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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.
| 1451. |
The rate of a gaseous reaction is given by the expression `k[A][B]`. If The volume of reaction vessel is suddenly reduced to one-fourth of the initial volume, the reaction rate relative to the original rate will be :A. `1//16`B. `1//8`C. 8D. 16 |
| Answer» Correct Answer - D | |
| 1452. |
The rate of a gaseous reaction is given by the expression K[A][B]. If the volume of the reaction vessel is suddenly reduced to 1/4 th of the initial volume, the reaction rate relating to original rate will beA. `1//10`B. `1//8`C. 8D. 16 |
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Answer» Correct Answer - d When volume is reduced to `(1)/(4)` , concentration become four times and reaction rate become 16 times because it is 2 order reaction . |
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| 1453. |
Consider a gaseous reaction, the rate of which is given by `k[A][B]`. The volume of the reaction vessel containing these gases is suddenly reduced to `1//4th` of the initial volume. The rate of the reaction as compared with original rate isA. `1//10`B. `1//8`C. `8`D. `16` |
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Answer» Correct Answer - D If volume is decreased by `(1)/(4)` of initial then conc. Of `A` and `b` Both are increased by `4` times of initial so rate incereased `16` times. |
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| 1454. |
The temperature coefficient of a reaction is determined by comparing velocity constants at two temperatures separated by………. . |
| Answer» Correct Answer - `10^(@)C` | |
| 1455. |
The temperature coefficient of a reaction is determined by comparing velocity constant at two temperatures differing by……………… . |
| Answer» Correct Answer - `10^(@)C` | |
| 1456. |
In the given graph the activation energy , `E_(a)` for the reverse reaction will be A. 150 kJB. 50 kJC. 200 kJD. 100 kJ |
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Answer» Correct Answer - b `DeltaH = E_(a) f = E_(a) b` `100 = 150-E_(b)` `E_(a) b = 50 `kJ . |
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| 1457. |
Catalyst in a reactionA. Lowers the activation energyB. Increases the rate of reactionC. Both (a) and (b)D. Initiates the reaction |
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Answer» Correct Answer - c A catalyst lowers the activation energy both of forward and backward reaction and increase the rate of reaction . |
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| 1458. |
The value of temperatuer coefficient is usually………………. . |
| Answer» Correct Answer - two | |
| 1459. |
The reactions having low activation energy are……………reactions. |
| Answer» Correct Answer - fast | |
| 1460. |
The energies of activation for forward and reverse reactions for `A_(2) + B_(2) hArr 2AB ` are `180 kJ mol^(-1)` and 200 `kJ mol^(-1)` respectively . The presents of catalyst lowers the activation energy of the both (forward and reverse ) reactions by `100 kJ mol^(-1)` . The enthalpy change of the reaction `(A_(2) + B_(2) to 2AB)` in the presence of catalyst will be (in kJ `mol^(-1)`)A. 300B. 120C. 280D. 20 |
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Answer» Correct Answer - d So , `DeltaH_("Reaction") = E_(f) - E_(b) = 80 - 100 = -20` . |
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| 1461. |
Which of the following are correct for redioactivity ?A. Half life is independent of initial amount of redioactive substance.B. On increasing temperature half life will decrease.C. The half life of `C-14` in `CO_(2)` is different form than in `C_(6)H_(12)O_(6.)`D. Half life of `U-238`and 235 are different. |
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Answer» Correct Answer - A::D |
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| 1462. |
There are four nuclei A,B.C and D having mass numbers `9, 10,11` and , `12` respectively. Their binding gnergies are ``54, 10, 11` and `12` respectively. Their binding energies are `54,70,66` and `78` MeV, respectively. Which of the following statement is/are true ?A. The most stable nuclei is B.B. A and C have equal stabilities.C. The relative order of stability is : `Dgt Bgt C= A.D. D is more stable than C but less stable than B. |
