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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
For a single step reaction `X+2Y rarr` Products, the molecularity isA. ZeroB. 1C. 2D. 3 |
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Answer» Correct Answer - D Elementary reaction definition |
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| 552. |
For the reaction: `2HIrarr H_(2)+I_(2)`, the expresison `-d(HI)//2dt` representsA. The rate of formation of `HI`B. The rate of disappearance of `HI`C. The instantaneous rate of the reactionD. The average rate of reaction |
| Answer» Correct Answer - C | |
| 553. |
The reaction `2AX(g)+2B_(2)(g)rarr A_(2)(g)+2B_(2)X(g)` has been studied kinetically and on the baiss of the rate law following mechanism has been proposed. I. `2A X hArr A_(2)X_(2) " " ("fast and reverse")` II. `A_(2)X_(2)+B_(2)rarrA_(2)X+B_(2)X` III. `A_(2)X+B_(2)rarrA_(2)+B_(2)X` where all the reaction intermediates are gases under ordinary condition. form the above mechanism in which the steps (elementary) differ conisderably in their rates, the rate law is derived uisng the principle that the slowest step is the rate-determining step (RDS) and the rate of any step varies as the Product of the molar concentrations of each reacting speacting species raised to the power equal to their respective stoichiometric coefficients (law of mass action). If a reacting species is solid or pure liquid, its active mass, i.e., molar concentration is taken to be unity, the standard state. In qrder to find out the final rate law of the reaction, the concentration of any intermediate appearing in the rate law of the RDS is substituted in terms of the concentration of the reactant(s) by means of the law of mass action applied on equilibrium step. Let the equilibrium constant of Step I be `2xx10^(-3) mol^(-1) L` and the rate constants for the formation of `A_(2)X` and `A_(2)` in Step II and III are `3.0xx10^(-2) mol^(-1) L min^(-1)` and `1xx10^(3) mol^(-1) L min^(-1)` (all data at `25^(@)C)`, then what is the overall rate constant `(mol^(-2) L^(2) min^(-1))` of the consumption of `B_(2)`?A. `6xx10^(-5)`B. `1.2xx10^(-4)`C. `3xx10^(-5)`D. `1.5xx10^(-5)` |
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Answer» Correct Answer - B Rate of overall reacation `=`Rate of step II `(+(d[N_(2)O])/(dt))=k_(II)[A_(2)X_(2)][B_(2)]` form equilibrium step, i.e., step I `k_(c )=([A_(2)X_(2)])/([AX]^(2)), k_(c )=` equilibrium constant of step I `[A_(2)X_(2)]=k_(c )[AX]^(2)` (`k_(II)=` rate of constant of formation of `A_(2)X)` `=k[AX]^(2)[B_(2)]` `k=`Rate constant of overall reacation Since for the rate of `A_(2)X` consumed `2mol` of `B_(2)` are formed so rate constant for the consmption of `B_(2)` will twice that of overall rate constant, i.e., `2k_(II)k_(c )` `2xx3xx10^(-2)xx2xx10^(-3)= 1.2xx10^(-4)mol^(-2)L^(2)min^(-1)` |
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| 554. |
For a single step reaction `X+2Y rarr` Products, the molecularity isA. ZeroB. `2`C. `3`D. `1` |
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Answer» Correct Answer - C Since theee molecules are colliding ismulaneously, so molecularity is `3`. It is determine theoretically form the number of reactant molecules in a balanced chemical reaction.Correct Answer - C Since theee molecules are colliding ismulaneously, so molecularity is `3`. It is determine theoretically form the number of reactant molecules in a balanced chemical reaction. |
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| 555. |
