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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.
| 1401. |
Find out two-third `(2//3)` life of a first order reaction in which `k=5.48 xx 10^(-14)s^(-1)` |
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Answer» For the first order reaction, `t=2.303/k log ([A]_(0))/[A]` Let `[A]_(0) = a, [A] = (a-2//3a) = 1//3a, k=5.48 xx 10^(-14)s^(-1)` `t_(2//3) = 2.303/k log ([A]_(0)/[A]) = 2.303/(5.48 xx 10^(-14)s^(-1))log a/(1//3a) = 2.303 / (5.48 xx 10^(4)s^(-1))log3` `(2.303)/(5.48 xx 10^(-14)s^(-1)) xx 0.4771= 2.0 xx 10^(13)`s |
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| 1402. |
Find out two-third `(2//3)` life of a first order reaction in which `k=5.48 xx 10^(-14)s^(-1)`A. `2.01xx10^(11)s`B. `2.01xx10^(13)s`C. `8.08xx10^(13)s`D. `16.04xx10^(11)s` |
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Answer» Correct Answer - B For two- third of a reaction, `[A]_(o)=a,[A]=a-(2a)/(3)rArra=(a)/(3)` `t_(2//3)=(2.303)/(k)"log"[A]_o/[A]=(2.303)/(k)"log"(a)/(a)=(2.303)/(k)log3` `t_(2//3)=(2.303xx0.4771)/(5.48xx10^(-14))=2.01xx10^(13)s` |
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| 1403. |
The time required for 100% completion of a zero order reaction isA. akB. `(a)/(2k)`C. `(a)/(k)`D. `(2k)/(a)` |
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Answer» Correct Answer - C For a zero order reaction, `x = kt` For 100% completion of the reaction, x = a Therefore, a = kt or `t = (a)/(k)` |
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| 1404. |
The rate constant for a zero order reaction isA. `k=(C_(0))/(2t)`B. `k=(C_(0)-C_(t))/(t)`C. `k=ln.(C_(0)-C_(t))/(t)`D. `k=(C_(0))/(C_(t))` |
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Answer» Correct Answer - B For a zero order reaction, `C_(0)-C_(1)=ktimplies k=(C_(0)-C_(t))/(t)` |
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| 1405. |
For the first order reaction, `k=5.48 xx 10^(-4)s^(-1)`, the two-third life time for this reaction isA. 2005 sB. 1000 sC. 2000 sD. 3005 s |
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Answer» Correct Answer - A For first order reaction, we know that `k=(2.303)/(t).log.(a)/((a-x))` Hence, `t=(2.303)/(5.48 xx 10^(-4))xxlog .(a)/(a-(2)/(3)a)` `=(2.303xx10^(4))/(5.48)xxlog3` `=(2.303xx10^(4))/(5.48)xx0.4771=2005s` |
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| 1406. |
In accordance to Arrhenius equation, the plot of log k against `(1)/(T)` is a straight line. The slope of the line is equal toA. `-E_(a)//R`B. `+E_(a)//R`C. `-(E_(a))/(2.303R)`D. `+(E_(a))/(2.303R)` |
| Answer» Correct Answer - C | |
| 1407. |
In accordance to Arrhenius equation, the plot of log k against `(1)/(T)` is a straight line. The slope of the line is equal toA. `(-E_(a))/(R )`B. `(-2.303)/(E_(a).R)`C. `(-E_(a))/(2.303R)`D. `(-E_(a))/(2.303)` |
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Answer» Correct Answer - C Arrhenius equation can be written as `logk=logA-(E_(a))/(2.303R)`, On comparing this equation with the straight line equation, `y=mx+c` we get,, `y=logk, m=-(E_(a))/(2.303R),x=(1)/(T),C=logA` |
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| 1408. |
If we plot a graph between log K and `(1)/(T)` by Arrhenius equation , the slope isA. `-(E_(a))/(R)`B. `+(E_(a))/(R)`C. `-(E_(a))/(2.303R)`D. `+(E_(a))/(2.303 R)` |
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Answer» Correct Answer - c ln k = ln `- (E_(a))/(RT)` is Arrhenius equation . Thus plots of ln k vs 1/T will give slope = `-E_(a)//RT` or `-E_(a)// 2.303R`. |
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| 1409. |
Half-life of a hypothetical reaction is found to be inversely proportional to the cube of initial concentration. The order of reaction isA. 4B. 3C. 5D. 2 |
