94420 + Interview Questions in Chemistry in Class 12 Page 498 InterviewSolution

24851.	The rate constant of a first-order reaction of the type 2AtoP" is "1.5xx10^(-1)s^(-1). The half-life period of the reaction is
Answer» `2.31xx10^(3)s` `4.62xx10^(3)s` `9.24xx10^(3)s` `1.5xx10^(-4)s` ANSWER :A

Discussion

24852.	The rate constant of a particular reaction doubles when the temperature changes from 300K to 310 K, calculate the energy of activation.
Answer» Solution :DATA: constant at 300 K is `K_(1)` Rate constant at 310 K is `K_(2), "" K_(2) = 2K, "" R = 8.314 J K^(-1)MOL^(-1)` Formula: `(k_(2))/(k_(1)) = (Ea)/(2.303R) {(T_(2) - T_(1))/(T_(1)T_(2))}` Substitution: `log ""(2k_(1))/(K_(1)) = (E_(a))/(8.314 xx 2.303) [(310 - 300)/(300 xx 310)]` `log 2 = (E_(a))/(8.314xx2.303)xx(10)/(300xx310)` `E_(a) = (0.3010xx8.314xx2.303xx300xx310)/(10)` `E_(a) = 53.598 kJ mol^(-1)`

Discussion

24853.	The rate constant of a first order reaction is given by K=2.1xx10^(10)"exp"(-2700//T). It meas that
Answer» Greater the activation ENERGY, greater will be the TEMPERATURE coefficient `((K_(T+10))/(K_(T)))` Freququency factor of the reaction is `2.1xx10^(10)` Half life increases with increase of temperature Activation energy of reaction is 5.4 is K.Cal Solution :a. `K=2.1xx10^(10)XXE^(((-2700)/T)),K=Axxe^(((-EA)/(RT))),logk=logA-((Ea)/(2.303Rt))` b. `A=2.1xx10^(10)` c. If `Tuarr(t_(1//2))darr` so `(t_(1/2)alpha1/T)` d. `(-2700)/T=(-epsilona)/(RT),epsilona=2700xxR=(2700xx2)/1000=5.4` Kcal

Discussion

24854.	The rate constant of a first order reaction is k =7.39 xx 10^(-5) " s"^(-1). Find the half-life of the reaction.
Answer» SOLUTION :If `t_(1/2)=0.93/K=0.693/(7.39xx10^-5)=(0.693xx10^5)/7.39=9.35xx10^3sec.`

Discussion

24855.	The rate constant of a first order reaction is doubled when the temperature is increased from 20^(@)C to 25^(@)C . How many times the rate constant will increase if the temperature is raised from 20 ^(@)Cto 40^(@)C
Answer» 4 8 16 32 Solution :`5^(@)C` rise CAUSES increase of 2 time . `20^(@)C` will cause `(2)^(4) = 16` TIMES .

Discussion

24856.	The rate constant of a first order reaction is given by
Answer» `K=(2.303)/(t//2)"log"([R]_(0))/([R])` `k=(2.303)/(t//2)"log"([R])/([R]_(0))` `k=(2.303)/(t)"log"([R]_(0))/([R])` `k=(2.303)/(t)"log"([R])/([R]_(0))` ANSWER :C

Discussion

24857.	The rate constant of a first order reaction is 60s^(-1). How much time will it take to reduce the intial concentration of the reactant to its 1//16th value ?
Answer» SOLUTION :`t=(2303)/(k)log(a)/(a//16)=(2303)/(60S^(-1))log16=4*62xx10^(-2)s.`

Discussion

24858.	The rate constant of a first order reaction is 3 xx 10^(-5) sec^(-1)M If initial concentration is 0.10M, the initial rate is (Ms^-1)
Answer» `3 xx10^(-6)` `3 xx 10^(-5)` `3 xx 10^(-8)` `3 xx 10^(-7)` Solution :RATE `=K xx[A]= 3xx 10^(-6)xx 0.1=3xx10^(-7)Ms^(-1)`