A reacts to form P , A plot of the reciprocal of the concentration of A vs time is a straight line . When the initial concentration of A is `1.0 xx 10^(-2) M` , its half-life is found to be 20 min . When initial concentration of A is `3.0 xx 10^(-3) M` , the half -life will beA. 20 minB. 40 minC. 56 minD. 67 min |
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Answer» Correct Answer - d `A to P` A plot of 1/[A] vs time is a straight line of second order . For second order , `t_(1//2) = (1)/(k[A]_(0))` where , k = rate constant `[A]_(0) ` = initial concentration of A. when , `[A]_(0) = 1 xx 10^(-2)` M , then `t_(1//2) = 20` min `20 = (1)/(k[1 xx 10^(-2)]) implies k = (1)/(20) [ 1 xx 10^(-2)] " " ... (i)` when , `[A]_(0) = 3 xx 10^(-3)` M , then `t_(1//2) = `? `t_(1//2) = (1)/(k[3 xx 10^(-3)])` substituting value of k from equation (i) `t_(1//2) = (1 xx 20 [1 xx 10^(-2)])/([3 xx 10^(-3)]) = 6.66 xx 10` min = 67 min . |
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| 556. |
For the reaction: `X(g) rarr Y(g)+Z(g)`, the following data were obtained at `30^(@)C`: `{:("Experiment",[X](mol L^(-1)),"Rate" (mol L^(-1) hr^(-1))),(I,0.17,0.05),(II,0.34,0.10),(III,0.68,0.20):}` The equilibrium constant for the reaction is `0.50`. Assuming that the reaction proceeds by one-step mechanism. Find the rate constant of reverse reaction?A. `0.294 hr^(-1)`B. `0.588 hr^(-1)`C. `0.123 hr^(-1)`D. `0.117 hr^(-1)` |
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Answer» Correct Answer - B `X(g) underset(k_(b))overset(k_(f))hArr Y(g) + Z(g)` `k_(eq)=(k_(f))/(k_(b))` `impliesK_(b)=(0.294hr^(-1))/(0.50)= 0.588hr^(-1)` |
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| 557. |
For the reaction: `X(g) rarr Y(g)+Z(g)`, the following data were obtained at `30^(@)C`: `{:("Experiment",[X](mol L^(-1)),"Rate" (mol L^(-1) hr^(-1))),(I,0.17,0.05),(II,0.34,0.10),(III,0.68,0.20):}` The rate constant of the above reaction isA. `0.588 hr^(-1)`B. `0.294 hr^(-1)`C. `0.123 hr^(-1)`D. `0.210 hr^(-1)` |
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Answer» Correct Answer - B In experiments I and II, [X] becomes double and rate is also doubled. Thus it is first, order reaction. `r=k_(f)[X]` `0.05mol L^(-1)hr^(-1)=k_(f)xx0.17molL^(-1)` `:. k_(f)= 0.29hr^(-1)` |
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| 558. |
For the reaction: `X(g) rarr Y(g)+Z(g)`, the following data were obtained at `30^(@)C`: `{:("Experiment",[X](mol L^(-1)),"Rate" (mol L^(-1) hr^(-1))),(I,0.17,0.05),(II,0.34,0.10),(III,0.68,0.20):}` In experiment `(I)`, what time will reactant `(X)` take to be reduced into `0.0425 M`?A. `2.4 hr`B. `0.48 hr`C. `3.1 hr`D. `4.7 hr` |
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Answer» Correct Answer - D `a=0.17M, (a-x)= 0.0425M` `t=(2.3)/(0.294hr^(-1))log.(0.17)/(0.0425)` `=(2.3)/(0.294)log4(2.3)/(0.294)xx0.6=4.7hr` |
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| 559. |
For which of the following represents the expresison for theee/fourth life reaction?A. `k/2.303 log4//3`B. `2.303/k log 3//4`C. `2.303/k log 4`D. `2.303/k log 3` |
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Answer» Correct Answer - C `t_(3//4) = (2.303)/(k)log.(1)/((1-(3)/(4))) = (2.303)/(k)log4` Correct Answer - C `t_(3//4) = (2.303)/(k)log.(1)/((1-(3)/(4))) = (2.303)/(k)log4` |
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| 560. |
If the rate of a reaction is 5 at `10^(@)C ` then on increasing the temperature to `30^(@)C` , new rate isA. 20B. 10C. 50D. 40 |
