))" or "t_(1//2)=(sqrt(2)"("sqrt(2)-1")")/(k)[A_(0)]^(1//2)`

Discussion

24760.	The rate law experession for four reactions are give below . In which case the percentage increase in the rate of reaction would be maximum when the concentration of A is doubled ?
Answer» rate = k[A] rate `= k[A]^(3//2)` rate `k[A]^(@)` rate =` k[A]^(2)` SOLUTION :(D) LARGER the order of REACTION more the percent increase in the rate of reaction for a given increase in concentration .

Discussion

24761.	The rate for the reaction, RCl+NaOH(aq)rarrROH+NaCl is given by rate=K_1[RCl]. The rate of the reaction is:
Answer» DOUBLED on doubling the CONCENTRATION of NaOH Halved on reducing the concentration of RCl to half Decreased on increasing the temperature of the reaction Unaffected by increasing the temperature of the reaction Answer :B

Discussion

24762.	The rate for a first order reaction is 0.6932 xx 10^(-2) mol l^(-1) "min"^(-1) and the initial concentration of the reactants is 1 M , T_(1//2) is equal to
Answer» 6.932 MIN 100 min `0.6932 xx 10^(-3)` min `0.6932 xx 10^(-2)` min Solution :`R = K["reactant"]^(-1) therefore k = (0.693 xx 10^(-2))/(1)` also `t_(1//2) = (0.693)/(k) = (0.693)/(0.693 xx 10^(-2)) = 100 ` min

Discussion

24763.	The rate for a first order reaction is 0.6932 xx 10^(-2) mol " "L^(-1) and the initial concentration of the reactant is 1M, t_(1//2) is equal to
Answer» `6.932` minute `0.6932 XX 10^(-3)` minute 100 minute `0.6932 xx 10^(-2)` minute ANSWER :C

Discussion

24764.	The rate for a first order reaction is 0.6932 xx 10^(-2)mol L^(-1) min^-1 and the initial concentration of the reactant is 1 M, t_(1//2) is equal to:
Answer» `0.6932 X 10^(-2)MINUTE` `0.6932 x x10^(-2)` 100minute 6.932minute Answer :C

Discussion

24765.	The rate for the first order reaction is 0.0069 mol L^(-1) min^(-1) and theinitial concentration is 0.2 mol L^(-1) . The half life period is
Answer» 638 s 1205 s 690 s `0.635 `s Solution :(B) Rate = K ( C) `0.0069 mol L^(-1) min^(-1) = (0.2 "mol L"^(-1))` k `k = (0.0069 )/(0.2)= (3.45xx10^(-2))/(60)` `= 5.75 xx10^(-4) s^(-1)` `t_(1//2)=(0.693)/(5.75xx10^(-4)) = 1205 ` s

Discussion

24766.	The rate expression for the reaction A_((g)) + B_((g)) rarr C_((g))" is rate" = k[A]^(2)[B]^(1//2). What changes in the initial concentrations of A and B will cause the rate of reaction to increase by a factor of eight?
Answer» [A`]=[A] : [B`] = 2[B] [A`]=2[A] : [B`] = 4[B] [A`]=[A] : [B`] = 4[B] [A`] = 4[A] : [B`] = [B] SOLUTION : Given `r= k[A]^2[B]^(1//2)` new RATE will be 8 times when [A]. = 2[A] and [B]. = 4[B] now `r. = [2A]^(2)[48]^(1//2)` = 8[A][B]

Discussion

24767.	The rate expression gives the relation between rate of reaction and
Answer» CONC. of reactants conc. of products RATE CONSTANT rate LAW Answer :A

Discussion

24768.	The rate expression for the reaction A_((g))+B_((g))toC_((g)) is rate =kC_(A)^(2)C_(B)^(1//2) what changes in the initial concentrations of A and B will cause the rate of reaction to increase by a factor of eight?
Answer» `C_(A)xx2,C_(B)xx2` `C_(A)xx2,C_(B)xx4` `C_(A)XX1,C_(B)xx4` `C_(A)xx4,C_(B)xx1` ANSWER :B

