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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.
| 1. |
Rate law for the reaction `A + 2B rarr C`, is found to be Rate = k [A] [B]. If the concentration of reactant B is doubled keeping the concentration of A constant, the value of rate constant will beA. the sameB. doubledC. quadrupledD. halved |
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Answer» Correct Answer - A |
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| 2. |
Which of the following statements is incorrect about the collision theory of chemicacl reaction ?A. It considers reacting molecules or atoms to be hard spheres and ignores their structrual features.B. Number of effective collisions determines the rate of reaction.C. Collision of atoms or molecules processig sufficient thereshold energy results into the product formation.D. Molecules should collide with sufficient threshold energy andproper orientatin for the collision to be effective. |
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Answer» Correct Answer - C |
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| 3. |
A chemical reaction was carried out at 300 K and 280 K. The rate constants were found to be `k_(1) and k_(2)` respectively. ThenA. `k_(2)=0.25 k_(1)`B. `k_(2)=0.5k_(1)`C. `k_(2)=4k_(1)`D. `k_(2)=2k_(1)` |
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Answer» Correct Answer - A By the increase in temperature the activation energy decrease. So the rate of reaction increases. Here temperature difference is 20K. At 10K the rate constant is doubled. Therefore, here it is four times. Hence, `k_(2)~~0.25k_(1)` |
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| 4. |
The half-life of a reaction is halved as the initial concentration of the reaction is doubled. The order of the reaction isA. 0.5B. 1C. 2D. 0 |
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Answer» Correct Answer - C Relation Between half-life time and initial concentration is `t_(1//2)prop(1)/((a_(0))^(n-1)` Then, `(t_(1//2))_(1)prop(1)/((a_(0))_(1)^(-1))" "...(i)` `(t_(1//2))/(2)prop(1)/((2a_(0))^(n-1)" "...(ii)` Divide by Eq(ii) `(t_(1//2)xx2)/t_(1//2)=((2)^(n-1)(a_(0))^(n-1))/((a_(0))^(n-1))` `2=(2)^(n-1)rArr -1=1` `n=2` |
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| 5. |
75 % of first order reaction is complete in 30 minutes. What is the time required for 93.75 % of the reaction (in minutes) ?A. 45B. 120C. 90D. 60 |
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Answer» Correct Answer - D `t_(1//2)=1/2xx30=15`min93.75% reaction can be completed in `4xxt_(1//2)=4xx15 =60`min |
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| 6. |
A reaction is 50 % complete in 2 hours and 75 % complete in 4 hours the order of reaction is |
| Answer» Since the amount left after one half period (2 hrs) is one half (50%) of the original amount and the amount left after two half period (4hrs) is one-fourth (25%), this suggests that the reaction is of rist order. | |
| 7. |
A certain zero order reaction has `k = 0.025 M s^(-1)` for the disappearance of `A`. What will be the concentration of `A` after `15` seconds if the initial concentration is `0.5 M` ?A. `0.5M`B. `0.32M`C. `0.12M`D. `0.06M` |
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Answer» `x=kt=0.025xx15=0.375M` Reamaning conc. `=0.5-0.375=0.125M` |
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| 8. |
If the concentration is mentioned in mol `L^(-1)` units and time in seconds, what are the units of k for zero order reaction and first order reaction? |
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Answer» For zero order reaction, k has units : mol `L^(-1)s^(-1)`. For first order reaction, k has units : `s^(-1)` |
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| 9. |
Average rate of reaction does not give the true picture of the reaction rate. Explain. |
| Answer» Average reaction rate obtained by dividing the change in molar concentration by the time taken for the change does not give the true pictue of the reaction rate, Actually, the change in molar concentration of the species keeps on changing with time, so is the change in molar concentration. Therefore, the molar concentration does not change at a uniform rate and average rate of reaction fails to give the true picture of the reaction rate. | |
| 10. |
Rate constant k of a reaction is dependent on temperatur: `k=Ae^(Ea//RT)` K has the least value atA. high T and high `E_(a)`B. high T and small `E_(a)`C. low T and low `E_(a)`D. low T and high `E_(a)` |