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Answer» Correct Answer - a For every `10^(@)` rise of temperature rate constant doubles . For `20^(@)` rise of temperature , rate constant becomes 4 times , i.e., `k_(2) = 4k_(1) = 4 xx 5 = 20.` |
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| 561. |
Which of the following is the correct statementA. Order of a reaction has always an integral valueB. Mechanism of a reaction proposed is always finalC. Zero order reaction are multi-step reactionsD. Order of reactions can be predicted even without knowing the rate law |
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Answer» Correct Answer - d It is experimentally determined . |
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| 562. |
For which of the following reactions, the units of rate constant and rate of reaction are same?A. First order reactionB. Second order reactionC. Third order reactionD. Zero order reaction |
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Answer» Correct Answer - D Unit of rate `= mol L^(-1) t^(-1)` Unit of zero order `= mol L^(-1) t^(-1)` Correct Answer - D Unit of rate `= mol L^(-1) t^(-1)` Unit of zero order `= mol L^(-1) t^(-1)` |
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| 563. |
Rate of the given reaction (i) ` A + B overset (r_(1) = 0.05)(to) X ` (ii) ` X + B overset(r_(2) = 0.89)(to) Y` (iii) `Y + A overset(r_(2) = 0.001)(to) (AY)` (iv) `AY + B overset(r_(4) = 0 .10)(to) AYB` will be determined byA. Step(i) Because the reaction starts with the formation of XB. Step (ii) because it is fastest stepC. Step (iii) because it is the slowest stepD. Step (iv) because it ends the reaction |
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Answer» Correct Answer - c Slowest step is the rate determining step . |
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| 564. |
Burning of coal is represented as `C(s)+O_(2)(g)rarr CO_(2)(g)`. The rate of this reaction is increased byA. Decrease in the concentration of oxygenB. Powdering the jump of coalC. Decreaisng the temperature of coalD. Providing inert atmosphere |
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Answer» Correct Answer - B Surface area increases.Correct Answer - B Surface area increases. |
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| 565. |
The slowest step of a particular reaction is found to be `1//2X_(2) + Y_(2) rarr XY_(2)`. The order of the reaction isA. `2`B. `3`C. `3.5`D. `1.5` |
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Answer» Correct Answer - D Slow step is: `(1)/(2)X_(2) + Y_(2) rarr XY_(2)` Hence, `r = k[X_(2)]^(1//2)[Y_(2)]` `OR = (1)/(2)+1 = 1(1)/(2)`Correct Answer - D Slow step is: `(1)/(2)X_(2) + Y_(2) rarr XY_(2)` Hence, `r = k[X_(2)]^(1//2)[Y_(2)]` `OR = (1)/(2)+1 = 1(1)/(2)` |
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| 566. |
(a) Write the rate expression. (b) Find the rate constant for `{:([A]_(0),[B]_(0),,,"`R_(0)` "Rate" (mol litre^(-1) s^(-1))),(0.1,0.2,,,0.05),(0.2,0.2,,,0.10),(0.1,0.1,,,0.05):}` |
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Answer» Rate `=K[A]^(m)[B]^(n)` where `m, n` are order of reaction with respect to `A` and `B` respectively. `K` is rate constant. `:. 0.05=K[0.1]^(m)[0.2]^(n) …(1)` `0.10=K[0.2]^(m)[0.2]^(n) …(2)` `0.05=K[0.1]^(m)[0.1]^(n) …(3)` By eqs. (1) and (2), `m=1` By eqs. (1) and (3), `n=0` `:. Rate=K[A]^(1)[B]^(0)` or `0.05=K[0.1]^(1)` or `K=0.5 sec^(-1)` |
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| 567. |
Calculate the overall order of a reaction which has the rate expresison. (a) Rate`= k[A]^((1)/(2))[B]^((3)/(2))` , (b) Rate `= k[A]^((3)/(2))[B]^(-1)`A. `3//2`B. `1//2`C. ZeroD. None of these |
