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InterviewSolution
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.
| 401. |
For the reaction, `2NO(g) + 2H_(2)(g) rarr N_(2)(g) + 2H_(2)O(g)` The rate expression can be written in the following ways:`(d[N_(2)])/(d t) = k_(1)[NO][H_(2)], (d[H_(2)O])/(d t) = k_(2)[NO][H_(2)]` `- (d[NO])/(d t) = k_(3)[NO][H_(2)], -(d[H_(2)])/(d t) = k_(4)[NO] [H_(2)]` The relationship between `k_(1), k_(2), k_(3), k_(4)` isA. `k_(2) = k_(1) = k_(3) = k_(4)`B. `k_(2) = 2k_(1) = k_(3) = k_(4)`C. `k_(2) = 2k_(3) = k_(1) = k_(4)`D. `k_(2) = k_(1) = k_(3) = 2k_(4)` |
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Answer» Correct Answer - B `2NO +2H_(2) rarr N_(2) + 2H_(2)O` `r_(NO)/2=r_(H_(2))/2=r_(N_(2))/1=r_(H_(2)O)/2` `(K_(3) xx [NO] xx [H_(2)])/(2) = (K_(4) xx [NO] xx [H_(2)])/(2)` `= (K_(1) xx [NO] xx [H_(2)])/(1)` `= (K_(2) xx [NO] xx [H_(2)])/(2)` `K_(3) = K_(4) = 2K_(1) = K_(2)`. |
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| 402. |
`2NO+2H_(2)rarr N_(2)+2H_(2)O`. The experimental rate law for above reaction is , Rate `=k[NO]^(2)[H_(2)]`. When time is in minutesand the concentration is in moles `//L`, the units for `k` areA. `("moles"^(3))/(L^(3)-"min")`B. `("moles")/(L-"min")`C. `("moles"^(2))/(L^(2)-"min")`D. `(L^(2))/("moles"^(2)-"min")` |
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Answer» Correct Answer - 4 Rate`=k[NO]^(2)[H_(2)]` `("conc.")/("time")=k("conc.")^(3)" "rArr" "k=(1)/("time(conc.)"^(2))` `k=(1)/(("moles"^(2))/(L^(2))("min"))=(L^(2))/("moles"^(2)-"min")` |
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| 403. |
`k_(34^(@)),k_(35^(@)) lt 1`, thenA. Rate increase with the rise in temperatureB. Rate decreases with rise in temperatureC. Rate does not change with rise in temperatureD. None of these |
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Answer» Correct Answer - A `(k_(34^(@)))/(k_(35^(@))) lt 1` or `(k_(35^(@)))/(k_(34^(@))) gt 1` Also, `k prop T`. Hence, rate increase with rise of temperature. |
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| 404. |
For a reaction, the rate constant is expressed as `k = Ae^(-40000//T)`. The energy of the activation isA. `40000 cal`B. `88000 cal`C. `80000 cal`D. `8000 cal` |
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Answer» Correct Answer - C `k = Ae^(-E_(a)//RT)` `:. (-E_(a))/(R ) = -40000` `:. E_(a) = 40000 xx 2 = 80000 cal` |
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| 405. |
For the reaction `2NO(g) + H_(2)(g) rarr N_(2)O(g) + H_(2)O(g)`, at `900 K` following data are observed. `{:("Initial pressure of NO (atm)","Initial pressure of" H_(2)o (atm),"Initial rate of pressure decrease" (atm min^(-1))),(0.150,0.40,0.020),(0.075,0.40,0.005),(0.150,0.20,0.010):}` Find the order of reaction.A. `3`B. `2`C. `1`D. `0` |
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Answer» Correct Answer - A form experiments `I` and `II, [H_(2)]` is constant and `[NO]` is doubled. Then rate becomes `4` times. So order of reaction w.r.t. `(NO) = 2`. form experiments `I` and `III, [NO]` is constant, and `[H_(2)]` is doubled then rate becomes `2` times. So order of reaction w.r.t. `(H_(2)) = 1`. Total order `= 3`. Hence answer (a) is correct. |
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| 406. |
A reaction takes place in theee steps: the rate constant are `k_(1), k_(2), and k_(3)`. The overall rate constant `k=(k_(1)k_(3))/(k_(2))`. If `E_(1), E_(2)`, and `E_(3)` (energy of activation) are `60, 30` and `10kJ`, respectively, the overall energy of activation isA. `40`B. `30`C. `400`D. `60` |