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Answer» Correct Answer - D We have the arrhenius equation, `k=Ae^(E_a//RT)` `=Ae^(ET//E_(a))` K becomes least when T is low and `E_(a)` is high. |
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| 11. |
How does the value of rate constant vary with reactant concentrations? |
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Answer» For a general nth order reaction, Rate constant(k) `propto 1/(C^(n-1))` |
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| 12. |
For the reaction:`N_(2)O_(4)(g)rarr2NO_(2)(g)` the number of moles of `N_(2)O_($)(g)` with time given as : `|{:("Time, min","0","5","10"),("Moles N"_(2)O_(4)(g),"0.200","0.170","0.140"):}|` What is the number of moles of `NO_(2)(g)` at `t=10` min? (Assume moles of `NO_(2)(g)=0` at t = 0)A. `0.280`B. `0.120`C. `0.110`D. `0.060` |
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Answer» Correct Answer - B |
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| 13. |
The initial rate of reaction `3 A + 2 B + C to ` Products , at different initial concentrations are given below The order with respect to the reactants , A , B and C are respectivelyA. 3 , 2 , 0B. 3, 2 , 1C. 2 , 2 , 0D. 2,1,0 |
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Answer» Correct Answer - d From `1^(st)` and `2^(nd)` sets of data- no change in rate with the change in concentration of C . So the order with respect to C is zero. From `1^(st)` and `4^(th)` sets of data Dividing eq. (4) by eq. (1) `(1.25 xx 10^(-3))/(5.0 xx 10^(-3)) = [(0.005)/(0.010)]^(x)` or `0.25 = (0.5)^(x)` or `(0.5)^(2) = (0.5)^(x)` `therefore x = 2` The order with respect to A is 2 . From the `1^(st)` and `3^(rd)` sets of data Dividing eq. (1) by eq. (3) `(5.0 xx 10^(-3))/(1.0 xx 10^(-2)) = [(0.005)/(0.010)]^(y)` or `(0.5)^(1) = (0.5)^(y) implies y = 1` The order with respect to B is 1 So the order with respective the reactants A , B and C is 2, 1 and 0. |
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| 14. |
A compound decomposes with a first-order rate constant of `0.00854s^(-1)`. Calculate the concentation after `5.0` minutes for an initial concentration of `1.2` M.A. `0.010` MB. `0.093` MC. `0.92` MD. `1.1` M |
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Answer» Correct Answer - B |
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| 15. |
The rate of the reaction `2NO+O_(2)rarr2NO_(2),"at" 25^(@)C` is `r=k[NO]^(2)[O_(2)]` If the initial concentrations of the reactant are `O_(2)=0.040 "mol"L^(-1) and NO==0.01 "mol" L^(-1)` , the rate constant of the reaction is :A. `7.0xx10^(-2)L"mol"^(-1)s^(-1)`B. `7.0xx10^(-4)L^(2)"mol"^(-2)s^(-1)`C. `7.0xx10^(2)L^(2)"mol"^(-2)s^(-1)`D. `7.0xx10^(3)L^(2)"mol"^(-2)s^(-1)` |
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Answer» Correct Answer - D `0.028=k[0.01]^(2)[0.040]` |
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| 16. |
`C^(14)` is a beta active nucleus. A sample of `C^(14)H_(4)` gas kept in a closed vessel shows increase in pressure with time . This is due to:A. the formation of `N^(14)H_(3) "and" H_(2)`B. the formation of `B^(11)H_(3) "and" H_(2)`C. the formation of `C_(2)^(14)H_(4) "and"H_(2)`D. the formation of `C^(12)H_(3), N^(14)H_(2) "and"H_(2)` |
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Answer» Correct Answer - a |
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| 17. |
The unit of first order rate constant when concentration is measured in terms of pressure and time in minutes is ………………. . |
| Answer» Correct Answer - `min^(-1)` | |
| 18. |
For a reaction between A and B, the initial rate of reaction is measured for various initial concentrations of A and B . The data provided are : The overall order of the reaction is :A. one(1)B. two (2)C. two and a half (2.5)D. between 1 and 2 |
| Answer» Correct Answer - D | |
| 19. |
For a reaction between A and B , the initial rate of reaction is measured for various initial concentrations of A and B . The data provided are The overall order of the reaction isA. OneB. TwoC. Two or halfD. Three |
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Answer» Correct Answer - a If the order of reaction w.r.t. A is n and the order of reaction w.r.t. B is m rate law becomes Rate = `k [A]^(n)[B]^(m)` From (i) ` 5 xx 10^(-5) = [0.20]^(n) [0.30]^(m) " " … (i)` From (ii) `5 xx 10^(-5) = [0.20]^(n) [0.10]^(m) " " ... (ii)` From (iii) ` 1 xx 10^(-5) = [0.40]^(n) [0.05]^(m) " "... (iii)` or `10 xx 10^(-5) = [0.40]^(n) [0.05]^(m)` From equations (i) and (ii) , `(5 xx 10^(-5))/(5 xx 10^(-5)) = [(0.20)/(0.20)]^(n) [ (0.30)/(0.10)]^(m)` `1 = (3)^(m) implies (3)^(0) = (3)^(m) implies m = 0` From equations (ii) and (iii) , `(5 xx 10^(-5))/(10 xx 10^(-5)) = ((0.20)/(0.40))^(n) ((0.10)/(0.05))^(m)` `(1)/(2) = ((1)/(2))^(n) implies ((1)/(2))^(1) = ((1)/(2))^(n) implies n = 1`s Overall order of the reaction = n + m = 1 + 0 = 1. |