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Answer» Correct Answer - B `OR = (3)/(2) - 1 = (1)/(2)`Correct Answer - B `OR = (3)/(2) - 1 = (1)/(2)` |
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| 568. |
For hypothetical chemical reaction `A rarr I`, it is found that the reaction is third order in `A`. What happens to the rate of reaction when the concentration of `A` is doubled?A. Rate increases by a factor of `2`.B. Rate decreases by a factor of `3`.C. Rate increases by a factor of `8`D. Rate remains unaffected. |
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Answer» Correct Answer - C `r_(1) = k[A]^(3)` …(i) `r_(2) = k[2A]^(3)` …(ii) Dividing Eq. (ii) by Eq. (i), `r_(2) = 8r_(1)`Correct Answer - C `r_(1) = k[A]^(3)` …(i) `r_(2) = k[2A]^(3)` …(ii) Dividing Eq. (ii) by Eq. (i), `r_(2) = 8r_(1)` |
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| 569. |
The initial concentration of `N_(2)O_(5)` in the following first order reaction: `N_(2)O_(5)(g) rarr 2NO_(2)(g)+(1)/(2)O_(2)(g)` was `1.24 xx 10^(-2) mol L^(-1)` at `318 K`. The concentration of `N_(2)O_(5)` after `60 min` was `0.20 xx 10^(-2) mol L^(-1)`. Calculate the rate constant of the reaction at `318 K`.A. `0.0104" min"^(-1)`B. `0.0204" min"^(-1)`C. `0.0304" min"^(-1)`D. `0.0404" min"^(-1)` |
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Answer» Correct Answer - C For a first order reaction , `"log"([R]_(1))/([R]_(2))=(k(t_(2)-t_(1)))/(2.303)` `k=(2.303)/((t_(2)-t_(1)))"log"([R]_(1))/[R]_(2)` `=(2.303)/((60min-0min))"log"(1.24xx10^(-2)"mol "L^(-1))/(0.20xx10^(-2)"mol "L^(-1))` `= (2.303)/(60)log6.2" min"^(-1)=0.0304" min"^(-1)` |
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| 570. |
The rate of a chemical reaction becomes double for every `10^(@)` rise in temperature. If the tempeature is raised by `50^(@)`C, the rate of reaction increases by about:A. 10 timesB. 24 timesC. 32 timesD. 64 times |
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Answer» Correct Answer - C For every `10^(@)C` rise in temperature, rate is doubled thus, temperature coefficient of the reaction=2 , when the temperature is increased by `50^(@)C,` rate becomes `=2^((50//10))=2^(5)=32"times."` |
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| 571. |
For a reaction `2Ararr3B,` if the rate of formation of B is `x"mol"//L,` the rate of consumption of A isA. xB. `(3x)/(2)`C. 3xD. `(2x)/(3)` |
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Answer» Correct Answer - D For reaction, `2Ararr3B` `-(1)/(2) (d[A])/(dt)=(1)/(3)(d[B])/(dt)` `(d[B])/(dt)=xx"mol"//L` `(d[A])/(dt)=-(2)/(3)rArr(d[B])/(dt)=-(2)/(3)x` Hence, the rate of consumption of `A=[(-d[A])/(dt)=(2)/(3)x]` |
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| 572. |
Two radioactive material A and B have disintegration constants `10lambda and2lambda` respectively. If initially they have same number of nuclei, then the ration of number of nuclei of A and B will be `(1)/(e)` after a time of :A. `(1)/(10lambda)`B. `(1)/(11lambda)`C. `(11)/(10lambda)`D. `(1)/(8lambda)` |
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Answer» Correct Answer - D |
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| 573. |
Which species is most likely to lose a position `(beta^(+))?`A. `._(7)^(12)N`B. `._(8)^(18)O`C. `._(9)^(20)F`D. `._(10)^(20)Ne` |
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Answer» Correct Answer - A |
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| 574. |