Discussion

24859.	The rate constant of a first-order reaction is 3 xx 10^(-6) per second. If the initial concentration is 0.10 m, the initial rate of reaction is
Answer» `3 XX 10^(-5) ms^(-1)` `3 xx 10^(-6) ms^(-1)` `3 xx 10` `3 xx 10^(-7) ms^(-1)` Solution :Given: Rate constant of the first-order reaction `(K) = 3 xx 10^(-6)` per sec and INITIAL concentration `[A] = 0.10 M`. We know that initial rate constant `K[A]=3 xx 10^(-6) xx 0.10 = 3 xx 10^(-7) ms^(-1)`

Discussion

24860.	The rate constant of a first order reaction is 4X10^-3sec^-1Ata reactant concentration of 0.02 M, the rate of reaction would be:
Answer» `8X10^-5Msec^-1` `4X10^-3Msec^-1` `2X10^-1Msec^-1` `4X10^-1Msec^-1` ANSWER :A

Discussion

24861.	The rate constant of a first order reaction is 3 xx 10^(-6) per second . If the initial concentration is 0.10 M , the initial rate of reaction
Answer» `3 XX 10^(-5) M s^(-1)` `3 xx 10^(-6) M s^(-1)` `3 xx 10^(-8) Ms^(-1)` `3 xx 10^(-7) M s^(-1)` Solution :Given : Rate constant of the FIRST order reaction `(k) = 3 xx 10^(-6)` per SEC and initial concentration `[A] = 0.10` M . We know that initial rate constant `k [A] = 3 xx 10^(-6) xx 0.10 = 3 xx 10^(-7) Ms^(-1)`.

Discussion

24862.	The rate constant of a first order reaction is 1.15 xx 10^(-3)s^(-1). Calculate its half life period (t1//2).
Answer» SOLUTION :`T_((1)/(2)) = (0.693)/(K) = (0.693)/(1.15 xx 10^(-3)) = 602.65 s`

Discussion

24863.	The rate constant of a first order reaction increases from 2xx10^(-2) to 8xx10^(-2) when the temperature changes from 300 K to 320 K. Calculate the energy of activation.
Answer» SOLUTION :We shall use the Arrhenius EQUATION `"log"(k_(2))/(k_(1))=(E_(a))/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]` Given : `k_(1)=2xx10^(-2), k_(2)=8xx10^(-2), T_(1)=300K, T_(2)=320K and R=8.314JK^(-1)"MOL"^(-1)` SUBSTITUTING these values in the above equation, we get `"log"(8xx10^(-2))/(2xx10^(-2))=(E_(a))/(2.303xx8.314)[(320-300)/(320xx300)]` `or log 4=(E_(a))/(19.147)xx(20)/(96000) or 0.6021=(E_(a))/(19.147)xx(1)/(4800)` `or E_(a)=0.6021xx19.147xx4800 or E_(a)=55.34"kJ mol"^(-1)`

Discussion

24864.	The rate constant of a first order reaction at 500 K is 2.35 xx 10^(-5) s^(-1). At what temperature, the half-life of the reaction is 256 min?
Answer» ANSWER :513 K

Discussion

24865.	The rate constant of a first order reaction of 0.0693"min"^(-1). What is the time (in min) required for reducing an initial concentration of 20 mole "lit"^(-1) to 2.5 mole "lit"^(-1)?
Answer» 40 10 20 30 Answer :D

Discussion

24866.	The rate constant of a first order reaction at 300 K and 310 K are respectively 1.2 xx 10^(3) s^(-1) and 2.4 xx 10^(3) s^(-1). Calculate the energy of activation. (R = 8.314 J K^(-1) mol^(-1))
Answer» SOLUTION :`LOG.(K_(2))/(K_(1)) = (E_(a))/(2.303 R) [(T_(2)-T_(1))/(T_(1)T_(2))]` `log.(2.4 XX 10^(-3))/(1.2 xx 10^(-3)) = (E_(a))/(2.303 xx 8.314) [(310 - 300)/(300 xx 310)]` `E_(a) = 53598 J` or `53.598 K J`.

