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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.
| 1851. |
The rate of reaction `2NO + Cl_(2) to 2NOCl` is doubled when the concentration of `Cl_(2)` is doubled and becomes eight times when the concentration of both NO and `Cl_(2)` are doubled. Predict the order of reaction. |
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Answer» Rate ( r) = `k[NO]^(x)[Cl_(2)]^(y)`…………..(i) Rate (2r)`= k[NO]^(x)[2Cl_(2)]^(y)`………..(ii) Rate (8r) = `k[2NO]^(x)[2Cl_(2)]^(y)`…………(iii) `(8r)/(2r) = (k[2NO]^(x)[2Cl_(2)]^(y))/(k[NO]^(x)[2Cl_(2)]^(y))` `4= [2]^(x) or x=2` `(2r)/(r)=(k[NO]^(x)[2Cl_(2)]^(y))/(k[NO]^(x)[Cl_(2)]^(y))` `2= [2]^(y)` or y=1 Order of reaction= 2+1=3. |
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| 1852. |
The rate of reaction: `2NO + Cl_(2) rarr 2NOCl` is given by the rate, equation rate `= k[NO]_(2)[Cl_(2)]`. The value of the rate constant can be increased byA. increasing the temperatureB. increasing the concentration of NOC. increasing the concentration of `Cl_(2)`D. Doing all these. |
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Answer» Correct Answer - A a) The value of rate constant depends upon temperature and not upon the initial concentration of reactants and products. |
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| 1853. |
In Arrhenius equation `k=Ae^(-E//RT)`, factor `e^(-E_(a)//RT)` is known as:A. frequency factorB. activation factorC. pre-exponential factorD. Boltzmann factor |
| Answer» Correct Answer - D | |
| 1854. |
The oxidation of ammonia takes place as, `4NH_(2)(g)rarr3O_(2)(g)rarr2N_(2)(g)+6H_(2)O(g)` If the rate of formation of `N_(2) "is" 0.7 M//s`, detemine the rate at which `NH_(3)` is consumed.A. `1.4 "mol" L^(-1)s^(-1)`B. `0.7 "mol" L^(-1)s^(-1)`C. `1.5 "mol" L^(-1)s^(-1)`D. none of these |
| Answer» Correct Answer - A | |
| 1855. |
For a reaction, `Cl_(2)(g) + 2NO(g) to 2NOCl(g)` ,the rate law is expressed as rate = `k[Cl_(2)][NO]^(2)`. What is the order of the reaction? |
| Answer» The rate law for the reaction ( r) = `k[AB_(2)][Cl_(2)]^(1//2)` | |
| 1856. |
The rate constant is given by the equation `k = P.Ze^(-E_(a)//RT)`. Which factor should register a decrease for the reaction to proceed more rapidly?A. TB. ZC. ED. pressure. |
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Answer» Correct Answer - C Lower the activation energy, faster is the reaction. |
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| 1857. |
The rate constant is given by the equation `K=Ae^(-E_(a)//RT)` which factor should register a decrease for the reaction to proceed more rapidly:A. `T`B. `A`C. `A` and `T`D. `E_(a)` |
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Answer» Correct Answer - d A decrease in `E_(a)` will increase rate constant `K` and thus rate of reaction of reaction increases. |
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| 1858. |
At a certain temperature, the first order rate constant `k_(1)` is found to be smaller than the second order rate constant `k_(2)`. If `E_(a)(1)` of the first order reaction is greater than `E_(a)(2)` of the second order reaction, then as temperature is raised:A. `k_(2)` will increase faster than `k_(1)`B. `k_(1)` will increase faster than `k_(2)` but will always remain less than `k_(2)`C. `k_(1)` will increase faster than `k_(2)` and become equal to `k_(2)`.D. `k_(1)` will increase faster than `k_(2)` and become greater than `k_(2)`. |
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Answer» Correct Answer - A Since `k_(2) gt k_(1)`. |
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| 1859. |
The rate constant is given by the equation `k = P.Ze^(-E//RT)`. Which factor should register a decrease for the reaction to proceed more rapidly ?A. `E_(a) `B. `T `C. `Z`D. `P` |
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Answer» Correct Answer - A Lower activation energy , faster will be reaction . |