Discussion

24769.	The rate expression for a reaction is rate =k[A]^(3//2)[B]^(-1), the order of reaction is
Answer» `0` `1//2` `3//2` `5//2` ANSWER :B

Discussion

24770.	The rateexpression fora reactionis Rate =K[A][B]^(2) unitfor rateconstant K is
Answer» `mol dm^(-3) SEC^(-1)` `(dm^(3)mol^(-1))^(2)sec^(-1)` `(dm^(3)mol^(-1)sec^(-1))^(2)` `dm^(3) mol^(-1)sec^(-1)` Answer :B

Discussion

24771.	For the reaction A + B rarr P, the rate is given by Rate = K [A]^(1) [B]^(2). By how many times does the rate of reaction increase when concentrations of A & B are doubled?
Answer» SOLUTION :`K_(1)=[A]^(1)[B]^(2)""rarr(1)` `K_(2)=[2A]^(1)[2D]^(2)` `K_(2)=[2A]4[D]^(2)` `K_(2)=8[A]^(1)[D]^(2)` `(K_(2))/(K_(1))=(8cancel([A]^(1)[D]^(2)))/(CANCEL([A]^(1)[D]^(2)))` `K_(2)=8K_(1)` The rate of reaction increase 8 times when concentration of A and B are doubled.

Discussion

24772.	The rate equation for the reaction 2A+BtoC is found to be : rate =K[A][B]. The correct statement in relation to this reaction is
Answer» unit of k must be `"sec"^(-1)` value of k is independent of the initial concentrations of A and B RATE of formation fo C is TWICE the rate of disappearance of A `t_(1//2)` is a constant Answer :B

Discussion

24773.	The rate equation for the decomposition of N_(2)O_(5) in "CC"l_(4) is rate =K[N_(2)O_(4)] where K=6.3x10^(-4)s^(-1) at 320 K. what would be the initial rate of decompositioni of N_(2)O_(5) in a 1.10M solution of N_(2)O_(4)?
Answer» `6.3xx10^(-5)` MOL `"LITRE"^(-1)s^(-1)` `0.63xx10^(-6)` mol `"litre"^(-1)s^(-1)` `6.3xx10^(-5)` mol `"litre"^(-1)s^(-1)` `0.63xx10^(-4)` mol `"litre"^(-1)s^(-1)` Solution :`r=K.[N_(2)O_(5)]=6.3xx10^(-4)xx(0.10)^(1)=6.3xx10^(-5)` `1N_(2)O_(5)to2NO_(2)+1/2O_(2)`

Discussion

24774.	The rate equation for a reactionA to Bis r = k[A]^(0) . If the initial concentration of the reaction is a mol dm^(-3) , the half-life period of the reaction is
Answer» `(k)/(a)` `(a)/(k)` `(2a)/(k)` `(a)/(2k)` SOLUTION :According to the rate equation GIVEN , the reaction is of ZERO ORDER . For a zero order reaction , `t_(1//2) PROP a` or `t_(1//2) = (a)/(2k)`.

Discussion

24775.	The rate equation for a reaction 3A +2B to C is given by (dx)/(dt) =k[A]^2[B]^1 which of the following statements is false
Answer» the reaction is overall a second order reaction the rate will not be ALTERED by doubling the concentration of B temerature will have no EFFECT on the rate of reaction all are false Solution :All the statements are flase. The overall order of the GIVEN reaction is THREE. The rate will be altered by doubling the conc. Of B as it apears in the rate law,. The change in TERMPERATURE will also affect the rate.

Discussion

24776.	The rate equation for a reaction AtoB is r=k[A]^(0). If the initial concentration of the reactant is a mol dm^(-3), the half life period of the reaction is
Answer» k/a a/k 2a/k a/2k ANSWER :D

Discussion

24777.	The rate determining step of a reaction is the step
Answer» which involves the minimum number of molecules which involves the MAXIMUM number of molecules which is the SLOWEST which is the fastest Solution :Slowest step is the RATE DETERMINING step.

