94420 + Interview Questions in Chemistry in Class 12

1.	The activity of transition elements is related to their -
Answer» variable OXIDATION states surface area complex forming ability magnetic moment ANSWER :A::B::C

Discussion

2.	The activity of the hair of an Egyptian mummy is 7 disintegrations minute^(-1). Find the age of mummy. Given t_(1//2) of .^(14)C is 5770 yearand disintegration rate offresh sample of .^(14)C is 14 disintegrations minute^(-1).
Answer» SOLUTION :`5770` YEARS

Discussion

3.	The activity of the radioactive sample drops to(1/64)^(th)of its original value in 2 hr find the decay constant (lambda).
Answer» Solution :`LAMBDA=(1)/(2) LN. (1)/((1//64))=2.079 HR^(-1)`

Discussion

4.	The activity of the enzymes pepsin is maximum at the pH value of
Answer» 1.5 6.5 8.5 10.5 Solution :Pepsin has highest enzymatic ACTIVITY at highly acidic CONDITION hence CHOICE (a) is correct which is higly acidic medum choices(b),(c) and (d) are basic in nature so are RULED out.

Discussion

5.	The activity of radioisotope changes with:
Answer» temperature pressure chemical enviorment none of these Answer :d

Discussion

6.	The activity of radioactive isotope decreases with:
Answer» temperature pressure chemical environment none of these. Answer :D

Discussion

7.	The activity of carbon -14 in a piece of an ancient wood is only 12.5 %. If the half-life period of carbon -14 is 5760 years, the age of the piece of wood will be (log 2 = 0.3010)
Answer» `17.281 xx 10^(2)` years `172.81 xx 10^(2)` years `1.7281 xx 10^(2)` years `1728.1 xx 10^(2)` years Solution :`t_(1//2) " of " C - 14 = 5760` years, `lamda = (0.693)/(5760)` Now, `t = (2.303)/(lamda) "log" (.^(14)C " original")/(.^(14)C "after time" t)` `= (2.303 xx 5760)/(0.693) "log" (100)/(12.5) = (2.303 xx 5760 xx 0.9030)/(0.693)` ` = 17281 = 172.81 xx 10^(2)` years

Discussion

8.	The activity of an old piece of wood is just one fourth of a fresh piece of wood. If half life period of ""^(14)C is about 6000 years, the age of old piece of wood is:
Answer» 6000 years 3000 years 9000 years 12000 years Answer :D

Discussion

9.	The activity of alkaline earth metalsas reducing agents:
Answer» INCREASES from Be to CA and DECREASES from Ca to BA Increases from Be to Ba Decreases from Be to Ba Decreases from Be to Ca and increases from Ca to Ba Answer :B

Discussion

10.	The activity of a radioactive nuclide (X^(100) is 6.023 curie at a certain time 't' . If its disintegration constant is 3.7 xx 10^(4)s^(-1) the mass of X after t sec is :
Answer» `6.022 xx 10^(6)` g `10^(-13) g` `10^(-15)` g `10^(-17)` g Answer :C

Discussion

11.	The activity of a radioactive sample decreases ot 1//3 of the original activity (A_(0)) in a period of 9 years. After 9 years more ,its activity is (A_(0))/(x). Find the value of X.
Answer» ANSWER :9

Discussion

12.	The activity of a radionuclide (X^(100)) is 6.023 curie . If the disintegration constant is 3.7 xx 10^(4) sec^(-1) , the mass of the radionuclide is :
Answer» `10^(-14) `g `10^(-6)` g `10^(-15)` g `10^(-3)` g Solution :`A = lambda . N , 6.023 xx 3.7 xx 10^(10) = 3.7 xx 10^(4) xx N_(t)` `N_(t)=6.023 xx 10^(6) ` atom = `("wt")/(GAW) = (N_(t))/(6.023 xx 10^(23)) = 10^(-15)` GMS