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| 1860. |
A sample contanining a radioactive isotope prduces 2000 counts per minutes in a Giege counter. Afte 120hours, the sample produces 250 counts per minutes. What is the half-life of the isotope?A. 15hB. 30hC. 40hD. 60h |
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Answer» Correct Answer - C |
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| 1861. |
The reaction `A+OH^(-)rarr` Products, obeys rate law expression as : `(-d[A])/(dt)=k[A][OH^(-)]` If initial concentrations of [A] and `[OH^(-)]` are 0.02 M and 0.3 M respectively and if it takes 30 sec for `1%` A to react at `25^(@)C`, caclutate the rate constant for the reaction |
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Answer» `A" "+ " "OH^(-)" "rarr` Products `t=0 " " 0.002 " "0.3` `k=(2.303)/(30xx(0.002-03))log_(10).(0.3[0.002-(0.002xx1)/(100)])/(0.002[0.3-(0.002xx1)/(100)])` `k=1.12xx10^(-3)"litre" "mol"^(-1)s^(-1)` |
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| 1862. |
Why in general a reaction does not proceed with a uniform rate throughout? |
| Answer» Since the change in moalr concentration of the reactants is not the same throghout therefore, the rate of reactants which is change in molar concentration per unit time does not remain constant. | |
| 1863. |
A Geigger Muller counter is used to study the radioactive process. In the absence of radioactive substances `A`, it counts `3` disintegration per second (dps). At the start in the presence of `A`, it records `23` dps, and after `10 m` in `13 dps`, i. What does it count after `20 m` in ?A. `8 dps, 10 mi n`B. `5 dps, 10 mi n`C. `5 dps, 20 mi n`D. `5 dps, 5 mi n` |
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Answer» Correct Answer - A In the absence of `A,3 dps` is zero error, hence Initial count`=23-3=20 dps` After `10 min= 13-3=10 dps` After `20 min= 5 dps` ( recorded `=5+3=8 dps)` (`50%` fall in `10 min, T_(50)=10min)` |
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| 1864. |
In Arrhenius equation `k = Ae^(-E_(a)//RT))`, A is the value of the rate constantA. when `E_(a)gtRT`B. `at 0^(@)C`C. at absolute zero `(0K)`D. when the temperature is infinite |
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Answer» Correct Answer - D In Arrhenius equation, `k=A` , when `e^(-E_(a)//RT)=1` . This happenes only when `E_(a)//RT=0` which is possible if either `E_(a)=0` or `T` is infinite. |
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| 1865. |
Two first order reactions proceed at `25^(@)C` at the same rate. The temperature coefficient of the rate of the first reaction is `2` and that of second reaction is `3`. Find the ratio of the rates of these reactions at `75^(@)C`. |
| Answer» Correct Answer - `7.5937;` | |
| 1866. |
Which of the following expression is correct for the rate of reaction given below ? `5Br^(-)(aq) + BrO_(3)^(-)(aq) + 6H^(+) (aq) to 3Br_(2) (aq) + 3H_(2) O(l)` .A. `(Delta[Br^(-)])/(Deltat)= 5(Delta[H^(+)])/(Deltat)`B. `(Delta[Br^(-)])/(Deltat)= (6)/(5)(Delta[H^(+)])/(Deltat)`C. `(Delta[Br^(-)])/(Deltat)= (5)/(6)(Delta[H^(+)])/(Deltat)`D. `(Delta[Br^(-)])/(Deltat)= 6(Delta[H^(+)])/(Deltat)` |
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Answer» Correct Answer - C Given, chemical reaction is `5Br^(-) (aq) + BrO_(3)^(-) (aq) + 6H^(+) (aq) to 3Br_(2)(aq) +3H_(2)O(l)` Rate law expression for the above equation can be written as `-(1)/(5)(Delta[Br^(-)])/(Deltat) = -(Delta[BrO_(3)^(-)])/(Deltat) = (-1)/(6)(Delta[H^(+)])/(Deltat) = (+1)/(3)(Delta[Br_(2)])/(Deltat)` `rArr" "(Delta[Br^(-)])/(Deltat)= -(Delta[BrO_(3)^(-)])/(Deltat) =(Delta[H^(+)])/(Deltat)` `rArr" "(Delta[Br^(-)])/(Deltat) = (5)/(6) (Delta[H^(+)])/(Deltat)` |
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| 1867. |
