94420 + Interview Questions in Chemistry in Class 12

1.	Rateof a reaction
Answer» increases withincreasesin temperature decreaseswithincreases in temperature doesnotdependontemperature Doesnotdependon CONCENTRATION Solution :Increase in termperaturefavoursfavourscollisionof MOLECULES

Discussion

2.	Rate of a reaction :
Answer» INCREASES with INCREASE in TEMPERATURE Decreases with increase in temperature Does not DEPEND on temperature Does not depend on concentration Answer :A

Discussion

3.	Rate of a chemical reactiondependson concentrationof reactant isindependenton concentration. Why?
Answer» Solution :ACCORDING to law of mass action rate of a REACTION is proportional to the concentration of each reactant involved in the reaction. Rate constant is the rate of reaction at unit concentration of each reactant. HENCE rate constant is independent of CONCENTRATIONS of reactants. Therefore rate is dependent and rate constant is independent on the concentration.

Discussion

4.	Rate of a chemical reaction can be kept constant by
Answer» stirring the compounds keeping the temperature constant both (a) and (b) NONE of these Solution :The rate of REACTION depends UPON conc. of REACTANT, surface area of reactant, PRESSURE, temperature, presence of light and catalyst.

Discussion

5.	Rate of a chemical reaction 2A (g) to B(g) is defined as : 'r_B'=1/V(dn_B)/(dt) Where n_B=number of moles of B formed and C_B =Concentration of B then which of the following relation correct .
Answer» `r_B=(dC_B)/(dt)+(C_B)/V(DV)/(dt)` (If volume V is not CONSTANT ) `r_B=(dC_B)/(dt)`(If volume V is constant) `r_B=(dC_B)/(dt)-(C_B)/V(dV)/(dt)` (If volume V is not constant ) `r_B=-(dC_B)/(dt)+(C_B)/V(dV)/(dt)` (If volume V is not constant ) SOLUTION :`r_B=(dC_B)/(dt)=1/V(dn_B)/(dt)=1/V(d(C_BV))/(dt)=V/V(dC_B)/(dt)+C_B/V(DV)/(dt)=(dC_B)/(dt)+C_B/V(dV)/(dt)`

Discussion

6.	Rate of a chemical reaction can be kept constant by:
Answer» STIRRING the compounds Keeping the TEMPERATURE constant Both (a) and (B) None Answer :B

Discussion

7.	Rate law has to be determined __________ and cannot be predicted.
Answer» SOLUTION :EXPERIMENTALLY

Discussion

8.	Rate law for the reactionA+2B to C is found to be Rate =k[A][B] Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be ________.
Answer» the same DOUBLED QUADRUPLED HALVED ANSWER :B

Discussion

9.	Rate law for the reaction A+2BtoC is found to be Rate =k[A][B] Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be ………………. .
Answer» the same doubled quadrupled halved Solution :RATE constant of a REACTION does not depend upon CONCENTRATIONS of the REACTANTS.

Discussion

10.	Rate law can not be determined from balanced chemical equation if ………………….. .
Answer» reverse REACTION is involved it is an ELEMENTARY reaction it is a SEQUENCE of elementary REACTIONS any of the reactants is in excess. Answer :A::C::D

Discussion

11.	Rate law for 2A+BtoC+D from following data: "S.No."""[A](M)""[B](M)"""Rate"(M//s) 1""0.01""0.01""2.5 2""0.01""0.02""5 3.""0.03""0.02""45
Answer» `r=K[A]^(1//3)[B]` `r=K[A]^(2)[B]` `r=K[A][B]^(1//3)` `r=K[A]^(2//3)[B]^(1//3)` ANSWER :B

Discussion

12.	Rate law cannot be determined from balanced chemical equation if ________
Answer» Reverse reactions is not involved It is an ELEMENTARY reaction It is a sequence of elementary reactions All of the reactants is in EXCESS. RATE LAW can be determined from balanced chemical equation if it is an elementary reaction. Answer :C

Discussion

13.	Rate law cannot be determined from balanced chemical equation if……
Answer» reverse REACTION is involved it is an elementary reaction it is a sequence of elementary reactions any of the REACTANTS is in excess. Solution :Only for IRREVERSIBLE elementary reaction RATE law is directly obtained.