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Answer» Correct Answer - A `k_(1) = Ae^(-E_(a_(1))//RT)` `k_(2) = Ae^(-E_(a_(2))//RT)` `k_(3) = Ae^(-E_(a_(3))//RT)` Overall rate: `k = (k_(1)k_(3))/(k_(2))` `:.` Overall `E_(a) = E_(a_(1)) + E_(a_(3))+E_(a_(2))` `= 60 + 10 - 30` `= 40 kJ` |
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| 407. |
A reaction takes place in theee steps: the rate constant are `k_(1), k_(2), and k_(3)`. The overall rate constant `k=k_(1)k_(3)//k_(2)`. If the energies of activation are `40,30, and 20 KJ mol^(-1)`, the overall energy of activation is (assuming `A` to be constant for all)A. 10B. 5C. 30D. 60 |
| Answer» Correct Answer - C | |
| 408. |
For the reaction: `2A+3BrarrC,[A]` is found to decrease at a rate of `2.0M.s^(-1)`. If the rate law is rate = k[A], how fast does [B] decrease under the same conditions?A. `0.66M.s^(-1)`B. `1.3M.s^(-1)`C. `2.0M.s^(-1)`D. `3.0M.s^(-1)` |
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Answer» Correct Answer - D |
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| 409. |
A reaction takes place in three steps. The rate constant are `k_(1), k_(2)` and `k_(3)`. The overall rate constant `k=(k_(1)k_(3))/(k_(2))`. If `E_(1), E_(2)` and `E_(3)` (energy of activation) are 60, 30 and 10 kJ, respectively, the overall energy. Of activation isA. 40 kJB. 30 kJC. 400 kJD. 300 kJ |
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Answer» Correct Answer - A `k_(1)=Ae^(-E_(a_(1))//RT),k_(2)=Ae^(-E_(a_(2))//RT),k_(3)=Ae^(-E_(a_(3))//RT)` Overall rate = `k=(k_(1)k_(3))/(k_(2))` Therefore, overall `E_(a)=E_(a_(1))+E_(a_(3))-E_(a_(2))` `=60 +10 -30 =40 kJ` |
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| 410. |
For the reaction `A+3BrarrC,` select the corrct statement :A. `(d[C])/(dt)=(-d[A])/(dt)`B. `(3d[C])/(dt)=(d[A])/(dt)`C. Rate law must be `r=k [A] [B]^(3)`D. Units for rate of reaction are independent of order of reaction |
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Answer» Correct Answer - A::D |
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| 411. |
If `a` is the initial concentration then time required to decompose half of the substance for nth order is inversely proportional to:A. `a^(n+1)`B. `a^(n-1)`C. `a^(n-2)`D. `a^(n)` |
| Answer» Correct Answer - B | |
| 412. |
The general expression for half-life period of an nth order reaction `(for n != 1)` isA. `t_(1//2)= (2^(n)-1)/((n-1)[A]_(0)^(n-1)k)`B. `t_(1//2)= (2^(n-1)-1)/((n+1)[A]_(0)^(n-1)k)`C. `t_(1//2)= (2^(n+1)+1)/((n+1)[A]_(0)^(n-1)k)`D. `t_(1//2)= (2^(n-1)-1)/((n-1)[A]_(0)^(n-1)k)` |
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Answer» Correct Answer - D For zero order reaction `t_(1//2)=([A]_(0))/(2k)` Or `t_(1//2)=(2^(0-1)-1)/((0-1)[A]_(0)^(0-1)k)` ltbr4gt For a second order reaction `t_(1//2)=(1)/(k[A]_(0))` Or, `t_(1//2)=(2^(2-1)=-1)/((2-1)[A]_(0)^(2-1)k)` and so on |
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| 413. |
From the rate expression for the following reactions, determine their order of reaction and dimensions of the rate constants. `a. 3NO(g) rarr N_(2)O(g),` Rate`=k[NO]^(2)` `b. H_(2)O_(2)(aq)+3I^(c-)(aq)+2H^(o+) rarr 2H_(2)O(l)+I_(3)^(c-),` Rate`=k[H_(2)O_(2)][I^(c-)]` `c. CH_(3)CHO(g)rarr CH_(4)(g)+CO(g),` Rate`=k[CH_(3)CHO]^(3//2)` `d. C_(2)H_(5)Cl(g) rarr C_(2)H_(4)(g)+HCl(g),` Rate `k[C_(2)H_(5)Cl]` |
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Answer» (a) Second order, litre `mol^(-1) time^(-1)` (b) `-do-, -do-` (c ) `3//2` order, `litre^(1//2) mol^(-1//2) time^(-1)` (d) `-do-, -do-` (e) First order, `time^(-1)` |
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| 414. |
The rate constant of a reaction increases by 5% when its temperature is raised from `27^@c` to `28^@c.` The activation energy of the reaction isA. `36.6kJ//mol`B. `16.6kJ//mol`C. `46.6kJ//mol`D. `26.6kJ//mol` |