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Answer» Correct Answer - A::B::D |
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| 1463. |
Which of the following statements are applicable to a balanced chemical equation of an wlwmwntry reaction ?A. Order is same as molecularicty.B. Order is less than the molecularity.C. Order is grater than the molecularity.D. Molecularity can never be zero. |
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Answer» Correct Answer - A::D |
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| 1464. |
For a complex reaction :A. order of overall reactionis same as molcularity of the slowest step (provided slowest step is having no reaction intermediate)B. order of overall reaction is less than the molecularity of the slowest step.C. order of overall reaction is greater than molecularity of the slowest step.D. molecularity of the slowest step is never zero or non interger. |
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Answer» Correct Answer - A::D |
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| 1465. |
Which of the following statements are in accordance with the Arrhenium equation?A. Rate of a reaction increases with increase in temperature.B. Rate of a reactrionincreases with dectease in activation energy.C. Rate constant decteases exponentially with increase in temperatureD. Rate of reaction decreases with decrease in activation energy. |
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Answer» Correct Answer - A::B |
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| 1466. |
Which of the following statements are in accordance with the Arrheenius equation ?A. Rate of a reaction increases with increase in temperatureB. Rate of a reaction increases with decreases in activation energyC. Rate constant decreases exponentially with increase in temperatureD. Rate of reaction decreases with decreases in activation energy |
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Answer» Correct Answer - A::B Arrhenius equations can be written as `K = A.e ^((-E_(a))/(RT))` `k prop ^(-E_(a))` i.e, rate of reaction increases with increase in temperature. |
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| 1467. |
In the reaction, `A to B`, the value of the rate constant was found to be `1.0 xx 10^(-2) mol^(-1) L s^(-1)`. What is the order of the reaction? How will the catalyst affect of the value of the rate constant? |
| Answer» Self explanatory | |
| 1468. |
The rate constant of a reaction is `3 xx 10^(2) hr^(-1)`. What is the order of the reaction? |
| Answer» From the units of rate constant, the reaction is of first order. | |
| 1469. |
a) write dimensions of first order reaction. b) Give reason for no change in concentration of `H_(2)` and `Cl_(2)` with respect to time in the reactions. `H_(2)(g) + Cl_(2) overset(hv)to 2HCl(g)` |
| Answer» a) `sec^(-1)` b) the reaction is photochemical reaction and zero order in nature. | |
| 1470. |
A first order of reaction is `75%` complete in `18` minute. What is the half-life period for the reaction ? |
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Answer» Correct Answer - 9 `K=2.303/t"log"a/((a-x))` `K=2.303/18"log"100/((100-75))=2.303/18xx"log" 4` `=7.7xx10^(-2)` Now `t_(1//2)=0.693/K=0.693/(0.77xx10^(-2))=9 min` |
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| 1471. |
The half life of a first order reaction having rate constant `k=1.7xx10^(-5) sec^(-1)` is :A. `12.1` hrsB. `9.7` hrsC. `11.3` hrsD. `1.8` hrs |
| Answer» `t_(1//2)=(0.693)/(k)=(0.693)/(1.7xx10^(-5))=40764 sec = 11.3 hrs ` | |
| 1472. |
For the first order reaction, the rate constant is `7.7xx10^(-2) sec^(-1)`. Calculate the time required for the initial concentration `1.5` mole of the reactant to be reduced to `0.75` mole. |
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Answer» Correct Answer - 9 `t=2.303/K"log"([A_(0)])/([A])` `=2.303/(7.7xx10^(-2))xx"log"1.5/0.75` `=2.303/(7.7xx10^(-2))xx"log"2 ~~9 sec` |
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| 1473. |
The rate constant for a chemical reaction has unit litre `mol^(-1) sec^(-1)`. Find the order of the reaction. |