Discussion

14.	Rate= k[NO]^(2) [O_(2)]^(1). By how many times does the rate of reaction change when the volume of the reaction vessel is reduced to 1//3^(rd) of its original volume ? Will ther he any change in the order of the reaction.
Answer» Solution :`r_(1) = k[NO]^(2) [O_(2)]^(1), r_(2) = k[(NO)/(3)]^(2) [(O_(2))/(3)]^(1)` `r_(2) = 27r_(1)` Rate of reaction increases by 27 TIMES Order of the reaction REMAINS same.

Discussion

15.	Rate expressionfor two reaction are : (a) "rate"_(a) = k_(a)[A] and (b) "rate"_(b) = k_(b) [B]^(2) . When[A] = [B] = 1 mol l^(-1) , k_(a) = k_(b) mol L^(-1). If [A] = [B] = 2 mol L^(-1) , write therelation between"rate"_(a)and "rate"_(b) .
Answer» Solution :`"Rate"_(a)=k_(a)[2" mol L"^(-1)]` `"Rate"_(B)=k_(b)["2 mol L"^(-1)]^(2)=k_(1)["2 mol L"^(-1)]^(2)` The ration of the RATES of two REACTIONS (a) and (b) is given as, `("Rate"_(a))/("Rate"_(b))=(k_(a)["2 mol L"^(-1)])/(k_(a)["2 mol L"^(-1)]^(2))=1/2` The ration of the rates = `1:2`.

Discussion

16.	Rate law.
Answer» Solution :It is defined as an EXPERIMENTALLY determined MATHEMATICAL equation which EXPRESSES the rate of a chemical reaction in terms of MOLAR concentrations of the reactants which influence the rate of the reaction.

Discussion

17.	Rate if a reaction can be expressed by Arrhenius equations as : k = Ae^(-E//RT) In this equation , E represents
Answer» The energy above which all the colliding MOLECULES will REACT The energy below which colliding molecules will not react The total energy of the reacting molecules at a temperature , T The fractionof molecules with energy greater than the activation energy of the REACTION Solution :None of the given options is PERFECTLY correct . The closest choice is option (b) .

Discussion

18.	Rate expression of a chemical change is (dx)/(dt) =k[A]^1[B]^(1//2)[C]^(3//2). The order of the reaction is
Answer» 2 3 1 zero Solution :Order of REACTION is sum of POWER raised on concentration terms in expression RATE `= 1+1/2+3/2=3` Thus order `= 1+ 1/2 + 1/2=3`.

Discussion

19.	Rate equation is the expression that gives the relation between rate of reaction and :
Answer» Temperature Concentration of products Concentration of reactants None of the above Answer :C

Discussion

20.	Rate expression for xA+yBto prodcuts is Rate =k[A]^(m)[B]^(n). Units of K with respect to A and B respectively are s^(-1) and M^(-1).s^(-1) when concentration of A and B are increased by 4 times, then
Answer» `R_(F)=16R_(i)` `R_(i)=16R_(f)` `R_(f)=8R_(i)` `R_(f)=64R_(i)` ANSWER :D

Discussion

21.	Rate equation for a second order reaction is :
Answer» `K=(2.303)/TLOG((a)/(a-X))` `K=(1)/tlog((a)/(a(a-x)))` `K=(1)/t.(x)/(a(a-x))` `K=(1)/t^2.(a)/((a-x))` Answer :C

Discussion

22.	Rate constnat K=2.303"min"^(-1) for a particular reactio the initial concentration of there reactant is 1 mole/litre then rate of reaction after 1 minute is
Answer» `2.303M "min"^(-1)` `0.2303 M "min"^(-1)` 0.1M "min"^(-1)` NONE of these Solution :`K=2.303/t "log"(C_(o))/(C_(t))` `C_(t)=0.1` RATE `=k_(1)[c_(t)]` Rate `=0.2303M "min"^(-1)`

Discussion

23.	Rate contant k_2 of a reaction at 310 Kis two time of its rate constant k_1at 300 K.Calculate activation energy of the reaction.(log 2=0.3010,log 1=0)
Answer» SOLUTION :LOG `k_2/k_1=(EA)/(2.303R)((T_2-T_1)/(T_1T_2))` LOG2=`(Ea)/(2.303xx8.314)[((310-300))/(310xx300)] Ea=(0.3010xx2.303xx8.314xx300)/10=53598J`

Discussion

24.	Rate constant of a reaction (k) is 175 "litre"^(2)mol^(-2)sec^(-1). What is the order of reaction ?
Answer» first second third zero Solution :( C ) : On the BASIS of GIVEN UNITS of k, the reaction is of 3rd order.