Discussion

24867.	The rateconstantof a firstorderreaction at 25^(@) C is 0.24 s^(-1) ifthe energyof activation ofthe reactionis 88kJ mol^(-1) atwhattemperaturewouldthisreactionhave rateconstantof 4xx 10^(-2)s^(-1)
Answer» SOLUTION :Temperature=283 . 6 K

Discussion

24868.	The rate constant of a first order reaction at 27^@C is 10^(-3)mi n^(-1). The temperature coefficient of this reaction is 2. What is the rate constant (in min^(-1)) at 17^@C for this reaction?
Answer» `10^(-3)` `5xx10^(-4)` `2XX10^(-3)` `10^(-2)` Answer :B

Discussion

24869.	The rate constant of a certain first order reaction is 200S^(-1). What is its half life period ?
Answer» SOLUTION :`t_(1//2) = (0.693)/(K) [K=200S^(-1)] = (0.693)/(200s^(-1))` `t_(1//2) =0.003465` SECONDS.

Discussion

24870.	The rate constant of a a reaction depends on
Answer» EXTENT of reaction Time of reaction TEMPERATURE Initial COCENTRATION of the reactants Solution :RATE DEPENDS on temperature.

Discussion

24871.	The rate constant of 1st order is 0.0005min^(-1) . Find its half life period.
Answer» SOLUTION :`t_(1//2)=0.6931/K=0.6931/0.0005`=1386minutes

Discussion

24872.	The rate constant K_a of one reaction is found to be double than that of rate constant K_(a)'' of another reaction. Then the relation between the corresponding activation energies of the two reactions E_(a) ' and E_(a) '' can be represented as,
Answer» `E_1gtE_2` `E_1ltE_2` `E_1=E_2` NONE of the above Answer :D

Discussion

24873.	The rate constant K_(1) of a reaction is found to be double that of rate constant K_(2) of another reaction. The relationship between corresponding activation energies of the two reaction at same temperature (E_(1) "and" E_(2)) can be represented as:
Answer» `E_1 GT E_2` `E_1 LT E_2` `E_1 = E_2` NONE of these Answer :D

Discussion

24874.	The rate constantK_(1) of a reaction is found to be double that of rate constant K_(2) of another reaction. The relationship between corresponding activation energies of the two reactions E_(1) and E_(2) can be represented as
Answer» `E_(1)gtE_(2)` `E_(1)ltE_(2)` `E_(1)=E_(2)` `E_(1)=2E_(2)` Answer :B

Discussion

24875.	The rate constant K_(1) and K_(2) for two different reactions are are 10^(16)e^(-2000//T) and 10^(15)e^(-1000//T), respectively. The temperature at which K_(1)=K_(2) is
Answer» `1000K` `2000K` `(1000)/(2.303)K` `(2000)/(2.303)K` ANSWER :C

Discussion

24876.	The rate constant k , for the reaction N_(2)O_(5) to 2NO_(2) (g) + (1)/(2) O_(2)(g) correspond to concentration of N_(2)O_(5) initially and at time t .
Answer» `[N_(2)O_(5)]_(t) [N_(2)O_(5)]_(0) + KT` `[N_(2)O_(5)]_(0) = [N_(2)O_(5)]_(t) e^(kt)` `log_(10) [N_(2)O_(5)]_(t) = log_(10)[N_(2)O_(5)]_(0) - kt` ln `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))= kt` Solution :Rate constant = `2.3 xx 10^(-2) sec^(-1)` It means it is a FIRST order reaction (because UNIT of rate constant is `sec^(-1)`) For first order reaction k = `(1)/(t)` ln `(a)/(a-x)` Kt = ln `(a)/(a-x)` = ln `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))`.

Discussion

24877.	The rate constant (K') of one reaction is double of the rate constant (K") of another reaction. Then the relationship between the corresponding activation energies of the two reactions (E_(a)^(') "and" E_(a)^('')) will be
Answer» `E_(a) GT E_(a)` `E_(a) = E_(a)` `E_(a) lt E_(a)` `E_(a) = 4E_(a)` Solution :As `K' gt K'' . E'_(a) lt E''_(a)` (GREATER the rate CONSTANT , less is the activation energy).

