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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.
| 451. |
The variation of `K` and `k_(f)/K_(b)` with temperature shows the following effects:A. (a) For endothermic reactions `K` and `K_(f)/K_(b)` both decreasesB. (b) For endothermic reactions `K` and `K_(f)/K_(b)` both increasesC. (c ) For endothermic reactions `K` and `K_(f)/K_(b)` both decreasesD. (d) For endothermic reactions `K` and `K_(f)/K_(b)` both increases |
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Answer» Correct Answer - b, c `E_(a)` is always positive thus `K` increases with temperature. `DeltaH` is -ve, `K_(f)/K_(b)` decreases with temperature. `DeltaH` is +ve, `K_(f)/K_(b)` increases with temperature. |
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| 452. |
For the reaction `A+B+C rarr D`, the following observations were made: (i) When the concentration of `A` was doubled, the rate of formation of `D` was doubled (ii) When the concentration of `B` was halved, the rate of formation of `D` becomes one fourth (iii) Doubling the concentration of `C` had no effect on rate. Slect the correct statement(s):A. (a) Rate equation, `r=k[A][B]^(1//2)`B. (b) Reactant `C` must involve after rate determining stepC. (c ) Only `A` and `B` participate in the rate determining stepD. (d) Order of `C` is `1` |
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Answer» Correct Answer - b, c Since the rate of raection is independent of concentration of `C`, the rate determining step must not involve `C`. |
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| 453. |
Select the correct statements:A. (a) A transient state cannot be isolatedB. (b) A transient state is a unstable stateC. (c ) The degree of dissociation for a reaction obeying first order kinetics is `1-e^(-KT)`D. (d) A large value of half-life implies for a fast reaction |
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Answer» Correct Answer - a, b, c Higher is `t_(1//2)` lesser is rate constant, slower is rate. |
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| 454. |
The rate constant of a reaction is given as `k=2.1xx10^(10)e^(-2700//RT)` It means that `"log "k " vs "1//T` will be a straight line with intercept on `"log "k" axis"=log2.1xx10^(10).` Number of effective collisions of temperature are `2.1xx10^(10)cm^(-3)s^(-1).` Half-life of a reaction increases of temperature. `"log"k" vs "1//T` will be a straight line with `slope=-(2700)/(2.303R).` Which of the above statements are true? Choose the correct option.A. I and IIB. II and IIIC. III and IVD. I and IV |
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Answer» Correct Answer - d `k=2.1xx10^(10)e^(-2700//RT)` `" ""In " k="In "2.1xx10^(10)+"In "e^(-2700//RT)` `"log "k="log "2.1xx10^(10)-(2700)/(2.303RT)` `"log"k=-(2700)/(2.303RT)(1)/(T)+"log "2.1xx10^(10).` On comparing it with straight line equation, y=mx+c, we get `therefore" "m(slope)=-(2700)/(2.303R)` and `C("intercept")="log "2.1xx10^(10)` |
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| 455. |
For a reaction `pA +qB rarr` products, the rate law expression is `r=k[A]^(l)[B]^(m)` thenA. `(p+l)lt(l+m)`B. `(p+q)gt(l+m)`C. (p+q) may or may not be equal to (l+m)D. (p + q)=(l + m) |
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Answer» Correct Answer - C |
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| 456. |
The rate constant is numerically the same for I order, II order and III order. Select the correct statements:A. (a) If `[A] lt 1,` then `r_(1) gt r_(2) gt r_(3)`B. (b) If `[A] gt 1,` then `r_(1) lt r_(2) lt r_(3)`C. (c ) If `[A] gt 1,` then `r_(1) gt r_(2) gt r_(3)`D. (d) If `[A] = 1,` then `r_(1) = r_(2) = r_(3)` |
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Answer» Correct Answer - a, b, d `r_(1)=K[A] …(i)` `r_(2)=K[A]^(2) …(ii)` `r_(3)=K[A]^(3) …(iii)` |