Discussion

25.	Rate constant value depends on:
Answer» Temperature Time Initial concentration None of these Answer :A

Discussion

26.	Rate constant of a reaction is K=3.4 xx 10^-4 mol^-1 LS^-1 What is the order of the reaction ?
Answer» SOLUTION :SECOND ORDER.

Discussion

27.	Rateconstantof a reaction is 10^(-3) s^(-1). Calculatethepercentagecompletionof thereaction in .
Answer» ANSWER :0.1821

Discussion

28.	Rate constant of a reaction is k = 3.14 xx 10^(-4) mol L^(-1)s^(-1). What is the order of the reaction.
Answer» SOLUTION :SECOND ORDER.

Discussion

29.	Rate constant of a first order reaction is 6.93 xx 10^(-3) mi n^(-1). Calculate the half-life period.
Answer» Solution :GIVEN `K = 6.93 xx 10^(-3) mi N^(-1)` `t_(1/2) = `? `t_(1/2) = 0.693/K = 0.693/6.93 xx 10^(-3)mi n^(-1) = 100 min`

Discussion

30.	Rate constant of a reaction at 290K was found to be 3.2 xx 10^-3 . At 300K , it will be :
Answer» `1.28xx10^-2` `9.6xx10^-3` `6.4xx10^-3` `3.2xx10^-4`. ANSWER :C

Discussion

31.	Rate constant of a first order reaction at 298 K is 15.5 s^(-1). What is the approximate rate constant for this reaction at 308 K?
Answer» Solution :The EXPECTED RATE CONSTANT for this REACTION is `31s^(-1)`.

Discussion

32.	Rate constant of a first order reaction is 0.0693 min^(-1). Calculate the percentage of the reactant remaining at the end of 60 minutes.
Answer» Solution :FORMULA : `K = (2.303)/(t) log""([R]_(0))/([R])` ` 0.0693 = (2.303)/(60) log""(100)/([R]) rArr 1.8 = log (100)/([R])` ANTILOG `1.8 = (100)/([R])` `63.1 = (100)/([R]) rArr [R] = 1.584%`

Discussion

33.	Rate constant of a first order reaction A to B is 0.0693min^(-1). Calculate rate (i) at start and (ii) after 30 minutes. Initial concentration of A is 1.0M.
Answer» Solution :`k=0.0693min^(-1)` `t_((1)/(2))=(0.693)/(k_(1))=(0.693)/(0.0693)=10`min Since `C=C_(@)((1)/(2))^(n)` `n=(t)/(t_(1//2))` `n=(30)/(10)=3``C_(@)=1M` `:.C=1xx((1)/(2))^(3)=(1)/(8)M` Rate of the reaction at the START of the reaction `=k_(1)xxC_(@)`. `=10xx0.0693xx1=0.693M min^(-1)` Rate after `30`min `=k_(1)C=00693xx(1)/(8)=8.66xx10^(-8)M min^(-1)`

Discussion

34.	Rate constant of a first order reaction A products is 0.016 min^(-1). Calculate the time required for 80% of the reaction to be completed.
Answer» SOLUTION :`K = (2.303)/(t) LOG.(a)/(a-x)` `t = (2.303)/(0.016)log.(100)/(100-80)` `t = (2.303)/(0.016)log 5` `t = 100.6 MIN`

Discussion

35.	Rate constant of a 1st order reaction is 0.5 s^(-1).What is the half-life period ?
Answer» SOLUTION :`t_(1//2)=0.693/K=(0.693/0.5)s=1.386s`

Discussion

36.	Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation. log k log A(E_a)/(2.303R)(1/T) where E_a is the activation energy . When a graph is plotted is or logk Vs 1/T a straight line with a slope of - 400K is obtained . Calculate the activation energy .
Answer» SOLUTION :LOG k = log A - `E_a/(2.303R)(1/T)` y = c + mx `m=-E_a/(2.303R)` `E_a=-2.303Rm` `E_a=-2.303xx8.314"J K "^(-1)"MOL"^(-1)XX(-400K)` `E_a=76,589"J mol"^(-1)` `E_a = 76 589 "kJ mol"^(-1)` .