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Answer» Correct Answer - A `"log"(k_(2))/k_(1)=E_(a)/(2.303R)xx(T_(2)-T_(1))/(T_(1).T_(2))` log`(105)/(100)=E_(a)/(2.303xx8.314)xx(1)/(300xx301)` `E_(a)=36.65kJ" mol"` |
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| 415. |
The activation energy for the reaction `:` ltbr. `2Hl(g) rarr H_(2)(g)+I_(2)(g)` is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?A. `1.82xx10^(-18)`B. `1.47xx10^(19)`C. `2.67xx10^(16)`D. `3.89xx10^(19)` |
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Answer» Correct Answer - B Fraction of molecules (x) having energy equal to or greater than activation energy may be calculated as follows: `x=n//N=e^(-Ea)"//RT"` In `x=-E_(a)/(RT)or logx=-E_(a)/(2.303RT)` or `logx=-(209xx10^(3)J" mol"^(-1))/(2.303xx(8.314Jk^(-1)" mol"^(-1))xx581k)` `=-18.8323` `x=Antilog(-18.8323)` `=1.471xx10^(-19)` Fraction of molecules `=1.47xx10^(-19` |
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| 416. |
`A + B rarr` Product, `(d x)/(d t) = k[A]^(a)[B]^(b)` If `((d x)/(d t)) = k`, then order is:A. `4`B. `2`C. `1`D. `0` |
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Answer» Correct Answer - D Given `(dx)/(d t) = k[A]^(a)[B]^(b)` and if `(d x)/(d t) = k` then `a + b = 0` i.e. zero order reaction. |
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| 417. |
Number of natural life times `(T_(av))` required for a first order reaction to achieve `99.9%` level of comletion isA. `6.9`B. `1.5`C. `0.105`D. `9.2` |
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Answer» Correct Answer - A Time required for `99.9%` reaction is `10` times of half-life period. `T_(99.9) = 10T_(50)` But `T_(50) = (0.693)/(k)` `:. T_(av) = (1)/(k)` and `T_(50) = 0.693 T_(av)` `:. T_(99.9) = 10 xx 0.693 T_(av) = 6.93 T_(av)` |
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| 418. |
For the first-order reaction `(C=C_(0)e^(-k_(1)^(t)))` and `T_(av)=k_(1)^(-1)`. After two average lives concentration of the reactant is reduced to :A. `25%`B. `75%`C. `(100)/(e)%`D. `(100)/(e^(2))%` |
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Answer» Correct Answer - D |
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| 419. |
For the first-order reaction `T_(av)` (average life), `T_(50)` and `T_(75)` in the increasing order are :A. `T_(50)ltT_(av)ltT_(75)`B. `T_(50)ltT_(75)ltT_(av)`C. `T_(av)ltT_(50)ltT_(75)`D. `T_(av)=T_(50)ltT_(75)` |
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Answer» Correct Answer - A |
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| 420. |
which is not true for a second order reaction?A. It can have rate constant `1xx10^(-3)"L mol"^(-1)s^(-1)`B. Its half-life is inversely proportional to its initial concentrationC. Time to complete `75%` reaction is twice of half-lifeD. `T_(50)=(1)/(Kxx"initial conc".)` |
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Answer» Correct Answer - C |
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| 421. |
For the Ist order reaction : A(g) `rarr` B(g) + C(g) + D(s) taking place at constant pressure and temperature condition. Initially volume of container containing only A was found to be 10 L and after `0.693` hrs it was `17.5` L. The rate canstant for the reaction is :A. `(1)/(0.693)"ln" (3)/(2)hr^(-1)`B. 2 `hr^(-1)`C. `(1)/(0.693)"ln"(4)/(3)hr^(-1)`D. 1 `hr^(-1)` |
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Answer» Correct Answer - B |
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| 422. |