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| 457. |
For the reaction `H_(2) + Br_(2) rarr 2 HBr` overall order is found to be `3//2`. The rate of reaction can be expressed as:A. `[H_(2)][Br_(2)]^(1//2)`B. `[H_(2)]^(1//2)[Br_(2)]`C. `[H_(2)]^(3//2)[Br_(2)]^(0)`D. all of these |
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Answer» Correct Answer - D |
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| 458. |
If rate constant is numerically the same for three reaction of first, second and third order respectively, then which of the following is correct?A. If [A]=1 then `r_(1)=r_(2)=r_(3)`B. If [A]lt1 then `r_(1)gtr_(2)gtr_(3)`C. If [A]gt1 then `r_(3)gtr_(2)gtr_(1)`D. All of the above |
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Answer» Correct Answer - D |
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| 459. |
If rate constant is numerically the same for three reaction of first, second and third order respectively, then which of the following is correct?A. `if[A]gt1,r_(3)gtr_(2)gtr_(1)`B. `if[A]=1,r_(1)=r_(2)=r_(3)`C. `if[A]lt1,r_(1)gtr_(2)gtr_(3)`D. All of these |
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Answer» Correct Answer - D Consider a reaction `ArarrB"products"` `r_(1)=k[A]^(1)" for first order"` `r_(2)=k[A]^(2)"for second order"` `r_(3)=k[A]^(3)"for third order"` `because "k is same all the three cases".` `"If"[A]=1,r_(1)=r_(2)=r_(3)` `"If"[A]lt1,r_(1)gtr_(2)gtr_(3)` `"If"[A]gt1,r_(3)gtr_(2)gtr_(1)` |
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| 460. |
For the first order reaction, calculate the ratio of the time taken to complete `99%` of the reaction to the time taken to complete `90%` of the reaction. |
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Answer» `t=(2.303)/k log a/(a-x)` `t_(99%) = (2.303)/k log (100)/1=2.303/k xx 2 = 4.606/k` `t_(90%) = (2.303)/k log 100/10 = 2.303/k xx 1=2.303/k` `t_(99%)/t_(90%)=4.606/2.303 = 2`. |
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| 461. |
If 50% of a radioactive substance dissociates in 15 min, then the time taken by substance to dissociate 99% will beA. 50 minB. 100 minC. 99 minD. 150 min |
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Answer» Correct Answer - C Radioactive disintegration is a first order reaction, `k=(0.693)/(t_(1//2))=(0.693)/(15)" "(Given, t_(1//2)=15min)` =0.0462 Given, a=100, x=99 t=? `or" "0.0462=(2.303)/(t)"log"(100)/(100-99)` `=(2.303)/(t)" log "100` `or" "0.0462xxt =2.303xx2` `therefore" "t=(2.303xx2)/(0.0462)=99min` |
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| 462. |
For a reaction carried at 400 K, `NO_(2)(g) + CO_(2)(g) to CO_(2)(g) + NO_(2)(g)`, the proposed mechanism is as follows: `NO_(2) + NO_(2) overset("Slow")to NO + NO_(3) NO_(3) + CO overset("Fast")to CO_(2) + NO_(2)` What is the rate law for the reaction? |
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Answer» The slow step for the reaction is: `NO_(2) + NO_(2) to NO + NO_(3)` (Slow) The rate law equation is: `Rate( r) = k[NO_(2)][NO_(2)]= k[NO_(2)]^(2)`. |
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| 463. |
For a reaction A + 2B `to C`, rate =`k[A]^(x)[B]^(y)`. What is the order of the reaction? |
| Answer» Order of reaction = x+y | |
| 464. |
What are the units of rate constant for zero order reactions? |
| Answer» Units of k for zero order reaction: mol `L^(-1)s^(-1)` | |
| 465. |
The rate constant for the decomposition of hydrocarbons is `2.418xx10^(-5)s^(-1)` at `546K`. If the energy of activation is `179.9kJ mol^(-1)`, what will be the value of pre`-`exponential factor? |
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Answer» `k=2.418xx10^(-5)s^(-1)` ` T = 546 k ` `E_(a)=179.9KJ mol^(-1) =179.9xx10^(3) J mol ^(-1)` According to the Arrhenius equation, `k=Ae^(-E)` `rArr "In "k=In A-E_(0)/(RT)` `rArr"log"k=log A-E_(0)/(2.303RT)` `rArr"log"A=log K+E_(0)/(2.303RT)` `="log"(2.418xx10^(-5)S^(-1))+(179.xx10^(3) J "mol"^(-1))/(2.303xx8.314 "JK"^(-1)"mol"^(-1)xx546K)` =`(0.3835 - 5) + 17.2082` 12.5917 Therefore, A = antilog `(12.5917)` `= 3.9 × 10^(12) s^(−1)` (approximately) |