| Answer» From the units of rate constant, the reaction is of second order. | |
| 1474. |
State the units of rate constant in zero order reaction. |
| Answer» Units of k = mol `L^(-1)s^(-1)` | |
| 1475. |
Identify the reaction order from the rate constant value =`3.2 xx 10^(-5)` litre `mol^(-1)sec^(-1)` |
| Answer» The reaction of second order. | |
| 1476. |
To know the numerical value of a rate expression, we muyst know the concentration or the pressure of a substance at two different times during the course of a reaction. This information can be obtained experimentally, if we have a method ofA. measuring timeB. concentration or pressureC. keeping the conditions, especially the temperature constantD. all of these |
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Answer» Correct Answer - D The rate of a chemical reaction can be expressed as a given change in a given interval of time. The rate usually is expressed in termp of the change in the concentration or the pressure of one component. It may be expressed as a decrease in the concentration ( or pressure ) of a reactant or an increase in the concentration (or pressure ) of a product. |
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| 1477. |
For the reaction, `H_(2)+I_(2)overset(k_(1))underset(k_(2))hArr2HI`. The rate law expression is :A. `[-(1)/(2)(d[HI])/(dt)]=k_(1)[H_(2)][I_(2)]`B. `[-(1)/(2)(d[HI])/(dt)]=(k_(1)[HI]^(2))/(k_(2)[H_(2)][I_(2)])`C. `[-(1)/(2)(d[HI])/(dt)]=k_(1)[H_(2)][I_(2)]-k_(2)[HI]^(2)`D. `[-(1)/(2)(d[HI])/(dt)]=k_(1)k_(2)[H_(2)][I_(2)]` |
| Answer» Correct Answer - C | |
| 1478. |
For a Ist order decomposition of `A` as given Therefore rate constant `(K)` for the overall decomposition of `A` isA. `k=k_(1)+k_(2)+.....+k_(n)`B. `k=(k_(1)+k_(2)+.....+k_(n))//n`C. `k=k_(1)xxk_(2)xx.....xxk_(n))`D. none of these |
| Answer» Correct Answer - A | |
| 1479. |
For the reaction `NH_(4)^(+)+OCN^(-)rarrNH_(2)CONH_(2),` rthe probable mechanism is, `NH_(4)^(+)+OCN^(-)hArrNH_(4)OCN" "` (fast) and `NH_(4)OCNrarrCONH_(2)" "` (slow) The rate law will be :A. rate =`k[NH_(2)CONH_(2)]`B. rate =`k[NH_(4)]^(+)[OCN]^(-)`C. rate =`k[NH_(4)OCN]`D. None of these |
| Answer» Correct Answer - B | |
| 1480. |
The rate expresison for the reaction: `NH_(4)CNO hArr NH_(2)CONH_(2)` can be derived form the mechanism: (i) `NH_(4)CNO underset(k_(2))overset(k_(1))hArr NH_(4)NCO ("Fast")` (ii) `NH_(4)CNO overset(k_(3))rarrNH_(3) + HNCO` ("fast") (iii) `NH_(3)+HNCo overset(k_(4))rarr NH_(2)CONH_(2)` (Slow) Which of the following statement `(s)` is/are correct about the rate expresison?A. `(d_(["urea]))/(dt) = (k_(1)k_(3))/(k_(2))[NH_(4)NCO]`B. `(d_(["urea]))/(dt) = (k_(1)k_(3))/(k_(2)k_(4))[NH_(4)NCO]`C. `(d_(["urea]))/(dt) = k[NH_(4)NCO]`D. `(d_(["urea]))/(dt) = (k_(1) xx k_(2))/(k_(3) xx k_(4))[NH_(4)NCO]` |
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Answer» Correct Answer - A::C `(d)/(dt)[NH_(2)CONH_(2)] = k_(4)[NH_(3)][HNCO]` from (iii) Applying steady state approximation to `HNCO` or `NH_(3)` `(d[HNCO])/(dt) = O = k_(3) [NH_(4)NCO]-k_(4)[NH_(3)][HNCO]` `:. (k_(3))/(k_(4)) = ([NH_(3)][HNCO])/([NH_(4)CNO])` `(d["urea"])/(dt) = k_(4) xx [NH_(3)][HNCO] = k_(4) xx (k_(3))/(k_(4))[NH_(4)NCO]` Also, `[NH_(4)NCO] = (k_(1))/(k_(2)) xx [NH_(4)CNO]` `:. (d["urea"])/(dt) = k_(3) xx (k_(1))/(k_(2)) xxx [NH_(4)CNO]` `= k[NH_(4)CNO]` |
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| 1481. |
For the reaction `NH_(4)^(+)+OCN^(-)rarrNH_(2)CONH_(2),` rthe probable mechanism is, `NH_(4)^(+)+OCN^(-)hArrNH_(4)OCN" "` (fast) and `NH_(4)OCNrarrCONH_(2)" "` (slow) The rate law will be :A. rate =`k[NH_(2)CONH_(2)]`B. rate =`k[NH_(4)]^(+)[OCN]^(-)`C. rate =`k[NH_(4)OCN]`D. None of these |
| Answer» Correct Answer - B | |
| 1482. |
In some cases, it is found that a large number of colliding molecules have energy more than thereshold value, yet the reaction is slow. Why? |
| Answer» The reactant molecules may not be properly oriented at the time of collisions. | |
| 1483. |
`E_(1)` and `E_(2)` are the reaction is inversely proportional to initial concentration of the reactant, what is the order of reaction? |
| Answer» As `E_(2)gtE_(1)`, this means that the reaction is exothermic in nature. | |