Discussion

37.	Rate constant (k)of a reaction is 5xx10^(-2)s^(-1).Find the half life (t_(1/2)) of the reaction.
Answer» SOLUTION :`t_(1/2)=0.693/5xx10^(-2)=13.86` SEC

Discussion

38.	Rate constant K varies with temperature by equation log K("min"^(-1))=5-((2000))/T.We can conclude that (R=8.314J"mol"^(-1)K^(-1) (or) cal "mol"^(-1)K^(-1))
Answer» Pre exponential factor A is 5 Ea is 4 K cal/mol Pre exponential factor, A is `10^(5)` Ea is 9.212 Kcal/mol Solution :`K=A.e^(-Ea//RT),logK=logA-(epsilon_(a))/(2.303R)xx1/T` `logA=5,A=10^(5),(E_(a))/(2.303RT)=2000/T,E_(a)=(2000xx2.303xx2)/1,E_(a)=9.212KCal`

Discussion

39.	Rate constant 'k' of a reaction varies with temperature 'T' according to the equation: log k= logA-(E_(a))/(2.303R)((1)/(T)) where E_(a) is the activation energy. When a graph is plotted for log k vs (1)/(T), a straight line with a slope of -4250 K is obtained. Calculate 'E_(a)'for the reaction. [R=8.314JK^(-1)"mol"^(-1)]
Answer» SOLUTION :The equation `log k=log A -(E_(a))/(2.303R) ((1)/(T))` is of the type `y=mx +C` where m is the slope of the line. `therefore (-E_(a))/(2.303R)= -4250 K or E_(a)=2.303R xx 4250 K` or`E_(a)=2.303xx 8.314 JK^(-1)"MOL"^(-1) xx 4250K` `=81375"J mol"^(-1)=81.375 "kJ mol"^(-1)`

Discussion

40.	Rate constant k for first order reaction has been found to be 2.54xx10^(-3)s^(-1). Calculate its three-fourth time.
Answer» Solution :For the first ORDER reaction, use the relation `k=(2.303)/(t)"log"([R]_(0))/([R])` For three-foursth of the reaction to take place, `[R]=([R]_(0))/(4)or ([R]_(0))/([R])=4` Substituting the values in the rate equation, we have `2.54xx10^(-3)=(2.303)/(t)LOG4` or `t=(2.303)/(2.54)xx10^(3)xx0.6021=5.46xx10^(2)s`

Discussion

41.	Rate constant (k) for x Ato products is 0.04" M - min"^(-1). When [A] changes form 0.08 M to 0.02M, ratio of initial to final rate is
Answer» `1//4` `1//16` 1 4 Answer :C

Discussion

42.	The rate constant 'k'. For a reaction varies with temperature 'T' according to the question. log k = log A-(E_(a))/(2.303R)(1/T) Where E_(a) is the activation energy. When a graph is plotted for log k vs 1//T, a straight line with a slope of -4250 K is obtained. Calcualte E_(a) for this reaction.
Answer» SOLUTION :SLOPE =`E_a/(2.303R)`=-4250 K So, `E_a=-2.303xxRxx"Slope"` `=-2.303xx8.314 J K^(-1) "mol"^(-1) xx4250` `=81375.3 "J mol"^(-1)` `=81.375 "KJ mol"^(-1)`

Discussion

43.	Rate constant k for a reaction varies with temperature according to the equation log k ="constant"-(E_(a))/(2.303R)*(1)/(T) where E_(a) is the energy of activation for the reaction. When a graph is plotted for log k vs 1//T, a straight line with a slope of -6670 K is obtained. Calculate energy of activation for this reaction. State the units. ""[R=8.314JK^(-1)"mol"^(-1)]
Answer» Solution :The following data is provided SLOPE `= -6670K, R=8.314 JK^(-1)"mol"^(-1)` Slope `= -(E_(a))/(2.303R) or E_(a)= -"Slope" xx 2.303 xx R` SUBSTITUTING the VALUES, we have `E_(a)= -(-6670)K xx 2.303 xx 8.314JK^(-1)"mol"^(-1)=+127.714"KJ mol"^(-1)`