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| 466. |
Consider a certain reaction `A rarr` Products with `k=2.0xx10^(-2)s^(-1)`. Calculate the concentration of `A` remaining after `100s` if the initial concentration of `A` is `1.0 mol L^(-1)`. |
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Answer» `k = 2.0xx10^(-2) s^(-1)` `T = 100s` ` [A]_(0)=1.0mol^(-1)` Since the unit of k is `s^(−1)`, the given reaction is a first order reaction. `implies 2.0xx10^(-2)s^(-1)=(2.303)/(100s)"log"(1.0)/[A]` `implies 2.0xx10^(-2)s^(-1)=(2.303)/(100s)(-"log"[A])` `implies -log[A]= (2.0xx10^(-2)xx100)/(2.303)` `implies [A]= anti"log"(-(2.0xx10^(-2)xx100)/(2.303))` `0.135"mol"L^(-1)` (approximately) Hence, the remaining concentration of A is` 0.135 "mol"^(-1)` |
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| 467. |
The rate constant for the decompoistion of a certain reaction is described by the equation: `log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)` A two-step mechanism has been suggested for the reaction of nitric oxide and bromine: `NO(g) + Br_(2)(g) overset(k_(1))rarr NOBr_(2)(g)` `NOBr_(2)(g)+NO(g) overset(k_(2))rarr 2NOBr(g)` The observed rate law is, rate `= k[NO]^(2)[Br_(2)]`. Hence, the rate-determining step isA. `NO(g)+Br_(2)(g) rarr NOBr_(2)(g)`B. `NOBr_(2)(g)+NO(g) rarr 2NOBr(g)`C. `2NO(g) + Br_(2)(g) rarr 2NOBr(g)`D. None of these |
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Answer» Correct Answer - B `NO(g)+Br_(2)(g) overset(k_(1))hArr NOBr_(2)(g)` …(i) `NOBr_(2)(g)+NO(g)underset(RDA)overset(k_(2))rarr 2NOBr(g)` …(ii) `:. Rate = k_(2)[NOBr_(2)][NO]` …(iii) Since `NOBr_(2)` is the reactive intermediate, so its concentration is determined form step (i). `:. k_(1) or k_(eq) = ([NOBr_(2)])/([NO][Br_(2)])` or `[NOBr_(2)] = k_(1)[NO][Br_(2)]` Substitute the `[NOBr_(2)]` in step (iii), `:. Rate = k_(1)k_(2)[NO]^(2)[Br_(2)] = K[NO]^(2)[Br_(2)]` |
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| 468. |
The rate constant for the decompoistion of a certain reaction is described by the equation: `log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)` Energy of activation (in `kcal`) isA. `57.6 kcal`B. `1.25 xx 10^(4) kcal`C. `14.0 kcal`D. `14 xx 10^(4) kcal` |
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Answer» Correct Answer - A `E_(a)//2.303RT = (1.25 xx 10^(4) K)/(T)` `:. E_(a) = 1.25 xx 10^(4) xx 2.303 xx 10^(-3) kcal` `= 57.6 kcal` |
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| 469. |
The rate constant for the first order decompoistion of a certain reaction is described by the equation `log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)` (a) What is the energy of activation for the reaction? (b) At what temperature will its half-life periof be `256 min`? |
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Answer» Step I. Calculation of activation energy `(E_(a))` According to Arrhenius equation , k=`Ae^(-Ea//RT)`. log k `= log A - (E_(a))/(2.303 RT)` The given equ. Is log k `= 14.34 - (1.25 xx 10^(4)(K))/(T)` On comparison: `E_(a)/(2.303 RT) = (1.25 xx 10^(4)(K))/(T)` `E_(a) = 1.25 xx 10^(4) xx 2.303 xx 8.314 (J mol^(-1))` `=239,339 J mol^(-1) = 239.339 kJ mol^(-1)` Step II. Calculation of desired temperature For the first order reaction , k `= 0.693/t_(1//2) = 0.693/(256 min) = 0.693/(256 xx 60s) = 4.51 xx 10^(-5) s^(-1)` According to Arrheneius theory, log k `= 14.34 - (1.25 xx 10^(4))/T` , `-4.35 = 14.34 -(1.25 xx 10^(4))/(18.69) = 669 K` |
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| 470. |