Discussion

13.	The activity of a radioactive nuclide is 2 xx 10^(7) disintegrations per minute (dpm). After 23.03 minutes, its activity is reduced to 2 xx 10^(6) dpm. What is the average life (in min) of this nuclide
Answer» 1000 10 1 0.1 Solution :`t = (2.303)/(lamda) "log"(N_(0))/(N)` or, `lamda = (2.303)/(t) "log" (N_(0))/(N) = (2.303)/(23.03) "log" (2XX 10^(7))/(2 xx 10^(6))` or, `lamda = 0.1 "MIN^(-1)` Now, average lige `= (1)/(lamda) = (1)/(0.1) = 10` min

Discussion

14.	The activity of a catalyst has been explained in terms of spontaneous adsorption of reactant moleculse on the surface of catlyst and ther by forming an activated adsorbed complex along with evolution of heat . The adorbed activated complex undergoes to decompositio to regencerate catlyst and give products . adorpton is of two types : (i) Physical adsorptio due to van der Waal'sforces of attraction between reactant and catalyst molecules , weak , less exothermic, mulitlay and non-dirctional . (ii) Chemical dasorpton due to free valencies , strong, more exothermic, unilayer. The extent of adormpotion ( x//m) vs temperature at constant pressure gives isobars. Freundilish studied influence of pressure on the phusical adorptons and reported that log x/m log K + 1/n log P. Select the correct statements .
Answer» I, II,III I, II, IV II, III, Iv I, II Solution :All these are FACTS .

Discussion

15.	The activity of a catalyst has been explained in terms of spontaneous adsorption of reactant moleculse on the surface of catlyst and ther by forming an activated adsorbed complex along with evolution of heat . The adorbed activated complex undergoes to decompositio to regencerate catlyst and give products . adorpton is of two types : (i) Physical adsorptio due to van der Waal'sforces of attraction between reactant and catalyst molecules , weak , less exothermic, mulitlay and non-dirctional . (ii) Chemical dasorpton due to free valencies , strong, more exothermic, unilayer. The extent of adormpotion ( x//m) vs temperature at constant pressure gives isobars. Freundilish studied influence of pressure on the phusical adorptons and reported that log x/m log K + 1/n log P. Select the correct satements .
Answer» I, III, IV I,II, III II, III, Iv I,II , IV Solution :All these are FACTS .

Discussion

16.	The activity ofa radioactive isotope falls to 12.5% in 90 days. Calculate the half life and decay constant.
Answer» Solution :`N_(0)=100 N=12.5 t=90 days` DECAYCONSTANT `LAMBDA=(2.303)/(t) log (N_(0))/(N)` `=0.02558 log 8` `=2.311xx10^(-2) days^(-1)` `t^(1//2)=(0.693)/(lambda)=(0.963)/(2.311xx10^(-2))=29.99 days`

Discussion

17.	The active mass of solid is generally taken as
Answer» `GT 1` `=1` ` LT 1` 0 Answer :B

Discussion

18.	The active species produced in Hofmann's bromide reaction is :
Answer» BR `Br_2` OBR `Obr_2` ANSWER :C

Discussion

19.	The activity and selectivity of zeolites as catalyst is based on :
Answer» Their PORE size Size of their CAVITIES on the surface Both (a) and (B) None Answer :C

Discussion

20.	The activation energy ofr a reaction can be determined by measuring the reaction rate at different:
Answer» temperature catalyst concentrations reactant concentrations times on the REACTION curve. Answer :D

Discussion

21.	The active component of dynamite is
Answer» keiselghur nitro GLYCERINE nitro BENZENE trinitro TOLUENE Answer :B

Discussion

22.	The activation energy of reaction is equal to
Answer» THRESHOLD ENERGY for the reaction threshold energy + energy of the reactants threshold energy - energy of the reactants threshold energy+ energy of the products SOLUTION :ACTIVATION energy =Threshold energy - Energy of the reacants

Discussion

23.	The activation energy of the reaction is 75.2 kJ mol^(-1) in the absence of a catalyst and 50.14" kJ mol"^(-1) with a catalyst. How many times will the rate of reaction grow in the presence of the catalyst if the reaction proceeds at 25^(@)C? (R=8.314" JK"^(-1)" mol"^(-1)).
Answer» ANSWER :`24660" TIMES"`