Discussion

24778.	The rate equation for a certain reaction is: r=k[Cl_(2)][NO]^(2). What is the overall order of the reaction?
Answer» SOLUTION :Rate LAW ( r) `=k[Cl_(2)][NO]^(2)` Order of REACTION= 1+2=3

Discussion

24779.	The rate determining step for the preparation of nitrobenzene from benzene is
Answer» Removal `OVERSET(+)NO_(2)` Removal of `overset(+)NO_(3)` Formation of `overset(+)NO_(2)` Formation of `overset(+)NO_(3)` Answer :C

Discussion

24780.	Therate cosntantof firstorderreactionis 6.8 xx 10^(-4)s^(-1) Ifthe initialconcentrationof the reactantis0.04 M, Whatis itsmolarityafter 20minutes? Howlongwillit takefor 25%of thereactanttoreact ?
Answer» Solution :Given :Rateconstant`=k = 6.8xx 10^(-4) s^(-1)` `[A]_(0)=0.04 M` (i)Aftert= 20min= 20x 60 s= 1200s`[A]_(t)= ?` `k= (2.303)/(t) log_(10).([A]_(0))/([A]_(t))` `:. log_(10).[[A]_(0)]/[[A]_(t)] = (kxxt)/(2.303) - (6.8 XX 10^(-4) xx 1200)/(2.303 ) = 0.3543` `:. [[A]_(0)]/[[A]_(t)]=AL 0.3543 =2.26` `:. [A]_(t) =[[A]_(0))/(2.26) =(0.04)/(2.26) =0.0177 M` (ii)For 25%reactantto reactt= ? REACTANT reacted=25%of0.04 M `=(25)/(100) xx 0.04` `=0.01 M` `:.` Reactantleft `=[A]_(t)= 0.04- 0.01 =0.03 M` `t= (2.303)/(k) log_(10) .([[A]_(0)])/([[A]_(t)])` `=(2.303)/(6.8 xx 10^(-4)) log_(10) .(0.04)/(0.03)` `=(2.303)/(6.8xx 10^(-4))` `=423 s` `=(423)/(60) MIN = 7.05 min`

Discussion

24781.	The rate constants of a reaction at 500K and 700K are 0.02s^(-1) and 0.07s^(-1), respectively. Calculatevalue of activation energy for the reaction.[ Given R=8.314JK^(-1)MI^(-1)].
Answer» SOLUTION :Use the Arrhenius EQUATION `"LOG"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` Given : `T_(1)=500K, T_(2)=700K, k_(1)=0.02s^(-1) and k_(2)=0.07s^(-1)` Substituting the values in the equation above, we have `"log"(0.07)/(0.02)=(E_(a))/(2.307xx8.314)[(700-500)/(500xx700)] or 0.5441 = (E_(a))/(19.1804) xx(200)/(350000)` or `E_(a)=(0.5441xx19.1804xx3500)/(2)o=18263J=18.263kJ`.

Discussion

24782.	The rate constants of a reaction at 500 k and 700 k are 0.02"sec"^(-1) and 0.07 "sec"^(-1) respectively. Calculate the value of E_(a)
Answer» 18230.8J 190.83J 185.02J 172.02J Solution :`(K_(2))/(K_(1))=(Ae^(-Ea_(2)//RT_(2)))/(Ae^(-EA//RT_(1))),0.07/0.02=e^((-epsilon_(a))/(RT_(2))+(epsilon_(a))/(RT_(1)))` `2.303log(7/2)=(E_(a))/R(1/(T_(1))-1/(T_(2))),2.303xx0.544=(E_(a))/R(1/500-1/700)` `E_(a)=((2.303xx0.544xx8.314xx500xx700)/200)=18228` Joule

