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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1551. |
What is the half-life of `""_(6)C^(14),` if its disintegration constant is `2.31xx10^(-4) yr^(-1)` ?A. `0.3 xx 10^(4)` yearsB. `0.3 xx 10^(3)` yearsC. `0.3 xx 10^(8)` yearsD. `0.3 xx 10^(2)` years |
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Answer» Correct Answer - a `t_(1//2) = (0.693)/( 2.31 xx 10^(-4)) = 0.3 xx 10^(4)` years . |
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| 1552. |
The decomposition of dimethyl ether is a fractional order reaction. The rate of reaction is givenby rate `= k (p_(CH_(3)OCH_(3)))^(3//2)`. If the pressure is measured in bar and time in minutes, then what are the units of rate and constant?A. bar `"min"^(-1), "bar"^(2) "min"^(-1)`B. bar `"min"^(-1), "bar"^(2), "min"^(-1)`C. `"bar"^(-1//2) "min"^(-1),"bar"^(2),"min"^(-1)`D. bar `"min"^(-1), "bar"^(1//2) "min"^(-1)` |
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Answer» Correct Answer - B In terms of pressure Rate `k = (P_(CH_(3)OCH_(3))^(3//2))` Unit of rate = bar `"min"^(-1)` Unit of rate constant = `("rate")/((p_(CH_(3)OCH_(3))^(3//2))` `= ("bar" "min"^(-1))/("bar"^(3//2)) = "bar"^(-1//2) "min"^(-1)` |
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| 1553. |
The order of reaction may be a fractional number. |
| Answer» Correct Answer - T | |
| 1554. |
In the above problem if concentration of reactant is less than `1 M` then:A. `r_(1)=r_(2)=r_(3)`B. `r_(1)gt r_(2) gtr_(3)`C. `r_(1)lt r_(2)lt r_(3)`D. All of these |
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Answer» Correct Answer - b If `[A] lt 1,` `r_(1) gt r_(2) gt r_(3)` |
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| 1555. |
The rate constant is unmerically the same for three reactions of first, second and third order respectively, the unit of concentration being in moles per litre. If `R_(1), R_(2) and R_(3)` are the rates of three reactions of first, second and third order respectively aqnd K is the rate constant, which of the following relationships is valid for the three reactions.A. `r_(1)=r_(2)=r_(3)`B. `r_(1)gtr_(2)gtr_(3)`C. `r_(1)ltr_(2)ltr_(3)`D. All of these |
| Answer» Correct Answer - C | |
| 1556. |
The chemical in a lightsticks give off light as they react. When the lightsticks is placed in warm water the glow increases . This is because the :A. activation energy for the process is lowered.B. average kinetic energy of the reactants increases.C. higher temperature catalyzes the reaction.D. higher temperature changes the wavelength of light emitted. |
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Answer» Correct Answer - B |
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| 1557. |
The molecularity and order of reaction can be `0, 1, 2`, etc. |
| Answer» Correct Answer - T | |
| 1558. |
In the Q.No , 108 if the concentration of the reactant is less than 1 M , then, :A. `r_(1)=r_(2)=r_(3)`B. `r_(1)gtr_(2)gtr_(3)`C. `r_(1)ltr_(2)ltr_(3)`D. all of these |
| Answer» Correct Answer - B | |
| 1559. |
The hydrolysis of ester in alkaline medium is a |
| Answer» Correct Answer - F | |
| 1560. |
Which of these factors affect the value of the rate constant for a reaction? (P) Temperature (Q) Reactant concentration (R) Use of catalystA. P onlyB. Q onlyC. P and Q onlyD. P,Q and R |
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Answer» Correct Answer - C |
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| 1561. |
All first order reactions are unimolecular. |
| Answer» Correct Answer - F | |
| 1562. |
Define rate of reaction, Write two factors that affects the rate of reactions. |
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Answer» for the definition of rate of reactions, For the factors affecting rate of reactions |
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| 1563. |
All radioactive decompoistion reactions are…………..order reactions. |
| Answer» Correct Answer - T | |
| 1564. |
Negative catalyst stop chemical reactions. |
| Answer» Correct Answer - F | |