Discussion

24.	The activation energy of reactant molecules in a reaction depends upon
Answer» Temperature Nature of the REACTANTS Colliison PER UNIT time Concentration of reactants Answer :B

Discussion

25.	The activation energy of a reaction is zero . The rate constant of the reaction :
Answer» a) INCREASES with INCREASE in TEMPERATURE b) decreases with increase in temperature c) decreases with DECREASE in temperature d) is nearly independent of temperature ANSWER :D

Discussion

26.	The activation energy of a reaction is zero. The rate constant of the reaction:
Answer» Increases with INCREASE of temperature Decreases with decrease of temperature Decreases with increase of temperature Independent of temperature Solution :`K=Ae^(-Ea//RT),Ea=0` then `((-E_(a))/(RT))=0`

Discussion

27.	The activation energy of a reaction is 94.14" kJ mol"^(-1) and the value of rate constant at 313 K is 1.8xx10^(-1)sec^(-1). Calculate the frequencyfactor, A.
Answer» Solution :Here, we are given that `E_(a)=94.14" kJ mol"^(-1)=94140" J mol"^(-1)` `T=313" K",k=1.8xx10^(-5)sec^(-1)` Substituting the VALUES in the EQUATION : `LOGK=(E_(a))/(2.303" RT")+log" A"` or `""log" A"=logk+(E_(a))/(2.303" RT")=log(1.8xx10^(-5)s^(-1))+(94140" J mol"^(-1))/(2.303xx8.314" JK"^(-1)mol^(-1)xx313"K")` `=(log" "1.8)-5+15.7082=0.2553-5+15.7082=10.9635` `"A"=" antilog "(10.9635)=9.194xx10^(10)s^(-1).`

Discussion

28.	The activation energy of a reaction is 94.14 KJ/mol and the value of rate constant at 40^@ C is 1.8 xx 10^(-1) sec^(-1). Calculate the frequency factor A.
Answer» SOLUTION :Given, `E_a=94.14xx10^(-3) J MOL^(-1), T=40+273 =313` K , `K=1.8xx10^(-1) "SEC"^(-1)` By using , `K=Ae^(-E_a//RT) rArr ` In K = In A -`E_a/(RT)` Or log K= log A -`E_a/(2.303RT)` Or `log(1.8xx10^(-1))+(94.19xx10^3)/(2.303xx8.314xx313)`=log A Or A = ANTILOG (10.9635) `=9.194xx10^10 "sec"^(-1)`

Discussion

29.	The activation energy of a reaction is 75.24 kJ "mol"^(-1) in the absence of a catalyst and 50.14 "kJ mol"^(-1) with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25^(@)C?
Answer» SOLUTION :Using Arrhenius equation in the following form and substituting the values, we GET In `k_(1)="In A"-(E_(a_(1)))/(RT) and "In "k_(2)="In A"-(E_(a_(2)))/(RT)` ` "orIn "k_(2)-"In "k_(1)=(1)/(RT) (E_(a_(1))-E_(a_(2)))` ` "orlog"(k_(2))/(k_(1))=(1)/(2.303xxRT) (75.24"kJ mol"^(-1)-50.14"kJ mol"^(-1))` `=(1)/(2.303xxRT) (25.10" kJ mol"^(-1))` `=(1)/(2.303xx8.314JK^(-1)"mol"^(-1)xx298K) xx25100 "J mol"^(-1)=4.40` or`(k_(2))/(k_(1))="ANTILOG "4.40=2.5xx10^(4)=25000` The rate or reaction will grow by 25000 times in the presence of catalyst. The decrease in activation energy TAKES place by`25.1 "kJ mol"^(-1)`.