Which of the following statements is correct for a possible order reaction?A. The rate of a reaction decreases with passage of time as the concentration of reactants decreases.B. The rate of reaction is same at any time during the reaction.C. The rate rate of a reaction is independent of temperature change.D. The rate of a reaction decreases with increase in concentration of reactant(s). |
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Answer» Correct Answer - A |
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| 423. |
For the reaction : A `rarr ` Product, the order of reaction is equal to :A. `("ln"(r_(0))_(1)+"ln"(r_(0))_(2))/("ln"[A_(0)]_(1)+"ln"[A_(0)]_(2))`B. `1-("ln"(t_(1//2))_(1)-"ln"(t_(1//2))_(2))/("ln"[A_(0)]_(1)-"ln"[A_(0)]_(2))`C. `("ln"(t_(1//2))_(1)-"ln"(t_(1//2))_(2))/("ln"[A_(0)]_(1)-"ln"[A_(0)]_(2))`D. `1+("ln"(t_(1//2))_(1)-"ln"(t_(1//2))_(2))/("ln"[A_(0)]_(1)-"ln"[A_(0)]_(2))` |
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Answer» Correct Answer - B |
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| 424. |
For an acid catalysed hydrolysis of an ester which of the following options regarding initial rate of reaction is correct?A. There will be no effect of concentration of `[H^(+)]` on rate of reaction.B. Rate of reaction will be same if `0.05` M HCl is taken or `0.05"M H"_(2)SO_(4)` is taken.C. Rate of reaction will be faster when 1 M HCl is taken as compared to when 1 M `CH_(3)COOH` is taken.D. The reaction proceeds following third order kinetics. |
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Answer» Correct Answer - C |
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| 425. |
The time required for the decompoistion of `99.9%` fraction of a first order reaction is………….to that of its half-life time.A. 5B. 10C. 15D. 20 |
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Answer» Correct Answer - B For a first order reaction, time required to complete any particular fraction is given as `t_(f)=(2.303)/(K)log((1)/(1-f))` Substituting the value of `f=(99.9)/(100)` , we get `t_(99.9%)=(2.303)/(K)log(1)/(1-0.999)` `=(2.303)/(K)log((1)/(0.001))` `=(2.303)/(K)log((10^(3))` `=(2.303)/(K)3log(10)` `=(6.903)/(K)(log(10=1))` Half life of reaction is given as `t_(1//2)=(0.693)/(K)` Dividing the two result we get `(t_(99.9%))/(t_(1//2))=(6.909//K)/(0.693//K)` `=(6.909)/(0.693)` `=9.97~~10` Alternatively, we can calculate `yt_(99.9%)` as follows |
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| 426. |
Which of the following graphs represents zero order if `A rarr P` At `t = 0 rArr [A]_(0)` At `t = t rArr [A]_(t)`A. B. C. D. |
| Answer» Correct Answer - C | |
| 427. |
How many half-lives are needed to complete the zeroth order reaction ?A. TwoB. FourC. InfiniteD. Eight |
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Answer» Correct Answer - A The half-life of a reaction, symbolized by `t_(1//2)` , is the time required for the reactant concentration to drop to one half of its intial value. Suppose a zero order reaction of the from `a0` initial concentration `ArarrB` `(a-x)x` concentration at time `t` Integrated rate law is `K_(0)=(x)/(t)` where `K_(0)` is the rate constantl of a zero order reaction. When `t=t_(1//2)` , half of the reatant is converted into product. thus `x=a//2` . Using integrated rate law, we can write `t_(1//2)=(a)/(2K)` Multiplying both sides by `2` , we get `2t_(1//2)=(a)/(K)=t_(oo)` where `t_(oo)` is the time taken for the completion of the reaction. Thus, two half lives are required to complete the zero order reaction. For first order reaction, the integrated rate law is `t=(2.303)/(k)log((a)/(a-x))` Substituting the value of `K` in terms of half-life we get `t=(2.303)/(0.693//t_(1//2))log((a)/(a-x))` When reaction is complete, `t=t_(oo)` and `x=a` . Therefore `t_(oo)=(2.303t_(1//2))/(0.693)log((a)/(a-a))` `=(2.303)/(0.693)t_(1//2)log((a)/(0))` `=(2.303)/(0.693)t_(1//2)(oo)` Therefore, infinite half lives are needed to complete a first order reaction. |