| 1565. |
A catalyst affects the rate of a chemical reaction by:A. increasing the average kinetic energy of the reactants.B. increasing the number of collisions between the reactants.C. decreasing the energy difference between the reactants and products.D. providing an alternative reaction pathway with a lower activation energy. |
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Answer» Correct Answer - D |
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| 1566. |
An iron catalyst is used in the Haber process in which process `N_2 and H_2` react to produce `NH_3` What is the role of this catalyst ?A. It provides a pathway with a lower activation energyB. It increases the equilibrium constant of the reactionC. It rasises the kinetic energies of the reactantsD. It interacts with the `NH_3` |
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Answer» Correct Answer - C |
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| 1567. |
Catalyst in a chemical reaction :A. increases the activation energyB. does not change activation energyC. does not change `DeltaH`D. none of the above |
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Answer» Correct Answer - B |
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| 1568. |
All of the following are expected to affect the rate of an irreversible chemical reactions except :A. adding a catalystB. removing some productsC. increasing the temperatureD. decreasing the reactant concentration |
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Answer» Correct Answer - D |
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| 1569. |
A catalyst speeds up a chemical reaction by :A. shifting the equilibrium.B. increasing the activation energy.C. decreasing the reaction enthalpy.D. providing an alternative enthalpy. |
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Answer» Correct Answer - A |
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| 1570. |
For the reaction `2N_(2)O_(5) rarr 4NO_(2)+O_(2)` rate of reaction and rate constant are `1.02 xx 10^(-4)` and `3.4 xx 10^(-5) sec^(-1)` respectively. The concentration of `N_(2)O_(5)` at that time will beA. 1.732B. 3C. `1.02 xx 10^(-4)`D. `3.4 xx 10^(5)` |
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Answer» Correct Answer - B b) `[N_(2)O_(5)]= ("Rate")/k = (1.2 xx 10^(-4) mol L^(-1)s^(-1))/(3.4 xx 10^(-5)s^(-1))`=3 |
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| 1571. |
For the reaction `2N_(2)O_(5) rarr 4NO_(2)+O_(2)` rate of reaction and rate constant are `1.02 xx 10^(-4)` and `3.4 xx 10^(-5) sec^(-1)` respectively. The concentration of `N_(2)O_(5)` at that time will beA. `1.732`B. `3`C. `1.02xx10^(-4)`D. `3.4xx10^(5)` |
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Answer» Correct Answer - B The reaction `2N_(2)O_(5)rar4NO_(2)+O_(2)` is a first order reaction (Look at the unit of the unit of rate constant). Thus Rate `=k[N_(2)O_(5)]` `[N)_(2)O_(5)]=(Rate)/(k)` `=(1.02xx10^(-4)molL^(-1_s^(-1))/(3.4xx10^(-5)s^(-1))` `=3.0molL^(-1)` |
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| 1572. |
If `3A rarr 2B`, then the rate of reaction of `+ (dB)/(d t)` is equal toA. `-(3)/(2)(d[A])/(dt)`B. `-(2)/(3)(d[A])/(dt)`C. `-(1)/(3) (d[A])/(dt)`D. `+2(d[A])/(dt)` |
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Answer» Correct Answer - B For reaction `3A to 2B` `rate =-(1)/(3) (d[A])/(dt)`[ Rate of disappearance ] `=+(1)/(2)(d[B])/(dt)`[ rate of appearance ] `therefore + (d[B])/(dt) =-(2)/(3) (d[A])/(dt)` |
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| 1573. |
If `3A rarr 2B`, then the rate of reaction of `+ (dB)/(d t)` is equal toA. `- (3)/(2) (d[A])/(dt)`B. `- (2)/(3) (d[A])/(dt)`C. `- (1)/(3) (d[A])/(dt)`D. `+ 2(d[A])/(dt)` |
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Answer» Correct Answer - B For the reaction `3Ararr2B` `Rate `=-(d[A])/(3dt)=+(d[B])/(2dt)` Thus, `+(d[B])/(dt)=-(2)/(3)=-(2)/(3)(d[A])/(dt)` |
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| 1574. |