Discussion

24878.	The rate constant (k') of one of the reaction is found to be double that of the rate constant (k") of another reaction. Then the relationship between the corresponding activation energies of the two reactionscan be represented as
Answer» `(E_agt E_a )` `(E_a LT E_a )` `(E_a = E_a)` none of these. Solution :As `k= Ae^(-E_a//RT)` THEREFORE if k. GT k" than `E_a. lt E_a...`

Discussion

24879.	The rate constant (K) for the reaction 2Ararr Product was found to be 2.5X10^-5Lmol^-1sec^(-1) after15 sec, 2.5X10^-5 L mol^(-1)sec^(-1) after 30 sec and2.55X10^-5 L mol^-1 sec^-1 after 50sec. The order of reaction is:
Answer» 2 3 Zero 1 Answer :A

Discussion

24880.	The rate constant (K) for the reaction2A + B toProduct was found to be 2.5 xx 10^(-5) litre mol^(-1) sec^(-1) after 15 sec , 2.60 xx 10^(-5) litre mol^(-1) sec^(-1) after 30 sec and 2.55 xx 10^(-5) litre mol^(-1) sec^(-1) after 50 sec . The order of reaction is
Answer» Solution :By OBSERVING the UNIT of RATE constant `MOL^(1-N) "litre"^(n-1) "sec"^(-1) = mol^(-1) "litre" sec^(-1)` 1 - n= `-1 implies n = 2`

Discussion

24881.	The rate constant (K) for the reactio. 2A+Bto Products was found to be 2.5xx10^(-5)litre"mol"^(-1)"sec"^(-1) after 15 sec, 2.60xx10^(-5) litre"mol"^(-1)"sec"^(-1) after 30 sec and 2.55xx10^(-5) litre "mol"^(-1)"sec"^(-1) after 50 sec. The order of reaction is
Answer» 2 3 zero 1 Solution :It is 2ND ORDER

Discussion

24882.	The rate constant is numerically the same for three reactions of first, second and third order respectively. Which of the following is correct ?
Answer» if [A]= 1 then `r_1 =r_2 = r_3` if [A] lt 1 then `r_1 gt r_2 gt r_3` if [A] gt 1 then `r_1 lt r_2 lt r_3` all SOLUTION :`r_1=k[A]` … (i) for 1st ORDER reaction. `r_2=k[A]^2`…(ii) for 2nd order reaction. `r_3=k[A]^3`…(iii) for 3rd order reaction. When RATE constant k is the same for these reaction, it is clear that, `r_1=r_2=r_3` when [A] = 1 mol `L^(-1)` `r_1 gt r_2 gt r_3` When [A] `lt 1` `r_1 lt r_2 lt r_3` When [A] `gt 1`.

Discussion

24883.	The rate constant is numerically same for three reactions of first second and third order respections of first second and third order respectively .Which of the following is true for the rate of these three reactions if concentration of the reactant is same and larger than 1 M ?
Answer» `r_(1) = r_(2)=r_(3)` `r_(1) lt r_(2) lt r_(3)` `r_(1) gtr_(2) gt r_(3) ` no definte relation Solution :(B) `r_(1)=k_(1)[A] , r_(2) = k_(2)[A]^(2) , r_(3) =k_(3) [A]^(3)` if `k_(1)= k_(2) = k_(3) ` and `[A] gt 1 M` `r_(3) gt r_(2) gt r_(1)` or `r_(1)lt r_(2) lt r_(3)`

Discussion

24884.	The rate constant is numerically same for three reactions of 1st, 2nd and 3rd order respectively. If conc, of the reactant is more than 1M, which one is true for the rates of the three reactions ?
Answer» `r_1=r_2=r_3` `r_1ltr_2ltr_3` `r_1gtr_2gtr_3` NONE of these Answer :B

Discussion

24885.	The rate constant is given by the equation, k=P.Ze^(-E_a//RT) Which factor should register a decrease for the reaction to proceed more rapidly?
Answer» T Z `E_a` P Solution :`E_a`

Discussion

24886.	The rate constant is given by the equation K=Ae^(-Ea//RT)which factor should register a decrease for the reaction to proceed rapidly:
Answer» T Z A `E_a ` ANSWER :D

Discussion

24887.	The rate constant is given by the equation k = pze^(-E//RT) . Which factor should register a decrease for the reaction to proceed more rapidly
Answer» T Z E P Solution :When k increases , rate of REACTIONALSO increases , `k = (pz)/(e^(E//RT))` for k to increase p , z , T should increase and E should decrease . (e = 27).