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| 428. |
How much faster would a reaction proceed at `25^(@)C` than at `0^(@)C` if the activation energy is `65 kJ` ?A. `2` timesB. `5` timesC. `11` timesD. `16` times |
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Answer» Correct Answer - C `"log"((K_(2))/(K_(1))) = (E_(a))/(2.303R)((T_(2) - T_(1)))/(T_(1)T_(2))` `= (65 xx 10^(3)xx(298-273))/(2.303 xx 8.3 xx 298 xx 273)` By calculation we fins `(K_(2))/(K_(1)) = 11` |
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| 429. |
Writte the rate law and order for the following reaction: `AB_(2)+C_(2)to AB_(2)C+C` (slow) `AB_(2)+C to AB_(2)C` (fast) |
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Answer» The rate law for the reaction (r) = `k [AB_(2)][C_(2)]` Order of reaction = 1+1 = 2 |
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| 430. |
The temperature at which the average speed of perfect gas molecules is double than at `17^(@)C` isA. `34^(@)C`B. `68^(@)C`C. `162^(@)C`D. `887^(@)C` |
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Answer» Correct Answer - D Average speed `(V_(av)) prop (T)^(1/2)` `:. (V_(av))_(2)/(V_(av))_(1) = sqrt((T_(2))/(T_(1)))` `("Given" ((V_(av))_(2))/(V_(av))_(1) = 2)` `(2)^(2) = (T_(2))/(T_(1)) rArr T_(2) = 4T_(1)` `T_(1) = 273 + 17 = 290 K` `:. T_(2) = 4 xx 290 = 1160 K = 1160 - 273 = 887^(@)C` |
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| 431. |
when concentration of reactant in reaction `A to B ` is increased by 4 times , the rate increase only 2 times The order of the reaction would beA. 2B. `(1)/(3)`C. `4`D. `(1)/(2)` |
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Answer» Correct Answer - D Rate `=2, [A]=4` `[4]^(1//2) =2 therefore n=1//2` |
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| 432. |
The temperature at which the average speed of perfect gas molecules is double than at `17^(@)C` isA. `mol L^(-1) min^(-1)`B. `L^(2) mol^(-2) min^(-1)`C. `L mol^(-1) min^(-1)`D. `min^(-1)` |
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Answer» Correct Answer - B `k = ["conc"]^(1-n)min^(-1)` For third order reaction `= [mol L^(-1)]^(1-3)min^(-1)` `= L^(2) mol^(-2) min^(-2)` |
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| 433. |
Magnesium (Z=12) has isotopes that range form `Mg-20, "to"Mg-31, "Only" Mg-24, Mg-25,andMg-26` are not radioactive. What mode of radioactive deacy would convert. What mode of radioavtive deacy would convert `Mg-20, Mg-21, Mg-22, and Mg-23` into stable isotopes most quickly?A. Electron emissionB. Alpha particle emissionC. Gamma emissinD. Positron emission |
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Answer» Correct Answer - D |
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| 434. |
The rate of a reaction increases four-fold when the concentration of reactant is increased `16` times. If the rate of reaction is `4 xx 10^(-6) mol L^(-1) s^(-1)` when the concentration of the reactant is `4 xx 10^(-4) mol L^(-1)`. The rate constant of the reaction will beA. `2 xx 10^(-4) mol^(1//2) L^(1//2) s^(1//2)`B. `1 xx 10^(-2) s^(-1)`C. `2 xx 10^(-4) mol^(-1//2) L^(1//2) s^(-1)`D. `25 mol^(-1) L min^(-1)` |
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Answer» Correct Answer - A Rate`prop sqrt("Concentration") = ksqrt("Concentration")` `k = (Rate)/("Concentration")^(1//2)` `= (4 xx 10^(-6))/(4 xx 10^(-4))^(1//2) = (4 xx 10^(-6))/(2 xx 10^(-2)) = 2 xx 10^(-4) mol^(1//2) L^(-1//2) s^(-1)` |
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| 435. |