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| 20. |
The following rate data were obtained at `303 K` for the following reaction: `2A + B rarr C + D` What is the rate law? What is the order with respect to each reactant and the overall order? Also calculate the rate constant and write its units. |
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Answer» Let the order of the reaction with respect to A be x and with respect to B be y . Therefore , rate of the reaction is given by , Rate = `k[A]^(x) [B]^(y)` According to the question , `6.0 xx 10^(-3) = k [0.1]^(x) [0.1]^(y) " " (i)` `7.2 xx 10^(-2) = k [0.3]^(x) [0.2]^(y) " " (ii)` `2.88 xx 10^(-1) = k [0.3]^(x) [0.4]^(y) " " (iii) ` `2.40 xx 10^(-2) = k [0.4]^(x) [0.1]^(y) " " (iv)` Dividing equation (iv) by (i) , we obtain `(2.40 xx 10^(-2))/(6.0 xx 10^(-3)) = (k [0.4]^(x)[0.1]^(y))/(k [0.1]^(x) [0.1]^(y))` `implies 4 = ([0.4]^(x))/([0.1]^(x))` `implies 4 = ((0.4)/(0.1))^(x)` `implies (4)^(1) = 4^(x)` `implies x = 1 ` Dividing equation (iii) by (ii) , we obtain `(2.88 xx 10^(-1))/(7.2 xx 10^(-2)) = (k [0.3]^(x) [ 0.4]^(y))/(k[0.3]^(x) [ 0.2]^(y))` `implies 4 = ((0.4)/(0.2))^(y)` `implies 4 = 2^(y)` `implies 2^(2) = 2^(y)` `implies y = 2 ` Therefore , the rate law is Rate = k `[A][B]^(2)` `implies k = ("Rate")/([A][B]^(2))` From experiment l , we obtain `k = (6.0 xx 10^(-3) "mol" L^(-1) "min"^(-1))/((0.1 "mol" L^(-1)) (0.1 "mol" L^(-1))^(2))` `6.0 L^(2) "mol"^(-2) "min"^(-1)` From experiment lll , we obtain `k = (2.88 xx 10^(-1) "mol" L^(-1) "min"^(-1))/((0.3 "mol" L^(-1)) (0.4 "mol" L^(-1))^(2))` `6.0 L^(2) mol^(-2) "min"^(-1)` From experiment IV , we obtain `k = (2 . 40 xx 10^(-2) "mol" L^(-1) "min"^(-1))/((0.4 "mol" L^(-1)) (0.1 "mol" L^(-1))^(2))` `= 6.0 L^(2) mol^(-2) "min"^(-1)` Therefore , rate constant , K = `6.0 L^(2) mol^(-2) "min"^(-1)` |
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| 21. |
The radioactivity of sample is `R_(1)` at a time `T_(1)` and `R_(2)` at a time `T_(2)` . If the half -life of thespeciman is T, the number of atoms that have disintegrated in the time `(T_(2)-T_(1))` is equal to :A. `(R_(1)T_(1)-R_(2)T_(2))`B. `(R_(1)-R_(2))`C. `(R_(1)-R_(2))//T`D. `(R_(1)-R_(2))T//0.693` |
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Answer» Correct Answer - d |
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| 22. |
A sample of radiative substance is found `90%` of it’s initial amount after one day. What `%` of the original sample can be found after 3 days ?A. 81B. 72.9C. 25D. `65.61` |
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Answer» Correct Answer - 2 Equal fraction decay in equal periods of time, fraction of sample remaining after 3 days `rArr(0.9)^(3)=0.729` |
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| 23. |
In a reaction between A and B, the initial rate of reaction was measured for different initial concentration of A and B as given below: `{:(A//M,0.20,,0.20,0.40),(B//M,0.30,,0.10,0.05),(r_(0)//M s^(-1),5.07xx10^(-5),,5.07xx10^(-5),7.6xx10^(-5)):}` |
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Answer» Let the order of the reaction with respect to A be x and with respect to B be y . Therefore , `r_(0) = k[A]^(x) [B]^(y)` `5.07 xx 10^(-5) = k [0.20]^(x) [0.30]^(y) " " (i)` `5.07 xx 10^(-5) = k[0.20]^(x) [0.10]^(y) " " (ii)` `1.43 xx 10^(-4) = k [0.40]^(x) [0.05]^(y) " " (iii)` Dividing equation (i) by (ii) , we obtain `(5.07 xx 10^(-5))/(5.07 xx 10^(-5)) = (k[0.20]^(x) [0.30]^(y))/(k[0.20]^(x) [0.10]^(y))` `implies 1 = ([0.30]^(y))/([0.10]^(y))` `implies ((0.30)/(0.10))^(0) = ((0.30)/(0.10))^(y)` `implies y = 0` Dividing equation (iii) by (ii) , we obtain `(1.43 xx 10^(-4))/(5.07 xx 10^(-5)) = (k[0.40]^(x)[0.50]^(y))/(k [0.20]^(x)[0.30]^(y))` `implies (1.43 xx 10^(-4))/(5.07 xx 10^(-5)) = ([0.40]^(x))/([0.20]^(x)) " " {:[("Since" y = 0),( [0.05]^(y) = [0.30]^(y) = 1)]:}` ` implies 2.821 = 2^(x)` `implies "log" 2.821 = x "log" 2 " " ` (Taking log on both sides ) `implies x = ("log" 2 .821)/("log" 2)` = `1.496` = `1.5` (approximately ) Hence, the order of the reaction with respect to A is `1.5` and with respect to B is zero . |
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| 24. |
Radioactivity of a sample (Z =22) decrease 90% after 10 years . What will be the half -life of the sampleA. 5 yearsB. 2 yearsC. 3 yearsD. 10 years |