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Answer» Correct Answer - A `t_(1//2)prop(1)/(a^(n-1))` Hence, `t_(1//2)prop(1)/(a^(3))`, only when n = 4 |
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| 1410. |
Half life of a reaction is found to be inversely proportional to the cube of its initial concentration . The order of reaction isA. 2B. 5C. 3D. 4 |
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Answer» Correct Answer - d Half-life of a reaction is found to be inversely proportional to the cube of initial concentration . The order of reactions is 4 . |
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| 1411. |
Which among the following is a false statementA. Half life of a third order reaction is inversely proportional to the square of initial concentration of the reactantB. Molecularity of a reaction may be zero or fractionalC. For a first order reaction `t_(1//2) = (0.693)/(K)`D. Rate of zero order reaction is independent of initial concentration of reactant |
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Answer» Correct Answer - b Molecularity of a reaction never become zero or fraction . |
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| 1412. |
`2N_(2)O_(5) rarr 4NO_(2) + O_(2)` If `(-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)]` What is the relation between `k_(1), k_(2)`, and `k_(3)`?A. `k_(1)=k_(2)=k_(3)`B. `2k_(1)=k_(2)=4k_(3)`C. `2k_(1)=4k_(2)=k_(3)`D. None |
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Answer» Correct Answer - B `-(1)/(2)(d[N_(2)O_(45)])/(dt)=(1)/(4)(d[NO_(2)])/(dt)=(d[O_(2)])/(dt)` |
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| 1413. |
`N_(2)O_(5)` decomposes according to equation, `2N_(2)O_(5) rarr 4 NO_(2)+O_(2)` (a) What does `-(d[N_(2)O_(5)])/(dt)` denote? (b) What does `(d[O_(2)])/(dt)` denote? (c) What is the unit of rate of this reaction? |
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Answer» (a) Rate of decomposition of `N_(2)O_(5)`. (b) Rate of formation of `O_(2)`. (c ) Unit of rate`= mol litre^(-1) time^(-1)`. |
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| 1414. |
Which of the following will react fastest (in terms of amount of product formed in a give time) and which will react at the highest rate? (a) `1` mole of `A` and `1` mole of `B` in `1` litre vessel. (b) `2` mole of `A` and `2` mole of `B` in `2` litre vessel. (c ) `0.2` mole of `A` and `0.2` mole of `B` in `0.1` liter vessel. |
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Answer» (b) will react faster because of more reactant present there will be more product produced per unit time. (c ) will react at highest rate because `[A]` and `[B]` are highest and thus rate will be more. |
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| 1415. |
Derive the relationship between rate of reaction, rate of disappearence of `X, Y` and rate of formation of `X_(2)Y_(3)` for the reaction: `2X+3Yrarr X_(2)Y_(3)` |
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Answer» Average rate of reaction `=-1/2(Delta[X])/(Deltat)=-1/3(Delta[Y])/(Deltat)=(Delta[X_(2)Y_(3)])/(Deltat)` Instantaneous rate of reaction `=-1/2(d[X])/(dt)=-1/3(d[Y])/(dt)=(d[X_(2)Y_(3)])/(dt)` |
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| 1416. |
State one condition under which a bimolecular reaction may be kinetically of first order reactions. |
| Answer» This is possible in case of the reactants is taken in large excess so that its concentration changes very slightly in the reaction. The reaction is pseudo first order reaction. For example, hydrolysis of an ester in acidic medium. | |
| 1417. |
