94420 + Interview Questions in Chemistry in Class 12 Page 1756 InterviewSolution

87751.	A reaction rate constant is given by k = 1.2 xx 10^(14) e^(-(25000//RT)) sec^(-1) . It means
Answer» log k versus log T will GIVE a straight line with slope as `- 25000` log k versus T will give a straight line with slope as `- 25000` log k versus log 1/T will give a straight line with slope `-25000` log k versus 1/T will give a straight line Solution :ACCORDING to the Arrhenius equation a straight line is to be obtained by plotting the logarithm of the rate CONSTANT of a chemical reaction (log k) against 1/T.

Discussion

87752.	A reaction proceeds with the energy of activation of 55.3 kcal/mole. If AH of the reaction is 1 kcal, what would be the energy of activation of the reverse reaction?
Answer» SOLUTION :54.3kcal

Discussion

87753.	A reaction proceeds in three stages. The first stage is a slow and involves two molecules of reactants. The second and third stage are fast. The overall order of the reaction is
Answer» first order SECOND order third order zero order SOLUTION :Rate is DETERMINED only is slow STEP of mechanism.

Discussion

87754.	A reaction proceeds in three stages. The first stage is a slow and involves two molecules of reactants .The second and third stage are fast .The overall order of the reaction is :
Answer» FIRST order Second order Third order Zero order Answer :B

Discussion

87755.	A reaction proceeds by a two step mechanism A_2underset(k_2)overset(k_1)hArr2A (fast reaction) A+B to Products (slow reaction) What is the rate law for the overallreaction ?
Answer» rate `=k[A_2][B]` rate `=k[A_2]^2[B]` rate `=k[A_2]^(1/2)` rate `=k[A_2]^(1/2)[B]` Solution :From `A_(2) to 2 A` (FAST reaction ) `k = ([A]^(2))/([A_(2)]) or [A]^(2) = k XX [A_(2)]` or `A = sqrt(k xx [A_(2)]) "" … (i)` Form ` A + B to` Products (slow reaction ) , the rate LAW is :` (dx)/(dt) = k [A] [ B]` Substituting the value of [A] from (i) in (II) , `(dx)/(dt) = "rate" = k sqrt(k xx [ A_(2)]) xx [B]` `= k xx sqrtk xx [A_(2)]^(1//2) xx [ B]= k [A_(2)]^(1//2) xx [B]`

Discussion

87756.	A reaction P to Q is completed 25% in 25 min, 50% completed in 25 min if [P] is halved, 25% completed in 50 min if [P] is doubled. The order of reaction is
Answer» 1 2 0 3 Solution :By hit and trial method, if reaction is of zero order, `K = (1)/(t){[A]_(0)-[A]}` Case I. `k = (1)/(25) XX (25)/(100) = 10^(-2)` Case II. `k = (1)/(25) xx (1)/(2) {[A]_(0)-[A]}` (as[P] is HALVED) `= (1)/(25) xx (1)/(2) xx (50)/(100) = 10^(-2)` Case III. `k = (1)/(50) xx 2{[A]_(0)-[A]}` `""` (as [P] is doubled ) ` = (1)/(50)k xx 2 xx (25)/(100) = 10^(-2)` As k comes out to be constant, hence the given reaction is of zero order.

Discussion

87757.	(A) Reaction of HI with ter-butylethylether produces ter.butyliodide & ethyl alcohol(R) It follows SN^1reaction
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :A

Discussion

87758.	A reactionof firstorderwithrespect to reactantA andsecondorderwithrespectto reactantB. the rate lawforthe reactionis givenby
Answer» `"Rate"=K[A][B]^(2)` Rate `=[A][B]^(2)` Rate `=[A]^(2)[B]` Rate `=k [A]^(0)[B]^(2)` ANSWER :A

Discussion

87759.	(A) Reaction of Cl_(2) with hot and concen -trated NaOH is disproportionation reaction (R ) Oxidation state of chlorine changed from 0"to"-1and+5
Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is FALSE Both (A) and (R ) are false Answer :A