A hypothetical reaction `A_(2) + B_(2) rarr 2AB` follows the mechanism as given below: `A_(2) hArr A+A ("fast")` `A+B_(2) rarr AB+B` (slow) `A+B rarr AB` (fast) The order of the overall reaction isA. `2`B. `1`C. `3//2`D. Zero |
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Answer» Correct Answer - c `r=K[A][B_(2)]` and `K_(c )=([A][A])/([A_(2)])` or `[A]=K_(c )^(1//2)[A_(2)]^(1//2)` Thus `r=K.K_(c )^(1//2)[A_(2)]^(1//2)[B_(2)]` `=K^(1)[A_(2)]^(1//2)[B_(2)]` or order is `3//2`. |
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| 1575. |
The possible mechanism for the reaction `2NO + 2H_(2) to N_(2) + 2H_(2)O` is i) `2NO ltimplies N_(2)O_(2)` ii) `N_(2)O_(2) + H_(2) overset("slow")to N_(2)O+H_(2)O` iii) `N_(2)O + H_(2)O overset("fast")to N_(2) + H_(2)O` a) What is rate law for the reaction? b) What is the order of the reaction?A. Rate =`[N_(2)O_(2)],"order"=1`B. Rate`=[N_(2)O_(2)][H_(2)][H_(2)]"order"=2`C. Rate `=[N_(2)O_(2)]^(2),"order"=2`D. Rate `=[N_(2)O_(2)]^(2),"order"=3` |
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Answer» Correct Answer - B The slowest step of the reaction is rate determining step `N_(2)O_(2) +H_(2)to N_(2)O+H_(2)O` Rate of reaction ` =[N_(2)O][H_(2)]` hence order og reaction = 2 |
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| 1576. |
The reaction, `2NO+Br_(2)rarr 2NOBr`, is supposed to follow the following mechanism, (i) `NO+Br_(2) overset("fast")(hArr) NOBr_(2)` (ii) `NOBr_(2) +NO overset(slow)(rarr)2NOBr` suggest the rate law expression.A. `r=k[NO]^(2)[Br_(2)]`B. `r=k[NO][Br_(2)]`C. `r=k[NO][Br_(2)]^(2)`D. `r=k[NOBr_(2)]` |
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Answer» Correct Answer - A (a) For slowet step : rate `=k[NO][Br_(2)]" "...(i)`For equilibrium, `k=([NOBr_(2)])/([NO][Br_(2)])" "...(ii)`by(i) and(ii), `r=k dot"K_(c)[NO]^(2) [Br_(2)]` rate `=k[NO]^(2)[Br_(2)]` |
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| 1577. |
Conisder the reaction mechanism: `A_(2) overset(k_(eq))hArr 2A ("fast")` (where `A` is the intermediate.) `A+B underset(k_(1))rarr P (slow)` The rate law for the reaction isA. `k_(1)[A][B]`B. `k_(1)k^(1//2)[A_(2)]^(1//2)[B]`C. `k_(1)k^(1//2)[A][B]`D. `k_(1)k^(1//2)[A]^(2)[B]` |
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Answer» Correct Answer - B `A_(2) overset(k_(eq))hArr 2A ("fast")` `rArr k_(eq) = [A]^(2)//[A_(2)] rArr [A] = sqrt(k_(eq)[A_(2)])` `A+B overset(k_(1))rarrP (slow) rarr RDS` `rArr Rate_(RDS) = k_(1)[A][B] = k_(1)sqrt(k_(eq)) sqrt([A_(2)]).[B]` Note: Overall reaction is : `A_(2)+2B rarr 2P` |
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| 1578. |
For the reaction, `2H_(2)(g)+2NO(g) rarr N_(20(g) +2H_(2)O(g)` Rate `=k[H_(2)][NO]^(2)`. This mechanism has been proposed: Step 1: `H_(2)+NO rarr H_(2)O+N` Step2 : `N=NO rarr N_(2)+O` Step 3 : `O+H_(2) rarr H_(2)O` Which statement about this rate law and mechanism is correct?A. The mechanism is consistent with the rate law if step 1 is the rate determineing step.B. This mechanism is consistent with the rate law if step 2 is the rate detemining step.C. The mechanism is consistent with the rate law if step 3 is the rate determining step.D. This mechanism can not be consistent with the rate law, regardless of which step is rate-determining. |
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Answer» Correct Answer - B |
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| 1579. |
The reaction,`2NO+2H_(2)rarr N_(2)+2H_(2)O` has been assigned to follow the following mechanism I. `NO+NO hArr N_(2)O_(2)" "` (fast) II.`N_(2)O_(2)+H_(2) rarr N_(2)O+H_(2)O " "` (slow) III. `N_(2)O +H_(2) rarr N_(2)+H_(2)O " "` (fast) The rate constant of step II is `1.2 xx 10^(-4) "mole"^(-1)L"min"^(-1)` while equilibrium constant of step I is `1.4 xx 10^(-2)` . What is the rate of reaction when concentration of NO and `H_(2)` each is 0.5 mole `L^(-1)`?A. `2.1 xx 10^(-7)` mol`L^(-1) "min"^(-1)`B. `3.2 xx 10^(-6)` mol`L^(-1) "min"^(-1)`C. `3.5 xx 10^(-4)` mol`L^(-1) "min"^(-1)`D. None of above |
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Answer» Correct Answer - A |