Pieces of wood burn faster than a log of wood of the same mass becauseA. surface area of log of wood is larger and needs more time to burnB. pieces of wood have lager surface areaC. All pieces of wood catch fire at the same timeD. Block of wood has higher density than pieces of the same wood |
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Answer» Correct Answer - B Factors affecting rate of reaction . |
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| 436. |
When the concentration of a reactant in reaction `A rarr B` is increased by `8` times but rate increases only `2` times, the order of the reaction would beA. 2B. `(1)/(3)`C. `4`D. `1/2` |
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Answer» Correct Answer - B `r_(1) prop [A]^(n) (8[A])^(n)prop 2r_(1)` `((8[A])^(n))/([A]^(n))=2implies 8^(n) = 2 therefore =(1)/(3)` |
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| 437. |
For an imaginary reaction `2X+3Y to "products"` Given: rate of disappearance of `X = r_1` rate of disappearance of `Y = r_(2)` `r_1 and r_2` are related as:-A. `3r_(1) = 2r_(2)`B. `r_(1)=r_(2)`C. `2r_(1)= 3r_(2) `D. `r_(1)=2r_(2)` |
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Answer» Correct Answer - A Rate `=-(1)/(2)r_(1)=-(1)/(3)r_(2)` |
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| 438. |
A first orde reaction is 75% completed after 32min. When was 50% of the reaction completed?A. 16 minutesB. 8 minutesC. 4 minutesD. 32 minutes |
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Answer» Correct Answer - A 75 % completed in 32 minutes I .e., 2 half - life has passed `therefore ` 1 half - life = 16 minutes |
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| 439. |
Statement: Temperature coefficient is the ratio of two rate constants preferably `35^(@)C` and `25^(@)C`. Explanation: It can also be given as `e^((-E_(a))/R[(T_(2)-T_(1))/(T_(1)T_(2))])`A. (a) `S` is correct but `E` is wrongB. (b) `S` is wrong but `E` is correctC. (c ) Both `S` and `E` are correct and `E` is correct explanation of `S`D. (d) Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - D Both statement and explanation are correct. |
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| 440. |
For the reaction `2NO_(2)+F_(2)rarr 2NO_(2)F`, following mechanism has been provided: `NO_(2)+F_(2) overset("slow")(rarr) NO_(2)F+F` `NO_(2)+Foverset("fast")(rarr) NO_(2)F` Thus rate expression of the above reaction can be writtens as:A. `r=K[NO_(2))]^(2)[F_(2)]`B. `r=K[NO_(2)][F_(2)]`C. `r=k [NO_(2)]`D. `r=k [F_(2)]` |
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Answer» Correct Answer - B Slowest step is the rate determining step , `r=K[NO_(2)][F_(2)]` |
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| 441. |
For the reaction `2NO_(2)+F_(2)rarr 2NO_(2)F`, following mechanism has been provided: `NO_(2)+F_(2) overset("slow")(rarr) NO_(2)F+F` `NO_(2)+Foverset("fast")(rarr) NO_(2)F` Thus rate expression of the above reaction can be writtens as:A. (a) `S` is correct but `E` is wrongB. (b) `S` is wrong but `E` is correctC. (c ) Both `S` and `E` are correct and `E` is correct explanation of `S`D. (d) Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 442. |
For the non-equilibrium process, `A + B rarr` Product, the rate is first-order w.r.t. `A` and second order w.r.t. `B`. If `1.0 mol` each of `A` and `B` were inrofuced into `1.0 L` vessel and the initial rate was `1.0 xx 10^(-2) mol L^(-1) s^(-1)`, calculate the rate when half the reactants have been turned into Products.A. `1.25 xx 10^(-3) mol L^(-1)s^(-)`B. `1.0 xx 10^(-2) mol L^(-1)s^(-)`C. `2.50 xx 10^(-3) mol L^(-1)s^(-)`D. `2.0 xx 10^(-2) mol L^(-1)s^(-)` |