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Answer» Correct Answer - c `k= (2.303)/(t) "log" (a)/(a-x) = (2.303)/(10) "log" (100)/(10)` `k = (2.303)/(10)` `t_(1//2) = (0.693)/(2.303) xx 10 = 3 ` yrs. |
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| 25. |
75% of a first order reaction was completed in 32 min . When would 50% of the reaction completedA. 24 minB. 16 minC. 8 minD. 64 min |
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Answer» Correct Answer - b The first order rate constant is given as `k = (2.303)/(t) "log" (a_(0))/(a_(0) - x) ` _______(i) also , half life `t_(1//2) =(2.303 "log"2)/(K)` ______(ii) Equating K from equation (i) and (ii) `therefore (2.303"log"2)/(t_(1//2)) = (2.303)/(32) "log" (100)/(100-75)` or , `("log"2)/(t_(1//2)) = (1)/(32)` log 4 . or `("log"2)/(t_(1//2)) = (1)/(32) xx 2` log 2 `therefore t_(1//2) = 16 ` minutes |
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| 26. |
A reaction is first order in `A` secod order in `B`: (i) write differential rate equation. (ii) How is the rate affected when the concentration of `B` is tripled ? (iii) How is the rate affected when the concentration of both `A` and `B` is doubled? |
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Answer» (i) The differential rate equation will be `-(d [R])/("dt") = k [A] [B]^(2)` (ii) If the concentration of B is increased three times , then `-(d[R])/("dt") = k[A] [3B]^(2)` = `9* k [ A] [B]^(2)` Therefore , the rate of reaction will increase 9 times (iii) 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 . |
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| 27. |
Acid catalysed hydrolysis of ethyl acetate follows a pseudo-first order kinetics with respect to ester . If the reaction is carried out with large excess of ester , the order with respect to ester will beA. 1.5B. 0C. 2D. 1 |
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Answer» Correct Answer - b With large excess of ester the rate of reaction is independent of ester concentration . |
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| 28. |
Radioactive elements were incorporated into the earth when the solar system formed All rocks and minerals contain tiny amount of these radioactive elements which breakdown spontaneously into more stable atoms overtime A commonly used radiometric dating technique relies on the breakdown of `._(19)K^(40)` to `._(18)Ar ^(40)` precise measurements of the amount of `K^(40)` relative to `Ar^(40)` in an igneous rock can tell the age of rock An igneous rock sample was found to contain `0.2` gm potassium and `0.6` gm of Ar The age of the igneous rock is `(lambda of K^(40)=6.93xx10^(-10)year^(-1))`A. `1.2xx10^(8)` yearsB. `2xx10^(9)` yearsC. `2xx10^(10)` yearsD. `2.4xx10^(9)` years |
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Answer» Correct Answer - B |
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| 29. |
In a pseudo first order hydrolysis of ester in water the following results were obtained: `{:(t//s,0,30,60,90),(["Ester"],0.55,0.31,0.17,0.085):}` (i) Calculate the average rate of reaction between the time interval `30` to `60` seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. |
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Answer» (i) Average rate of reaction between the time interval , 30 to 60 seconds , = `(d ["Ester"])/("dt")` = `(0.31 - 0.17)/(60 - 30)` `= (0.14)/(30)` `= 4. 67 xx 10^(-3) "mol" L^(-1) s^(-1)` (ii) For a pseudo first order reaction , `k = (2.303)/(t) "log" ([R]_(0))/([R])` For t = 30 s , `k_(2) = (2.303)/(30) "log" (0.55)/(0.31)` `= 1.911 xx 10^(-2) s^(-1)` For t = 60 s `k_(2) = (2.303)/(60) "log" (0.55)/(0.17)` `= 1.957 xx 10^(-2) s^(-1)` For t = 90 s , `k_(3) = (2.303)/(90) "log" (0.55)/(0.085)` =`2.075 xx 10^(-2) s^(-1)` Then , average rate constant , `k = (k_(1) + k_(2) + k_(3))/(3)` = `((1.911 xx 10^(-2)) + (1.957 xx 10^(-2)) + (2.075 xx 10^(-2)))/(3)` `= 1.98 xx 10^(-2) s^(-1)` |
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| 30. |
In a pseudo first order hydrolysis of ester in water the following result6s were obtained: `{:(t//s,0,30,60,90),(["Ester"],0.55,0.31,0.17,0.085):}` (i) Calculate the average rate of reaction between the time interval `30` to `60` seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.A. `1.91 xx 10^(-2) s^(-1)`B. `4.67 xx 10^(-3) mol L^(-1) s^(-1)`C. `1.98 xx 10^(-3) s^(-1)`D. `2.07 xx 10^(-2) s^(1)` |
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Answer» Correct Answer - B Average rate during the time interval 30-60s. Rate `= - (C_(2) - C_(1))/(t_(2) - t_(1)) = - ((0.17 - 0.31))/(60 - 30) = (0.14)/(30)` `= 4.67 xx 10^(-3) mol L^(-1) s^(-1)` |
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| 31. |