A reaction involiving two different reactants can never be:A. (a) Unimolecular reactionB. (b) I order reactionC. (c ) II order reactionD. (d) Bi-molecular reaction |
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Answer» Correct Answer - a Two reactants leads to bimolecular reaction may be of I or II order. |
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| 1868. |
Which of the following curves represents a second order reaction ? [x = product oncentration `(a-x)` = reactant concentration)A. B. C. D. |
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Answer» Correct Answer - C For the simple type of second order reaction `Ararr`products the integrater rate law is `(1)/([R]_(t))=Kt+(1)/([R]_(0))` Since the integrated rate law has the form `y=mx+b,a` graph of `1//[R]_(t)` or `1//a-x` versus time is a straigh line if the reaction is second order: `overset((1)/([R]_(t))=Kt+(1)/([R]_(0)))` `({:(uarr,uarr,uarr,uarr),(y,m,x,b):}` The slope of the straght line is rate constant `K` , and the intercept is `1//[R]_(0)` . First graph corresponds to a first order reaction, second graph corresponds to a third order reation. while the fourth graph corresponds to a third order reaction. |
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| 1869. |
Which of the following curves represents a first order reaction.A. B. C. D. All of these |
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Answer» Correct Answer - D A first order reaction exhibits an exponential decay of the reaction concentration `[R_(t)=R_(0)e^(Kt)]` and a linear decay of the logarithm of the reactant concentration `[logR_(1)=logR_(0)-(k)/(2.303).t]` Integrated rate law for a first order reaction is `K=(2.303)/tlog(([R]_(0))/([R]_(t)))` or `log({:(([R]_(0))/([R]_(t)),=,((K)/(2.303)),t),(overset(uarr,)(y),=,overset(uarr)(m),overset(uarr)(x)):})` |
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| 1870. |
For a first order reactionA. `t_(0.75)=4t_(0.5)`B. `t_(0.75)=2t_(0.5)`C. `t_(0.75)=3t_(0.5)`D. `t_(0.75)=1.5t_(0.5)` |
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Answer» Correct Answer - B For a first order reaction, the time required for `75%` of the reaction to complete is twice that required for `50%` of the reaction to complete because `75%` of the reaction `(100%overset(t_(1//2))(rarr)50%overset(t_(1//2))(rarr)25%)`involves two half lives while `50%` of the reaction `(100%overset(t_(1//2))to50%)` involves just one half life. |
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| 1871. |
The half life for a reaction between fixed conecntration of reactants varies with temperature as follows: `{:(t^(@)C,520,533,555,574),(t_(1//2)sec,1288,813,562,477):}` Calcutate the activation energy of this reaction. |
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Answer» Correct Answer - `86.47KJ^(-1)` Use: `log((k_(2))/K_(1))=(E)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))` `k_(2)=0.693//t_(2)` `k_(1)=0.693//t_(1)` `:.log((t_(1))/(t_(2)))=(E)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))` |
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| 1872. |
The following data wer obtained for a given reaction at 300 K Calculate by what factor the rate of catalyst reaction is increases ? |
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Answer» Correct Answer - 2000 Use: `(k_(p))/(k_(a))="antilog"((DeltaE)/(2.303RT))` `k_(p)` = rate constant in presence of catalyst `k_(a)`= rate constant in presence of catalyst `DeltaE=76-57=19 kJ//"mol"` |
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| 1873. |
A catalyst increases the rate of reaction by :A. decreasing the free energy change for reaction.B. decreasing the enthalpy change of reaction.C. decreasing the activation energy for reaction.D. all of the above |
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Answer» Correct Answer - C |
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| 1874. |