Certain bimolecular reactions which follow the first order kinetics are calledA. First order reactionsB. Unimolecular reactionsC. Bimolecular reactionsD. Pseudounimolecular reactions |
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Answer» Correct Answer - d It is the characteristic of pseudo-unimolecular reactions . |
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| 1418. |
The half life for the reaction `N_(2)O_(5) hArr 2NO_(2) + (1)/(2) O_(2)` is 24 hrs at `30^(@)C` . Starting with 10 g of `N_(2)O_(5)` how many grams of `N_(2)O_(5)` will remain after a period of 96 hoursA. 1.25 gB. 0.63 gC. 1.77 gD. 0.5 g |
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Answer» Correct Answer - b `k = (0.063)/(24) hr^(-1) = (2.303)/(96)` log `(10)/(a -x)` or log `(10)/(a-x) = 1.2036` or 1 - log ( a-x) = `1.2036` or log (a-x) = `-0.2036 = 1.7964` or (a-x) = antilog 1.7964 or (a-x) = antilog 1.7964 = 0.6258 gm . |
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| 1419. |
Half-life of a reaction is found to be inversely proportional to the cube of its initial concentration. The order of reaction isA. `2`B. `5`C. `3`D. `4` |
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Answer» Correct Answer - D half-life of a reaction is found to be inversely proportional to the cube of initial concentration. The order of reaction is `4`. `t_(1//2) = (1)/([a_(0)]^(3))` Also, `t_(1//2) prop [a_(0)]^(1-n)` `[a_(0)]^(1-3) = [1]^(1-3)` `rArr -3 = 1-n` `rArr n = 4` |
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| 1420. |
The half life for the reaction `N_(2)O_(5) hArr 2NO_(2)+1/2 O_(2)` in `24 hr` at `30^(@)C`. Starting with `10 g` of `N_(2)O_(5)` how many grams of `N_(2)O_(5)` will remain after a period of `96` hours ?A. `1.25 g`B. `0.63 g`C. `1.7 g`D. `0.5 g` |
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Answer» Correct Answer - B `k = (0.693)/(24)hr^(-1) = (2.303)/(96)"log"(10)/(a-x)` or `"log"(10)/(a-x) = 1.2036` or `1 - log(a-x) = 1.2036` or `log(a-x) = -0.2036 = 1.7964` or `(a-x) = "anitlog" 1.7964 = 0.6258g` |
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| 1421. |
For reaction `A+2BtoC`. The amount of C formed by starting the reaction with 5 mole of A and 8 mole of B is :A. 5 molesB. 8 molesC. 16 molesD. 4 moles |
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Answer» Correct Answer - D d) `A + 2B to C`. 5 moles of A require 10 moles of B. But No. of moles of B are 8 only (limiting reactant) `therefore` 8 moles of B will react with 4 moles of A to form 4 moles of C. |
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| 1422. |
The active mass of solid is generally taken asA. `gt 1`B. `= 1`C. `lt 1`D. 0 |
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Answer» Correct Answer - B The active mass of solids is usually taken as unity.. |
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| 1423. |
What is the energy of activation of a reaction is its rate doubles when the temperature is raised from 290 K to 300 K ?A. 12 kcalB. 15 kcalC. 10 kcalD. 20 kcal |
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Answer» Correct Answer - A ` log.(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `log2=(E_(a))/(2.303xx2)[(300-290)/(290xx300)]` `E_(a)`=12062 cal =12.0 cal |
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| 1424. |
Which of the following statements is correct?A. The rate of a reaction decreases with passage of time as the concentration of reactants decreases.B. The rate of a reaction is same at any time during the reaction.C. The rate of a reaction is independent of temperature changeD. The rate of a reaction decreases with increases in concentration of reactants(s). |
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Answer» Correct Answer - A |
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| 1425. |