Discussion

87760.	A: Reaction of carboxylic acid with hydrazoic acid in the presence of conc. H_(2)SO_(4) to form primary amine is called Schmidt reaction. R: In this reaction isocyanate intermediate is formed.
Answer» If both Assertion & REASON are TRUE and the reason is the CORRECT explanation of the assertion, then mark (1). If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2) If Assertion is true STATEMENT but Reason is false, then mark (3) If both Assertion and Reason are false statements, then mark (4) Answer :2

Discussion

87761.	(A) Reaction of benzaldehyde with formaldehyde in presence of base is crossed Cannizzaro's reaction. (R) Formaldehyde is always oxidized in this reaction because it is better oxidising agent.
Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) is false Answer :C

Discussion

87762.	A reaction of aryl diazonium salts that does not involved loss of nitrogen takes place when they react with phenol and aromatic amines. Aryl diazonium ion relatively is weak electrophile but has sufficient reactivity to attack strongly activated aromatic ring. The reaction is known as azo coupling. The coupling of diazonium ions with phenols or other electron rich aromatic compounds is useful commercial reaction as azo compounds are highly coloured and many of them are used as dyes. Coupling between arenediazonium cation and amines takes place most rapidly at pH
Answer» <P>`10-14` `0-2` `5-7` `5-8` Solution :Basic coupling takes place in weakly acidic MEDIUM `(P^(N)=5" to "7)`

Discussion

87763.	A reaction of aryl diazonium salts that does not involved loss of nitrogen takes place when they react with phenol and aromatic amines. Aryl diazonium ion relatively is weak electrophile but has sufficient reactivity to attack strongly activated aromatic ring. The reaction is known as azo coupling. The coupling of diazonium ions with phenols or other electron rich aromatic compounds is useful commercial reaction as azo compounds are highly coloured and many of them are used as dyes. Which of the folllowing is responsible for the colour of diazo compounds ?
Answer» Benzene RING `-N = N-`group `-OH` group `-NH_(2)` group Solution :`-N=N-` group

Discussion

87764.	A reaction occurs spontaneously if :
Answer» `TtriangleSgttriangleH and TRIANGLEH` is +ve and `TRIANGLES is -ve` `TtriangleS=triangleH`and both `triangleH `and` triangleS` are +ve `TtriangleSlttriangleH`and both `triangleH` and `triangleS` is +ve `TtriangleSgttriangleH`and both `triangleH` and `triangleS`are is +ve Answer :D

Discussion

87765.	A reaction is spontaneous if free energy change DeltaG , for the reaction is negative . The question arises, how we can explain , the spontaneous of reversible reaction taking the example of dissociation equilibrium N_2O_4 hArr 2NO_2(g) the variation of free energy with the fractionof N_2O_4 dissociated under standard conditions is shown in the figure below. When 1 mole of N_2O_4 changes into the equilibrium mixture the volume of DeltaG^@ is
Answer» 0.84 kJ `-0.84 kJ` 5.40 0.84 kJ 0.84-5.40 kJ Solution :The lowest POINT of the plot REPRESENTS EQUILIBRIUM mixture. When 1 mole of `N_2O_4` changes into equilibrium mixtture , free ENERGY is LOWERED by 0.84 kJ, `DeltaG=-0.84 KJ`.

Discussion

87766.	A reaction mixture for the combustion of SO_2 was prepared by opening a stopcockconnecting two separate chambers, one having a volume of 2.125 litres filled with SO_2 at 0.75 atm and the other having 1.5 litres volume filled with oxygen at 0.50 atm, both gases were at 80^@C.(i) What were the mole fractions of SO_2 and O_2in the mixture, and the total pressure?(ii) If the mixture was passed over a catalyst that promoted the formation of SO_3and was then returned to the original two connected vessels, what were the mole fractions in the final mixture and what was the final total pressure? Assume that the conversion of SO_2 is complete to the extent of the availability of O_2
Answer» SOLUTION :(i) 0.68,0.32, 0.64 ATM (II) 0.06, 0.94,0.44 atm

Discussion

87767.	A reaction is spontaneous if free energy change DeltaG , for the reaction is negative . The question arises, how we can explain , the spontaneous of reversible reaction taking the example of dissociation equilibrium N_2O_4 hArr 2NO_2(g) the variation of free energy with the fractionof N_2O_4 dissociated under standard conditions is shown in the figure below. The standard free energy change for the conversion of 1 mole of N_2O_4 into 2 moles of NO_2 is
Answer» 5.40 KJ `-5.40 KJ` (5.4+0.84 )KJ (5.40 +2 `XX`0.84)kJ Solution :`DeltaG=G` for 2 MOLES of `NO-2-G` for 1 moleof `N_2O_4` =5.40 kJ