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| 1580. |
What is the rate law for the hypothetical reaction with the mechanism shown? `2A hArr` intermediate 1 `" "`(fast equilibrium) intermediate `1+B rarr ` intermediate 2`" "` (slow) intermediate`2 +B rarr A_(2)B_(2)" "` (fast)A. Rate=`k[A]^(2)`B. Rate`=[B]^(2)`C. Rate=`k[A][B]`D. Rate =`k[A]^(2)[B]` |
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Answer» Correct Answer - D |
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| 1581. |
For the reaction `2A_(3)hArr 3B_(2)` Mechanism `A_(3)hArr B_(2) +I^("*"), " "k_(c)` (fast) `A_(3)+I^("*") rarr 2B_(2)," "k` (slow) For the net reaction, correct statement is :A. order is one but molcularity 2B. order is 2 but molecularity 1C. both order and molecularity is not definedD. order is one but molecularity is not defined |
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Answer» Correct Answer - D |
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| 1582. |
Why does it take more time to boil an egg or cook rice at higher altitudes? |
| Answer» The boiling point of water decreases at higher altitudes since the atmospheric pressure decreases. The decrease in temperature is excepted to decrease the reaction rate or the cooking proces. Time, taken is generally more as compared to the plains. | |
| 1583. |
In the rate-limiting approximation for a two-step reaction, the overall rate of the reaction is always equal to the rate of the ……….. Step in the reaction mechanism.A. firstB. secondC. fastestD. slowest |
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Answer» Correct Answer - D |
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| 1584. |
In order to determine order/rate constant of any gaseous reaction pressure data at constant volume and temperature can be analysed. For a gaseous reaction occurring in a rigid vessel at 300 K following data was observed. `2A(g) rarr 3B(g) +2C(g)` `{:(,"Time (min)",,"10 min",,"30 min",,infty time),(,"Pressure increase (mm of Hg)",,"30 mm",,52.5 min,,"60 min"):}` What will be the total increase in pressure 20 min after the reaction?A. 45 mm of HgB. 50 mm of HgC. 55 mm of HgD. 40 mm of Hg |
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Answer» Correct Answer - A |
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| 1585. |
The solvolysis of cinnamyl chloride can be studied spectrophotometrically by observing the decrease in Absorbance of the adsorption maximum at 260 nm. The following observation were made in ethanoic NaOH at 298 k. `{:(,"Time (min)",,"0",,"10",,"20"),(,"Absorbance at 260 nm",,0.4,,0.36,,0.324):}` Absorbance is directly proportional to the concentration of cinnamyl chloride. [Given: l n `10/9, l n 2.5=0.9, l n 5/3=0.5]` The order of reaction (solvolysis) is: |
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Answer» Correct Answer - B |
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| 1586. |
Fluorine - 18 is a radioactive isotope that decays by positron emission to form oxygen - 18 with a half-life of 100 min. Physicians use `^(18)F` for the study of brain by injecting a quantity of fluoro substituted glucose into the blood of a patient The glucose accummulates in the region where the brain is active and needs nourishment. If a sample of glucose that contains `.^(18)F` is injected into the blood what percentange will remain after 5 hours?A. `12.5`B. 50C. 25D. `6.25` |
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Answer» Correct Answer - A |
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| 1587. |
Fluorine - 18 is a radioactive isotope that decays by positron emission to form oxygen - 18 with a half-life of 100 min. Physicians use `^(18)F` for the study of brain by injecting a quantity of fluoro substituted glucose into the blood of a patient The glucose accummulates in the region where the brain is active and needs nourishment. How long does it take for `99.9%` decay of `.^(18)F` ?A. Infinite timeB. `16.67` hourseC. 10 hrsD. `33.33` hours |
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Answer» Correct Answer - B |