In the sequences of the following `._(92)Xe^(238) overset(-alpha)to Y overset(-beta)toZ overset(-beta)toL overset(-nx)to._(84)M^(218)` the value of n will beA. 5B. 4C. 3D. 6 |
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Answer» Correct Answer - B |
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| 575. |
For the reaction, `R-X overset(Slow)rarrR^(o+)+X^(o+)`, `R^(o+) + overset(ɵ)(OH) overset("Fast") rarr ROH` Which type of reaction coordinate diagram represents the above reaction mechanism?A. B. C. D. d. This is `S_(N)l` reaction where no tranistion state will be present |
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Answer» Correct Answer - B Since the reaction is `S_(N)1`, so diagram (b) represents `S_(N)1` reaction, with two tranistion states and intermediate carbocation. Diagram (A) represents `S_(N)2` reaction and has one `TS` and no intermediate. |
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| 576. |
3A `rarr` B + C It would be a zero order reaction when :A. the rate of reaction is proportional to square of concentration of AB. the rate of reaction remains same at any concentration of AC. the rate remains unchanged at any concentration of B and CD. the rate of reaction doubles it concentration of B is increased to double |
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Answer» Correct Answer - B |
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| 577. |
If for a reaction `t_(x//y)` represents time required for `(x)/(y)`th fraction of reactant to react then identify the correct option for a first order reaction : Statement-I : `t_(3//4)=t_(1//2)xx2` Statement-II : `t_(15//16)=t_(1//2)xx4` Statement-II : `t_(7//8)=t_(3//4)xx2` Statement-IV : `t_(15//16)=t_(3//4)xx3`A. All statements are correctB. Only Statement-I is correctC. Statement-III and Statement-IV are incorrectD. Statement-II and Statement-IV are correct |
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Answer» Correct Answer - C |
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| 578. |
For a certain reaction of order n, the time for half change, `t_(1//2)` is given by `t_(1//2)=((2-sqrt2))/(k)xxC_(0)^(1//2)` where k is constant and `C_(0)` is the initial concentration. What is n?A. 1B. 2C. 0D. `0.5` |
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Answer» Correct Answer - D |
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| 579. |
Which of the following statement is incorrect for a photochemical reaction?A. Photochemical reactions are complex reactions.B. Overall order of photochemical reactions is always zero.C. Only the first step of such reactions follows zero order kinetics.D. The rate of such reactions depends on the intensity of radiations absorbed. |
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Answer» Correct Answer - B |
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| 580. |
In a zero order reaction half life is 100 sec. After how much time `(7)/(8)` fraction of reactant will be reacted?A. 300 sec.B. 200 sec.C. 175 sec.D. 25 sec. |
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Answer» Correct Answer - C |
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| 581. |
For zero order reaction which is true?A. The rate constant is dimensionless.B. Amount of reactant remain same throughout.C. `t_(1//2)prop` initial concentration of reactant.D. A plot of concentration of reactant vs time is a straight line with slope equal to K. |
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Answer» Correct Answer - C |
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| 582. |