The decomposition of a drug in human was found to be a first order process. The activation energy for the decomposition is `(3100RxxIn2)` and pre-exponential factor `A=4096hr^(-1)` . How long will it take the concentration of the drug In the blood to fall to half of its intial value at 310K? (Given: In `2=0.7`)A. `10.5 hr`B. `0.175` minC. `10.5` minD. `0.175` sec |
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Answer» Correct Answer - C |
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| 1875. |
In a reversible reaction `2NO_(2)underset(k_(2))overset(k_(1))iffN_(2)O_(4)`, the rate of disappearance of `NO_(2)` is equal toA. `(2k_(1))/(k_(2))[NO_(2)]^(2)`B. `2k_(1)[NO_(2)]-2k_(2)[N_(2)O_(4)]`C. `2k_(1)[NO_(2)]-2k_(2)[N_(2)O_(4)]`D. `(2k_(1)-k_(2))[NO_(2)]` |
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Answer» Correct Answer - C `2NO_(2)underset(k_(2))overset(k_(1))ltimpliesN_(2)O_(4)` Rate of reaction, `-1/2(d[NO_(2)])/(dt) = k_(1)[NO_(2)]^(2) - k_(2)[N_(2)O_(4)]` Rate of disappearance of `NO_(2)` `(-d[NO_(2)])/(dt) = 2k_(1)[NO_(2)]^(2)-2k_(2)[N_(2)O_(4)]` |
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| 1876. |
In a reversible reaction `2NO_(2)underset(k_(2))overset(k_(1))iffN_(2)O_(4)`, the rate of disappearance of `NO_(2)` is equal toA. `(2k_(1))/(k_(2))[NO_(2)]^(2)`B. `2k_(1)[NO_(2)]^(2)-2k_(2)[N_(2)O_(4)]`C. `2k_(2)[NO_(2)]^(2)-k_(2)[N_(2)O_4]`D. `(2k_(1)-k_(2))[NO_(2)]` |
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Answer» Correct Answer - B `2NO_(2)overset(k_(1))underset(k_(2))hArrN_(2)O_(4)` `Rate=-(1)/(2)(d[NO_(2)])/(dt)` `=k_(1)[NO_(2)]^(2)-k_(2)[N_(2)O_(4)]` `"therefore Rate of disappearance of " NO_(2)` `i.e, -(d[NO_(2)])/(dt)=2k_(1)[NO_(2)]^(2)-2k_(2)[N_(2)O_(4)]` |
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| 1877. |
On mixing 1 `dm^(3)` of 3M ethanol with 1 `dm^(3)` of 2 M ethanoic acid, an ester is formed. `C_(2)H_(5)OH+CH_(3)COOHrarrCH_(3)COOC_(2)H_(5)+H_(2)O` If each solution is diluted with an equal volume of water , the decrease in the initial rate would beA. 0.5 timesB. 4 timesC. 0.25 timesD. 2 times |
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Answer» Correct Answer - C `r=k[CH_(3)COOH][C_(2)H_(5)OH]` `or" "k=k[(n]/[V)][(n]/[V)]` Increasing volume (V) by twice (due to dillution) will reduce the rate by 4 times. `i.e." "r=0.25k[CH_(3)COOH][C_(2)H_(5)OH]` |
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| 1878. |
Cyclopropane rearranges to form propane : `DeltararrCH_(3)-CH=CH_(2)` This follows first order kinetics. The rate constant is `2.174xx10^(-3) "sec"^(-1)` . The initial concentration of cyclopropane is `0.29 M` . What will be the concentration of cyclopropane after 100 secA. `0.035 M`B. `0.22 M`C. `0.145 M`D. `0.0018 M` |
| Answer» Correct Answer - B | |
| 1879. |
If the amount of radioactive substance is increased three times and simultaneously temperature also increases thrice, the number of atoms disintegrating per unit time would be:A. `(1)/(3)rd "of original atoms" `B. constantC. tripleD. 9 times |
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Answer» Correct Answer - C |
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| 1880. |
The bromination of acetone that occurs in acid solution is represented by this equation. `CH_(3)COCH_(3) (aq) + Br_(2) (aq) rarr` `CH_(3)COCH_(2) Br(aq) + H^(+) (aq) + Br(aq)` These kinetic data were obtained for given reaction concentrations. Initial concentration, `M` `{:([CH_(2)COCH_(3)],[Br_(2)],[H^(+)],("Initail rate) (disappearance of "Br_(2)),),(0.30,0.05,0.05,5.7 xx 10^(-5),),(0.30,0.10,0.05,5.7xx10^(-5),),(0.30,0.10,0.10,1.2xx10^(-4),),(0.40,0.5,0.20,3.1xx10^(-4),):}`A. `rate =K [CH_(2) COCH_(3)][H^(+)]`B. `rate= K [CH_(2)= COCH_(3)[Br_(2)]`C. `rate= K[CH_(3)COCH_(3)][Br_(2)][H^(+)]^(2)`D. `rate = k [CH_(3)COCH_(3)][BR_(2)][H^(+)]` |