Discussion

44.	Rate constant for the decomposition of ethylene oxide into CH_(4) and CO may be described by the equation logl(s^(-1))=14.34-(1.25xx10^(4))/(T) (a) What is the energy of activation of this reaction ? (b) What is the value of k at 670K ?
Answer» SOLUTION :`(a) 239.34kj MOL^(-1)` `(B) 4.82xx10^(-5)s^(-1)`

Discussion

45.	Rate constant for a reaction varies with temperature as, ln K(sec^(-1))=14.34-(1.25xx10^(4))/(T) Which statement(s) is (are) correct?
Answer» (a) The graph plotted `log_(10) K` vs. `1/T` is straight LINE with `E_(a)=24.83 KCAL` (B) Pre-exponential factor`=13.34` (c ) The rate constant at `500 K` is `2.35xx10^(-5) sec^(-1)` (d) `E_(a)=30.63 kcal` Solution :N//A

Discussion

46.	Rate constant K = 2.303 "min"^(–1) for a particular reaction the initial concentration of the reactant is 1 mole/litre then rate of reaction after 1 minute is :-
Answer» 2.303 M `"MIN"^(-1)` 0.2303 M `"min"^(-1)` 0.1 M `"min"^(-1)` None Solution :`K=2.303/t "LOG" C_o/C_t` `c_t`=0.1 RATE= `k_1[c_t]` Rate = 0.2303 M `"min"^(-1)`

Discussion

47.	Rate constant k for a first order reactions has been found to be 2.54 xx 10^(-3) sec^(-1). Calculate its 3//4^(th) life. (log 4 = 0.6020)
Answer» SOLUTION :`K = (2.303)/(t) log.(a)/(a-x)` `k -2.54 xx 10^(-3)sec^(-1), t_(3//4) = ?`

Discussion

48.	Rast method is based upon the use of …………………..as solvent whose molal depression constant is ……………… .
Answer» Solution :camphor, `"39.7 K kg MOL"^(-1)`

Discussion

49.	Rate constant for a reaction is lamda. Average life is representant by
Answer» `1//LAMDA` `LN 2//lamda` `(lamda)/(SQRT2)` `(0.693)/(lamda)` ANSWER :D

Discussion

50.	Rate constant for a reaction is 1.85 xx 10^(2)s^(-1). Give the order of reaction.
Answer» SOLUTION :FIRST ORDER REACTION.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Rateof a reaction

Rate of a reaction :

Rate of a chemical reactiondependson concentrationof reactant isindependenton concentration. Why?

Rate of a chemical reaction can be kept constant by

Rate of a chemical reaction 2A (g) to B(g) is defined as : 'r_B'=1/V(dn_B)/(dt) Where n_B=number of moles of B formed and C_B =Concentration of B then which of the following relation correct .

Rate of a chemical reaction can be kept constant by:

Rate law has to be determined __________ and cannot be predicted.

Rate law for the reactionA+2B to C is found to be Rate =k[A][B] Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be ________.

Rate law for the reaction A+2BtoC is found to be Rate =k[A][B] Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be ………………. .

Rate law can not be determined from balanced chemical equation if ………………….. .

Rate law for 2A+BtoC+D from following data: "S.No."""[A](M)""[B](M)"""Rate"(M//s) 1""0.01""0.01""2.5 2""0.01""0.02""5 3.""0.03""0.02""45

Rate law cannot be determined from balanced chemical equation if ________

Rate law cannot be determined from balanced chemical equation if……

Rate= k[NO]^(2) [O_(2)]^(1). By how many times does the rate of reaction change when the volume of the reaction vessel is reduced to 1//3^(rd) of its original volume ? Will ther he any change in the order of the reaction.

Rate expressionfor two reaction are : (a) "rate"_(a) = k_(a)[A] and (b) "rate"_(b) = k_(b) [B]^(2) . When[A] = [B] = 1 mol l^(-1) , k_(a) = k_(b) mol L^(-1). If [A] = [B] = 2 mol L^(-1) , write therelation between"rate"_(a)and "rate"_(b) .

Rate law.