| 1484. |
Express the relation between the half-life period of a reaction and initial concentration of the reaction of second order. |
| Answer» For the second order reaction, `t_(1//2) alpha 1/[A]_(0)` | |
| 1485. |
In a reaction, `2A to` Products, the concentration of A decreases from 0.5 mol `L^(-1)` to 0.4 mol `L^(-1)` in 10 minutes. Calculate the rate during this interval.A. `0.012`B. `0.024`C. `2 xx 10^(-3)`D. `2 xx 10^(-4)` |
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Answer» Correct Answer - D d) Rate =`(Delta[A])/(dt)= (0.12M)/(600s) = 2 xx 10^(-4)` |
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| 1486. |
In a reaction, `2A to` Products, the concentration of A decreases from 0.5 mol `L^(-1)` to 0.4 mol `L^(-1)` in 10 minutes. Calculate the rate during this interval. |
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Answer» For the reaction, `2A to `products. Avergage rate = `-1/2(Delta[A])/(dt) = -1/2((0.4-0.5) mol^(-1))/(10 min)` `=((0.1)molL^(-1))/(20 min) = 0.005 mol L^(-1)min^(-1)` |
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| 1487. |
In a reaction `2A rarr B` the concentration of A decreases from `0.3 mol L^(-1) to 0.2 mol L^(-1)` in 10 minutes. The reaction rate during trhis interval is mol `L^(-1) minute^(-1)`A. `0.01`B. `-0.005`C. `-0.01`D. `0.005` |
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Answer» Correct Answer - D Reaction rate=rate of disappearance of `A` per mole `=-(1)/(2)((Delta[A])/(Deltat))=-(1)/(2)((0.2-0.3)molL^(-1))/(10min)` `=0.005molL^(-1)min^(-1)` Reaction rate is always a positive equantity. The negative sign simply indicates the fall in concentration of reaction `A` . |
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| 1488. |
Rate of which reaction increase with temperature:A. of anyB. of exothermic reactionsC. of endothermic reactionD. of none |
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Answer» Correct Answer - A |
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| 1489. |
For a zero order reaction. Which of the following statement is false:A. The rate is independent of the temperature of the reactionB. The rate is independent of the concentration of the reactantsC. The half life depends upon the concentration of the reactantsD. The rate constant has the unit mole `L^(-1)sec^(-1)` |
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Answer» Correct Answer - A |
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| 1490. |
The first order rate constant k is related to temperature as log `k=15.0-(10^(6)//T)`. Which of the following pair of value is correct?A. `A=10^(15)"and E"=1.9xx10^(4)KJ`B. `A=10^(-15)"and E"=40KJ`C. `A=10^(15)"and E"=40KJ`D. `A=10^(-15)"and E"=1.9xx10^(4)KJ` |
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Answer» Correct Answer - A |
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| 1491. |
For a given reaction, energy of activation for forward reaction `(E_(af)` is 80 kJ `mol^(-1).DeltaH=-40"kJ mol"^(-1)` for the reaction. A catalyst lowers `E_(af)` to 20 kJ `mol^(-1)`. The ratio of energy of activation for reverse reaction before and after addition of catalyst is :A. `1.0`B. `0.5`C. `1.2`D. `2.0` |
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Answer» Correct Answer - D |
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| 1492. |
Which of the following statement is correct regarding enthalpy of reaction?A. It does not depend on physicall state of reactant or products.B. It always increases with increase in temperature.C. It is equal to ratio of activation energy of forward reaction to activation energy of backward reaction.D. It does not depend on presence of catalyst. |
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Answer» Correct Answer - D |
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| 1493. |
Consider the following reactions at 300 K. A `rarr` B (uncatalysed reaction) `Aoverset("catalyst")rarrB` (catalyst reaction) The activation energy is lowered by `8.314"KJ mol"^(-1)` for the catalysed reaction. How many times the rate of this catalysed reaction greater than that of uncatalysed reaction?(Given `e^(3.33)=20)`A. 15 timesB. 38 timesC. 22 timesD. 28 times |