Discussion

30.	The activation energy of a reaction is 58.3kJ/"mole". The ratio of the rate constants at 305 K and 300 K is about (R=8.3Jk^(-1)mol^(-1)) ("Antilog" 0.1667=1.468)
Answer» `1.25` `1.75` `1.5` `2.0` ANSWER :C

Discussion

31.	The activation energy of a reaction is 9 kcal/mol. The increase in the rate constant when its temperature is raised from 395 to 300K is approximately
Answer» 0.1 0.5 1 0.25 Solution :`LOG. (k_2)/(k_1)= 9000/ (2.303xx 2)(5/(395 xx 300))= 0.1103` Hence, `k_2/k_1 = 1.288` or `k_2 = 1.288 k_1, i.e., ` INCREASE = 38.8 %`~~`25%

Discussion

32.	The activation energy of a reaction is 58.3 kJ/mole. The ratio of the rate constnat at 205 K and 300 K is about (R=8.31Jk^(-1)"mole"^(-1))(Antilog 0.1667=1.468)
Answer» 1.25 1.75 1.5 `2.0` ANSWER :C

Discussion

33.	The activation energy of a reaction is 56 .2 kJ mole . The ratio of rate constants at 305 K and 300 K is ( R=8 J mol^(-1) K^(-1)) about :
Answer» `1.25` `1.47` `1.10` `1.60` ANSWER :B

Discussion

34.	The activation energy of a reaction is 225 k cal mol"^(-1)and the value of rate constant at 40^@C "is "1.8xx10^(-5)s^(-1).Calculate the frequency factor , a , .
Answer» Solution :Here, we are given that `E_a = 22.5"k CAL mol"^(-1) = 22500 "cal mol"^(-1)` `T = 40^@C = 40+273=313K` `k=1.8xx10^(-5) "sec"^(-1)` Substituting the value in the equation `LOGA = log k + ((E_a)/(2.303RT))` `log A = log (1.8xx10^(-5))+((22500)/(2.303xx1.987xx313))` `logA =log (1.8)-5+(15.7089)` A = antilog (10.9642) `A = 9.208 xx10^(10) "collisions s"^(-1)`

Discussion

35.	The activation energy of a reaction can be determined from the slope of which of the following graphs ?
Answer» `1nk"VS"(1)/(T)` `(T)/(1nk)"vs"(1)/(T)` `1nk"vs"T` `(1nk)/(T)"vs"T` ANSWER :A

Discussion

36.	The activation energy of a reaction can be determined from the slope of which of the following graphs
Answer» `(lnK)/(T)` vs. T ln K vs. `(1)/(T)` `(T)/("ln"K) "vs" . (1)/(T)` ln K vs. T Solution :`K = Ae^(-Ea//RT)` ln K = ln `A - (Ea)/(RT) ln_(E) e ` ln K = ln `A - (Ea)/(R) xx (1)/(T) `

Discussion

37.	The energy of activation for a reaction is 100 KJ mol^(-1). The peresence of a catalyst lowers the energy of activation by 75%. What will be the effect on the rate of reaction at 20^(@)C, other things being equal?
Answer» SOLUTION :`2.35xx10^13` TIMES LOG`K_2/K_1=(E_a-E_p)/(2.303RT) RARR log K_2/K_1=(75xx10^3)/(2.303xx8.314xx293)`

Discussion

38.	The activation energy of a reaction can be determined from the slope of which of the following graphs?
Answer» In `K " VS " 1/T` `T/In k " vs " 1/T` In `k " vs " T` In `(k)/T " vs " T` SOLUTION :a) It is the CORRECT ANSWER.

Discussion

39.	The activation energy of a certain uncatalysed reaction at 300 K is 76 k) per mole. The activation energy is lowered to 57 kJ per mole by the lise of a catalyst. By what factor is the rate of the catalysed reaction increased?
Answer» SOLUTION :2031 TIMES

Discussion

40.	The activation energy of a reaction can be determined by
Answer» changing the concentration of REACTANTS evaluating rate constant at standard TEMPERATURE evaluating rate constants at two different temperatures by doubling CONC. of reactants Answer :C