Consider the graph under question 8. Which of the following options does not show instaneous rate of reactiono at `40^(th)` second?A. `(V_(5)-V_(2))/(50-30)`B. `(V_(4)-V_(2))/(50-30)`C. `(V_(3)-V_(2))/(40-30)`D. `(V_(3)-V_(1))/(40-20)` |
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Answer» Correct Answer - B |
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| 1426. |
A first order reaction is `15%` compelte in 20 minutes. How long it take to be `60%` complete? |
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Answer» For the first order reaction, `t=(2.303)/k log (a)/(a-x)` `20 min=2.303/(k) log 100/85 = 2.303/k log a/(a-x)` 20 min=`(2.303)/(k) log (100/85) = (2.303)/k log 1.1765`...........(i) Iind case: `a=100%, (a-x)= 40%` `t=(2.303)/k log 100/40 = 2.303/k log 2.5`.............(ii) Dividing eqn (ii) by eqn. (i), `t/(20 "min")=(log 2.5)/(log 1.1765)= (0.3979)/(0.0706)` or `t=(0.3979)/(0.0706) xx (20"min") = 112.7 min` |
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| 1427. |
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):}` Rate law of the above experiment is:A. `r=k[A][B]`B. `r=k[A]^(2)[B]`C. `r=k[A][B]^(2)`D. `r=k[A]^(2)[B]^(0)` |
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Answer» Correct Answer - B |
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| 1428. |
Given the following two mechanisms, one with catalyst and the other without catalyst. (i) `A + B rarr C` (slow) (ii) `C +B rarr F + A` (fast) (iii) `B + B rarr F` (slow) Which mechanism use the catalyst and what is it ?A. Step (i), `A`B. Step (ii), `B`C. Step (iii), `F`D. Steps (i) and (ii), `C` |
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Answer» Correct Answer - A Since `(A)` is recovered back in the reaction, so it catalyst in the step `(i)`.Correct Answer - A Since `(A)` is recovered back in the reaction, so it catalyst in the step `(i)`.Correct Answer - A Since `(A)` is recovered back in the reaction, so it catalyst in the step `(i)`. |
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| 1429. |
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):}` Rate constant of the above experiment is:A. `1.3xx10^(-1)`B. `2.6xx10^(-2)`C. `2.6xx10^(-1)`D. `1.3xx10^(-2)` |
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Answer» Correct Answer - C |
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| 1430. |
Study the two photochemical reactions and answer the questions given below: For the overall reaction between A and B to yield C and D two mechanisms are proposed: (I) `A+B rarr AB* rarr C+D, k_(1)=1xx10^(-5)M^(-1)s^(-1)` (II) `A rarr A* rarr E, k_(1)=1xx10^(-4)M^(-1)s^(-1)` `E+B rarr C+D, k_(2)=1xx10^(10)M^(-1) s^(-1)` (species with * are short-lived) Rate according to mechanism II when concentration of each reactant is 1 M will be:A. `1xx10^(-4) M s^(-1)`B. `1xx10^(10)M s^(-1)`C. `1xx10^(-6)M s^(-1)`D. `1xx10^(-10) M s^(-1)` |
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Answer» Correct Answer - A |
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| 1431. |
Study the two photochemical reactions and answer the questions given below: For the overall reaction between A and B to yield C and D two mechanisms are proposed: (I) `A+B rarr AB* rarr C+D, k_(1)=1xx10^(-5)M^(-1)s^(-1)` (II) `A rarr A* rarr E, k_(1)=1xx10^(-4)M^(-1)s^(-1)` `E+B rarr C+D, k_(2)=1xx10^(10)M^(-1) s^(-1)` (species with * are short-lived) At what concentration of B, rates of two mechanism are equal?A. 1 MB. 5 MC. 7 MD. 10 M |
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Answer» Correct Answer - D |
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| 1432. |