Discussion

87768.	Areaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ?
Answer» SOLUTION :For a second order reaction, `R=k[A]^(2)` where A is a reactant. (i) If the concentration of A is doubled then, `R_(1)=k[2A]^(2)=4K[A]^(2)` `THEREFORE (R_(1))/(R)=(4k[A]^(2))/(k[A]^(2))=4` `therefore ` RATE will increase 4 time original rate. (ii) When the concentration of A is reduced to half, `R_(2)=k[(A)/(2)]^(2)=(1)/(4)k[A]^(2)` `therefore (R_(2))/(R)=(1//4xxk[a]^(2))/(k[A]^(2))=(1)/(4)` `therefore ` Rate will reduce to `1//4` th initial value.

Discussion

87769.	A reaction is spontaneous at low temperature but non-spontaneous at high temperature. Which of the following is true for the reaction
Answer» `DeltaH gt 0, DeltaS gt 0` `DeltaH LT 0, DeltaS gt 0` `DeltaH gt 0, DeltaS = 0` `DeltaH lt 0, DeltaS lt 0` SOLUTION :We know that `DeltaG=DeltaH-TDeltaS` When `DeltaHlt0` and `DeltaSlt0` then `DeltaG` will be NEGATIVE at LOW TEMPERATURES and the reaction will be spontaneous.

Discussion

87770.	A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Answer» SOLUTION :Rate `=k[A]^(2)=KA^(2)`. [Let [A] be REPRESENTED by a] If `[A]=2a, " rate"=k((a)/(2))^(2)=(1)/(4)ka^(2)`. The rate becomes one-fourth.

Discussion

87771.	A reaction is secondorder with respect to a reactant . How is the rate of reaction affected if the concentration of the reactant is : (i) doubled (ii) reduced to half.
Answer» Solution :(i) Reaction is SECOND ORDER with respect to the reactant `:. "Rate" = K [ A] ^2 = ka ^2` When [A]=2a Rate = `k (2a)^2=4 ka ^2` = 4 times Therefore , when concentration of the reactant is doubled the rate will become 4 times (ii) when `[A]= 1/2 a` Rate `=k(1/2a)^2=1/4ka^2=1/4k` Therefore , rate will be reduced to one - fourthe the initial rate.

Discussion

87772.	A reaction is said to be first order if it's rate is proportional to the concentration of reactant Let us consider a reaction A(g) rarr B(g) + c(g) At t=0 a 00 At time t a-x xx The rate of reaction is given by the expression (dx)/(dt)=k(a-x) and integrated rate equation for a given reaction is represted as k=1/t l n(a/(a-x)) where a=initial concentration and (a-x) = concentration of A after time t. Consider a reaction A(g) rarr 3B(g) + 2C(g) with rate constant 1.386xx10^(-2)min^(-1) starting with 2 moles of A in 12.5 litre vessel initially if reaction is allowed to take place at constant pressure and at allowed to take place at constant pressure and at 298 K then find the concentration of B after 100 min.
Answer» `0.04` MIN `0.36` M `0.09` M None of these Answer :C

Discussion

87773.	A reaction is said to be first order if it's rate is proportional to the concentration of reactant Let us consider a reaction A(g) rarr B(g) + c(g) At t=0 a 00 At time t a-x xx The rate of reaction is given by the expression (dx)/(dt)=k(a-x) and integrated rate equation for a given reaction is represted as k=1/t l n(a/(a-x)) where a=initial concentration and (a-x) = concentration of A after time t. Thermal decomposition of compound X is a first order reaction If 75% of X is decompossed in 100 min, how long will it take for 90% of the compound to decompose? [Given :log 2=0.30]
Answer» 190 MIN `176.66` min `166.66` min `156.66` min Answer :C

Discussion

87774.	A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) Doubled (ii) reduced to half ?
Answer» Solution :(i) ` r = K[A]^(2) rArr r = k[2A]^(2) rArr r4k[A]^(2)` Rate increases by four times. (ii) ` r = k[A]^(2)` `r = k[(A)/(2)]^(2) rArr = (1)/(4) k[A]^(2)`. Rate increases by four times.