Discussion

41.	The activation energy of a chemical reaction can be determined by,
Answer» evaluating rate CONSTANT at two different temperatures changing the concentration of reactants evaluating the concentration of reactants at two different temperatures evaluating rate consntant at standard TEMPERATURE Solution : `:LOG""(K_(2))/(K_(1))=(E_(a))/(2.303R)(T_(2)-T_(1))/(T_(1)T_(2))`

Discussion

42.	The activation energy for the reaction,2HI(g)toH_2(g)+I_2(g)is 209.5kj mol^(-1)at 581 K.Calculate the fraction of molecules of reactant having energy equal to or greater than activation energy.
Answer» SOLUTION :Let the fraction be X `E^(E//RT)` i.e.,x=`e^(E//RT)` In `x= -E_a/(RT)log x=-E_a/(2.303RT)=(-209.5xx10^3J)/(2.303xx8.314xx581)=-18.8323` `thereforex=Antilog` of -18.8323=Antilog of `bar19 .1677=1.471xx10^(-19)`

Discussion

43.	The activation energy for the reaction 2HI_((g))toH_(2(g))+I_(2(g)) is 209.5 KJ mol^(-1) at 581 K.Calculate the fraction of molecules of reactants having energy equal to or freater than activation energy.
Answer» SOLUTION :FRACTION of molecules =`e^(-E_(a))/(RT)` `X=e^((209.5xx1000)/(8.314xx581))` `X=e^(-43.37)` `=1.461xx10^(-19)` OR `therefore K=A e^(-(E_(a))/(RT))` `therefore (k)/(A)=e^(-(E_(a))/(RT))` [But ,`(k)/(A)`=Fraction of molecule=X] `therefore X=-e^(-(E_(a))/(2.303RT)` `therefore log X=-(209.5xx10^(3)J mol)/(2.303xx8.134 J K^(-1)mol^(-1)xx581K)` `therefore` X=Antilog (-18.8323)=`1.4713xx10^(-19)`

Discussion

44.	The activation energy for the reaction, 2HI_((g))rarr H_(2(g))+I_(2(g)) is 209.5 KJ mol^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Answer» Solution :In the given case `E_(a)=209.5 "KJ mol"^(-1)=209500J mol^(-1)` T = 581 K `R = 8.314 JK^(-1) mol^(-1)` Now the FRACTION of molecules of reactants having energy equal to or greater than activation energy is given as. `x = e^(-E_(a)//RT)` `rArr log x = - E_(a//RT)` `rArr log x =-(E_(a))/(2.303 RT)` `rArr log x = (209500 "J mol"^(-1))/(2.303xx8.314 JK^(-1)mol^(-1)xx581)` `= 18.8323` x = Antilogs (18.8323) = Antilogs `BAR(19).1677` `= 1.471xx10^(-19)`.

Discussion

45.	The activation energy for the reaction, 2" HI "(g)toH_(2)(g)+I_(2)(g), is 209.5" kJ mol"^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Answer» Solution :FRACTION of molecules having energy equal to or GREATER than ACTIVATION energy `=x=(n)/(N)=e^(-E_(a)//RT)` `:.LNX=-(E_(a))/(RT)" or "logx=-(E_(a))/(2.303"RT")` or `logx=-(209.5xx10^(3)" J mol"^(-1))/(2.303xx8.314" JK"^(-1)mol^(-1)xx581" K")=-18.8323` `:.x=" Antilog"(-18.8323)="Antilog "bar(19).1677=1.471xx10^(-19)`

Discussion

46.	The activation energy for the reaction 2HI (g) to H_(2)(g) +I_(2)(g), is 209.5 kJ "mol"^(-1) at 581 K. Calculate the fraction of molecules of reactants hving energy equal to or greater than activation energy.
Answer» Solution :Fraction of molecules having energy equal to or greater than activation energy, X is GIVEN by: `x=(n)/(N)=e^(-E_(a)//RT) or "In"x= -(E_(a))/(RT) or " log"x= -(E_(a))/(2.303RT)` SUBSTITUTING the values, we get `log x= -(209.5xx10^(3)"J mol"^(-1))/(2.303xx8.314JK^(-1)"mol"^(-1) XX 581K)= -18.8323` `therefore x="Antilog" (-18.8323)="Antilog" bar(19).1677=1.471xx10^(-19)`.