Discussion

24888.	The rate constant is given by the equation K=Ae^(-Ea//RT). Which factor should register a decrease for the reaction to proceed more rapidly:
Answer» T Activated complex can be isolated A and T `E_(a)` SOLUTION :rate `ALPHA(1/("ACTIVATION ENERGY"))`

Discussion

24889.	Therate constant for two parallel reactions were found to be 1.0xx10^(-2)dm^(3)mol^(-1)and 3.0xx10^(-2)dm^(3)mol^(-1)s^(-1). Ifthe corresponding energies of activation of the correspoing energies of activation of the parallel reaction are 60.0 Kj mol^(-1) and 70.0 KJ mol^(-1) respectively ,what is the apparent overall energy of activation ?
Answer» `130.0KJmol^(-1)` `67.5KJmol^(-1)` `100.0KJmol^(-1)` `65.0KJmol^(-1)` ANSWER :B

Discussion

24890.	The rate of a reaction triples when temperature changes from 20^(@)C to 50^(@)C. Calculate the energy of activation.
Answer» `50` `54` `100` `34` Answer :B

Discussion

24891.	The rate constant for the second order reaction was shown by log k=12-(3163)/(T) . Concentration and time were in molL^(-1) and min. respectively. Calculate the half-life at 43.3^(@)C. The initial concentration of each reactant was 0.001molL^(-1)
Answer» ANSWER :`10` MINUTES

Discussion

24892.	The rate constant for the reaction , 2N_(2)O_(5) to 4NO_(2) + O_(2) is 3 xx 10^(-5) sec^(-1) . If the rateis 2.40 xx 10^(-5) mol "litre"^(-1) sec^(-1) . Then the concentration of N_(2)O_(5) (in mol litre^(-1) ) is
Answer» `1.4` 1.2 `0.04` `0.8` Solution :Rate = `k (N_(2)O_(5))` HENCE `2.4 xx 10^(-5) = 3.0 xx 10^(-5) (N_(2)O_(5))` or `(N_(2)O_(5)) = 0.8` mol `L^(-1)`

Discussion

24893.	The rate constant for the reaction 2N_2O_5rarr2N_2O_4 +O_2 is 3xx10^-5 sec^-1. If the rate is 2.4xx10^-5 M sec^-1, the concentration of N_2O_5 is
Answer» 1.4 M 0.04M 1.2 M 0.8 M Answer :D

Discussion

24894.	The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.2 - (1.0xx10^(4))/(T) Calculate E_(a) for this reaction and rate constant k if its half-life period be 200 minutes. [Given : R=8.314 JK^(-1)"mol"^(-1)]
Answer» Solution :A general equation relating log k and `E_(a)` is GIVEN by `log k =log A-(E_(a))/(2.303RT) ""…(i)` The given equation is `log k=14.2 -(1.0xx10^(4))/(T)""…(ii)` On comparing equations (i) and (ii), we have `(E_(a))/(2.303RT)=(1.0xx10^(4))/(T)` or `E_(a)=1.0xx10^(4)xx2.303 xx R "" ...(III)` Substituting the value of R in equation (iii), we have `E_(a)=1.0xx10^(4)xx2.303xx8.314=191471.4"J mol"^(-1)` Half-life PERIOD for a first order reaction `t_(1//2)=(0.693)/(k)` or `ko=(0.693)/(t_(1//2))=(0.693)/(200"min")=0.0034" min"^(-1)`

Discussion

24895.	The rate constant for the first order decomposition of N_(2) O_(5) is given by the following equation: k=(2.5 xx 10^(14) s^(-1) ) e^((-25000K)//T) Calculate E_a for this reaction and rate constant if itshalf-life period be 300 minutes.
Answer» Solution :`K=AE^(-E_(a) //RT)""…(i)` Also `k=(2.5 xx 10^(14) s^(-1) ) e^((-25000K)//T)""…(ii)` From (i) and (ii), we have `Ae^(-E_(a)//RT)=(2.5 xx 10^(-14) s^(-1) ) e^((-25000K)//T)` Equating similar terms in the above equation, we have `(-E_(a))/(RT) = (-25000K)/( T)` or `E_(a) = R xx 25000 K` or `E_(a) =8.314 J K^(-1) mol^(-1) xx 25000 K` or `E_(a) = 207.85 kJ "mol"^(-1)` For a first ORDER reaction `t_(1//2) = (0.693)/(k)` or `k=(0.693)/( t_(1//2))= (0.693)/(300 "minutes") = 2.31 xx 10^(-3) "minutes"^(-1)` or `k=(0.693)/( 300 xx 60S) = (0.693)/(18000 s) = 3.85 xx 10^(-5) s^(-1)`.