Discussion

87775.	A reaction is represented by A overset(k_1)rarr B(slow ) and A+B overset(k_2)rarrC(fast) where k_1and k_2are the ratge constants of two steps. The rate of production of C will be given by
Answer» `k_1[A][B]` `k_1[A]` `k_1k_2[A]` `k_2[A][B]` Solution :The slow step `A overset(k_1)TOB`, is the rate determining step of the reaction,HENCE rate of production of C, i.e., `(d[C])/(dt)=k_1[A]`.

Discussion

87776.	A reaction is of the first order reaction to A and is of second order relative to B .what will be the effect on rateif the concentration of A and B are doubled
Answer» Velocity REMAINS constant 4 times 2 times 8 times Solution :`R = k[A] [B]^(2)` r' = k `[2A][2B]^(2) = 8k [A] [B]^(2) ` or r' = 8r .

Discussion

87777.	A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half ? What is the unit of rate constant of such a reaction ?
Answer» SOLUTION :RATE=`K[A]^2` Unit of `k=(MOL L^(-1))/S=k(mol^(-1))^2` `k=mol^(-1) L S^(-1)`

Discussion

87778.	A reaction is of second order in A and first order in B. How is the rate affected when the concentration of both A and B is doubled ?
Answer» SOLUTION :A reaction is SECOND order in A and first order in B When concentration of both A and B is doubled then, Rate `= k[2A]^(2)[2B]=8k [A]^(2)[B]=8` (INITIAL rate) This shows that rate will increase 8 times to the initial rate.

Discussion

87779.	A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Answer» SOLUTION :RAT `=k[R]^(2)` (i) If R is INCREASED to 2R, then Rate `=k[2R]^(2)=4k[R]^(2)` Thus, the rate increases four times. (II) If R is reduced to `(R )/(2)`, then Rate `=k[(R )/(2)]^(2)= (1)/(4) k[R]^(2)` Thus, the rate is reduced to one-fourth.

Discussion

87780.	A reaction is of second order in A and first order in B. How is the rate affected on increasing the concentration of A three times ?
Answer» Solution :A reaction is second order in A and first order in B When the concentration of A is increased three TIMES, (i.e) 3A, then Rate `= K[3A]^(2)[B]` `= 9k [A]^(2)[B]=0` (initial rate) This shows the rate will INCREASE 9 times to the initial rate.

Discussion

87781.	A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times. (ii) the concentrations of A as well as B are doubled?
Answer» Solution :`(dx)/(DT)= k[A][B]^(2) " Let "[A]=a, [B]=b, (dx)/(dt)=k.a.b^(2)` (i) When the concentration of B is increased to THREE TIMES, `(dx)/(dt)= ka(3b)^(2) or (dx)/(dt)=9kab^(2)` Thus the rate increase 9 times. (ii) When the concentration of A as well as B are DOUBLED, `(dx)/(dt)= k(2a)(2b)^(2)" or "(dx)/(dt)=k.2a.4b^(2) " or "(dx)/(dt)=8kab^(2)` Thus the rate increases 8 times.

Discussion

87782.	A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times, (ii) the concentrations of A as well as B are doubled ?
Answer» Solution :According to QUESTION the given reaction is rate EQUATION `:.` Rate = `k[A]^(1)[B]^(2)` (i) When concentration of B is tripled it means conc. Of B becomes 3 times Rate `= k[A][3B]^(2)` `:.` The rate of reaction becomes 9 times (ii) When conc. of A and B both is DOUBLED, then conc. of A becomes 2A and that of B becomes 2B. `:.` Rate `= k[2A][2B]^(2)` `= 8k[A][B]^(2)` Rate of the reaction becomes 8 times.

Discussion

87783.	A reaction is firt order in A and second order in B.
Answer» Solution :(i) Write the difference RATE equation . (ii)How is the rate affected on increasing the concentration of B three times ? (iii)How is the rate affected when the concentration of both A and B are doubled ? (i) `(dx)/(DT)=K[A]^1[B]^2` (ii) If concentration .B. is triple them the rate will become 9 times . (IV) When concentrationof both A and B are doubled , then the rate will become 8 times .