Discussion

47.	The activation energy for a reaction that doubles the rate when the temperature is raised from 300 K to 310 K is (log 2 = 0.3)
Answer» `50.6 KJ MOL^(-1)` `75.45 kJ mol^(-1)` `45.5 kJ mol^(-1)` `53.4 kJ mol^(-1)` ANSWER :D

Discussion

48.	The activation energy for most of the reaction is approximately 50 kJ mol^(-1). The rate for temperature coefficient for such reaction will be
Answer» `~~2` `~~3` `LT1` `gt 4` Answer :A

Discussion

49.	The activation energy for a simple chemical reactionA to Bis E_(a) in forward direction . The activation energy for reverse reaction
Answer» Is ALWAYS double of `E_(a)` Is negative of `E_(a)` Is always less than `E_(a)` Can be less than or more than `E_(a)` SOLUTION :In EXOTHERMIC and ENDOTHERMIC reactions will be more and less than `E_(a)` RESPECTIVELY .

Discussion

50.	The activation energy for a reaction when the temperature is raised from 300K to 310K is
Answer» `50.6kJ mol^(-1)` `53.6kJmol^(-1)` `56.6kJ mol^(-1)` `59.6 kJ mol^(-1)` Solution :b) ACCORDING to ARRHENIUS equation, `LOG (K_(2))/(K_(1)) = E_(a)/2.303[1/T_(1)-1/T_(2)]` `log(2K_(1))/(K_(1))=(E_(a))/(2.303R) [1/T_(1)-1/T_(2)]` `LOG2 = (E_(a))/(2.303 xx 8.314 Jk^(-1)mol^(-1))[1/(300K-1/310K)]` `0.30 = (E_(a))/(2.303 xx (8.314 J mol^(-1)) xx (9300)` `=53.6 kj mol^(-1)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The activity of transition elements is related to their -

The activity of the hair of an Egyptian mummy is 7 disintegrations minute^(-1). Find the age of mummy. Given t_(1//2) of .^(14)C is 5770 yearand disintegration rate offresh sample of .^(14)C is 14 disintegrations minute^(-1).

The activity of the radioactive sample drops to(1/64)^(th)of its original value in 2 hr find the decay constant (lambda).

The activity of the enzymes pepsin is maximum at the pH value of

The activity of radioisotope changes with:

The activity of radioactive isotope decreases with:

The activity of carbon -14 in a piece of an ancient wood is only 12.5 %. If the half-life period of carbon -14 is 5760 years, the age of the piece of wood will be (log 2 = 0.3010)

The activity of an old piece of wood is just one fourth of a fresh piece of wood. If half life period of ""^(14)C is about 6000 years, the age of old piece of wood is:

The activity of alkaline earth metalsas reducing agents:

The activity of a radioactive nuclide (X^(100) is 6.023 curie at a certain time 't' . If its disintegration constant is 3.7 xx 10^(4)s^(-1) the mass of X after t sec is :

The activity of a radioactive sample decreases ot 1//3 of the original activity (A_(0)) in a period of 9 years. After 9 years more ,its activity is (A_(0))/(x). Find the value of X.

The activity of a radionuclide (X^(100)) is 6.023 curie . If the disintegration constant is 3.7 xx 10^(4) sec^(-1) , the mass of the radionuclide is :

The activity of a radioactive nuclide is 2 xx 10^(7) disintegrations per minute (dpm). After 23.03 minutes, its activity is reduced to 2 xx 10^(6) dpm. What is the average life (in min) of this nuclide

The activity ofa radioactive isotope falls to 12.5% in 90 days. Calculate the half life and decay constant.