Discussion

24896.	The rate constant for the reaction : 2N_(2) O_(5) rarr 4NO_(2) +O_(2) is 3.0xx10^(-5) "sec"^(-1) . If the rate is 2.40xx10^(-5) mol "litre"^(-1) "sec"^(-1) then the concentration of N_(2)O_(5)(in mol "litre"^(-1) ) is :
Answer» `1.4` `1.2` `0.02` `0.8` Solution :(D) RATE`= K[N_(2)O_(5)]` or `[N_(2)O_(5)] = ("Rate")/(k)` `= (2.40xx10^(-5))/(3.0xx10^(-5))` =0.8

Discussion

24897.	The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.34 -1.25 xx 10^(4)K//T Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes?
Answer» Solution :According to Arrhenius EQUATION `K=Ae^(-E_(a)//RT)` Taking LOGARITHM of both sides, In k`="In A"-(E_(a))/(RT) or logk=logA-(E_(a))/(2.303RT)` Equating similar terms in the two equations `(E_(a))/(2.303RT)=(1.25xx10^(4)K)/(T) or E_(a)=2.303R xx 1.25 xx 10^(4)K` or `E_(a)=2.303 xx (8.314JK^(-1)"mol"^(-1))xx1.25xx10^(4)K=239.34"kJ mol"^(-1)` When `t_(1//2)=256"min", k=(0.693)/(256xx60s)=4.51xx10^(-5)s^(-1)` Substituting this value in the given equation, we GET `log(4.51xx10^(-5))=14.34-(1.25xx10^(4)K)/(T)` or `(-5+0.6542)=14.34-(1.25xx10^(4)K)/(T)` or `(1.25xx10^(4)K)/(T)=18.6858 or T=669K`.

Discussion

24898.	The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.34-1.25xx10^(4)"K"//"T" Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes ?
Answer» Solution :According to ARRHENIUS equation, `k="A e"^(-E_(a)//"RT")` or `lnk=lnA-(E_(a))/(RT)" or "logk=logA-(E_(a))/(2.303"RT")` Comparing with the given equation, `(E_(a))/(2.303" RT")=(1.25xx10^(4)"K")/(T)` or `E_(a)=2.303" R "xx1.25xx10^(4)" K"=2.303xx(8.314" JK"^(-1)" mol"^(-1))xx1.25xx10^(4)" K"=239.347" kJ mol"^(-1)` When `t_(1//2)=256" min",k=(0.693)/(256xx60"s")=4.51xx10^(-5)s^(-1)` Substituting this value in the given equation, `log(4.51xx10^(-5))=14.34-(1.25xx10^(4)" K")/(T),i.e.,(-5+0.6542)=14.34-(1.25xx10^(4)"K")/(T)` or `(1.25xx10^(4)K)/(T)=18.6858" or "T=669" K".`

Discussion

24899.	The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation: log k=14.34-1.25xx10^(4)K//T [Where unit of k is s^(-1)] Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes?
Answer» SOLUTION :Calculation of `E_(a)`:The REACTION is first order. The decomposition of `H_(2)O_(2)` occurs in reaction .It is given that, log k=`14.34-1.25xx10^(4)K//T` ….(a) log k=`log A-(E_(a))/(2.303 RT)` (Arrhenius(b)) Comparing equation (a) and (b), log k and 14.34 =log A is accepted and removing it, So,`(E_(a))/(2.303RT)=(1.25xx10^(4))/(T)k` `therefore E_(a)=1.25xx104 kxx2,303xx8.314 JK^(-1) mol^(-1)` Calculation of decomposition constant k on the base of `t_((1)/(2))`: `k=(0.693)/(t)` Where ,`t_((1)/(2))=256` MINUTE `=(0.693)/(256xx60s)` Calculation of `t_((1)/(2))` TEMPERATURE: Given equation :log k=14.34-`(1.25xx10^(4)K)/(T)` `therefore log(4.51xx10^(-5))=14.34-(1.25xx10^(4)K)/(T)` `therefore -4.3458-14.34=-(1.25xx10^(4)K)/(T)` `therefore-18.6858=-(1.25xx10^(4)K)/(T)` `therefore T=(1.25xx10^(4)K)/(18.6858)=668.96 K=669K`