Select the correct option for 1st order reaction `2Ararr`product having rate constant of reaction `1.386xx10^(-2)`min.A. Time required for `75%` completion of reaction is 100 min.B. Time required for `99.9%` completion of reaction is 250 min.C. Rate of decoposition of reactant at concentration `0.1` M is `1.386xx10^(-3)M//min`D. Rate of decoposition of reactant at concentration `0.1` M is `1.386xx10^(-3)M//sec`. |
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Answer» Correct Answer - B |
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| 583. |
The rate of the reaction `2 NO + Cl_(2) to 2 NOCl` is given by the rate equation rate = `k [NO]^(2) [Cl_(2)]` The value of the rate constant can be increased byA. Increasing the temperatureB. Increasing the concentration of NOC. Increasing the concentration of the `Cl_(2)`D. Doing all of these |
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Answer» Correct Answer - a Concentration do not affect rate constant |
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| 584. |
Calculate the rate of reaction for the change, `2Ararr` Products, when rate constant of the reaction is `2.1xx10^(-5) time^(-1)` and `[A]_(0)=0.2M`. |
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Answer» The unit of rate constant suggest it to be first order. Thus rate `=K[A]` `=2.1xx10^(-5)xx0.2` `=4.2xx10^(-6) mol litre^(-1) sec^(-1)` |
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| 585. |
The data given below are for the reaction of `NO` and `Cl_(2)` to from `NOCl` at `295 K`. `{:([Cl_(2)],[NO],,,"Initial rate" (mol litre^(-1) sec^(-1))),(0.05,0.05,,,1xx10^(-3)),(0.15,0.05,,,3xx10^(-3)),(0.05,0.15,,,9xx10^(-3)):}` (a) What is the order with respect to `NO` and `Cl_(2)` in the reaction? (b) Write the rate expression. (c ) Calculate the rate constant. (d) Determine the reaction rate when conc. of `Cl_(2)` and `NO` are `0.2 M` and `0.4 M` respectively. |
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Answer» For the reaction, `2NO+Cl_(2)rarr 2NOCl` Rate=`K[Cl_(2)]^(m)[NO]^(n) …(1)` Where, m and n are order of reaction w.r.t. `Cl_(2)` and `NO`, respectively. From the given data: `1xx10^(-3)= K[0.05]^(m)[0.05]^(n) …(2)` `3xx10^(-3)= K[0.15]^(m)[0.05]^(n) ...(3)` `9xx10^(-3)= K[0.05]^(m)[0.15]^(n) ...(4)` By eqs. (2) and (3), `m=1` By eqs. (2) and (4), `n=2` (a) `:.` Order w.r.t. `NO` is `2` and w.r.t. to `Cl_(2)` is `1`. (b) Also, rate expression `r=K[Cl_(2)]^(1) [NO]^(2)` (c ) And rate constant, `K=(r)/([Cl_(2)]^(1)[NO]^(2))` `=(1xx10^(-3))/([0.05]^(1)[0.05]^(2))` `=8 litre^(2) mol^(-2) sec^(-1)` (d) Further, `r=K[Cl_(2)]^(1)[NO]^(2)=8[0.2]^(1)[0.4]^(2)` `=0.256 mol litre^(-1) sec^(-1)` |
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| 586. |
The reaction between A and B is first order with respect to A and zero with respect to B. Fill in tha blanks in the following table: `{:("Experiment",,[A]//M,[B]//M,"Initial rate"//"M min"),(I.,,0.1,0.1,2.0xx10^(-2)),(II.,,....,0.2,4.0xx10^(-2)),(II.,,0.4,0.4,.......),(IV.,,....,0.2,2.0xx10^(-2)):}` |
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Answer» (II)- is `0.2`, (III)- is `8xx10^(-2)` , (IV)- is `0.1` |
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| 587. |