Rate if a reaction can be expressed by Arrhenius equations as : k = Ae^(-E//RT) In this equation , E represents

Rate expression of a chemical change is (dx)/(dt) =k[A]^1[B]^(1//2)[C]^(3//2). The order of the reaction is

Rate equation is the expression that gives the relation between rate of reaction and :

Rate expression for xA+yBto prodcuts is Rate =k[A]^(m)[B]^(n). Units of K with respect to A and B respectively are s^(-1) and M^(-1).s^(-1) when concentration of A and B are increased by 4 times, then

Rate equation for a second order reaction is :

Rate constnat K=2.303"min"^(-1) for a particular reactio the initial concentration of there reactant is 1 mole/litre then rate of reaction after 1 minute is

Rate contant k_2 of a reaction at 310 Kis two time of its rate constant k_1at 300 K.Calculate activation energy of the reaction.(log 2=0.3010,log 1=0)

Rate constant of a reaction (k) is 175 "litre"^(2)mol^(-2)sec^(-1). What is the order of reaction ?

Rate constant value depends on:

Rate constant of a reaction is K=3.4 xx 10^-4 mol^-1 LS^-1 What is the order of the reaction ?

Rateconstantof a reaction is 10^(-3) s^(-1). Calculatethepercentagecompletionof thereaction in .

Rate constant of a reaction is k = 3.14 xx 10^(-4) mol L^(-1)s^(-1). What is the order of the reaction.

Rate constant of a first order reaction is 6.93 xx 10^(-3) mi n^(-1). Calculate the half-life period.

Rate constant of a reaction at 290K was found to be 3.2 xx 10^-3 . At 300K , it will be :

Rate constant of a first order reaction at 298 K is 15.5 s^(-1). What is the approximate rate constant for this reaction at 308 K?

Rate constant of a first order reaction is 0.0693 min^(-1). Calculate the percentage of the reactant remaining at the end of 60 minutes.

Rate constant of a first order reaction A to B is 0.0693min^(-1). Calculate rate (i) at start and (ii) after 30 minutes. Initial concentration of A is 1.0M.

Rate constant of a first order reaction A products is 0.016 min^(-1). Calculate the time required for 80% of the reaction to be completed.

Rate constant of a 1st order reaction is 0.5 s^(-1).What is the half-life period ?

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation. log k log A(E_a)/(2.303R)(1/T) where E_a is the activation energy . When a graph is plotted is or logk Vs 1/T a straight line with a slope of - 400K is obtained . Calculate the activation energy .

Rate constant (k)of a reaction is 5xx10^(-2)s^(-1).Find the half life (t_(1/2)) of the reaction.

Rate constant K varies with temperature by equation log K("min"^(-1))=5-((2000))/T.We can conclude that (R=8.314J"mol"^(-1)K^(-1) (or) cal "mol"^(-1)K^(-1))

Rate constant k for first order reaction has been found to be 2.54xx10^(-3)s^(-1). Calculate its three-fourth time.

Rate constant (k) for x Ato products is 0.04" M - min"^(-1). When [A] changes form 0.08 M to 0.02M, ratio of initial to final rate is

The rate constant 'k'. For a reaction varies with temperature 'T' according to the question. log k = log A-(E_(a))/(2.303R)(1/T) Where E_(a) is the activation energy. When a graph is plotted for log k vs 1//T, a straight line with a slope of -4250 K is obtained. Calcualte E_(a) for this reaction.

Rate constant for the decomposition of ethylene oxide into CH_(4) and CO may be described by the equation logl(s^(-1))=14.34-(1.25xx10^(4))/(T) (a) What is the energy of activation of this reaction ? (b) What is the value of k at 670K ?

Rate constant for a reaction varies with temperature as, ln K(sec^(-1))=14.34-(1.25xx10^(4))/(T) Which statement(s) is (are) correct?

Rate constant K = 2.303 "min"^(–1) for a particular reaction the initial concentration of the reactant is 1 mole/litre then rate of reaction after 1 minute is :-

Rate constant k for a first order reactions has been found to be 2.54 xx 10^(-3) sec^(-1). Calculate its 3//4^(th) life. (log 4 = 0.6020)

Rast method is based upon the use of …………………..as solvent whose molal depression constant is ……………… .

Rate constant for a reaction is lamda. Average life is representant by

Rate constant for a reaction is 1.85 xx 10^(2)s^(-1). Give the order of reaction.