Study the two photochemical reactions and answer the questions given below: For the overall reaction between A and B to yield C and D two mechanisms are proposed: (I) `A+B rarr AB* rarr C+D, k_(1)=1xx10^(-5)M^(-1)s^(-1)` (II) `A rarr A* rarr E, k_(1)=1xx10^(-4)M^(-1)s^(-1)` `E+B rarr C+D, k_(2)=1xx10^(10)M^(-1) s^(-1)` (species with * are short-lived) Rate according to mechanism I when concentration of each reactant is `0.1` M will be:A. `1xx10^(-7)M s^(-1)`B. `1xx10^(-6) M s^(-1)`C. `1xx10^(-5) M s^(-1)`D. `1xx10^(-4)M s^(-1)` |
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Answer» Correct Answer - A |
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| 1433. |
Statement: For the reaction `2O_(3) rarr 3O_(2)`, the rate `=K[O_(3)]^(2)[O_(2)]^(-1)` Explanation: The reaction has -ve order for `O_(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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| 1434. |
The half-life of tritium is 12.3yrs. If 48.0 mg of tritium is released form a nuclear power plant during the course of an accident, calculate the mass (in mg) of the nuclide that will remain after 5.0yrs.A. 36.2B. 32C. 24D. 16 |
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Answer» Correct Answer - A |
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| 1435. |
Carbon-14 radioactively decays via the emission of a beta particle. Which of the following is the product of this decay?A. Beryllium-10B. Boron-14C. Carbon-13D. Nitrogen-14 |
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Answer» Correct Answer - D |
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| 1436. |
If half lives of a radioactive element, undergoing parallel path `alpha-"decay"` and `alpha-"decay"` are 4 years and 12 years respectively, then percentage of element that remains after 12 years will be:A. 0.5B. 0.125C. 0.0625D. 0.25 |
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Answer» Correct Answer - C |
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| 1437. |
A sample of `.^(14)CO_(2)` was to be mixed with ordinary `CO_(2)` for a biological tracer experiment. In order that `10 "cm"^(3)` of diluted gas should have `10^(4)` dis/min, what activity `("in"mu Ci)` of radioactive carbon is needed to prepare 60L of diluted gas at 1 atm and 273 K? `[1 Ci=3.7 xx10^(10) "dps"]`A. `270mu Ci`B. `27muCi`C. `2.7 muCi`D. `2700 muCi` |
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Answer» Correct Answer - b |
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| 1438. |
The activation energy of a reaction is 75.2 kJ `mol^(-1)` in the absence of a catalyst and 50.14 kJ `mol^(-1)` in the presence of a catalyst. How many times will the reaction grow in the presence of catalyst if the reaction proceeds at `25^(@)`C ? |
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Answer» Let the rate constant in the absence of catalyst = `k_(1)` Let the rate constant in the presence of catalyst = `k_(2)` Activation energy in the absence of catalyst `(E_(1)) = 75.2 kJ mol^(-1)` Activation energy in the presence of catalyst `(E_(2)) = 50.14 kJmol^(-1)` According to Arrhenius equation, `k_(1) = Ae^(-E1//RT), k_(2)= Ae^(-E2//RT)` `k_(2)/k_(1) = e^((E_(1)-E_(2)//RT))` or ln `k_(2)/k_(1)= (E_(1)-E_(2))/(RT)` `log K_(2)/K_(1) = (E_(1)-E_(2))/(RT)` `=((75.3-50.14) xx (10^(3)Jmol^(-1)))/(2.303 xx (8.314 JK^(-1)mol^(-1) xx 298K)) = 4.391` `k_(2)/k_(1)= "Antilog" 4.391 = 24604` or `k_(2) = 24604 k_(1)` Thus, the reaction rate has increased nearly by 24604 times. |
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| 1439. |
The ratio of rate constant at `27^(@)C` and `37^(@)C` is `Q_(10)`. What should be the energy of activation of a reaction for which `Q_(10) = 2.5` ?A. `71 kJ`B. `212 kJ`C. `35 kJ`D. `12.1 kJ` |