For the reaction `CH_(3)COOCH_(3) + H_(2)O overset(H^(+))rarr CH_(3)COOH + CH_(3)OH`. The progress of the process of reaction of reaction is followed byA. Finding the amount of methanol formed at different intervalsB. Finding the amount of acetic acid formed at different intervalsC. Using a voltmeterD. Using polarimeter |
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Answer» Correct Answer - B Because as reaction progresses the amount of acetic acid increases. |
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| 32. |
In a pseudo first order hydrolysis of ester in water the following result6s were obtained: `{:(t//s,0,30,60,90),(["Ester"],0.55,0.31,0.17,0.085):}` (i) Calculate the average rate of reaction between the time interval `30` to `60` seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. |
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Answer» (i) Average rate of reaction `=(Delta[ester])/(Deltatime)` `=(0.17-0.31)/(60-30)=-4.67xx10^(-3) mol L^(-1) s^(-1)` (ii) `K=2.303/t "log" a/((a-x))` Where `a` is initial conc. At time t. `K_(1)=2.303/30"log"0.55/0.31=1.91xx10^(-2)` `K_(2)=2.303/60"log"0.55/0.17=1.96xx10^(-2)` `K_(3)=2.303/90"log"0.55/0.085=2.01xx10^(-2)` Thus, `K=(K_(1)+K_(2)+K_(3))/(3)` `=(1.91xx10^(-2)+1.96xx10^(-2)+2.01xx10^(-2))/(3)` `=1.96xx10^(-2)` |
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| 33. |
Kinetic of acidic hydrolysis of ester is a pseudo Ist order reaction `CH_(3)COOCH_(3)(aq)+H_(2)_(excees)(l)` `overset [H^(+)] rarr CH_(3)COOH(aq) + CH_(3)OH(aq)` Rate law is given by `r=K[H^(+)][H_(2)O][CH_(3)COOCH_(3)]` `K=1.8xx10^(-3)M^(-2)sec^(-1)` where `H^(+)` ion concentration is given by acid catalyst in an experiment 1 M `CH_(3)COOCH_(3)` is hydrolyscd using `0.1` M HCI as catayst. Reaction mixture is titrated against standard NaOH solution at different times If initially 10 ml Of NaOH was required while after long time 100 ml of NaOH is required then find volume of NaOH required at 230.3 secA. 91 mlB. 81 mlC. 75 mlD. 50 ml |
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Answer» Correct Answer - A |
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| 34. |
Kinetic of acidic hydrolysis of ester is a pseudo Ist order reaction `CH_(3)COOCH_(3)(aq)+H_(2)_(excees)(l)` `overset [H^(+)] rarr CH_(3)COOH(aq) + CH_(3)OH(aq)` Rate law is given by `r=K[H^(+)][H_(2)O][CH_(3)COOCH_(3)]` `K=1.8xx10^(-3)M^(-2)sec^(-1)` where `H^(+)` ion concentration is given by acid catalyst in an experiment 1 M `CH_(3)COOCH_(3)` is hydrolyscd using `0.1` M HCI as catayst. Calculate the time at which ester concentration reduce to `0.25` M `[l n2=0.7]` (Given Density of pure `H_(2)O=1 gm //ml)`A. 70 secB. 140 secC. `700/9` secD. `350/9` sec |
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Answer» Correct Answer - B |
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| 35. |
The half life period of `C^(14)` is 5760 years. For a 200 mg of sample of `C^(14)`, the time taken to change to 25 mg isA. 11520 yearsB. 23040 yearsC. 5760 yearsD. 17280 years. |
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Answer» Correct Answer - D d) `R_(0)=200mg`, R=25mg, `t_(1//2)=5760` years, Suppose required years =x Suppose no. of half lifes=n `(R/R_(0)) = (1/2)^(n)` or `(25/250) = (1/2)^(n)` `(1/8) = (1/2)^(n)` or `(1/3)^(3) = (1/2)^(n)` n=3 Now, `n=x/(t_(1//2)) = 3 xx 5760 Y` `=17280` years. |
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| 36. |
The mathematical expression which describes the reaction rate in terms of molar concentrations of the reactants as determined experimentally is called rate law. The sum of the coefficients of the reacting species that are involved in the rate equation of a particular reaction is known as the order of the reaction, The reactions may be classified as first, second and third order reactions depending upon the number of of reacting species involved in the rate equation. It can be fractional in complex reations and even zero in certain reactions such as photochemicla reactions. The units of the rate constant(k) can help ini predicting the nature of a particular reaction. 1) A substance decomposes in a solution followig first order kinetics. Flask A contains 1L of 1M solution and flask B has 100 mL of 0.6 M solutions. After 8 hours, the concentrations of substance in flask A becomes 0.25 M. What will be the time taken for the concentration of the same substances in flask B to becomes 0.3 M?A. `0.4` hrB. `2.4`hrC. `4.0` hrD. unpredictable as rate constant is not given. |
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Answer» Correct Answer - C c) In flask A: `k=(2.303)/8 log 1.0/(0.25) =2.303/4 log2` In flask B: `t=2.303/k log 0.6/0.3 = (2.303 xx 4)/(2.303) = 4` Log 2 cancels out |