The active mass of solid is generally taken as

The active species produced in Hofmann's bromide reaction is :

The activity and selectivity of zeolites as catalyst is based on :

The activation energy ofr a reaction can be determined by measuring the reaction rate at different:

The active component of dynamite is

The activation energy of reaction is equal to

The activation energy of the reaction is 75.2 kJ mol^(-1) in the absence of a catalyst and 50.14" kJ mol"^(-1) with a catalyst. How many times will the rate of reaction grow in the presence of the catalyst if the reaction proceeds at 25^(@)C? (R=8.314" JK"^(-1)" mol"^(-1)).

The activation energy of reactant molecules in a reaction depends upon

The activation energy of a reaction is zero . The rate constant of the reaction :

The activation energy of a reaction is zero. The rate constant of the reaction:

The activation energy of a reaction is 94.14" kJ mol"^(-1) and the value of rate constant at 313 K is 1.8xx10^(-1)sec^(-1). Calculate the frequencyfactor, A.

The activation energy of a reaction is 94.14 KJ/mol and the value of rate constant at 40^@ C is 1.8 xx 10^(-1) sec^(-1). Calculate the frequency factor A.

The activation energy of a reaction is 75.24 kJ "mol"^(-1) in the absence of a catalyst and 50.14 "kJ mol"^(-1) with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25^(@)C?

The activation energy of a reaction is 58.3kJ/"mole". The ratio of the rate constants at 305 K and 300 K is about (R=8.3Jk^(-1)mol^(-1)) ("Antilog" 0.1667=1.468)

The activation energy of a reaction is 9 kcal/mol. The increase in the rate constant when its temperature is raised from 395 to 300K is approximately

The activation energy of a reaction is 58.3 kJ/mole. The ratio of the rate constnat at 205 K and 300 K is about (R=8.31Jk^(-1)"mole"^(-1))(Antilog 0.1667=1.468)

The activation energy of a reaction is 56 .2 kJ mole . The ratio of rate constants at 305 K and 300 K is ( R=8 J mol^(-1) K^(-1)) about :

The activation energy of a reaction is 225 k cal mol"^(-1)and the value of rate constant at 40^@C "is "1.8xx10^(-5)s^(-1).Calculate the frequency factor , a , .

The activation energy of a reaction can be determined from the slope of which of the following graphs ?

The activation energy of a reaction can be determined from the slope of which of the following graphs

The energy of activation for a reaction is 100 KJ mol^(-1). The peresence of a catalyst lowers the energy of activation by 75%. What will be the effect on the rate of reaction at 20^(@)C, other things being equal?

The activation energy of a reaction can be determined from the slope of which of the following graphs?

The activation energy of a certain uncatalysed reaction at 300 K is 76 k) per mole. The activation energy is lowered to 57 kJ per mole by the lise of a catalyst. By what factor is the rate of the catalysed reaction increased?

The activation energy of a reaction can be determined by

The activation energy of a chemical reaction can be determined by,

The activation energy for the reaction,2HI(g)toH_2(g)+I_2(g)is 209.5kj mol^(-1)at 581 K.Calculate the fraction of molecules of reactant having energy equal to or greater than activation energy.

The activation energy for the reaction 2HI_((g))toH_(2(g))+I_(2(g)) is 209.5 KJ mol^(-1) at 581 K.Calculate the fraction of molecules of reactants having energy equal to or freater than activation energy.

The activation energy for the reaction, 2HI_((g))rarr H_(2(g))+I_(2(g)) is 209.5 KJ mol^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

The activation energy for the reaction, 2" HI "(g)toH_(2)(g)+I_(2)(g), is 209.5" kJ mol"^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

The activation energy for the reaction 2HI (g) to H_(2)(g) +I_(2)(g), is 209.5 kJ "mol"^(-1) at 581 K. Calculate the fraction of molecules of reactants hving energy equal to or greater than activation energy.

The activation energy for a reaction that doubles the rate when the temperature is raised from 300 K to 310 K is (log 2 = 0.3)

The activation energy for most of the reaction is approximately 50 kJ mol^(-1). The rate for temperature coefficient for such reaction will be

The activation energy for a simple chemical reactionA to Bis E_(a) in forward direction . The activation energy for reverse reaction

The activation energy for a reaction when the temperature is raised from 300K to 310K is