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Answer» Correct Answer - A `log Q10` or `log 2.5 = (E_(a))/(2.303R)[(1)/(T_(1)) - (1)/(T_(2))]` Or `log 2.5 = (E_(a))/(2.303R)[(T_(2) - T_(1))/(T_(1)T_(2))]` Subsititute the value of `R` in `kJ`. `E_(a)` comes out to be `71 kJ` |
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| 1440. |
The rate constant for a first order reaction increases from `4 xx 10^(-2)` to `8 xx 10^(-2)` when the temperature changes from `27^(@)`C to `37^(@)`C. Calculate energy of activation for the reaction. |
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Answer» According to Arrhenius equation, `(log)k_(2)/k_(1) = E_(a)/(2.303R)[1/T_(1)-1/T_(2)]` `k_(2)/k_(1) = (8xx 10^(-2))/(4 xx 10^(-2)) = 2, T_(1) = 273 + 27 = 300K, T_(2)= 273 + 37 = 310 K` `R= 8.314 JK^(-1)mol^(-1)` `log2 = (E_(a))/(2.303 xx (8.314 J K^(-1)mol^(-1)))[1/300K - 1/310K]` `E_(a) = (0.3010 xx 2.303 xx (8.314 J mol^(-1)) xx (300) xx (310))/10` `53598.59 J mol^(-1) = 53.6 kJ mol^(-1)` |
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| 1441. |
Conisder the following reaction at `300 K`: `Ararr B` (uncatalyzed reaction) `ArarrB` (catalyzed reaction) The activation energy is lowered by `8.314 kJ mol^(-1)` for the catalyzed reaction. The rate of this reaction isA. `15` timesB. `38` timesC. `22` timesD. `28` times |
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Answer» Correct Answer - D `log.(k_(2))/(k_(1)) = (E_(a)-E_(ac))/(2.303 RT) = (8.314 xx 10^(3))/(8.314 xx 300) xx (1)/(2.303)` `rArr (k_(2))/(k_(1)) ~~ 28` |
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| 1442. |
For the given reaction: `H_(2) + I_(2) rarr 2HI` Given: `{:(T(K),1//T(K^(-1)),log k,),(769,1.3 xx 10^(-3),2.9,),(67,1.5 xx 10^(-3),1.1,):}` The activation energy will beA. `41.4 kcal mol^(-1)`B. `40 kcal mol^(-1)`C. `-41.4 kcal mol^(-1)`D. `-40 kcal mol^(-1)` |
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Answer» Correct Answer - B `k = Ae^(-E_(A)//RT)` `rArr log k = log A - (E_(a))/(2.303 RT)` `{:(log k_(1) = log A-(E_(a))/(2.303RT) ((1)/(T_(1)))),(log k_(2)=logA-(E_(a))/(2.303RT) ((1)/(T_(2)))):}} "Subtract to get" E_(a) = 40 lcal mol^(-1)` |
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| 1443. |
The rate constant for a first order reaction becomes six times when the temperature is raised from 350 to 400 K. Calculate the energy of activation for the reactions. |
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Answer» According to Arrhenius equation, `log k_(2)/k_(1)= E_(a)/(2.303 R) [1/T_(1)-1/T_(2)]` `k_(2)/k_(1) = 6, T_(1)=350K, T_(2)=400K , R= 8.314 JK^(-1)mol^(-1)` `log 6= log 6 = E_(a)/(2.303 xx (8.314 J mol^(-1)))[1/(350K)-1/(400K]]` `0.77815 = (E_(a))/(2.303 xx (8.314 J mol^(-1)))[(400-350)/(350 xx 400)]` `0.77815 = E_(a)/(2.303 xx (8.314 J mol^(-1)))[50/(350xx400)]` `E_(a) = ((0.77815 xx 2.303 xx (8.314 J mol^(-1)) xx 350 xx 400))/(50)` `41718 J mol^(-1) = 41.718 kJ mol^(-1)` |
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| 1444. |
The rate of formation of `C_(6)H_(6)+3H_(2) underset(k_(b))overset(k_(f))hArr C_(6)H_(12)` for the forward reaction is first order with respect to `C_(6)H_(12)` and `H_(12)` each. Which one of the options is/are correct?A. `k_(eq) = (k_(f))/(k_(b))`B. `k_(eq) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)]^(3))`C. `r_(f) =k_(f)[C_(6)H_(6)][H_(2)]`D. `r_(b) = k_(b)[C_(6)H_(12)][H_(2)]^(-2)` |
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Answer» Correct Answer - A::B::C::D `k_(eq) = (k_(f))/(k_(b)) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)]^(3)) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)][H_(2)]^(2))` `r_(f) = k_(f) xx[C_(6)H_(6)][H_(2)]` `r_(b) = k_(b) xx "Unknown"` At equilibrium, `r_(f) = r_(0)` `:. k_(f)[C_(6)H_(6)][H_(2)] = k_(b) xx " Unknown"` `r_(b) = k_(b) xx (k_(f))/(k_(b)) xx [C_(6)H_(6)][H_(2)]` `= k_(f)[C_(6)H_(6)][H_(2)]` `= k_(b)[C_(6)H_(12)][H_(2)]^(-2)` |