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| 37. |
For a complex reaction ……… .A. Order of overall reaction is same as molecularity of the slowest stepB. Order of overall reaction is less than the molecularity of the slowest stepC. order of overall reacticn is greater than molecularity of the slowest stepD. molecularity of the slowest step is never zero or non-integer |
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Answer» Correct Answer - A::D (a) For a complex reaction, order of overall reaction = molecularity of slowset step As rate of overall reaction depends upon total number of moleucles involved in slowest step of the reaction. Hence, moleculartiy of the slowest step is equal to order of overall reaction. (b) Since, the completion of any reaction is not possible in the absence of reactants. Hence, slowest step of any chemical reactions must contain at least one reactant. Thus, molecularity of the slowest step is never zero or non- integer. |
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| 38. |
The mathematical expression which describes the reaction rate in terms of molar concentrations of the reactants as determined experimentally is called rate law. The sum of the coefficients of the reacting species that are involved in the rate equation of a particular reaction is known as the order of the reaction, The reactions may be classified as first, second and third order reactions depending upon the number of of reacting species involved in the rate equation. It can be fractional in complex reations and even zero in certain reactions such as photochemicla reactions. The units of the rate constant(k) can help ini predicting the nature of a particular reaction. The rate constant is numerically the same for three reactions of first, second and third order respectively. Which one is true for rates of three reactions if concentrations of the reactant is 1M?A. `r_(1) = r_(2)=r_(3)`B. `r_(1) gt r_(2) lt r_(3)`C. `r_(1) lt r_(2) lt r_(3)`D. All the above. |
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Answer» Correct Answer - C c) `r_(1) = k[A]` `r_(2)=k[A]^(2)` `r_(3) = k[A]^(3)` For the same concentration of [A]=1 |
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| 39. |
The mathematical expression which describes the reaction rate in terms of molar concentrations of the reactants as determined experimentally is called rate law. The sum of the coefficients of the reacting species that are involved in the rate equation of a particular reaction is known as the order of the reaction, The reactions may be classified as first, second and third order reactions depending upon the number of of reacting species involved in the rate equation. It can be fractional in complex reations and even zero in certain reactions such as photochemicla reactions. The units of the rate constant(k) can help ini predicting the nature of a particular reaction. The concentration of a reactant in a solution falls from 0.2 M to 0.1 M in 2 hours and to 0.05 M in horus. The order of the reaction is:A. zeroB. twoC. oneD. half. |
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Answer» Correct Answer - C c) Since half-life period `(t_(1//2))` remains constant, order of reaction=1. |
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| 40. |
The mathematical expression which describes the reaction rate in terms of molar concentrations of the reactants as determined experimentally is called rate law. The sum of the coefficients of the reacting species that are involved in the rate equation of a particular reaction is known as the order of the reaction, The reactions may be classified as first, second and third order reactions depending upon the number of of reacting species involved in the rate equation. It can be fractional in complex reations and even zero in certain reactions such as photochemicla reactions. The units of the rate constant(k) can help ini predicting the nature of a particular reaction. For a reaction `N_(2)O_(5) to 2NO_(2) + 1/2O_(2)` Given: `(-d[N_(2)O_(5)])/(dt) = k_(1)[N_(2)O_(5)]` `(d[NO_(2)])/(dt) = k_(2)[N_(2)O_(5)]` `(d[O_(2)])/(dt) = k_(3)[N_(2)O_(5)]` the relation between `k_(1), k_(2)` and `k_(3)` is:A. `2k_(1)=k_(2)=4k_(3)`B. `k_(1)=k_(2)=k_(3)`C. `2k_(1)=4k_(2)=k_(3)`D. None of these. |
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Answer» Correct Answer - A a) `(-d[N_(2)O_(5)])/(dt) = 1/2d([NO_(2)])/(dt) = (2d[O_(2)])/(dt)` `k_(1)[N_(2)O_(5)]= k_(2)/k_(1) [N_(2)O_(5)] = 2k_(3)[N_(2)O_(5)]` or `2k_(1) = k_(2)=4k_(3)` |