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Answer» Correct Answer - A Let the order of reaction wrt `CH_(3)COOCH_(3) .Br_(2) and H^(+) are x,y,and z ` respectively thus . `Rate (I) =[CH_(3)COCH_(3)]^(x)[Br_(2)]^(y)[H^(+)]^(z) ` `4.7xx10^(-5)=(0.30)^(x)(0.05)^(y)(0.05)^(z)` `5.7xx10^(-5)=(0.30^(x)(0.10)^(y)(0.05)^(z)` `1.2xx10^(-4)=(0.30)^(x)(0.10)^(y)(0.10)^(z)` `3.1xx10^(-4)=(0.40)^(x)(0.05)^(y)(0.20)^(z)` from Eqs (i) and(ii) `1=((1)/(2))^(y)or 1^(@)=((1)/(2))^(y)` `y=0` from Eqs(ii) and (iii) `z=1` from Eqs.(i) and (iv) `x=1` thus rate law`prop |CH_(2)COCH_(3)|^(1)|Br_(2)|^(0)[H^(+)]^(1)` `=K[CH_(3)COCH_(3)][H^(+)]` |
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| 1881. |
The reaction obey I order with respect to `H_(2)` and `ICl` both. `H_(2) (g) + 2ICl(g) rarr 2 HCl(g) + I_(2) (g)` Which of the following mechanism is in consistent with the given fact ? Mechanism A: `H_(2) (g) + 2Cl rarr 2HCl(g) + I_(2)(g)` Mechanism B: (i) `H_(2)(g) + ICl(g) overset ("slow")rarr HCl(g) + HI(g)` (ii) `HI(g) + ICl(g) rarr HCl(g) + I_(2)`A. Neither A nor BB. A onlyC. B onlyD. A and B both |
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Answer» Correct Answer - C According to mechanism `A` , which proceeds through an elementary reaction, should be of first order with respect to `H_(2)(g)` and of second order with respect to `ICI` . Therefore, mechanism `A` is ruled out. On the other hand, there are two steos in mechanism `B` Rate of overall reaction is controlled by the slow step which involves one molecule of `H_(2)` and one molecule of `ICI` i.e., the law should be Rate `=k[H_(2)][ICI]` Therefore, mechanism `B` is consistant with the given in formation |
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| 1882. |
The reaction obey I order with respect to `H_(2)` and `ICl` both. `H_(2) (g) + 2ICl(g) rarr 2 HCl(g) + I_(2) (g)` Which of the following mechanism is in consistent with the given fact ? Mechanism A: `H_(2) (g) + 2Cl rarr 2HCl(g) + I_(2)(g)` Mechanism B: (i) `H_(2)(g) + ICl(g) overset ("slow")rarr HCl(g) + HI(g)` (ii) `HI(g) + ICl(g) rarr HCl(g) + I_(2)`A. `A` and `B` bothB. Neither `A` nor `B`C. `A` onlyD. `B` only |
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Answer» Correct Answer - D `I` step of mechanism `B` shows `I` order in both reactants. |
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| 1883. |
The reaction obey I order with respect to `H_(2)` and `ICl` both. `H_(2) (g) + 2ICl(g) rarr 2 HCl(g) + I_(2) (g)` Which of the following mechanism is in consistent with the given fact ? Mechanism A: `H_(2) (g) + 2Cl rarr 2HCl(g) + I_(2)(g)` Mechanism B: (i) `H_(2)(g) + ICl(g) overset ("slow")rarr HCl(g) + HI(g)` (ii) `HI(g) + ICl(g) rarr HCl(g) + I_(2)`A. only BB. Both A and BC. Neither A nor BD. Only A |
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Answer» Correct Answer - A in the reactions which take place in a number of steps , the slowest step is know as the rate determining step hence , rate reaction always depends on slow step . `H_(2)(g) +[C](g) to HCI(g)+HI(g)`is first order reaction with respect to `H_(2) and ICI` thus the mecheaniusm B will be more consistent with the given informataion . |
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| 1884. |
The rate law for a reaction between `A` and `B` is given by rate `= k[A]^(n)[B]^(m)`. On doubling the concentration of `A` and halving the concentration of `B`, the ratio of the new rate to the earlier rate of the reaction becomesA. (a) `n-m`B. (b) `2^(n-m)`C. (c ) `(1)/(2^(m+n))`D. (d) `m+n` |
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Answer» Correct Answer - b `r_(0)=K[A]^(n)[B]^(m)` `r_(1)=K[2A]^(n)[B//2]^(m)` `r_(1)=K 2^(n-m) [A]^(n)[B]^(m)` or `r_(1)=rxx2^(n-m)` |
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| 1885. |
One of the hazards of nuclear explosion is the generation of `.^(90)Sr` and its subsequent incorporation in bones. This nuclide has a half-life of `28.1` year. Suppose one micro-gram was absorbed by a new-born child, how much `Sr^(90)` will remain in his bones after 20 year? |