Discussion

24783.	The rate constants of a reaction at 500 K and 700 K are 0.02s^(-1)" and "0.07s^(-1) respectively. Calculate the values of E_(a) and A.
Answer» SOLUTION :`log""(k_(2))/(k_(1))=(E_(a))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))` `log""(0.07)/(0.02)=(E_(a))/(2.303xx8.314" JK"^(-1)MOL^(-1))xx(700"K"-500"K")/(700"K"xx500"K")`, `log3.5=(E_(a))/(2.303xx8.314" JK"^(-1)mol^(-1))xx(200"K")/(700"K"xx500"K")` or `""E_(a)=(0.5441xx2.303xx8.314xx700xx500)/(200)"J mol"^(-1)=18231.4" J mok"^(-1)=18.23" kJ mol"^(-1)` Further, `logk=-(E_(a))/(2.303"R")+LOGA" or"logA=logk+(E_(a))/(2.303"RT")` Substituting `T=500 K, k=0.02s^(-1)`, we get `logA=log0.02+(18231.4" J mol"^(-1))/(2.303xx8.314"JK"^(-1)mol^(-1)xx500"K")=-log50+12.4095=-1.699+1.904=0.205` `A=" Antilog "(0.205)=1.603`

Discussion

24784.	The rate constants of a reaction at 300K and 280 K respectively are K_(1) and K_(2). Then
Answer» `K_(1)=20K_(2)` `K_(2)=4K_(1)` `K_(1)=4K_(2)` `K_(1)=0.5K_(2)` Answer :C

Discussion

24785.	The rate constants k_1 and k_2 for two different reactions are 10^(16).e^(-2000//T)" and "10^(15).e^(-1000//T), respectively. The temperature at which kı = ky is
Answer» 2000 K `(1000 )/(2.303)`K 434.22 K `(2000)/(2.303)`K Solution : `K_1=10^(16)E^(-2000//T),K_2=10^(15)e^(-1000//T)` The temperature at which `K_1=k_2` will be `10^(6)e^(-2000//T)=10^(15)e^(-1000//T)` `implies (e^(-2000//T))/(e^(-1000//T))=(10^(15))/(10^(16))` `implies e^((-1000)/(T)) = 10^(-1) implies log_e e^((-1000)/(T)) = log_e10^(-1)` `implies 2.303 log_(10) e^((-1000)/(T)) = 2.303xxlog_(10)10^(-1)` `implies (-1000)/(T)xxlog_(10) e=-1 therefore T=(1000)/(2.303)K`

Discussion

24786.	The rate constants k_(1)" and "k_(2) for two different reactions are 10^(16). e^(-2000//T) and 10^(15).e^(-1000//T) respectively. The temperature at which k_(1)=k_(2) is
Answer» `1000 K` `(2000)/(2.303)K` `2000 K` `(1000)/(2.303)K` Solution :`k_1=10^16eF^(-2000//T), k_2=10^(15)e^(-1000//T` When `k_1=k_2,10^16e^(-2000//T)=10^15e^(-1000//T)` or`10E^(-2000//T)=e^(=-1000//T)` Taking natural logarithm of both SIDES, we get In `10-2000/T=(-1000)/T" or "2303-2000/T=(-1000)/T` or `1000/T=2303` or `T=1000/(2*303)K`

Discussion

24787.	The rate constant value for a reaction is 1.75xx10^(2)L^(2)mol^(-2)sec^(-1).The half life period t_((1)/(2))alpha……
Answer» `[R_(o)]^(-2)` `[R_(o)]` `[R_(o)]^(2)` `[R_(o)]^(-1)` Solution :`K=1.75xx10^(2)LI^(2)"mole"^(-2)SEC^(-1)` Accroding to the unit of K,it indicates `3^(RD)` ORDER reaction,so half life time. `t_((t)/(2))prop[R_(o)]^(1-n)therefore [R_(o)]^(-2)`

Discussion

24788.	The rate constants k_(1) and k_(2) for two different reactions are 10^(16) * e^(-2000//T) and 10^(15) * e^(-1000//T) , respectively . The temperature at which k_(1) = k_(2) is
Answer» 2000 K `(1000)/(2.303) K` 1000 K `(2000)/(2.303)K` SOLUTION :`10^(16) e^((-2000)/(T)) = 10^(15) e^((-1000)/(T))` `10 = e^((-1000)/(T))/(e^(-2000)/(T)) implies 10 e^((1)/(T) xx 1000) implies "ln" 10 = (1000)/(T)` `implies 2.303` log 10 = `(1000)/(T) implies T = (1000)/(2.303) K` .