Discussion

24900.	The rate constant for the decomposition of N_(2)O_(5) at various temperatures is given below : {:("T"//""^(@)C,,,0,,,20,,,40,,,60,,,80),(10^(5)xxk//s^(-1),,,0.0787,,,1.70,,,25.7,,,178,,,2140):} Draw a graph between ln k and 1//T and calculate the values of A and E_(a). Predict the rate constant at 30""^(@)C and 50""^(@)C.
Answer» SOLUTION :PROCEED as in Solved PROBLEM 5, page `4//119`.

Discussion

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The rate constant of a first order reaction is given by

The rate constant of a first order reaction is 60s^(-1). How much time will it take to reduce the intial concentration of the reactant to its 1//16th value ?

The rate constant of a first order reaction is 3 xx 10^(-5) sec^(-1)M If initial concentration is 0.10M, the initial rate is (Ms^-1)

The rate constant of a first-order reaction is 3 xx 10^(-6) per second. If the initial concentration is 0.10 m, the initial rate of reaction is

The rate constant of a first order reaction is 4X10^-3sec^-1Ata reactant concentration of 0.02 M, the rate of reaction would be:

The rate constant of a first order reaction is 3 xx 10^(-6) per second . If the initial concentration is 0.10 M , the initial rate of reaction

The rate constant of a first order reaction is 1.15 xx 10^(-3)s^(-1). Calculate its half life period (t1//2).

The rate constant of a first order reaction increases from 2xx10^(-2) to 8xx10^(-2) when the temperature changes from 300 K to 320 K. Calculate the energy of activation.

The rate constant of a first order reaction at 500 K is 2.35 xx 10^(-5) s^(-1). At what temperature, the half-life of the reaction is 256 min?

The rate constant of a first order reaction of 0.0693"min"^(-1). What is the time (in min) required for reducing an initial concentration of 20 mole "lit"^(-1) to 2.5 mole "lit"^(-1)?

The rate constant of a first order reaction at 300 K and 310 K are respectively 1.2 xx 10^(3) s^(-1) and 2.4 xx 10^(3) s^(-1). Calculate the energy of activation. (R = 8.314 J K^(-1) mol^(-1))

The rateconstantof a firstorderreaction at 25^(@) C is 0.24 s^(-1) ifthe energyof activation ofthe reactionis 88kJ mol^(-1) atwhattemperaturewouldthisreactionhave rateconstantof 4xx 10^(-2)s^(-1)

The rate constant of a first order reaction at 27^@C is 10^(-3)mi n^(-1). The temperature coefficient of this reaction is 2. What is the rate constant (in min^(-1)) at 17^@C for this reaction?

The rate constant of a certain first order reaction is 200S^(-1). What is its half life period ?

The rate constant of a a reaction depends on

The rate constant of 1st order is 0.0005min^(-1) . Find its half life period.

The rate constant K_a of one reaction is found to be double than that of rate constant K_(a)'' of another reaction. Then the relation between the corresponding activation energies of the two reactions E_(a) ' and E_(a) '' can be represented as,

The rate constant K_(1) of a reaction is found to be double that of rate constant K_(2) of another reaction. The relationship between corresponding activation energies of the two reaction at same temperature (E_(1) "and" E_(2)) can be represented as:

The rate constantK_(1) of a reaction is found to be double that of rate constant K_(2) of another reaction. The relationship between corresponding activation energies of the two reactions E_(1) and E_(2) can be represented as

The rate constant K_(1) and K_(2) for two different reactions are are 10^(16)e^(-2000//T) and 10^(15)e^(-1000//T), respectively. The temperature at which K_(1)=K_(2) is

The rate constant k , for the reaction N_(2)O_(5) to 2NO_(2) (g) + (1)/(2) O_(2)(g) correspond to concentration of N_(2)O_(5) initially and at time t .