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| 1588. |
Fluorine - 18 is a radioactive isotope that decays by positron emission to form oxygen - 18 with a half-life of 100 min. Physicians use `^(18)F` for the study of brain by injecting a quantity of fluoro substituted glucose into the blood of a patient The glucose accummulates in the region where the brain is active and needs nourishment. What is the average life for decomposition of `18_(F)`?A. 100 minB. 200 minC. `69.3` minD. 144 min |
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Answer» Correct Answer - D |
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| 1589. |
The following reaction was carried out in water : `Cl_(2) + 2I^(ɵ) rarr I_(2) + 2Cl^(ɵ)` The initial concentration of `I^(ɵ)` was `0.25 mol L^(-1)` and the concentration after `10 min` was `0.23 mol L^(-1)`. Calculate the rate of disappearance of `I^(ɵ)` and rate of appearance of `I_(2)`. |
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Answer» `Delta[I^(-)]=[I^(-)]_("final")-[I^(-)]_("initial")` `=0.23-0.25=-0.02"mol" L^(-1)` `Deltat=10-0=10` min `-(Delta[I^(-)])/(Deltat)=-((0.02))/(10)=0.02 "mol" L^(-1) "min"^(-1)` Rate of apperance of `I_(2)-(1)/(2)(Delta[I^(-)])/(Deltat)=(0.002)/(2)` `= 0.001 "mol" L^(-1) min ^(-1)` |
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| 1590. |
`99%` at a first order reaction was completed in `32 min`. When will `99.9%` of the reaction complete.A. `48 min`B. `46 min`C. `50 min`D. `45 min` |
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Answer» Correct Answer - A `K xx 32 = ln ((100)/(1))` …(i) `K xx t = ln((100)/(0.1))` …(ii) eq. `(ii)//(i)` `(t)/(32) = (3 ln 10)/(2 ln 10)` `t = 48 min` |
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| 1591. |
`99%` at a first order reaction was completed in `32 min`. When will `99.9%` of the reaction complete.A. 48 minB. 50 minC. 35 minD. 39 min |
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Answer» Correct Answer - A Time required to complete any particular fraction of first order reaction is `_(t//2)=(2.303)/(k)log(1)/(1-f)` therefore `t_(99%)=(2.303)/(k)log(1)/(1-0.99)` `=(2.303)/(k)log(1)/(0.01)` `=(2.303)/(k)log10^(2)` `t_(99.9%)=(2.303)/(k)log(1)/(1-0.999)` `=(2.303)/(k)log(1)/(0.001)` `=(2.303)/(k)log10^(3)` Dividing Eq. `(1)` by Eq. `(2)` , we get `(t_(99%))/(t_(99.9%))=(log10^(2))/(log10^(3))=(2log10)/(3log10)=(2)/(3)` Thus, `t_(99.9%)=(3)/(2)t_(99%)` `=(3)/(2)(32min)=48min` |
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| 1592. |
The following reaction was carried out in water `Cl_(2) + 2I^(-) to 2Cl^(-) + I_(2)` The initial concentration of `I^(-)` was `0.50 mol L^(-1)` and concentration after 10 minutes was 0.46 mol `L^(-1)`. Calculate the rate of disappearance of `I^(-)` and rate of appearance of `I_(2)`. |
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Answer» The chemicaal reaction is: `Cl_(2) + 2I^(-) to 2Cl^(-) +I_(2)` `Delta[I^(-)]=(0.046 -0.50) = -0.04 mol L^(-1)`, `Deltat=10min`. Rate of disappearance of `I^(-)` `-(Delta[I^(-)])/(Deltat) = -(0.04 mol L^(-1))/(10 "min")=0.004 mol L^(-1)min^(-1)` Rate of appearance of `I_(2)`. `(Delta[I_(2)])/(Deltat) = 1/2(Delta[I^(-)])/(Deltat) = 0.04/2 mol L^(-1)min^(-1) = 0.002 mol L^(-1)min^(-1)`. |
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| 1593. |
In the first order reaction, half of the reaction is completed in 100 seconds. The time for `99%` of the reaction to occur will beA. 664.64sB. 646.6sC. 660.9sD. 654.5 s |
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Answer» Correct Answer - A a) `k=0.693/100 s^(-1)` `t(99.9%) = 2.303/k log a/(a-x)` `=(2.303 xx 100s/(0.693)) xx log 100/1` `=(2.303 xx 100 xx 2)/(0.693) = 666.64` s |
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| 1594. |
The mechanism of the reaction `A+2B rarr C+D` is: Step I: `A+B underset(k_(2))overset(k_(1))hArr I` Step II: `B+I overset(k_(3))rarr C+D` In the first step is a fast equilibruim , then the incorrect statement is :A. Order of reacton with respect to A is 1.B. Order of reacton with respect to B is 2.C. Overall rate of reaction is ,`r=K_(3).[A][B]^(2)`D. Rates of forward and backward reaction of step I is much greater than the rate of reaction of step II. |