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Answer» Correct Answer - D |
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| 1494. |
The rate of a heterogeneous reaction (as iron (solid) any oxygen gas) does not depend on:A. Concentration of reactantsB. Surface area of reactantsC. pressure of reactant gasesD. potential energy of reactant |
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Answer» Correct Answer - d Rate of reaction depends upon conc., pressure of `O_(2)` and surface area of iron. |
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| 1495. |
The rate constant of a second order reaction is `10^(-2) "mol"^(-1) litre sec^(-1)`. The rate constant when expressed in `cm^(3) "molecule"^(-1) min^(-1)` is:A. `9.96xx10^(-22)`B. `9.96xx10^(-23)`C. `9.96xx10^(21)`D. `1.004xx10^(-24)` |
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Answer» Correct Answer - a `K=10^(-2) mol^(-1) litre sec^(-1)` `=(10^(-2)xx1000xx60)/(6.02xx10^(23)) c c "molecule"^(-1) min^(-1)` `=9.9618xx10^(-22) c c "molecule"^(-1) min^(-1)` |
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| 1496. |
The rate constant of reaction is `5xx10^(-2) "litre " ^(3) "mole"^(-3) "min"^(-1)` the order of reaction isA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - D `k =("rate")/([R]^(n))=(M//s)/(M^(n))=M^(1-n)S^(-1)` when `n= 4, K= "litre"^(3) "mole" ^(-3) min^(-1)` |
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| 1497. |
`A+2B to C +D ` for a reaction from following data correct rate law = ` A. `"Rate " = K [A]^(2) [B]`B. `"rate" =k [A]^(2) [B]^(2) `C. `"Rate " =K[A][B]^(2)`D. `"Rate " =K[A][B] ` |
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Answer» Correct Answer - C Rate `=k [A][B]^(2)` Keeping [B]constant ,[A],is made a 4 times , rate also become 4 times , hence rate `prop [A].` keeping [A] constant ,[B] is doubled , rate becomes 4 times , hence rate `prop [B]^(2)`. `therefore "Rate"=k[A][B]^(2)` |
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| 1498. |
The rate constant of a reaction has same dimensions as rate of reaction The reaction is ofA. third orderB. second orderC. first orderD. zero order |
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Answer» Correct Answer - D `k=("rate")/([R]^(n)),k = ` rate , when n=0 |
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| 1499. |
when concentration of reactant is incresed eighteen times the rate becomes two times , the rate of reaction isA. `1`B. `(1)/(2)`C. `(1)/(3) `D. `(1)/(4)` |
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Answer» Correct Answer - C `(k_(2))/(k_(1))[(A_(2))/(A_(1))]^(n) or 2=(8)^(n) therefore = (1)/(3)` |
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| 1500. |
Frequently a species can react in different ways to give a variety of products For example toluene can be nitrated at the ortho meta or para positions we shall consider the simplest case, that of two competing irreversible first- order reactions: `A overset(k_(1)) rarr C` and `A overset (k_(2)) rarr D` where the stoichimetric coefficients are taken as unity for simplicity. THe rate law is `((d[A])/(dt))=-k_(1)[A] -k_(2)[A] =-(k_(1)+k_(2))[A]` `Rightarrow [A] = [A]_(0)e^-(k_(1)+k_(2))t` For C, we have `((d[C])/(dt))=k_(1)[A] =k_(1)[A]_(0)e^-(k_(1)+k_(2))t` Multiplication by dt and integration from time 0(where `[C]_(0)=0` to an arbitary time t gives `[C] =(k_(1)[A]_(0))/(k_(1)+k_(2))(1-e^-(k_(1)+k_(2))t)` Similarly integration of `((d[D])/(dt))` gives `[D] = (k_(2)[A]_(0))/(k_(1)+k_(2))(1-e^-(k_(1)+k_(2))t)` The sum of the rate constants `k_(1)+k_(2)` appears in the exponentials for both [C] and [D] At any time, we also have, `[C]/[D] = k_(1)/k_(2)` At high temperature, acetic acid decomposition into `CO_(2)` and `CH_(4)` and simultaneously into `CH_(2)CO` (ketene) and `H_(2)O` (i) `CH_(3)COOH overset (k_(1)=3s^(-1)rarr CH_(4)+CO_(2)` (ii) `CH_(3)COOH overset(k_(2)=4s^(-1)rarr CH_(2)CO+H_(2)O` What is the fraction of acetic acid is reacting as per reaction(i)?A. `3/4`B. `3/7`C. `4/7`D. None of these |
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Answer» Correct Answer - B |
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