Discussion

24789.	The rate constant, the activation energy and the frequency factor of a chemical reaction at 25^(@)C are 3.0xx10^(-4)s^(-1),104.4"KJ mol"^(-1) and 6.0xx10^(14)s^(-1) respectively. The value of the rate constant as Trarroo is :
Answer» `2.0xx10^(18)s^(-1)` `6.0xx10^(14)s^(-1)` infinite `3.6xx10^(30)s^(-1)` ANSWER :B

Discussion

24790.	The rate constant , the activation energy and the arrhenius parameter of a chemical reaction at 25^(@)C and 3.0 xx 10^(4) s^(-1) , 104. 4 kJ mol^(-1) and 6.0 xx 10^(14) s^(-1) respectively . The value of the rate constant as T to oo is
Answer» `2.0 xx 10^(18) s^(-1)` `6.0 xx 10^(14) s^(-1)` INFINITY `3.6 xx 10^(30) s^(-1)` Solution :`T_(2) = T`(SAY) , T = `25^(@)C = 298 K` `E_(a) = 104.4 kJ mol^(-1) = 104.4 xx 10^(3) J mol^(-1)` `k_(1) = 3 xx 10^(-4), k_(2) = ? implies log (k_(2))/(k_(1)) = (E_(a))/(2.303R) [(1)/(T_(1)) - (1)/(T_(2))]` log `(k_(2))/(3 xx 10^(-4)) = (104.4 xx 10^(3) J mol^(-1))/(2.303 xx (8.314 J k^(-1) mol^(-1)))` `[(1)/(298K) -(1)/(T)]` As `T implies OO (1)/(T) to 0` `therefore` log `(k_(2))/(3 xx 10^(-4)) = (104.4 xx 10^(3) J mol^(-1))/(2.303 xx 8.314 xx 298)` log `(k_(2))/(3 xx 10^-4) = 18.297 , (k_(2))/(3 xx 10^(-4)) = 1.98 xx 10^(18)` or `k_(2) = (1.98 xx 10^(18)) xx (3 xx 10^(-4)) = 6 xx 10^(14) s^(-1)`.

Discussion

24791.	The rate constant, the activation energy and the arrhenius paremeter of a chemical reaction at 25^(@) are 3.0times10^(-4)s^(-), 104.4kJ*mol^(-1) and 6.0times10^(14)s^(-1) respectively. The value of the rate constant as T to infty is
Answer» `2.0times10^(18)s^(-1)` `6.0times10^(14)s^(-1)` Infinity `3.6times10^(30)s^(-1)` Solution :`T_(2)=T(SAY), T_(1)=25^(@)=298K, E_(a)=104.4kJmol^(-1)=104.4times10^(3)Jmol^(-1)` `k_(1)=3times10^(-4), k_(2)=?` `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `"log"(K_(2))/(3times10^(-4))=(104.4times10^(3)Jmol^(-1))/(2.303times(8.314JK^(-1)mol^(-1))[(1)/(298)-(1)/(T)]` As `T to INFTY, (1)/(T)to 0` `therefore "log"(K_(2))/(3times10^(-4))=(104.4times10^(3)Jmol^(-1))/(2.303times8.314times298)` `"log"(k_(2))/(3times10^(-4))=18.297, (k_(2))/(3times10^(-4)=1.98times10^(18)` `k_(2)=6times10^(14)s^(-1)`

Discussion

24792.	The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are 3.5 x 10^(-5)s^(-1), 110.5 kJ mol^(-1) and 6.0 x 10^(14) s^(-1) respectively The value of rate constant at T to is
Answer» `1.5 X 10^5 s^(-1)` `2.0 x 10^(10) s^(-1)` `6.0 x 10^(14) s^(-1)` Zero Answer :C