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| 1445. |
The rate law for a reaction `A + B to C + D` is given by the expression k[A]. The rate of reaction will beA. doubled on doubling the concentration of BB. halved on reducing the concentration of A to halfC. Decreased on increasing the temperature of the reactionD. unaffected by any change in concentration or temperature. |
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Answer» Correct Answer - B The rate of reaction depends upon concentration of only A. |
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| 1446. |
A first order reaction is `20%` complete in `10 min`. Calculate (a) the specific rate constant of the reaction and (b) the time taken for the reaction to reach `75%` completion. |
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Answer» Correct Answer - (a) `k = 0.0223 min^(-1)` (b) `62.16 min` (a) `k = (2.303)/(t) log.(a)/(a-x)` For `20%` completion, `k = (2.303)/(10)log.(100)/(80) = 0.0223 min^(-1)` (b) Half life `= (0.693)/(k) = (0.693)/(0.0023) = 31.08 min` For `75%` completion of the reaction, two half life lives are required, which is equal to `31.08 xx 2 = 62.16 min`.Correct Answer - (a) `k = 0.0223 min^(-1)` (b) `62.16 min` (a) `k = (2.303)/(t) log.(a)/(a-x)` For `20%` completion, `k = (2.303)/(10)log.(100)/(80) = 0.0223 min^(-1)` (b) Half life `= (0.693)/(k) = (0.693)/(0.0023) = 31.08 min` For `75%` completion of the reaction, two half life lives are required, which is equal to `31.08 xx 2 = 62.16 min`. |
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| 1447. |
A reaction in which reactants (R ) are converted into products (P) follows second order kinetics. If concentration of R is increased by four times, what will be the increase in the rate formation of p?A. 9 timesB. 4 timesC. 16 timesD. 8 times |
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Answer» Correct Answer - C `R to P` For a second order reaction rate `= k [R]^(2)` If conc. Of R is increased by four times, rate `= k [4R]^(2)` Hence, the rate of formation of P increases by 16 times. |
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| 1448. |
The rate of a gaseous reaction is given by the expresison `k[A]^(2)[B]^(3)`. The volume of the reaction vessel is suddenly reduced to one-half of the initial volume. The reaction rate relative to the original rate will beA. `(1)/(8) a`B. `(1)/(2)a`C. 2aD. 32a |
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Answer» Correct Answer - D Rate `= k [A]^(2) [B]^(3) = a` when volume is reduced to one half then conc. Of reactants will be doubled. Rate `= k[2A]^(2) [2B]^(3) = 32 [A]^(2) [B]^(3) = 32 a` |
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| 1449. |
For the reaction `N_(2)(g) + 2 H_(2) (g) to 2 NH_(3) (g)` under certain conditions of temperature and partial pressure of the reactants , the rate of formation of `NH_(3)` is `0.001 kg h^(-1)` . The rate of conversion of `H_(2)` under the same conditions isA. `1.82 xx 10^(-4)`kg/hrB. `0.0015` kg/hrC. `1.52 xx 10^(4)` kg/hrD. `1.82 xx 10^(-14)` kg/hr |
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Answer» Correct Answer - b `(-dN_(2))/(dt) = (-1)/(3) (dH_(2))/(dt) = (1)/(2) (dNH_(5))/(dt) implies (dH_(2))/(dt) = (3)/(2) xx 0.001 = 0.001 ` 5 kg 1`hr^(-1)`. |
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| 1450. |
The rate of a chemical reaction doubles for every `10^(@)C` rise of temperature. If the temperature is raised by `50^(@)C`, the rate of the reaction increases by aboutA. 12 timesB. 16 timesC. 32 timesD. 50 times |
| Answer» Correct Answer - C | |