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Answer» Correct Answer - `(6.1xx 10^(-7)g)` |
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| 1886. |
The rate law for a reaction between `A` and `B` is given by rate `= k[A]^(n)[B]^(m)`. On doubling the concentration of `A` and halving the concentration of `B`, the ratio of the new rate to the earlier rate of the reaction becomesA. `(1)/(2^(m+n))`B. m+nC. m-nD. `2^(n-m)` |
| Answer» Correct Answer - D | |
| 1887. |
The rate law for a reaction between `A` and `B` is given by rate `= k[A]^(n)[B]^(m)`. On doubling the concentration of `A` and halving the concentration of `B`, the ratio of the new rate to the earlier rate of the reaction becomesA. `(1)/(2^(m+n))`B. `2^((n-m))`C. `(m+n)`D. `(n-m)` |
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Answer» Correct Answer - B `Rate_(1)=kk[A]^(n)[B]^(m)` On doubling the concentration of `A` and having the concentration of `B` , we get `Rate_(2)=k[2A]^(n)[B//2]^(m)` Dividing `Rate_(2)` by `Rate_(1)` , we get `(Rate_(2))/(Rate_(1))=(k[2A]^(n)[B//2]^(m))/(k[A]^(n)[B]^(m))=(2)^(n)((1)/(2))^(m)` `=2^(n-m)` |
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| 1888. |
The rate law for a reaction between the substances A and B is given by Rate = `k[A]^(n)[B]^(m)` On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as:A. `(1)/(2^(m+n))`B. `(m=n)`C. `(n-m)`D. `2^((n-m))` |
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Answer» Correct Answer - D |
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| 1889. |
The rate law for a reaction between the substances A and B is given by Rate = `k[A]^(n)[B]^(m)` On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as:A. (m+n)B. (n-m)C. `2^(n-m)`D. `1/(2^(m+n))` |
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Answer» Correct Answer - C c) `Rate_(1) = k[A]^(n)[B]^(m)` `Rate_(2) = k[2A]^(n)[1//2B]^(m)` `(Rate_(2))/(Rate_(1))= (k[2A]^(n)[1//2B]^(m))/(k[A]^n)[B]^(m)` `=[2]^(n([1//2])^(m))= 2^(n).2^(-m)` `=2(n-m)` |
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| 1890. |
The rate of a first order reaction is `1.8 xx10 ^(-3 ) "Mol "L^(-1) "Min"^(-1)` . When the initial concentration is `0.3 "Mol "L^(-1) `. The rate constant in the units of second isA. `1xx10^(-2) S^(-1)`B. `1xx10^(-4)S^(-1)`C. `6xx10^(-2)S^(-1)`D. `6xx10^(-4)S^(-1)` |
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Answer» Correct Answer - B Rate =k [concentration ]` therefore k=("Rate ")/("[concentration ]")` `=(1.8xx10^(-3))/(0.3)=6xx10^(-3)xx(1)/(60)=10^(-4)` |
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| 1891. |
During nuclear explosion, one of the products is `.^(99)Sr` with half`-` life of `28.1 `years. If `1 mug` of `.^(90)Sr` was absorbed in the bones of a newly born baby instead of calcium, how much of its will remain after 10 years and 60 years if it is not lost metabolically. |
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Answer» `t_(1//2) = 28.1 Y: k = (0.693)/(t_(1//2))= 0.693/28.1 Y^(-1)` All radioactive explosions follow first order kinetics. `t= (2.303)/k log a/(a-x)` Ist case, a=1 mug, t= 10 Y, `k=0.693/28.1 Y^(-1)` `10Y = 2.303/(0.693//28.1 Y^(-1)) log a/(a-x)` `loga/(a-x) = (10Y)xx(0.693//28.1Y^(-1))/(2.303) = 0.107` `a/(a-x)` = Antilog 0.107 = 1.279 `(a-x) = a/1.279=(1"mug")/(1.279)=0.7819` mug IInd case, a=1mug, t=60Y, `k=0.693//28.1 Y^(-1)` 60 Y `= (2.303)/(0.693//28.1Y^(-1))= log a/(a-x)` `loga/(a-x) = (60Y) xx (0.693//28.1Y^(-1))/(2.303) = 0.642` `a/(a-x)`= Antilog 0.642 = 4.385 `(a-x) = a/4.385 = (1"mug")/(4.385) = 0.2280` mug |
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| 1892. |
For a second order reaction rate at a particular time is X . IF the initial concentration is tripled is tripled the rate will becomeA. `3x`B. `9x^(2)`C. `9x`D. `27x` |