Discussion

24793.	The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25^(@)C are 3.0xx10^(-4)s^(-1), 104.4kJmol^(-1) and 6.0xx10^(14)s^(-1) respectively, thevalue of the rate constant as T to oo is
Answer» `2.0xx10^(18)s^(-1)` `6.0xx10^(14)s^(-1)` `oo` `3.6xx10^(30)s^(-1)` SOLUTION :`K=Ae^(-E_(a)//RT)` When `T to oo` `k to A` `A=6xx10^(14)s^(-1)`

Discussion

24794.	The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25^(@)C are 3.0xx10^(-4)s^(-1), 104.4" kJ mol"^(-1)," and "6.0xx10^(14)s^(-1) respectively. The value of the rate constant as T to oo is
Answer» `2.0xx10^(18)s^(-1)` `6.0xx10^(14)s^(-1)` Infinity `3.6xx10^(30)s^(-1)` Solution :(B) : `K="A e"^(-E_(a)//"RT")." As "T to oo,k=A=6.0xx10^(14)s^(-1).`

Discussion

24795.	The rate constant, the activation energy and frequency factor of a chemical reaction at 25^(#)C are 3.0xx10^(-4)s^(-1), 104.4"kJ mol"^(-1) and 6.0xx10^(14)s^(-1) respectively. What is the value of the rate constant when T rarr oo ?
Answer» Solution :Formula : `k_(1)=Ae^(-E_(a//RT))` Given : Rate CONSTANT, `k_(1)-3.0xx10^(-4)s^(-1)` FREQUENCY factor, `A = 6.0xx10^(14)s^(-1)` Activation energy, `E_(a)=104.4 kJ` `= 104400 J` Temperatures, `T_(1)=25^(@)C , T_(2) = oo` `k_(2)=A` `k_(2)=A.e^(-(E_(a))/(RT))` `k_(2)=6.0xx10^(14)xx e^(-1(104400)/(8.314xx oo)` `k_(2)=6.0xx10^(14)xx1 (therefore e (-E_(a))/(R oo)=1)` `therefore` The rate constant at `T rarr oo` is `6.0xx10^(14)s^(-1)`.

Discussion

24796.	The rate constant, the activation and Arrhenius parameter of a chemical reaction at 25^(@)C are 3xx10^(-4)s^(-1), 104.4 kJ mol^(-1) and 6xx10^(14)s^(-1) respectively. The value of the rate constant at T to oo is
Answer» `2xx10^(18)s^(-1)` `6XX10^(14)s^(-1)` `OO` `3.6xx10^(30)s^(-1)` Answer :B

Discussion

24797.	The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25^@C are 3.0xx10^-4S^-1, 104.4 kj mol^-1 and 6.0xx10^14 s^-1 respectively. The value of the rate constant as Tto oo is:
Answer» `2.0xx10^18 s^-1` `6.0xx10^14 s^-1` INFINITY `3.6xx10^30 s^-1` ANSWER :B

Discussion

24798.	The rate constant of zero order reaction is 2xx10^(-2) mole L /sec. If the concentration of the reactant after 25 sec is 0.5 M, the initial concentration is
Answer» `2xx10^(-2)M` 1M 2M `2xx10^(2)M` SOLUTION :`C_(0)-C_(t)=Kxxt,C_(0)-0.5=2xx10^(-2)xx25,C_(0)=0.50+0.5,C_(0)=1M`

Discussion

24799.	The rate constant of the reaction Ato B is 0.6xx10^(-3) mole per second.If the concentration of A is 5 M ,then concentration of B after 20 minutes is :
Answer» 0.36 M 0.72 M 1.08 M 3.60 M Solution :For ZERO ORDER REACTION : x=k.t=`06xx10^(-3)xx20xx60=0.72m`

Discussion

24800.	The rate constant of the reaction at temperature 200 K is 10 times less than the rate constant at 400 K.What is the activation energy of the reaction ?
Answer» `1842.4" R"` `921.2" R"` `460.6" R"` `230.3" R"` SOLUTION :At `T_1=200 K,` if `k_1=k,` then at `T_2 = 400 K, k_2=20 k ` `log""(10k)/k=E_a/(2303R)(400-200)/(400xx200)" or "E_a=9212R`.

Discussion

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