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Answer» Correct Answer - C |
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| 1595. |
In the following first order reactions: `A+ "Reagent" overset(K_(1))(rarr) Product` `B+ "Reagent" overset(K_(2))(rarr) Product` The ratio of `K_(1)//K_(2)` when only `50%` of `B` reacts in a given time when `94%` of `A` has been reacted is:A. `4.06`B. `0.246`C. `2.06`D. `0.06` |
| Answer» Correct Answer - A | |
| 1596. |
For a non-equilibrium process `A + B to` Produts the rate is first order with respect to A and second order with respect to B. If 1.0 Mole each of A and B are introduced into a one liter vessel and the intial rate is `1.0 xx 10^(-2)`mol `L^(-1)s^(-1)`, the rate when half of the reaction have been eonsumed is:A. `1.2 xx 10^(-3)`B. `1.2 xx 10^(-2)`C. `1.2 xx 10^(-4)`D. None of these |
| Answer» Correct Answer - A | |
| 1597. |
The rate constant for a first order reaction is `60s^(-1)`. How much time will it take to reduce the initial concentrationof the reactant to its `1//16th` value ? |
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Answer» For the first order reaction `t=2.303/k log a/(a-x)` Let the initial concentration =a , Final concentration (a-x) `= a//16, k=60s^(-1)` `t=2.303/(60s^(-1)) log a/(a//16) = 2.303/(60s^(-1)) log 16` `=(2.303xx1.2041)/(60s^(-1)) = 0.0462s = 4.62 xx 10^(-2)s` |
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| 1598. |
Relating the different ways of expressing reaction rates: Ethanol `(CH_(3)CH_(2)OH)`, the avtive ingredient in alcoholic beverges and an octane booster in gasoline, is yielded by fermentation of glucose. The balance chemical equation is `C_(6)H_(12)O_(6)(aq.)overset("zymase")rarr 2CH_(3)CH_(2)OH(aq.)+2CO_(2)(g)` How is the rate of formation of ethanol related to the rate of consumption of glucose ? Strategy : The rate of disappearance of a reactant X is `-Delta[X]//Delta t` while the rate of appearance of a product Y is `Delta[Y]//Delta t`. We can express the rate of the reaction in terms of either rate expressoin. To find the relative rates, look at the coefficients in the balanced chemical equation. These rate expressions can be equated , if we first divide each by the coefficient of the corresponding substance in the chemical equation. |
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Answer» Rate of consumption of glucose `= - (Delta[C_(6)H_(12)O_(6)])/(Delta t)` Rate of formation of ethanol `= (Delta[CH_(3)CH_(2)OH])/(Delta t)` According to the balanced equation, 2 mol of ethanol are produced for each mole of glucose that reacts. Therefore, the rate of formation of ethanol is twice the rate of consumption of glucose. Divide each rate by the corresponding coefficent and then equate them : `- (Delta[C_(6)H_(12)O_(6)])/(Delta t) = (Delta[CH_(3)CH_(2)OH])/(2Delta t)` |
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| 1599. |
A radioactive element gets spilled over the floor of a room. Its half life period is 30 days. If its initial activity is ten times the permissible value, after how many days will it be safe to enter the room?A. 100 daysB. 1000 daysC. 300 daysD. 10 days. |
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Answer» Correct Answer - A a) Let the initial activity for safe working = A `therefore` Initial activity `[A]_(0) = 10 A` `k=0.693/30` days `t=(2.303)/t log ([A]_(0))/([A])` `t=(2.303 xx (30"days"))/(0.693) log 10` `=99.6 ~~ 100` days. |
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| 1600. |
A sample of rock from moon contains atoms of uranium and lead in the ration 1:3. Then age of rock will be `(t_(1//2)"for uranium"=4xx10^(10)years)` :A. `4xx10^(2)years`B. `1.2xx10^(10)years`C. `8xx10^(10)years`D. `2xx10^(9)years` |
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Answer» Correct Answer - C |
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