94420 + Interview Questions in Chemistry in Class 12 Page 497 InterviewSolution

24801.	The rate constant of the reaction A to B is 0.6 x 10^-3 Molar per second. If the concentration of A is 5 M then concentration of B after 20 minutes is
Answer» 1.08 M 3.60 M 0.36 M 0.72 M Solution :It is zero order reaction `THEREFORE 6 XX 10^(-4) = ("con")/(20 xx 60)` `therefore` Conc., of B = 0.72 M

Discussion

24802.	The rate constant of the reaction A to 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 Answer :B

Discussion

24803.	The rate constant of the reaction : 2H_(2)O(aq)to2H_(2)O(l)+O_(2)(g) is 3xx10^(-3)min^(-1). At what concentration of H_(2)O_(2), the rate of the reaction will be 2xx10^(-4)M s^(-1) ?
Answer» `6.67xx10^(-3)M` `2M` `4M` `0.08M` SOLUTION :Rate `=k[H_(2)O_(2)]` `2xx10^(-4)=(3XX10^(-3))/(60)xx[H_(2)O_(2)]` `[H_(2)O_(2)]=4M]`

Discussion

24804.	The rateconstantof reactionis 5xx10^(-2) "litre " ^(3)"mole"^(-3) "min"^(-1) theorderofreactionis
Answer» `1` `2` `3` `4` Solution :`k =("rate")/([R]^(N))=(M//s)/(M^(n))=M^(1-n)S^(-1)` when`n= 4, K= "litre"^(3)"mole" ^(-3) MIN^(-1)`

Discussion

24805.	The rate constant of reaction, 2A + B rarr C is 2.57xx10^(-5)"L mole"^(-1)sec^(-1) after 10 sec,, 2.65xx10^(-5)"L mol"^(-1)sec^(-1) after 20 sec., and 2.55xx10^(-5)"L mole"^(-1)sec^(-1) after 30 sec. The order of the reaction is :
Answer» 0 1 2 3 Answer :C

Discussion

24806.	The rate constant of raction is 0.2 "min"^(-1). The order of the reaction is
Answer» SECOND first zero third Solution :The unit of RATE constant of a first ORDER REACTION is `"TIME"^(-1)`.

Discussion

24807.	The rate constant of reaction is 3xx10^-3atm^-2sec^-1 . The order of reaction is
Answer» 1 2 3 0 Answer :B

Discussion

24808.	The rate constant of n^(th)order reaction has units :
Answer» `LITRE^(1-n)MOL^(1-n)sec^-1` `mol^(1-n)mol^(1-n)sec^-1` `mol^(1-n)litre^(n-1)sec^-1` NONE of these Answer :C

Discussion

24809.	The rate constant of nth order has units:
Answer» `"litre"^(1-n)MOL^(-1)sec^(-1)` `mol^(1-n)litre^(1-n)sec` `mol^(1-n^2)litre^(n^2)sec^(-1)` `mol^(1-n) litre^(n-1)sec^(-1)` Solution :For an nth ORDER reaction : RATE = `K [conc.]^n` `k = ("rate")/(["conc."]^n)` Units of `k = (mol L^(-1)s^(-1))/((molL^(-1))^n)= mol^(1-n) L^(n - 1) s^(-1)`

Discussion

24810.	The rate constant of first order reaction whose half-life is 480 s, is
Answer» `1.44 xx 10^(-3)s^(-1)` `1.44 s^(-1)` `0.72 xx 10^(-3)s^(-1)` `2.88 xx 10^(-3)s^(-1)` SOLUTION :For the first ORDER REACTION `implies ""k=(0.693)/(t_(1//2))=(0.693)/(480 sec)=1.44 xx 10^(-3)s^(-1)`

Discussion

24811.	The rate constant of a first order reaction is 0.60" sec"^(-1). What is its half-life period?
Answer» SOLUTION :`K=0.50sec^-1` HALF LIFE PERIOD, `t_(1/2)=0.693/(0.50sec^-1)=1.39 SEC.`

Discussion

24812.	Derive an expression for the rate constant of first order reaction. The rate constant of first order reaction is 0.346" min"^(-1). What is the half-life?
Answer» Solution :`K=0.346 min^-1` HALF life PERIOD`t_(1/2)=0.693/K` or,`t_(1/2)=0.693/(0346min^-1)`=2 MINUTES

Discussion

24813.	The rate constant of a second order reaction, 2A toProducts, is 10^(-4)" lit mol"^(-1)" min"^(-1). The initial concentration of the reactant is 10^(-2)" mol lit"^(-1). What is the half-life (in min) ?
Answer» 10 1000 100 `10^(6)` Solution :Half -life period of a second ORDER REACTION `t_(1//2) = (1)/("Rate CONSTANT" XX "Initial CONC.") = (1)/(ka)` `=(1)/(10^(-4) L mol^(-1) min^(-1) xx 10^(-2) mol L^(-1))` ` = 10^(6) min`

Discussion

24814.	The rate constant of a reaction of a reaction increases by 5% whenthe temperature of the reaction is increased from 300 to 301 K wheres equilibrium constant increases only by 2%. Calculate the activation energy for the forward as well as backward reaction.
Answer» Solution :ACCORDING to Arrhenius EQUATION, `log""(k_(2))/(k_(1))=(E_(a,f))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))` If `k_(1)=k" at "300" K, then at "301" K,"k_(2)=k+(5)/(100)k=1.05k` `:.log""(1.05k)/(k)=(E_(a,f))/(2.303xx8.314" JK"^(-1)MOL^(-1))((301" K"-300" K")/(300" K"xx301" K"))` `:.E_(a,f)=(log1.05)xx2.303xx(8.314" JK"^(-1)mol^(-1))xx300xx301" (K) "=366554" J mol"^(-1)=36.65" kJ mol"^(-1)` According to van't Hoff equation (giving the effect of temperature on equilibrium constant), `log""(K_(2))/(K_(1))=(DeltaH^(@))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))` If `K=K_(1)" at "300" K, then at "301" K","K"_(2)="K"+(2)/(100)"K"=1.02" K"` `:.log""(1.02"K")/("K")=(DeltaH^(@))/(2.303xx8.314" JK"^(-1)mol^(-1))((301"K"-300"K")/(300"K"XX301"K"))` or `DeltaH^(@)=" Activation energy for forward reaction-Activation energy for backward reaction"=E_(a,f)-E_(a,b)` `:.E_(a,b)=E_(a,f)-DeltaH^(@)=36.65-14.87" kJ mol"^(-1)=21.78" kJ mol"^(-1).`

Discussion

24815.	The rate constant of a zero order reaction in A is 0.003 "mol L"^(-1) "sec"^(-1). How long will it take for the initial concentration of A to fall from 0.10M to 0.075 M ?
Answer» SOLUTION :`t=([A]_0 -[A])/K=(0.10-0.075)/0.003`=8.3 SEC

Discussion

24816.	The rate constant of a reaction is equal to the rate of reaction when
Answer» Concentration of all REACTANTS are unity Concentration of all reactants and products are equal Concentration of reactants and products do not change with TIME The reaction takes INFINITE time for COMPLETION Answer :A

Discussion

24817.	The rate constant of a reaction is independent of
Answer» temperature catalyst cone. of reactant none of these Answer :C

Discussion

24818.	The rate constant of a reaction is equal to rate of reaction
Answer» when oncentrations of REACTANTS do not CHANGE with time when concentrations of all reactants and products are equal at time, t = 0 when oncentrations of all reactants are unity Solution :If `(dx)/(dt) = k [A]^(X) [ B]^(y) [C]^(z)` for a REACTION , then `(dx)/(dt) = k` When `[A] = 1 = [B]= [C]`

Discussion

24819.	The rate constant of a reaction is given byk=2.1xx10^10 exp(-2700//RT). It means that
Answer» log k versus `1/T` will be a straight LINE with SLOPE `=(-2700)/(2.303R)` log k versus `1/T` will be a straight line withintercepton axis = 10.32 When Tbecomes INFINITE , `k=2.1xx10^10` k is independent of temperature Solution :From Arrehenius equation `k=Ae^(-E_(a)//RT)` Given, `k=2.1xx10^10xxe^(-2700//RT)` INTERCEPT =log A - log `(2.1xx10^10)` = 10.32

Discussion

24820.	The rate constant of a reaction is given by k=2.1xx10^(10)C^(-(2700)/(2.303R) It means that
Answer» LOG K vs `(1)/(T)` will be straight line with slope `=(-2700)/(2.303R)` log k vs `(1)/(T)` will be straight line with intercept on log k axis `=log 2.1xx10^(10)` the NUMBER of effective collisions are `2.1xx10^(10)cm^(-3)sec^(-1)` half-life of the reaction increases with increase in temperature Solution :(A) and (B) are correct, (C ) is wrong because frequency factorgives total number of collisions and not the effective collision `cm^(-3)sec^(-1)`, (D) is wrong because half-life of the reaction decreases with increase in temperature (as reaction BECOMES faster)

Discussion

24821.	The rate constant of a reaction is given by k=2.1xx10^(10)" exp "(-2700/RT) It means that
Answer» log k vs. `1//T` will be a straight line with slope `=-(2700)/(2.303" R")` log k vs. `1//T` will be a straight line with intercept on log k axis `=2.1xx10^(10)` The number of effective collisions are `2.1xx10^(10)" cm"^(-3)" sec"^(-1)` Half life of the reaction INCREASES with increase of temperature. Solution :is wrong because frequency factor gives total number of collicions and not effective collisions `cm^-3 sec^-1`. (d) is wrng because half - life of the reactions decreases with incerease of temperature ( as reaction BECOMES FASTER ).

Discussion

24822.	The rate constant of a reaction is doubled when the temperature increased from400K to 410K. Calculate the activation energy (Ea) [R = 8.314 JK^(-1) " mol"^(-1)] .
Answer» SOLUTION :`log""(k_2)/(k_1) = (E_(a))/(2.303R) [(T_2 - T_1)/(T_1T_2)]` ` log 2 = (E_a)/(2.303 xx 8.314 ) [(410-400)/(400 xx 410)]` ` 0.310 = (E_a)/(19.147)[(10)/(164000)]` `E_a = (0.301 xx 19.147 xx 164000)/(10)` `= 94517 "Jmol"^(-1) OR 94.517 "kJmol"^(-1)`

Discussion

24823.	The rate constant of a reaction is 5.8xx10^(-2)s^(-1). The order of the reaction is ..........
Answer» First order zero order SECOND order Third order SOLUTION :The UNIT of RATE constant is `s^(-1)` and it indicates that the REACTION is first order.

Discussion

24824.	The rate constant of a reaction is 5.8xx10^(-2)s^(-1). The order of the reaction is
Answer» FIRST ORDER ZERO order Second order THIRD order SOLUTION :The unit of rate constant is `s^(-1)` and it indicates that the reaction is first order.

Discussion

24825.	The rate constant of a reaction is 2xx10^-5 sec^-1. If the initial concentration is 0.1 M, the initial rate is
Answer» `2xx10^-4 M sec^-1` `2xx10^-5M sec^-1` `2xx10^-6M sec^-1` `2xx10^-7M sec^-1` ANSWER :C

Discussion

24826.	The rate constant of a reaction is 2xx10^(-3)min^(-1) at 300 K temperature .By increase in temperature by 20K,its value becomes three time,then calculate the energy of activation of the reactuon .what will be its rate constant at 310 K temperature?
Answer» Solution :`E_(a)`=10480 cal,K(310 K)=`3.126xx10^(-3)MIN^(-1)`

Discussion

24827.	The rate constant of a reaction is 2xx10^(-3) min^(-1) at 300 K temperature .By increase in temperature by 10 K its value becomes doubl. Calculate energy of activation .Calculation the rate constant at 320 K.
Answer» SOLUTION :`E_(a)`=12810 cal,k(320 K)=`7.659xx10^(-3)min^(-1)`

Discussion

24828.	The rate constant of a reaction is 2.5xx10^(-2)" L mol"^(-1)s^(-1). Calculate the initial rate with concentration 0.2" mol L"^(-1).
Answer» Solution :From the units of rate constant, order of the reaction is 2. For a SECOND order reaction, rate = k `"[concentration]"^(2)` Initial rate of the reaction `=2.5xx10^(-2)(0.2)^(2)=1xx10^(-3)" MOL L"^(-1)s^(-1)`

Discussion

24829.	The rate constant of a reaction is 2.5xx10^-2mole^-1 lit sec^-1. The order of the reaction is
Answer» zero 1st 2nd 3rd Answer :C

Discussion

24830.	The rate constant of a reaction is 2.1xx 10^(-2)mol^(-2)litre^3 min^(-1). The order of reaction is
Answer» zero ORDER 1 2 3 Solution :UNIT of `K= (mol L^(-1))^(1-n)sec^(-1) or mi n^(-1)` for 3rd order, n = 3

Discussion

24831.	The rate constant of a reaction is 1.5xx10s^-1 at 100^@C. Calculate the value of activation energy for the reaction.
Answer» SOLUTION :From RELATION. `"LOG" K_2/K_1=E_a/(2.303R)[1/T_1-1/T_2]"implieslog"(4.5xx10^7)/(1.5xx10^7)` `=E_a/(2.303xx8.314)[1/323-1/373]IMPLIES"LOG3"=[E_a//(19.147)]xx50//(323xx373]` `E_a=(0.4771xx19.147xx323xx373)/50impliesE_a=22.01kj mol^-1`

Discussion

24832.	The rate constant of a reaction is 1.5xx10^7s^-1 at 50^@Cand 4.5×10^7s^−1 at 100^@C. Calculate the Arrhenius parameter for the reaction.
Answer» SOLUTION :From RELATION. `"LOG" K_2/K_1=E_a/(2.303R)[1/T_1-1/T_2]"implieslog"(4.5xx10^7)/(1.5xx10^7)` `=E_a/(2.303xx8.314)[1/323-1/373]IMPLIES"LOG3"=[E_a//(19.147)]xx50//(323xx373]` `E_a=(0.4771xx19.147xx323xx373)/50impliesE_a=22.01kj mol^-1`

Discussion

24833.	The rate constant of a reaction is 1.5 xx 10^(7)s^(-1) at 50^(@)C and 4.5 xx 10^(7) s^(-1) at 100^(@)C. Evaluate the Arrhenius parameters A and E_(a).
Answer» SOLUTION :Apply Arrhenius.s EQUATION. `2.19 XX 10^4 J//"mol" , 5.4 xx 10^20 s^(-1)`

Discussion

24834.	The rate constant of a reaction is 1.2xx10^(-5) "mol"^(-2) "litre"^(2) s^(-1) the order of the reaction is :
Answer» zero 1 2 3 Answer :D

Discussion

24835.	The rate constant of a reaction is 0.69 xx 10^(-1) "min"^(-1) and the initial concentration is0.2 mol l^(-1) . The half - life period is
Answer» 400 sec 600 sec 800 sec 1200 sec Solution :The unit of RATE constant SHOWS that REACTION is of first order . For first order reaction , half life is independent of initial CONC., of the reactant . THUS , `t_(1//2) = (0.693)/(0.69 xx 10^(-1)) = (0.693 xx 60)/(0.69 xx 10^(-1)) = 600` sec.

Discussion

24836.	The rate constant of a reaction is 10.8 xx 10^(-5) mol dm^(-3)s^(-1). The order of the reaction is
Answer» ZERO 1 2 3 Solution :On the BASISOF UNITS of K, ORDER = zero.

Discussion

24837.	The rate constant of a reaction has s^-1 units. The reaction is of …..
Answer» THIRD order Second order First order Zero order Answer :C

Discussion

24838.	The rate constant of a reaction has same units as the rate of reaction. The reaction is of …..
Answer» THIRD ORDER second order FIRST order Zero order Answer :D

Discussion

24839.	The rate constant of a reaction depends upon
Answer» temperature catalyst nature of reactant all of these Answer :D

Discussion

24840.	The rate constant of a reaction depend on.
Answer» Temperature Mass Weight time Answer :A

Discussion

24841.	The rate constants of a reaction at 700 K and 760 K are 0.011" M"^(-1)s^(-1) and 0.105" M"^(-1)s^(-1) respectively. Calculate the values of Arrhenius parameters.
Answer» SOLUTION :`2.824xx10^10`

Discussion

24842.	The rate constant of a reaction changes when
Answer» a catalyst is ADDED concentrations of the REACTANT are changed TEMPERATURE of changed both (a) and (c ) Solution :Rate CONSTANT of a reaction changes when either the temperature is changed or catalyst is added.

Discussion

24843.	The rate constant of a reaction changes with
Answer» Pressure Temperature INITIAL CONCENTRATION of reactants Extent of REACTION ANSWER :B

Discussion

24844.	The rate constant of a reaction at temperature 200K is 10 times than the rate constant at 400K . What is the activation energy (E_(a)) of the reaction (R=gas constant) ?
Answer» 1842.4 R 921.2 R 660.6 R 230.3 R Answer :B

Discussion

24845.	The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400K . What is the activation energy (E_(a)) of the reaction (R = gas constant )
Answer» 1842.4 R 921.2 R 460.6 R 230.3 R ANSWER :B

Discussion

24846.	The rate constant of a reaction at 500 K and 700 K are 0.02 s^(-1) and 0.07 s^(-1) respectively .Calculate the values of E_(a) and A.
Answer» SOLUTION :Calculation of `E_(a)`: `log(K_(2))/(k_(1))=(E_(a))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` `log (0.07 s^(-1))/(0.02 s^(-1))=(E_(a))/(2.303xx8.314 JK^(-1)mol^(-1))xx((700-500)/(700xx500))` `therefore log 3.5=(E_(a)xx200)/(2.303xx8.314xx700xx500)J mol^(-1)` `therefore E_(a)(0.5441xx2.303xx8.314xx700xx500)/(200) J mol^(-1)` Calculation of A: `K=Ae^(-(E_(a))/(RT))` Where,`E_(a)=18231 J mol^(-1)` 500 K temp.`K_(1)=0.02 s^(-1)`

Discussion

24847.	The rate constant of a reaction at 400 and 200 K are 0.04 and 0.02s^(-1) respectively. Calculate the value of activation energy .
Answer» Solution :According the ARRHENIUS equation `log(k_2/k_2)=E_a/(2.303R)((T_2-T_1)/(T_1T_2))` `T_2 =400K , k_2=0.04s^(-1)` `T_1=200K , k_1=0.02s^(-1)` `log(0.04s^(-1)/(0.02s^(-1)))=(E_a)/(2.303xx8.314"J K "^(-1) MOL^(-1))((400K-200K)/(200Kxx400K))` `log(2)=E_a/(2.303xx8.314" J K "^(-1)mol^(-1))((1)/(400K))` `E_a=log(2)xx2.303xx8.314"JK "^(-1)"mol"^(-1)xx400K` `E_a=2305"J mol"^(-1)`

Discussion

24848.	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 :`18.23 "KJ MOL"^(-1)`, 1.603

Discussion

24849.	The rate constant of a reaction at 200K is 10 times less than rate constant at 400K. Value of E_a? [R = gas constant]
Answer» `1842.4` R `921.2` R `460.6` R `230.3` R Answer :B

Discussion

24850.	The rate constant of a raction at 500 K and 700 K are 0.02 s^(-1) and 0.07s^(-1) respectively. Calculate the value of E_(a)
Answer» SOLUTION :18.230 KJ

Discussion

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