Discussion

87784.	A reaction is of first order in reactant A and of second order in reactant B. How is the rate of reaction affected when (i) concentration of B alone is increased to three times (ii) the concentration of A as well mass B is doubled ?
Answer» SOLUTION :`r=k[A][B]^(2)=kab^(2)"(i) "r=ka(3" b")^(2)=9" kab"^(2)" (ii) "r=k(2" a")(2" b")^(2)=8" kab"^(2)`.

Discussion

87785.	A reaction is first order with respect to the reactant A and Second roder with respect to the reactant B in a reaction. A + B to Product. (i) Write the differential rate equation. (ii) How is rate of reaction affected on increasing the concentration of B by 2 times.
Answer» Solution :(i) ` - (DX)/(DT) = k[A]^(1)[B]^(2)` (ii) ` - (dx)/(dt) = k[A][2B]^(2), (dx)/(dt) = 4k[A][B]^(2)` Rate of reaction INCREASES by four TIMES when concentration of B increases by 2 times.

Discussion

87786.	A reaction is first order w.r.t reactant A as well as w.r.t. B. Give the rate law. Also give one point of difference between average rate and instantaneous rate.
Answer» SOLUTION :RATE law `"rate = "k[A][B]` Difference between AVERAGE and instantaneous rate Average rate is equal to change in CONCENTRATION of reactants or products divided by the time taken for that change to occur. Instantaneous rate the rate of a reaction at a particular INSTANT.

Discussion

87787.	A reaction is first order with respect to the reactant A and second order with respect to the reactant B in a reaction A+Bto product How is the rate of the reaction affected on increasing tile concentration of B by two times.
Answer» SOLUTION :On increasing the concentration of B by two times the RATE of the REACTION increases by FOUR times.

Discussion

87788.	A reaction is first order with respect to the reactant A and second order with respect to the reactant B in a reaction A+Bto product Write the differential rate equation.
Answer» SOLUTION :`(DX)/(dx)=K[A][B]^2" "(dx)/(dt)=-K[A][B]^2`

Discussion

87789.	A reaction is first order with respect to reactant A and second order with respect to reactant B in a reaction A+Brarr product. i) Write the differential rate equation. ii) How is the rate of the reaction affected on increasing the concentration of B by two times?
Answer» Solution :i) The DIFFERENTIAL rate equation is `d[R]//dt=k[A][B]^(2)` ii) The rate increases by 4 TIMES by increasing the CONCENTRATIO B by 2 times.

Discussion

87790.	Reaction is first order in A and second in B. Write the different rate equation.
Answer» Solution :The differential rate EQUATION will be `(-d[R])/(DT)=K[A][B]^(2)`

Discussion

87791.	A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rte affected on increasing the concentration of B three times? (iii) How is the rate affected when concentrations of both A and B are doubled?
Answer» Solution :(i) `(dx)/(DT)=k[A][B]^(2)." Let"[A]=a and [B]=b`. Thus, `(dx)/(dt)=k ab^(2)`. (ii) If [B] is tripled, RATE `=k a(3b)^(2)=9kab^(2)`. The rate increases 9 times. (iii) If both [A] and [B] are doubled, Rate `=k(2a)(2B)^(2)=8 k ab^(2)`. The rate increases 8 times.

Discussion

87792.	A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is rate affected on increasing the concentration of B three times ? (iii) How is rate affected when the concentration of both A and B are doubled ?
Answer» Solution :(i) rate ` = - (DX)/(DT) = k[A]^(1) [B]^(2)` (ii)rate ` = k[A]^(!) [3B]^(2) = 9k[A]^(1) [B]^(2)`. Rate increases by 9 TIMES. (iii) `r = k[2A]^(1) [2B]^(2) = 8K [A}^(1) [B]^(2)` . Rate increases by 8 times.

Discussion

87793.	A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times ? (iii) How is the rate affected when the concentrations of both A and B are doubled ?
Answer» SOLUTION :(i) `(DX)/(dt) = k[A]^(1) [B]^(2)` (ii) If concentration of .B. is TRIPLED, then the rate will become 9 TIMES. (iii) When concentration of both A and B are doubled, then the rate will become 8 times.