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Answer» Correct Answer - A `A + B rarr` product `r = K[A]^(1) [B]^(2)` `r_(1) = K[1]^(1) [1]^(2) = 1 xx 10^(-2) (K = 1 xx 10^(-2))` `r_(1) = K[(1)/(2)][(1)/(2)]^(2) = 1 xx 10^(-2) xx (1)/(8)` = `1.25 xx 10^(-3) "mol L"^(-1) S^(-1)` |
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| 443. |
For the non-equilibrium process, `A + B rarr` Product, the rate is first-order w.r.t. `A` and second order w.r.t. `B`. If `1.0 mol` each of `A` and `B` were inrofuced into `1.0 L` vessel and the initial rate was `1.0 xx 10^(-2) mol L^(-1) s^(-1)`, calculate the rate when half the reactants have been turned into Products.A. `1.2xx10^(-3)`B. `1.2xx10^(-2)`C. `2.5xx10^(-4)`D. None of these |
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Answer» Correct Answer - a Rate `=K[A][B]^(2)` `:. 10^(-2)=K[1][1]^(2)` or `K=10^(-2) litre^(2) mol^(-2) sec^(-1)` New rate `=10^(-2)xx0.5xx(0.5)^(2)` `=1.2xx10^(-3) mol litre^(-1) sec^(-1)` |
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| 444. |
for the reaction, `2A + B rarr 3C + D`, which of the following does not express the reaction rateA. `(d[D])/(dt)`B. `-(d[A])/(2dt)`C. `-(d[C])/(3dt)`D. `-(d[B])/(dt)` |
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Answer» Correct Answer - c Rate of reaction `=-1/2(d[A])/(dt)` `=-(d[B])/(dt)=1/3(d[C])/(dt)=(d[D])/(dt)` |
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| 445. |
for the reaction, `2A + B rarr 3C + D`, which of the following does not express the reaction rateA. `(-d[C])/(3dt)`B. `(-d[B])/(dt)`C. `(d[D])/(dt)`D. `(-d[A])/(2dt)` |
| Answer» `(1)/(2)(d[A])/(dt)=(-d[B])/(dt)=+(1)/(3)(d[C])/(dt)=+(d[D])/(dt)="Rate"` | |
| 446. |
In which of the following, `E_(a)` for backward reaction is greater than `E_(a)` for forward reaction?A. (a) `A overset(E_(a)=50kcal)(rarr) B, DeltaH=-10 kcal`B. (b) `A overset(E_(a)=50kcal)(rarr) B, DeltaH=+10 kcal`C. (c ) `A+10 kcal rarr B, E_(a)=50 kcal`D. (d) `A-10 kcal rarr B, E_(a)=50 kcal` |
| Answer» Correct Answer - a, c | |
| 447. |
The rate of chemical reaction increases:A. (a) If the temperature is increasedB. (b) If the concentration of the reactants is decreasedC. (c ) If the concentration of the reactants is increasedD. (d) With time |
| Answer» Correct Answer - a, c | |
| 448. |
For a certain reaction the variation of rate constant with temperature is given by the equation `ln k_(t) = lnk_(0) + ((ln 3)t)/(10) (t ge 0^(@)C)` The value of temperature coefficient of the reaction isA. `0.1`B. `1`C. `10`D. `3` |
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Answer» Correct Answer - D `(k_(1))/(k_(0)) = mu^((t-0)/(10))` `log_(e) k_(t) -loge k_(0)= (t)/(10) log(mu)` `ln k_(t) = ln k_(0) + ((ln(mu))/(10)) t` `mu = 3`. |
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| 449. |
According to kinetic theory of gases:A. (a) collisions are always elasticB. (b) heavier molecules transfer more momentum to the wall of the containerC. (c ) only a small number of molecules have very high velocityD. (d) between collisions, the molecules move in straight lines with constant velocities |
| Answer» Correct Answer - a, b, c, d, | |
| 450. |
Which of the statement is (are) correct?A. (a) A pot of `log K_(p)` vs. `1/T` is linearB. (b) A plot of `log[X]` vs. time is linear for a first order reaction, `X rarr P`C. (c ) A plot of `log P` vs. `1/T` is linear at constant volumeD. (d) A plot of `log P` vs. `1/V` is linear at constant temperature |
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Answer» Correct Answer - a, b, d The relevant expressions are follows: (a) `logK_(p)=(DeltaH)/(RT)`= constant (b) `log[X]=log [X]_(0)+Kt` (c ) `P/T`= constant (V constant) (d) `PV`= constant (T constant) |
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