Using the concentration time equation for a first order reaction : The decomposition of `N_(2)O_(5)` to `NO_(2)` and `O_(2)` is first order, with a rate constant of `4.80xx10^(-4)//s att 45^(@)C` `N_(2)O_(5)(g)rarr2NO_(2)(g)+(1)/(2)O_(2)(g)` (a) If the initial concentration of `N_(2)O_(5) is 1.65xx10^(-2)mol//L`, what is its concentration after 825 s ? (b) How long would it take for the concentration of `N_(2)O_(5)` to decrease to `100xx10^(-2) mol L^(-1)` from its initiqal value, given in (a) ? Strategy : Since this reaction has a first order rate law, `d[N_(2)O_(5)]//dt = k[N_(2)O_(5)]` , we can use the corresaponding concentration time equation for a first order reaction : `k = (2.303)/(t) log ([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))` In each part, we substitute the know quantities into this equation and solve for the unkbnown. |
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Answer» (a) Substituting the values of k, t and `[N_(2)O_(5)]_(0)` into the concentration - time equation, we have `4.80xx10^(-4)s^(-1)= (2.303)/(825s)log(1.65xx10^(-2)mol L^(-1))/([N_(2)O_(5)]_(t))` or log `((1.65xx10^(-2)mol L^(-1))/([N_(2)O_(5)]_(t)) = ((4.80xx10^(-4)s^(-1))(825s))/((2.303)))` `= 0.172` or `log (([N_(2)O_(5)]_(t))/(1.65xx10^(-2)mol L^(-1)))= -0.172` To solve for `[N_(2)O_(5)]_(t)` , we take thew antilogarithm of both sides. This removes the long from the left and yield antilog `(-0.172)`, or `10^(-0.172)` , on the right, which equals `0.673`. Thus `([N_(2)O_(5)]_(t))/(1.65xx10^(-2)mol L^(-1)) = 0.673` Hence `[N_(2)O_(5)]_(t) = (1.65xx10^(-2)mol L^(-1)) (0.673)` `= 0.0111 mol L^(-1)` (b) Writing the integrated first order rate equation we have `k = (2.303)/(t) log ([N_(2)O_(5)]_(0)/([N_(2)O_(5)]_(t))) ` or `log (([N_(2)O_(5)]_(t))/([N_(2)O_(5)]_(0))) = (-kt)/(2.303)` Substituting the concentrations and the value of k, we get `log (((1.00xx10^(-2)mol L^(-1)))/((1.65xx10^(-2)mol L^(-1)))) = (-4.80xx10^(-4)s^(-1)xxt)/(2.303)` The left side equals `-0.217` , the right side equals `-2.08xx10^(-4)s^(-1)xxt` . Hence `0.217=2.08xx10^(-4)s^(-1)xxt` or `t = (0.217)/(2.08xx10^(-4)s^(-1))` `= 1.04xx10^(3)` |
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| 588. |
The experiment data for the reaction `2A + B_(2) rarr 2AB` is `|{:("Experiment",[A] M,[B_(2)] M,"Initial rate" (mol L^(-1) s^(-1)),),(I,0.50,0.5,1.6 xx 10^(-4),),(II,0.50,1.0,3.2 xx 10^(-4),),(III,1.00,1.0,3.2 xx 10^(-4),):}|` Write the most probable rate equation for the reacting giving reason for you answer. |
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Answer» Let rate expression be: rate `=K[A]^(m)[B_(2)]^(n)` Thus, `1.6xx10^(-4)=K(0.5)^(m)(0.5)^(n) …(i)` `3.2xx10^(-4)=K(0.5)^(m)(1.0)^(n) …(ii)` `3.2xx10^(-4)=K(1.0)^(m)(1.0)^(n) …(iii)` By eqs. (i) and (ii), `n=1` By eqs. (ii) and (iii), `m=0` Rate`=(dx)/(dt)K[B_(2)]^(1)` |
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| 589. |
A `22.4` litre flask contains `0.76 mm` of ozone at `25^(@)C`. Calculate: (i) the concentration of oxygen atoms needed so that the reaction `O+O_(3)rarr 2O_(2)` having rate constant equal to `1.5xx10^(7) litre mol^(-1) sec^(-1)` can proceed with a rate of `0.15 mol litre^(-1) sec^(-1)`. (ii) the rate of formation of oxygen under this condition. |
| Answer» Correct Answer - (i) `2.45xx10^(-4), (ii) 0.30 mol L^(-1) t^(-1);` | |
| 590. |
form the gaseous reaction `2A + B_(2) rarr 2AB`, the following rate data were obtained at `300 K`. Calculate the rate constant for the reaction and the rate of formation of `AB` when `[A]` is `0.02` and `[B_(2)]` is `0.04 mol L^(-1)` at `300K`. |
| Answer» Correct Answer - (a) `2`, (b) `1.28xx10^(-3)` | |
| 591. |
Assertion: For the reaction `RCl + NAOH(aq) rarr ROH + NACl`, the rate of reaction is reduced to half on reducing the concentration of `RCl` to half. Reason: The rate of the reaction is represented by `k[RCl]` i.e., it is a first-order reaction.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C Correct Reason The rate of reaction is represented by `k[RCl] [NaOH]`, i.e., it is a second order reaction. |