Discussion

87794.	A reaction is first order in A and second order in B. (i) Write differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times ? (iii) How is the rate affected when concentration of both A and B is doubled ?
Answer» Solution :(i) `(DX)/(DT)=k[A][B]^(2).` (ii) Rate `=k" ab"^(2)` If [B] is TRIPLED, Rate `=ka(3" b")^(2)=9" k ab"^(2)=9" times."` (iii) If both [A] and [B] are DOUBLED, Rate `=k(2" a")(2" b")^(2)=8" k ab"^(2)=8" times."`

Discussion

87795.	A reaction is first order in A and second order in B. (i) Write differential rate equation. (ii) How is rate affected when concentrated of B is tripled ? (iii) How is rate affected when concentration of both A and B is doubled ?
Answer» Solution :(a) A + B = Products. (i) The reaction is of Ist order with respect to A. Rate w.r.t. `[A] = k[A]` it is of SECOND order w.r.t. B. Rate w.r.t. `[B] = k[B]^(2)` `:.` Overall rate of reaction `= k[A][B]^(2)` `:. (DX)/(dt) = k[A][B]^(2)` (ii) When the concentration of B is tripled. `r^(2) = k[A][3B]^(2)`. `=9K[A][B]^(2)` `:.` Rate becomes 9 times. (iii) If the concentration of A and B is doubled, then `r = k[2A][2B]^(2)` `= 8k[A][B]^(2)` `:.` Rate becomes 8 times.

Discussion

87796.	A reaction is first order in A and second order in B. How the rate is affected (a) on increasing concentration of B three times and (b) on increasing concentration of both A and B twice ?
Answer» SOLUTION :Rateis given as rate `=K[A][B]^(2)` (a) If concentration of B is increaseed three times, NEW rate `=K[A][3B]^(2)` `9K[A][B]^(2)` Hence rate is increased by `9` times (b) If concentration of both A and B are doubled, `"new rate"=K[2A][2B]^(2)` `8K[A][B]^(2)` Hence rate is increased by `8` times

Discussion

87797.	Reaction is first order in A and second in B. How is the rate affected when the concentrations of both A and B are doubled ?
Answer» Solution :When the concentrations of both A and B are doubled `(-d[R])/(DT)=k[A][B]^(2)` `=k[2A][2B]^(2)` `= 8.k[A][B]^(2)` `THEREFORE` The RATE of reaction will increase 8 TIMES.

Discussion

87798.	Reaction is first order in A and second in B. How is the rate affected on increasing the concentration of B three times ?
Answer» Solution :If the concentration of B is increased three TIMES, then `(-d[R])/(DT)=k[A][3B]^(2)` `=9.k[A][3B]^(2)` `THEREFORE` The rate of reaction will increase 9 times.

Discussion

87799.	A reaction is catalysed by H^(+) ion, and in the rate law the dependence of rate is of first order with respect to the concentration of H^(+) ions, in presence of HA rate constant is 2xx10^(-3)"min"^(-1) and in presence of HB rate constant is 1xx10^(-10"min"^(-1). HA and HB have relative strength as :
Answer» `0.5` `0.002` `0.001` 2 Answer :D

Discussion

87800.	A reaction is catalysed by.X. Here .X.
Answer» DECREASES the rate CONSTANT of reaction Does not affect the EQUILIBRIUM constant of reaction Decreases the ENTHALPY of reaction Decreases the activation energy Answer :D

Discussion

Explore topic-wise InterviewSolutions in .

A reaction rate constant is given by k = 1.2 xx 10^(14) e^(-(25000//RT)) sec^(-1) . It means

A reaction proceeds with the energy of activation of 55.3 kcal/mole. If AH of the reaction is 1 kcal, what would be the energy of activation of the reverse reaction?

A reaction proceeds in three stages. The first stage is a slow and involves two molecules of reactants. The second and third stage are fast. The overall order of the reaction is

A reaction proceeds in three stages. The first stage is a slow and involves two molecules of reactants .The second and third stage are fast .The overall order of the reaction is :

A reaction proceeds by a two step mechanism A_2underset(k_2)overset(k_1)hArr2A (fast reaction) A+B to Products (slow reaction) What is the rate law for the overallreaction ?