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Answer» Correct Answer - C `r prop [x]^(2)`, concentration is 3 times ` therefore ` Rate will be `(3)^(2)` i.e., 9 times |
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| 1893. |
when the concentration of a particular reactant is tripled, the initial rate of the reaction increases by a factor of nine, what is the order of the reaction with respect to this reactant?A. ZeroB. OneC. TwoD. Three |
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Answer» Correct Answer - C |
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| 1894. |
In a reactioon if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of reaction? |
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Answer» For the reaction, let rate ( r) = `k[A]^(n)` According to available data, (27 r) = `k[3A]^(n)` `(27r)/ ( r) = (k[3A]^(n))/(k[A]^(n))` or `27 = (3)^(n)` or `(3)^(3) = (3)^(n)` n=3 and the order of reaction = 3 |
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| 1895. |
In a hypothetical reaction `2X+Y rarr M+N`. If the concentration of `Y` is kept constant but that of `X` is tripled, the rate of reaction then will beA. Increased by `3` timesB. Increased by `6` timesC. Increased by `9` timesD. Unpredictable |
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Answer» Correct Answer - D It requires more data e.g., different rate values are not given for different experiments.Correct Answer - D It requires more data e.g., different rate values are not given for different experiments. |
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| 1896. |
For a first order reaction velocity constant , `K = 10^(-3) s^(-1)` . Two third life for it would beA. 1100 sB. 2200 sC. 3300 sD. 4400 s |
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Answer» Correct Answer - a We know that `k = (2.303)/(t)` log `(a)/(a-x)` `10^(-3) = (2.303)/(t)` log `(a)/((a - (2a)/(3))) , 10^(-3) = (2.303)/(t)` log 3 `10^(-3) = (2.303)/(t) xx 0.4771 , t = (2.303 xx 0.4771)/(10^(-3))` `= 1098.7` sec = 1100 sec . |
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| 1897. |
A reaction is first order in `A` and second order in `B`. How is rate affected when concentration of `B` is tripled? |
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Answer» Correct Answer - 9 Rate `r_(1)=K[A]^(1)[B]^(2) …(1)` Now if concentration of `B` is triple So, `r_(2)=K[A]^(1)[3B]^(2) …(2)` Compare the eqs. (1) and (2), `r_(1)/r_(2)=1/9 :. R_(2)=9 r_(1)` `therefore` Rate becomes `9` times. |
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| 1898. |
The first order rate constant for the decomposition of `N_(2)O_(5)` is `6.2 xx 10^(-4) sec^(-1)`. The `t_(1//2)` of decomposition isA. `1117.7`B. `111.7`C. `223.4`D. `160` |
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Answer» Correct Answer - a `t_(1//2) = (0.693)/(k) = (0.693)/(6.2 xx 10^(-4)) = 1117.7 ` sec . |
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| 1899. |
The half-life period for the first order reaction is 693 seconds. The rate constant of this reaction would beA. `0.1 sec^(-1)`B. `0.01 sec^(-1)`C. `0.001 sec^(-1)`D. `0.001 sec^(-1)` |
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Answer» Correct Answer - c `t_(1//2) = (0.693)/(k)` Given `t_(1//2) = 693` sec `693 = (0.693)/(k) , k = (693 xx 10^(-3))/(693) , k = 10^(-3) = 0.001 sec^(-1)`. |
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| 1900. |
Rate constant of a reaction with a virsus is `3.1 xx 10^(-4) s^(-1)`. Time required for a virus to become `75%` inactivated isA. `17.5 min`B. `40 min`C. `74.5 min`D. `40 min` |
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Answer» Correct Answer - C Virsus inactivation follows first order kinetics based on unit of `k(s^(-1))`. `t_(50) = (0.693)/(k) = (0.693)/(3.1 xx 10^(-4)s^(-1)) = 2235s` `:. t_(75) = 2 xx t_(50) = 4470s = (4470)/(60) = 74.5 min` |
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