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| 592. |
The given reaction `2NO rarr O_(2) rarr 2NO_(2)` is an example ofA. first-order reactionB. second-order reactionC. third-order reactionD. none of these |
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Answer» Correct Answer - C It is a third-order reaction because Rate `= K[NO]^(2)[O_(2)]^(1) :. O.R. = 2+1 = 3` |
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| 593. |
Rate of a chemical reaction can be kept constant byA. Stirring the comoundsB. Keeping the tempature constantC. Both (a) & (b)D. Adding a catalyst |
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Answer» Correct Answer - B Factor affecting rate ocf chemical reaction . |
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| 594. |
The raction `2FeCl_(3) + SnCl_(2) rarr 2FeCl_(2) + SnCl_(4)` is an example ofA. fourth order reactionB. half order reactionC. third order reactionD. second order reaction |
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Answer» Correct Answer - C Experimentally determined rate law for the reaction is `Rate=k[FeCl_(3)]^(2)[SnCl_(2)]` Therefore, it is a third order reaction |
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| 595. |
Study the following experiment and answer the question at the end of it The following reactions were studied at `25^(@)` C in benzene solution containing `0.10` M pyridine `CH_(3)OH+(C_(6)H_(5))_(3)"CCl" rarr (C_(6)H_(5))_(3)C.OCH_(3)+HCI` A B C The following sets of data were observed: `{:("Set","Initial concentration",,"Time different",,"Final concentration[C]"),(,[A],[B]_(0),,,),("I",0.10M,0.05M,25 min,,0.0033 M),("II",0.10M,0.10 M,15 min,,0.0039 M),("III",0.20M,0.10M,7.5 min,,0.0077M):}` Rates `(d[C])/(dt)` in sets I, II and III are respectively (in M `min^(-1)`)A. `{:("I",,"II",,"III"),(1.30xx10^(-4),,2.6xx10^(-4),,1.02xx10^(-3)):}`B. `{:("I",,"II",,""III),(0.033,,0.0039,,0.0077):}`C. `{:("I",,"II",,"III"),(0.02xx10^(-4),,0.04xx10^(-4),,0.017):}`D. `{:("I",,"II",,"III"),("None of these",,,,):}` |
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Answer» Correct Answer - A |
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| 596. |
Which function of [X] . Polleted against time, will give a straight line for a second order reaction ? `Xrarr"Products"`A. [X]B. `[X]^(2)`C. ln [X]D. `(1)/([X])` |
| Answer» Correct Answer - D | |
| 597. |
Which of the reactions represented in these diagrams will show the greatest increase in rate for the same increase in temperature ? A. Reaction I forwardB. Reaction I reversedC. Reaction II forawardD. Reaction II reversed |
| Answer» Correct Answer - B | |
| 598. |
In which of the following reactions, the increase in the rate of reaction will be maximum?A. `{:(,E_(a),"Temperature rise"),((a),40 kJ//"mol",200-210K):}`B. `{:(,E_(a),"Temperature rise"),((b),90 kJ//"mol",300-320K):}`C. `{:(,E_(a),"Temperature rise"),((c),80 kJ//"mol",300-310K):}`D. All will have same rate |
| Answer» Correct Answer - B | |
| 599. |
Use the experimental data in the table to determine was studies What is the rate equation for the reaction ?A. Rate= k[A][B]B. Rate= `k[A]^(2)`C. Rate= `k[B]`D. Rate= `k[B]^(2)` |
| Answer» Correct Answer - D | |
| 600. |
For a gaseous reaction, the following data were recorded : The order of reaction is :A. secondB. firstC. zeroD. fractional |
| Answer» Correct Answer - B | |