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| 41. |
Rate law cannot be determined form balanced chemical equation if …….. .A. reverse reaction is involvedB. it is an elementary reactionC. it is a sequence of elementary reactionsD. any of the reactants is in excess |
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Answer» Correct Answer - A::C::D Rate law can be determined from balanced chermical equation if it is an elementary reaction. |
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| 42. |
Half life of a reaction becomes half when intial concentrations of reactants are made double. The order of the reaction will be:A. 1B. 2C. 0D. 3 |
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Answer» Correct Answer - B `t_(1//2)prop(1)/(a^(n-1))` `t_(1//2)prop(1)/(a)` where,n = order of reaction for second order reaction |
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| 43. |
The reaction `Aoverset(k)rarr` Product, is zero order while the reaction `Boverset(k)rarr` Product, is first order reaction. For what intial concentration of A are the half lives of the two reacions equal?A. `(log_(2)4) M`B. 2MC. 2 log 2 MD. ln 2 M |
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Answer» Correct Answer - A For zero order reaction, x=kt `:. (a)/(2)=kxxt_(1//2),i.e.,t_(1//2)=(a)/(2k)" ".....(i)` For frist order reaction, `t_(1//2)=(log_(e)2)/(k)" ".....(ii)` From eqs. (i) and (ii), `(a)/(2k)=(log_(e)2)/(k)` `a=log_(e)4M` |
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| 44. |
Interpration of the rate law: For a hydrothertical reaction `X+Y+Zrazrr` products the rate law is determined to be `"Rate" =k[X][Y]^(2)` What happens to the reaction rate when we make each of the following concentration changes ? (a) We dobule the concentration of X without changing the concentration of Y or Z. (b) We double the concentration of Y without changing the concentration of X or Z. (c ) We double the concentrations of Z without changing the concentration of X or Y. (d) We double all three concentrations simultaneously. Strategy: Interpret the law to predict the changes in reaction rate. Remember that changing concentrations does not change the value of k. |
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Answer» (a) According to the rate law expression, reaction is directly proportional to the first power of [X]. If we do not change [Y] or [Z], then doubling [X] (i.e., increasing [X] by a factor of 2) cause the reaction rate to increase by a factor of `2^(1)=2`, so the reaction rate doubles. (b) Rate is directly proportional to the second power of [Y]. If we do not change [X] or [Z], then doubling [Y] (i.e., increasing [Y] by a factor of 2) cause the reaction rate to increase by a factor of `2^(2)=4`, so the reaction rate quadruples. (c ) The reaction rate is indepent of [Z] so changing [Z] causes no change in reaction rate. (d) Doubling all concentrations would cause the changes decscribed in (a), (b), and (c ) simultaneouly. The rate would incerase by a factor of 2 due to the change in [X], by a factor of 4 due to the change in [Y], and be unaffected by the change in [Z]. The result is that the reaction rate increases by a factor of `2^(1)xx2^(2)=8`. |
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| 45. |
Method of intial rates: Given the following data, determine the rate law expression and the value of the rate constant for the reaction. `2X+Y+ZrarrU+V` Strategy: The rate law is of the from `"Rate" = k[X]^(a)[Y]^(b)[Z]^(c )` We must evaluate a,b,c, and k by using the reasoning outlined earlier. Thus, to determine order with respect to X, we must select those experiments in which concentration of X changes but the concentrations of Y and Z are kept onstant and so on. Note that the coefficient of U in the balenced equation is 1, so the rate of reaction is equal to the rate of formation of U. |
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Answer» Dependence on [Y] : In experiments 1 and 3, the intial concentrations of X and Z are unchanged. Thus, any change in the rate would be due to the change in concentration of Y. But we observethat the rate is the same in experiments 1 and 3, even though the concentration of Y is different. Thus, the reaction rate is independendent of `[Y]`, thus `b = 0([Y]^(@) =1)`. The rate law must be `"Rate" = k[X]^(a)[Z]^(c )` We can neglet changes in [Y] in the subsequent reasoning. Dependence on [Z]: Experiments 1 and 4 involve the same initial concentration of X, thus the observed [Z]. Thus, we must compare experiments 1 and 4 to find Z. [Z] has been multiplied by a factor of `(0.60M)/(0.20M) = 3.0 = [Z]` ratio Consequently, the rate changes by a factor of `(7.2xx10^(-6)Ma^(-1))/(2.4xx10^(-6)Ms^(-1)) = 3.0` = rate ratio The exponent C can be reduced from rate ratio `= ([Z] ratio)^(c )` `(3.0) = (3.0)^(c )` thus `c=1` i.e., the reaction is first order in Z. Now we know that the rate law is of the form . Rate `= k[X]^(a)[Z]` Dependence on [X]: We use experiments 1 and 2 to evaluate a, because [X] is changed, [Y] does not matter, and [Z] is unaltered. The observed rate change is due only to changed [X]. [X] has been multiplied by a factor of `(0.40M)/(0.20M)= 2.0= [X]` ratio Consequently, the rate changes by a factor of `(9.6xx10^(-6)Ms^(-1))/(2.4xx10^(-6)Ms(-1))= 4.0` = rate ratio The exponent a can be deduced from rate ratio `= ([X]ratio)^(a)` `4.0 = (2.0)^(a)` `(2.0)^(2)=(2.)^(a)` thus `a=2` i.e., the reaction is second order in X From these results, we can write the complete rate - law expression Rate `= k[X]^(2)[Y]^(@)[Z]^(1)` or Rate `= k[X]^(2)[Z]^(1)` To evalute the specific reaction rate , k, we can substitute any of the four sets of data into the rate-law expression we have just derived. Data from experiment 4 give `Rate_(4)= k[X]_(4)^(2)[Z]_(4)` `k = (Rate_(4))/([X]_(4)^(2)[Z]_(4))` `= (7.2x10^(-6)Ms(-1))/((0.20M)^(2)(0.60M))` `= 300xx10^(-6)M^(-2)s^(-1)` `= 3.0xx10^(-4)M^(-2)s^(-1)` The rate-law expression can also be written with the value of k incorporated. `Rate = 3.0xx10^(-4)M^(-2)s^(-1)[X]^(2)[Z]` This expression allows us to calculate the reaction rate at which this reaction occurs with any known concentrations of X and Z (provided some Y is present). |
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| 46. |
Two reactions `R_(2) and R_(2)` have identical pre - exponential factors. Activations enery of `R_(1)` exceeds that of `R_(2)` by 10 kJ `mol_(-1)` . If `k_(1) and k_(2)` are rate constants for rate constants for reactions `R_(1) and R_(2)` respectively at 300k , then In `(k_(2)/k_(1))`is equal to `(R=8.314 J mol^(-1)K^(-1))`A. 8B. 12C. 6D. 4 |
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Answer» Correct Answer - D We know `k_(2)=Ae^(-E_(a)//RT)` `(k_(2))/(k_(1))=e^((E_(a)^(1)-E_(b)^(2))//RT)` `ln.(k_(2))/(k_(1))=(E_(a)^(1)-E_(b)^(2))/(RT)` `=(10xx1000)/(8.314xx300)` `~~4` |
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| 47. |
The energy change accompanying the equilibrium reaction `A hArr B` is `-33.0 kJ mol^(-1)`. Assuming that pre-exponential factor is same for forward and backward reaction answer the following: The equilibrium constant `k` for the reaction at `300 K`A. `5.55 xx 10^(5)`B. `5.67 xx 10^(3)`C. `5.55 xx 10^(6)`D. `5.67xx 10^(2)` |
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Answer» Correct Answer - A `k_(f) = Ae^(-Ef//RT), k_(b) = Ae^(-E_(b)//RT)` `:. k = (k_(f))/(k_(b)) = (Ae^(-E_(f)//RT))/(Ae^(-E_(b)//RT)) = e^((E_(b)-E_(f))//RT)` or `log k = (E_(b) - E_(f))/(2.303 RT)` `= (33000 J mol^(-1))/(2.303 xx 8.314 J K^(-1) mol^(-1) xx 300 K) = 5.745` `k = "Antilog" (5.745) = 5.55 xx 10^(5)` |
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| 48. |
For the reaction: `aA + bB rarr cC+dD` Rate `= (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt)` In the following reaction, `xA rarr yB` `log.[-(d[A])/(dt)] = log.[(d[B])/(dt)] + 0.3` where negative isgn indicates rate of disappearance of the reactant. Thus, `x:y` is:A. `1:2`B. `2:1`C. `3:1`D. `3:10` |
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Answer» Correct Answer - B `xA rarr yB` `(-d[A])/(x dt) = (1)/(y) (d[B])/(dt)` `log.((-d[A])/(dt)) = log.(d[B])/(dt) + log.(x)/(y)` `:. log.((x)/(y)) = 0.3` `:. (x)/(y) = 2` or `x:y = 2:1` |
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| 49. |
For the reaction: `aA + bB rarr cC+dD` Rate `= (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt)` A reaction follows the given concentration-time graph. The rate for this reaction at `20 s` will be A. `4 xx 10^(-3) M s^(-1)`B. `1 xx 10^(-2) M s^(-1)`C. `2 xx 10^(-2) M s^(-1)`D. `7 xx 10^(-3) M s^(-1)` |
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Answer» Correct Answer - D If `H_(2)O` is in excess, the order of the reaction will be `1`. |
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| 50. |
For a hypothetical reaction `aA + bB rarr` Product, the rate law is: rate `= K[A]^(x)[B]^(y)`, thenA. `(a + b) = (x + y)`B. `(a + b) lt (x + y)`C. `(a + b) gt (x + y)`D. All of these |
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Answer» Correct Answer - D `aA + bB rarr` Products `r = k[A]^(x)[B]^(y)` `(a + b)` may be less, more equal to `(a + y)`. |
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