A reaction P to Q is completed 25% in 25 min, 50% completed in 25 min if [P] is halved, 25% completed in 50 min if [P] is doubled. The order of reaction is

(A) Reaction of HI with ter-butylethylether produces ter.butyliodide & ethyl alcohol(R) It follows SN^1reaction

A reactionof firstorderwithrespect to reactantA andsecondorderwithrespectto reactantB. the rate lawforthe reactionis givenby

(A) Reaction of Cl_(2) with hot and concen -trated NaOH is disproportionation reaction (R ) Oxidation state of chlorine changed from 0"to"-1and+5

A: Reaction of carboxylic acid with hydrazoic acid in the presence of conc. H_(2)SO_(4) to form primary amine is called Schmidt reaction. R: In this reaction isocyanate intermediate is formed.

(A) Reaction of benzaldehyde with formaldehyde in presence of base is crossed Cannizzaro's reaction. (R) Formaldehyde is always oxidized in this reaction because it is better oxidising agent.

A reaction occurs spontaneously if :

Areaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ?

A reaction is spontaneous at low temperature but non-spontaneous at high temperature. Which of the following is true for the reaction

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

A reaction is secondorder with respect to a reactant . How is the rate of reaction affected if the concentration of the reactant is : (i) doubled (ii) reduced to half.

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) Doubled (ii) reduced to half ?

A reaction is represented by A overset(k_1)rarr B(slow ) and A+B overset(k_2)rarrC(fast) where k_1and k_2are the ratge constants of two steps. The rate of production of C will be given by

A reaction is of the first order reaction to A and is of second order relative to B .what will be the effect on rateif the concentration of A and B are doubled

A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half ? What is the unit of rate constant of such a reaction ?

A reaction is of second order in A and first order in B. How is the rate affected when the concentration of both A and B is doubled ?

A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

A reaction is of second order in A and first order in B. How is the rate affected on increasing the concentration of A three times ?

A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times. (ii) the concentrations of A as well as B are doubled?

A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times, (ii) the concentrations of A as well as B are doubled ?

A reaction is firt order in A and second order in B.

A reaction is of first order in reactant A and of second order in reactant B. How is the rate of reaction affected when (i) concentration of B alone is increased to three times (ii) the concentration of A as well mass B is doubled ?

A reaction is first order with respect to the reactant A and Second roder with respect to the reactant B in a reaction. A + B to Product. (i) Write the differential rate equation. (ii) How is rate of reaction affected on increasing the concentration of B by 2 times.

A reaction is first order w.r.t reactant A as well as w.r.t. B. Give the rate law. Also give one point of difference between average rate and instantaneous rate.

A reaction is first order with respect to the reactant A and second order with respect to the reactant B in a reaction A+Bto product How is the rate of the reaction affected on increasing tile concentration of B by two times.

A reaction is first order with respect to the reactant A and second order with respect to the reactant B in a reaction A+Bto product Write the differential rate equation.

A reaction is first order with respect to reactant A and second order with respect to reactant B in a reaction A+Brarr product. i) Write the differential rate equation. ii) How is the rate of the reaction affected on increasing the concentration of B by two times?

Reaction is first order in A and second in B. Write the different rate equation.

A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rte affected on increasing the concentration of B three times? (iii) How is the rate affected when concentrations of both A and B are doubled?

A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is rate affected on increasing the concentration of B three times ? (iii) How is rate affected when the concentration of both A and B are doubled ?

A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times ? (iii) How is the rate affected when the concentrations of both A and B are doubled ?

A reaction is first order in A and second order in B. (i) Write differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times ? (iii) How is the rate affected when concentration of both A and B is doubled ?

A reaction is first order in A and second order in B. (i) Write differential rate equation. (ii) How is rate affected when concentrated of B is tripled ? (iii) How is rate affected when concentration of both A and B is doubled ?

A reaction is first order in A and second order in B. How the rate is affected (a) on increasing concentration of B three times and (b) on increasing concentration of both A and B twice ?

Reaction is first order in A and second in B. How is the rate affected when the concentrations of both A and B are doubled ?

Reaction is first order in A and second in B. How is the rate affected on increasing the concentration of B three times ?

A reaction is catalysed by.X. Here .X.