The rate constant (K') of one reaction is double of the rate constant (K") of another reaction. Then the relationship between the corresponding activation energies of the two reactions (E_(a)^(') "and" E_(a)^('')) will be

The rate constant (k') of one of the reaction is found to be double that of the rate constant (k") of another reaction. Then the relationship between the corresponding activation energies of the two reactionscan be represented as

The rate constant (K) for the reaction 2Ararr Product was found to be 2.5X10^-5Lmol^-1sec^(-1) after15 sec, 2.5X10^-5 L mol^(-1)sec^(-1) after 30 sec and2.55X10^-5 L mol^-1 sec^-1 after 50sec. The order of reaction is:

The rate constant (K) for the reaction2A + B toProduct was found to be 2.5 xx 10^(-5) litre mol^(-1) sec^(-1) after 15 sec , 2.60 xx 10^(-5) litre mol^(-1) sec^(-1) after 30 sec and 2.55 xx 10^(-5) litre mol^(-1) sec^(-1) after 50 sec . The order of reaction is

The rate constant (K) for the reactio. 2A+Bto Products was found to be 2.5xx10^(-5)litre"mol"^(-1)"sec"^(-1) after 15 sec, 2.60xx10^(-5) litre"mol"^(-1)"sec"^(-1) after 30 sec and 2.55xx10^(-5) litre "mol"^(-1)"sec"^(-1) after 50 sec. The order of reaction is

The rate constant is numerically the same for three reactions of first, second and third order respectively. Which of the following is correct ?

The rate constant is numerically same for three reactions of first second and third order respections of first second and third order respectively .Which of the following is true for the rate of these three reactions if concentration of the reactant is same and larger than 1 M ?

The rate constant is numerically same for three reactions of 1st, 2nd and 3rd order respectively. If conc, of the reactant is more than 1M, which one is true for the rates of the three reactions ?

The rate constant is given by the equation, k=P.Ze^(-E_a//RT) Which factor should register a decrease for the reaction to proceed more rapidly?

The rate constant is given by the equation K=Ae^(-Ea//RT)which factor should register a decrease for the reaction to proceed rapidly:

The rate constant is given by the equation k = pze^(-E//RT) . Which factor should register a decrease for the reaction to proceed more rapidly

The rate constant is given by the equation K=Ae^(-Ea//RT). Which factor should register a decrease for the reaction to proceed more rapidly:

The rate of a reaction triples when temperature changes from 20^(@)C to 50^(@)C. Calculate the energy of activation.

The rate constant for the second order reaction was shown by log k=12-(3163)/(T) . Concentration and time were in molL^(-1) and min. respectively. Calculate the half-life at 43.3^(@)C. The initial concentration of each reactant was 0.001molL^(-1)

The rate constant for the reaction , 2N_(2)O_(5) to 4NO_(2) + O_(2) is 3 xx 10^(-5) sec^(-1) . If the rateis 2.40 xx 10^(-5) mol "litre"^(-1) sec^(-1) . Then the concentration of N_(2)O_(5) (in mol litre^(-1) ) is

The rate constant for the reaction 2N_2O_5rarr2N_2O_4 +O_2 is 3xx10^-5 sec^-1. If the rate is 2.4xx10^-5 M sec^-1, the concentration of N_2O_5 is

The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.2 - (1.0xx10^(4))/(T) Calculate E_(a) for this reaction and rate constant k if its half-life period be 200 minutes. [Given : R=8.314 JK^(-1)"mol"^(-1)]

The rate constant for the first order decomposition of N_(2) O_(5) is given by the following equation: k=(2.5 xx 10^(14) s^(-1) ) e^((-25000K)//T) Calculate E_a for this reaction and rate constant if itshalf-life period be 300 minutes.

The rate constant for the reaction : 2N_(2) O_(5) rarr 4NO_(2) +O_(2) is 3.0xx10^(-5) "sec"^(-1) . If the rate is 2.40xx10^(-5) mol "litre"^(-1) "sec"^(-1) then the concentration of N_(2)O_(5)(in mol "litre"^(-1) ) is :

The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.34 -1.25 xx 10^(4)K//T Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes?

The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.34-1.25xx10^(4)"K"//"T" Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes ?

The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation: log k=14.34-1.25xx10^(4)K//T [Where unit of k is s^(-1)] Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes?