94420 + Interview Questions in Chemistry in Class 12 Page 495 InterviewSolution

24701.	The rate of a first order reaction is 1.5 xx 10^(-2) mol L^(-1) "min"^(-1) at 0.5 M concentration of the reactant . The half life of the reaction is
Answer» 8.73 min 7.53 min 0.383 min 23.1 min Solution :Rate `((DX)/(dt))` = k.c, `1.5 xx 10^(-2) = k xx 0.5` For FIRST order k = `(1.5 xx 10^(-2))/(0.5) = 3 xx 10^(-2) "MINUTE"^(-1)` `t_(1//2) = (.693)/(k) = (.693)/(3 xx 10^(-2)) = 23.1` minute.

Discussion

24702.	The rate of a first order reaction is 0.04" mol L"^(-1)s^(-1) at 10 seconds and 0.03" mol L"^(-1)s^(-1) at 20 seconds after initiation of the reaction. The half-life period of the reaction is
Answer» `44.1" s"` `54.1" s"` `24.1" s"` `34.1" s"` Solution :For a FIRST order reaction. `A to` product, and for concentration of the REACTANT at TWO different TIMES `k = (2303)/((2)-t_(1)) log""([A]_(1))/([A]_(2))` Also for first order reaction, rate `alpha` [A]. Hence `k=(2303)/(t_(2)-t_(1))log""(("Rate")_(1))/(("Rate")_(2))` `=(2303)/((20-10))log((004)/(003))=00287 s^(-1)` `t_(1//2)=(0693)/(k)=(0693)/(00287s^(-1))=2414s`

Discussion

24703.	The rate of a first order reaction is 0 04 mol L^-1s^-1 at 10 seconds and 003 molL^-1s^-1at 20 seconds after initiation of the reaction. The half life period of the reaction is
Answer» 24.1 s 34.1 s 44.1 s 54.1 s Answer :A

Discussion

24704.	The rate of a first order reaction is 0.69xx10^(-2) min^(-1) and the initial concentration is 0.5mol L^(-1) . The half life period is
Answer» 3000 s 0.33 s 50 s 100 s Solution :(A) Rate = K[C] `0.69xx10^(-2) = k[0.5]` or `k = (0.60xx10^(-2))/(0.5)` `t_(1//2) = (0.693)/(k) = (0.693xx0.5)/(0.693xx10^(2))` 50 min = 3000 sec

Discussion

24705.	The rate of a first order reaction is 0.04 mol l^(-1) s^(-1) at 10 seconds and 0.03 mol l^(-1) s^(-1) at 20 seconds after initiation of the reaction . The half-life period of the reaction is
Answer» 24.1 s 34.1 s 44.1 s 54.1 s Solution :`K = (2.303)/((t_(2) - t_(1))) "log" ((a-x_(1)))/((a-x_(2))) K = (2.303)/((20-10)) "log" ((0.04)/(0.03))` `K = (2.303 XX 0.1249 )/(10) (2.303 xx "log" 2)/(t_(1//2) = (2.303 xx "log" 2)/(10)` `t_(1//2) = (0.3010 xx 10)/(0.1249) = 24.1` SEC

Discussion

24706.	The rate of a first-order reaction is 0.04 mol l^(-1) s^(-1) at 10 seconds and 0.03 mol l^(-1)s^(-1) at 20 seconds after initation of the reaction.The half life perriod of the reaction is ……
Answer» 34.1 44.1 s 54.1s 24.1 s Solution :`K=(2.303)/(10)`log`(0.04)/(0.03)=(2.303xx0.124)/(10)` `t_((1)/(2))=(2.303xx0.3010xx10)/(2.303xx0.124)=24.27` second

Discussion

24707.	the rate ofa first- orderreactionis 0.04 molL^(-1) s^(-1)at 10sec and 0.03 mol L^(-1) S^(-1)at 20 sec afterinitiationof thereaction the half-lifeperiod of the reaction is
Answer» `34.1s` 44.1 s 54.1s 24.1 Solution :Givenorder of reaction =1 Rateof reaction at 10 s=0.04 MOL `L^(-1)S^(-1)` rateof reactionat 20s=0.03 `mol L^(-1)S^(-1)` ` therefore ` half - lifeperiod `(t_(1//2))=?` we have the equation forrate- constant 'K' in firstorderreaction . `K =(2303)/(t) log""(A_(t))/(A_(0))=(2.303)/(105 ) log""(0.04)/(0.03)` ` =(2.303)/(105)xx0.124` `k=0.028s^(-1)` weknow that `t_(1//2)=(0.693)/(k) =(0.693)/(0.028773391S^(-1))` `=24.14 s=24.1S`

Discussion

24708.	The rate of a first order reaction is 0.04" mil L"^(-1)s^(-1) at 10 minutes and 0.03" mol L"^(-1)s^(-1) at 20 minutes after initiation. Find the half life of the reaction.
Answer» Solution :For a FIRST order reaction, `ATO"PRODUCTS, for concentration of the reactant at two different TIMES,"` `k=(2.303)/(t_(2)-t_(1))log""([A]_(1))/([A]_(2))` But as rate `=k[A]," therefore, "(("rate")_(1))/((rate)_(2))=([A]_(1))/([A]_(2))` Hence, `k=(2.303)/(t_(2)-t_(1))log""(("rate")_(1))/(("rate")_(2))=(2.303)/((20-10)min)log""(0.04)/(0.03)=2.88xx10^(-2)min^(-1)` `:.t_(1//2)=(0.693)/(k)=(0.693)/(2.88xx10^(-2)min^(-1))=24.06" min "=1443.6 s.`

Discussion

24709.	The rate of a first order reaction has been found to be 2.45xx10^(-6) "mol L"^(-1) s^(-1) at concentration C_(1). The value of C_(1) is (rate constant = 3.5xx10^(-5)) :
Answer» 7.0 MOL `L^(-1)` 0.7 mol `L^(-1)` 0.07 mol `L^(-1)` 2.1 mol `L^(-1)` Solution :(C ) Rate = K(C ) `2.45 xx10^(-6) = 3.5 xx10^(-5) xxC_(1)` or `C_(1) = (2.45xx10^(-6))/(35xx10^(-5))= 0.07 "mol L"^(-1)` or `([A])/([A]_(0))= 1.77` `([A]_(0))/([A])= (1)/(1.77) = 56 .35 %` `:. ` Extent of reaction `= 100 - 56.3 = 43.75 %`

Discussion

24710.	The rate of a chemical reaction doubles for every 10^(@)C rise of temperature. If the temperature is raised by 50^(@)C, the rate of the reaction increases by about :
Answer» 24 times 32 times 64 times 10 times Solution :(B) : `roverset(10^(@)"rise")to2" r "OVERSET("Further "10^(@)" rise")to4" r "overset("Further "10^(@)" rise")to8" r "overset("Further "10^(@)" rise")to16" r "overset("Further "10^(@)" rise")to32" r"` Alternatively, as rate of reaction is doubled for EVERY `10^(@)` rise in temperature, increase in reaction rate `=2^(5)=32` times.

Discussion

24711.	The rate of a chemical reaction is increased in presence of a catalyst. This is because
Answer» activation ENERGY of the REACTION is less in the new path heat of reaction is decreased threshold energy is INCREASED activation energy of the new path is more ANSWER :A

Discussion

24712.	The rate of a chemical reaction doulets for every 10^(@)C rise of temperature. If the temperature is raised by 50^(@)C, the rate of the reaction by about,
Answer» 10 times 24 times 32 times 64times Solution :SINCE for every `10^(@)C` rise in TEMPERATURE increase in REACTION RATE`=2^(5)=3` times

Discussion

24713.	The rate of a chemical reaction doubles for every 10^(@)C rise in temperature. If the temperature is increased by 60^(@)C the rate of reaction increases by about
Answer» 20 times 32 times 64 times 128 times Solution :`x=60/10=6,(r_(2))/(r_(1))=MU^(x),r_(2)=r_(1)xx2^(6)=r_(1)xx64`

Discussion

24714.	The rate of a chemical reaction doubles for every 10^@C rise in temperature. If .the rate is increased by 60^@C, the rate of reaction increases by:
Answer» 29 times 32 times 64 times 128 times Answer :C

Discussion

24715.	The rate of a chemical reaction doubles for every 10^@C rise in temperature. If the rate is increased by 60^@C, the rate of reaction increases by:
Answer» 29 times 32 times 64 times 128 times Answer :C

Discussion

24716.	The rate of a chemical reaction doubles for every 10^(@) C rise of temperature . If the temperature is raised by 50^(@) C the rate of the reaction increases by about :
Answer» 32 times 64 times 10 times 24 times SOLUTION :(A) INCREASE in reaction rate `=2^(5)` = 32 times

Discussion

24717.	The rate of a chemical reaction doubles for an increases of 10 K in absolute temperature from 298 K. Calculate E_e.
Answer» Solution :log `k_2/k_1=E_1/(2.303R)[(T_2-T_1)/(T_1T_2)]` SUBSTITUTING we GET ,log 2=`(E_axx10)/(2.303xx8.314xx10^(-3)xx298xx308)` `thereforeE_a=(0.3010xx2.303xx8.314xx10^(-3)xx298xx308)/10=52089KJmol^(-1)`

Discussion

24718.	The rate of a chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_(a).
Answer» SOLUTION :52.898 KJ.

Discussion

24719.	The rate of a chemical reaction decreases as the reaction proceeds. This is because
Answer» The reactant concentration remain CONSTANT as the reaction proceeds. The PRODUCT concentration REMAINS constant as the reaction proceeds. Concentration of reactant decreases from time to time, as the reaction progresses Concentration of product decreases from time to time, as the reaction progresses Answer :C

Discussion

24720.	The rate of a chemical reaction…..
Answer» INCREASES as the REACTION proceeds. decreases as the reaction proceeds may INCREASE or decrease during the reaction. remains CONSTANT as the reaction proceeds. Answer :B

Discussion

24721.	The rate of a certain reaction is given by, rate = k[H^(+)]^(n). The rate increases 100 times when the pH changes from 3 to 1. The order (n) of the reaction is
Answer» 2 0 1 1.5 Solution : RATE (r) = `k[H^+]^N` When pH = 3 , `[H^+]` = `10^(-3)` and when pH = 1 , `[H^+]`=`10^(-1)` `THEREFORE (r_1)/(r_2)=(k(10^(-3))^n)/(k(10^(-1))^n) implies (1)/(100) =((10^(-3))/(10^(-1)))^n (therefore r_2=100 r_1)` `implies (10^(-2))^1=(10^(-2))^n implies n=1`

Discussion

24722.	The rate of a certain reaction increases by 2.5 times when the temperature is raised from 300K at 310K. If k is the rate constant at 300K then the rate constant at 310 K will be equal to
Answer» k 2K 2.5k 3k Solution :RATE of reaction `prop` Rate constant. As rate of reaction increases 2.5 times, the rate constant will BECOME 2.5 times. (ALTERNATIVELY, TEMPERATURE coefficient = 2.5).

Discussion

24723.	The rate of a certain reaction depends on concentration according to equation -(d[A])/(dt)=(k_(1)[A])/(1+k_(2)[A]) What will be the order of reaction when (i) concentration is very high (ii) very low ?
Answer» Solution :Given, `-(d[A])/(dt)=(k_(1)[A])/(1+k_(2)[A])` `IMPLIES(-d[A])/(dt)=(k_(1))/((1)/([A])+k_(2))` `(i)` When `[A]` is very high `(1)/([A])` is very SMAL , and thus negligible `:.-(d[A])/(dt)=(k_(1))/(k_(2))=` constant Thus, ORDER of reaction is ZERO. `(ii)` When `[A]` is very low. `[A+k_(2)[A]=k.` `-(d[A])/(dt)=(k_(1)[A])/(k.)=k.[A]` Thus order of reaction is one.

Discussion

24724.	The rate of a certain biochemical reaction catalyesd by an enzyme in human body is 10^4 times faster than when it carried out in the laboratory. The activation energy of this reaction :
Answer» Is zero Is DIFFERENT in TWO cases Is the same in both the cases None Answer :B

Discussion

24725.	The rate law of the reaction RCl+NaOHtoROH+NaCl is given by Rate =k[RCI]. The rate of this reaction
Answer» A & B B & C C & D B & D Answer :B

Discussion

24726.	The rate law of the reaction H_2 + Br_2to 2HBr is(d[HBr])/(dt) = k [H_2][Br_2]^(1//2)What is the order of the reaction? Is this an elementary reaction?
Answer» SOLUTION :`3/2, No`

Discussion

24727.	the rate law of the reaction A to Product is : rate = k[A]. It has been graphically represented. What is the rate constant for the reaction?
Answer» Solution :From the graph: Cast I: Rate = `K[A]` `1 xx 10^(-3) Ms^(-1) = k(0.1 M)` `k=(1 xx 10^(-3)Ms^(-1))/(0.1M) = 10^(-2)s^(-1)`

Discussion

24728.	The rate law of the reaction, A+2B toProduct , is given by d["Product"]//dt=K[A]^(2)*[B]. If A is taken in large excess, the order of the reaction will be
Answer» `0` `1` `2` `3` ANSWER :B

Discussion

24729.	The rate law of the reaction, 2A+ B to 2AC is represented as Rate = K [A]^(2) [B]. If A is taken in large excess, the order of the reaction will be,
Answer» Zero 1 2 3 Answer :B

Discussion

24730.	The rate law of nitration of benzene using a mixture of nitric acid and sulphuric acid is
Answer» RATE `=k ["BENZENE"] [NO_(2)^(+)] [H^(+)]` rate `= k [HNO_(3)] [H_(2)SO_(4)]` rate `= k ["benzene"] [NO_(2)^(+)]` rate `= k ["benzene"] [H_(2)SO_(4)]` Solution :Nitration of benzene is a second order reaction (FIRST order with respect to benzene and first order with respect to `NO_(2)^(+)` ) LTBR. `:.` Rate = `k` [Benzene] `[NO_(2)^(+)]`

Discussion

24731.	The rate law of a chemical reaction : 2NO+O_(2) to 2NO_(2), is given as rate =k[NO]^(2)[O_(2)]. How will the rate of reaction change if the volume of reaction vessel is reduced to 1//4th of its original value ?
Answer» Solution :For, `2NO+O_(2)to2NO_(2)` Rate `=k[NO]^(2)[O_(2)]` Let a MOLES of `NO` and `B` moles of `O_(2)` be taken to start a reaction at any time in a vessel of V litre `r_(1)=k[(a)/(V)]^(2)[(b)/(V)]`…………`(i)` If VOLUME of vessel is reduced to `V//4`, then for same moles of `NO` and `O_(2)` `r_(2)=k[(a)/(V//4)]^(2)[(b)/(V//4)]=64k[(a)/(V)]^(2)[(b)/(V)]`................`(ii)` By Eqs. `(i)` and `(ii)` , `(r_(2))/(r_(1))=64` `r_(2)` is `64` times of `r_(1)`

Discussion

24732.	The rate law from 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 as
Answer» (m + N) (n - m) `2^((n-m))` `1/(2^((m+n)))` ANSWER :C

Discussion

24733.	The rate law of a chemical reaction given below : 2NO(g) + O_(2)(g) to 2NO_(2)(g) is given as rate =k[NO]^(2).[O_(2)]. How will the rate of reaction change if the volume of reaction vessel is reduced to 1/4th of its original value ?
Answer» ANSWER :64

Discussion

24734.	The rate law of a reaction is : Rate = k[A]^(2)[B] On doubling the concentration of both A and B the rate x will become :
Answer» `4x^(2)` 8X `x^(3)` 9x Solution :(B) Let rate = `k [A]^(2) [B] =x ` rate `= k [2A]^(2) [2B] = 8 b[A]^(2) [B] =8x `

Discussion

24735.	The rate law for the reaction RCL + NaOH(aq) rarr ROH + NaCl is given by Rate = k[RCl]. The rate of the reaction will be
Answer» unaffected by increasing temperature of the reaction doubled on doubling the concentration of `NaOH` HALVED on REDUCING the concentration of `NaOH` to one HALF halved on reducing the concentration of `RCL` to one half Solution :`R_(1) = k[RCl]^(1)[NaOH]^(0)` `R_(2) = k[0.5 RCl]^(1)[NaOH]` `:. R_(2) = 0.5 R_(1)`

Discussion

24736.	The rate law for the reactionSucrose + Water overset([H^(+)])to Glucose + Fructose is given by :
Answer» Rate = K [sucrose] [water] Rate= K [sucrose] `["water"]^0` Rate = K `["sucrose"]^0` [water] Rate = K `["sucrose"]^(1//2) ["water"]^(1//2)` ANSWER :B

Discussion

24737.	The rate law for the reaction x A + y B to mP + nQis Rate = k[A]^(c) [D]^(d) . What is the total order of the reaction
Answer» (x+ y) (m + N) ( C + d) x/y Solution :Orderis the sum of the powers to which the concentration TERMS are RAISED in the rate equation.

Discussion

24738.	The rate law for the single step reaction 2A + B to 2C is given by
Answer» RATE=K[A].[B] `Rate=K[A]^2.[B]` Rate = K[2A].[B] `Rate=K[A]^2[B]^o` Answer :B

Discussion

24739.	The rate law for the reaction involving BrO_(3)^(-) and Br^(-) in acid mediumis : -(d[BrO_(3)^(-)])/(dt)=k[BrO_(3)^(-)][Br^(-)][H^(+)]^(2) Which of the following comments regarding the reaction are ture -
Answer» rate does not depend on the concentration of acid rate change with the variation of pH of the solution doubling `[H^(+)]` makes QUADRUPLES the rate it is a pseudo FIRST order REACTION Answer :A::B::C

Discussion

24740.	The rate law for the reaction RCl + NaOH (aq) to ROH + NaCl is given by , Rate = k_(1)[RCl]. The rate of the reaction will be
Answer» DOUBLED on doubling the concentration of sodium hydroxide HALVED on REDUCING the concentration of alkyl halide to one half Increased on increasing the temperature of the reaction Unaffected by increasing the temperature of the reaction. Solution :As RATE k = [RCl] , on decreasing the concentration of RCl to half , the rate will ALSO be halved . Rate will also increase with temperature .

Discussion

24741.	The rate law for the reaction 2H_(2(g))+2NO_((g)) to N_(2(g))+2H_(2)O_((g)) is given by rate =-=k[H_(2)][NO]^(2) The reaction occurs in the following two steps: (a) H_(2(g))+2NO_((g)) to N_(2)O_((g))_H_(2)O_((g)) N_(2)O_((g))+ H_(2(g)) to N_(2(g))+H_(2)O_((g)) What is the rate of N_(2)O in the mechanism ? What is the molecularity of each of the elementary stpes ?
Answer» SOLUTION :Rate of `N_(2)O=K[N_(2)O][H_(2)O]` Molecularity for 1ST STEP is 3 Molecularity for 2nd step is 2.

Discussion

24742.	Theratelaw for thereaction C_(2)H_(4)Br_(2)+ 3I^(-) to C_(2) H_(4) + 2Br^(-)+ I_(3)^(-) isRate = k[C_(2)H_(4) Br_(2)][I^(-) ]. Therate of the reactionis found to be1.1 xx 10^(-4) M//s whenthe concentrations of C_(2) H_(4) Br_(2)and I^(-) are0.12 Mand 0.18 M respectively. Calculatethe rateconstantof thereaction.
Answer» SOLUTION :Given :`C_(2) H_(4)Br_(2)+ 3I^(-) to C_(2) H_(4) + 2Br^(-) + I_(3)^(-)` Byrate LAW,Rateof REACTION `=R = k XX [ C_(2) H_(4) Br_(2) ][I^(-)]` `R= 1.1xx 10^(-4)Ms^(-1)` `[C_(2) H_(4)Br_(2) ]=0.12 M , [I^(-) ]= 0.18 M` Rateconstant = k = ? `R= k xx [C_(2)H_(4) Br_(2) ] xx [I^(-)]` `Ms^(-1)` `:. k= (R)/([C_(2) H_(4) Br_(2) ] xx [I^(-)])= (1.1 xx 10^(-4))/(underset(M)(0.12)xx underset(M)0.18) = 5.1 xx 10^(-3) M^(-1) s^(-1)`

Discussion

24743.	The rate law for the reaction: Ester +H^(+)overset("Fast")to Acid + alcohol is: -dx/dt = k["Ester"][H^(+)]^(@) What would be the effect on the rate if the concentration of acid is doubled?
Answer» SOLUTION :RATE of REACTION will REMAIN UNCHANGED.

Discussion

24744.	The rate law for the reaction : Ester +H^(+)to"Acid"+"Alcohol is ": (dx)/(dt)=k["Ester"][H^(+)]^(0) What would be the effect on the rate if (i) concentration of the ester is doubled ? (ii) concentration of H^(+) is doubled ?
Answer» SOLUTION :(i) The rate of REACTION will be doubled. (II) No effect on rate.

Discussion

24745.	The rate law for the decomposition of N_(2)O_(5) is : rate =k[N_(2)O_(5)]. What is the significance of k in this equation ?
Answer» Solution :K REPRESENTS the RATE constant or specific reaction rate for the reaction and is equal to the rate of reaction when concentration of `N_(2)O_(5)` is `1" mol L"^(-1)`.

Discussion

24746.	The rate law for the decomposition of N_(2)O_(5) is given as: Rate =k[N_(2)O_(5)]. What is the significance of k in the equation?
Answer» Solution :k represents the rate constant for the reaction. It becomes equal to the reaction rate when the molar concentration of `N_(2)O_(5)` is 1 MOL `L^(-1)`.

Discussion

24747.	The rate law for reaction A+B rarr C is Rate =K[A] [B] Given K=6.93xx10^(-4)M^(-1)sec^(-1) Find the time taken(sec) when concentration of [A] changes from 10^(-4) M to 5xx10^(-5) M Given: [B] =1M
Answer» 10 100 1000 10000 Answer :C

Discussion

24748.	The rare law for a reaction between th substances A and B is given by Rate =k[A]^(m)[B]^(n) 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 as
Answer» `(m+n)` `(n-m)` `2^((m-n))` `(1)/(2^((m+n)))` ANSWER :C

Discussion

24749.	The rate law for the chemical reaction : 2NO_(2)Clrarr 2NO_(2)+Cl_(2) is :The rate determining step is :
Answer» `2NO_(2)Cl RARR 2NO_(2)+2CL` `NO_(2)+Cl_(2) rarr NO_(2)Cl+Cl` `NO_(2)Cl+Clrarr NO_(2)+Cl_(2)` `NO_(2)ClrarrNO_(2)+Cl` ANSWER :D

Discussion

24750.	The rate law for a reaction of A,B and C has been found to be rate = k[A]^2[B][L]^(3//2) How would the rate of reaction Change when (i) Concentration of [L] is quadrupled (ii) Concentration of both [A] and [B] are doubled (iii) Concentration of [A] is halved (iv) Concentration of [A] isreduced to (1//3)and consideration of [L] is quadrupled .
Answer» SOLUTION :Rate `=k[A]^2[B][L]^(3//2)........(1) ` (i) when [L]=[4L] Rate `=k[A]^2[B][4L]^(3//2)` Rate = 8 `(k[A]^2[B][L]^(3//2))......(2)` Comparing (1) and (2) , rate is increased by 8 times. (ii) when [A] = [2A] and [B] = [2B] Rate = `k[2A]^2[2B][L]^(3//2)` Rate = `8(k[A]^2[B][L]^(3//2) )............(3)` Comparing (1) and (3) , rate is increased by 8 times . (iii) When `[A] = (([A])/(2))` Rate `k=(([A]]/2)[B][L]^(3//2)` Rate `=(1/4)(k[A]^2[B][L]^(3//2))...........(4)` Comparing (1) and (4) , rate is reduced to `1/4` times. (IV) when [A] `= (([A])/(3))` and [L]=[4L] Rate `=k[A/3]^2[B][4L]^(3//2)` Rate `=(8/9)(k[A]^2[L]^(3//2)).........(5)` comparing (1) and (5) , rate is reduced to 8/9 times .

Discussion

Explore topic-wise InterviewSolutions in .

The rate of a first order reaction is 1.5 xx 10^(-2) mol L^(-1) "min"^(-1) at 0.5 M concentration of the reactant . The half life of the reaction is

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the rate ofa first- orderreactionis 0.04 molL^(-1) s^(-1)at 10sec and 0.03 mol L^(-1) S^(-1)at 20 sec afterinitiationof thereaction the half-lifeperiod of the reaction is

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The rate of a chemical reaction doubles for every 10^(@)C rise of temperature. If the temperature is raised by 50^(@)C, the rate of the reaction increases by about :

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The rate of a chemical reaction doulets for every 10^(@)C rise of temperature. If the temperature is raised by 50^(@)C, the rate of the reaction by about,

The rate of a chemical reaction doubles for every 10^(@)C rise in temperature. If the temperature is increased by 60^(@)C the rate of reaction increases by about

The rate of a chemical reaction doubles for every 10^@C rise in temperature. If .the rate is increased by 60^@C, the rate of reaction increases by:

The rate of a chemical reaction doubles for every 10^@C rise in temperature. If the rate is increased by 60^@C, the rate of reaction increases by:

The rate of a chemical reaction doubles for every 10^(@) C rise of temperature . If the temperature is raised by 50^(@) C the rate of the reaction increases by about :

The rate of a chemical reaction doubles for an increases of 10 K in absolute temperature from 298 K. Calculate E_e.

The rate of a chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_(a).

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The rate of a chemical reaction…..

The rate of a certain reaction is given by, rate = k[H^(+)]^(n). The rate increases 100 times when the pH changes from 3 to 1. The order (n) of the reaction is

The rate of a certain reaction increases by 2.5 times when the temperature is raised from 300K at 310K. If k is the rate constant at 300K then the rate constant at 310 K will be equal to

The rate of a certain reaction depends on concentration according to equation -(d[A])/(dt)=(k_(1)[A])/(1+k_(2)[A]) What will be the order of reaction when (i) concentration is very high (ii) very low ?

The rate of a certain biochemical reaction catalyesd by an enzyme in human body is 10^4 times faster than when it carried out in the laboratory. The activation energy of this reaction :

The rate law of the reaction RCl+NaOHtoROH+NaCl is given by Rate =k[RCI]. The rate of this reaction

The rate law of the reaction H_2 + Br_2to 2HBr is(d[HBr])/(dt) = k [H_2][Br_2]^(1//2)What is the order of the reaction? Is this an elementary reaction?

the rate law of the reaction A to Product is : rate = k[A]. It has been graphically represented. What is the rate constant for the reaction?

The rate law of the reaction, A+2B toProduct , is given by d["Product"]//dt=K[A]^(2)*[B]. If A is taken in large excess, the order of the reaction will be

The rate law of the reaction, 2A+ B to 2AC is represented as Rate = K [A]^(2) [B]. If A is taken in large excess, the order of the reaction will be,

The rate law of nitration of benzene using a mixture of nitric acid and sulphuric acid is

The rate law of a chemical reaction : 2NO+O_(2) to 2NO_(2), is given as rate =k[NO]^(2)[O_(2)]. How will the rate of reaction change if the volume of reaction vessel is reduced to 1//4th of its original value ?

The rate law from 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 as

The rate law of a chemical reaction given below : 2NO(g) + O_(2)(g) to 2NO_(2)(g) is given as rate =k[NO]^(2).[O_(2)]. How will the rate of reaction change if the volume of reaction vessel is reduced to 1/4th of its original value ?

The rate law of a reaction is : Rate = k[A]^(2)[B] On doubling the concentration of both A and B the rate x will become :

The rate law for the reaction RCL + NaOH(aq) rarr ROH + NaCl is given by Rate = k[RCl]. The rate of the reaction will be

The rate law for the reactionSucrose + Water overset([H^(+)])to Glucose + Fructose is given by :

The rate law for the reaction x A + y B to mP + nQis Rate = k[A]^(c) [D]^(d) . What is the total order of the reaction

The rate law for the single step reaction 2A + B to 2C is given by

The rate law for the reaction involving BrO_(3)^(-) and Br^(-) in acid mediumis : -(d[BrO_(3)^(-)])/(dt)=k[BrO_(3)^(-)][Br^(-)][H^(+)]^(2) Which of the following comments regarding the reaction are ture -

The rate law for the reaction RCl + NaOH (aq) to ROH + NaCl is given by , Rate = k_(1)[RCl]. The rate of the reaction will be

The rate law for the reaction: Ester +H^(+)overset("Fast")to Acid + alcohol is: -dx/dt = k["Ester"][H^(+)]^(@) What would be the effect on the rate if the concentration of acid is doubled?

The rate law for the reaction : Ester +H^(+)to"Acid"+"Alcohol is ": (dx)/(dt)=k["Ester"][H^(+)]^(0) What would be the effect on the rate if (i) concentration of the ester is doubled ? (ii) concentration of H^(+) is doubled ?

The rate law for the decomposition of N_(2)O_(5) is : rate =k[N_(2)O_(5)]. What is the significance of k in this equation ?

The rate law for the decomposition of N_(2)O_(5) is given as: Rate =k[N_(2)O_(5)]. What is the significance of k in the equation?

The rate law for reaction A+B rarr C is Rate =K[A] [B] Given K=6.93xx10^(-4)M^(-1)sec^(-1) Find the time taken(sec) when concentration of [A] changes from 10^(-4) M to 5xx10^(-5) M Given: [B] =1M

The rare law for a reaction between th substances A and B is given by Rate =k[A]^(m)[B]^(n) 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 as

The rate law for the chemical reaction : 2NO_(2)Clrarr 2NO_(2)+Cl_(2) is :The rate determining step is :