The rate constant for the first order decomposition of a certain reaction is described by the equation `log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)` (a) What is the energy of activation for the reaction? (b) At what temperature will its half-life period be `256 min`? |
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Answer» Arrhenius equation is given by , `k = Ae^(-E_(a) //RT)` `implies ` ln k = ln A `- (E_(a))/(RT)` `implies` ln k = log A `-(E_(a))/(RT)` `implies ` log k = log A `- (E_(a))/(2.303 RT) " " (i)` The given equation is log k = `14. 34 - 1.25 xx 10^(4) K//T " " (ii)` From equation (i) and (ii) , we obtain `(E_(a))/(2.303RT) = (1.25 xx 10^(4)K)/(T)` `implies E_(a) = 1.25 xx 10^(4) K xx 2.303 xx R` =` 1.25 xx 10^(4) K xx 2.303 xx 8.314 J K^(-1) mol^(-1)` = `239339.3 J mol^(-1)` (approximately) = `239. 34 kJ mol^(-1)` Also , when `t_(1//2)` = 256 minutes `k = (0.693)/(t_((1)/(2)))` = `(0.693)/(256)` = `2.707 xx 10^(-3) "min"^(-1)` =`4.51 xx 10^(-5) s^(-1)` It is also given that log k = `14.34 - 1.25 xx 10^(4) K//T` `implies` log `(4.51 xx 10^(-5)) = 14.34 - (1.25 xx 10^(4)K)/(T)` `implies "log" (0.654 - 05) = 14.34 - (1.25 xx 10^(4)K)/(T)` `implies (1.25 xx 10^(4)K)/(T) = 18. 686` `implies T = (1.25 xx 10^(4) K)/(18.686)` = 668. 85 K = 669 K (approximately) |
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| 471. |
Rate constant for the decomposition of ethylene oxide into `CH_(4)` and CO may be described be the equation. `logk(s)^(-1)=14.34-(1.25xx10^(4))/(T)` (a) What is the energy of activation of this reaction ? (b) What is the value of k at 670K ? |
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Answer» (a) We know that, `log_(10)k=log_(10)A-(E)/(2.3030RT)" "....(i)` `log _(10)(s^(-1))=14.34-(1.25xx10^(4))/(T)" "....(ii)` Comparing eq. (i) and eq. (ii), we get `(E)/(2.303R)=1.25xx10^(4)` `E=1.25xx10^(4)xx2.303xx8.314xx10^(-3)` `E=239.339KJ//mol` (b) Substituting the value of T, i.e., at 670 K , in equation (ii), we get, `logk(s^(-1))=14.34-(1.25xx10^(4))/(670)=-4.3167` `k=4.82xx10^(-5)s^(-1)` |
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| 472. |
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with `t_(1//2)=3.00hr.` What fraction of sample of sucrose remains after `8 hr` ? |
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Answer» For a first order reaction , `k = (2.303)/(t) "log" ([R]_(0))/([R])` It is given , `t_(1//2) = 3.00` hours `k = (0.693)/(t_((1)/(2)))` Therefore , = `(0.693)/(3) h^(-1)` = `0.231 h^(-1)` Then , `0.231 h^(-1) = (2.303)/(8h) "log" ([R]_(0))/([R])` `implies "log" ([R]_(0))/([R]) = (0.231 h^(-1) xx 8 h)/(2.303)` `implies ([R]_(0))/([R])`= antilog (`0.8024`) `implies ([R]_(0))/([R]) = 6.3445` `implies ([R])/([R]_(0)) = 0.1576` (approx) =`0.158` Hence , the fraction of sample of sucrose that remains after 8 hours is `0.158` |
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| 473. |
The half-life for a reaction at initial concentration of `0.5` and `1.0 "mole litre"^(-1)` are `200 sec` and `100 sec` respectively. The order of the reaction isA. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - C `t_(1/(2)) prop (1)/(a^(n-1`))" "(200)/(100) = ((1)/(0.5))^(n-1)` `2 = 2^(n-1) n-1=1" " n = 2` |
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| 474. |
For the zero order reaction : A `rarr` P, `K=10^(-2)("mol"//"litre")sec^(-1)` If initial concentration of A is `0.3` M, then find concentration of A left after 10 sec.A. 0 MB. `0.2` MC. `0.1` MD. `0.15` M |
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Answer» Correct Answer - B |
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| 475. |
What are the units for the rate constant of a zero-order reaction?A. timeB. `"time"^(-1)`C. M.timeD. `"M.time"^(-1)` |
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Answer» Correct Answer - D |
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| 476. |
For a first order reaction A `rarr` B, find `[(t_(7//8))/(t_(1//2))]`.A. 3B. 4C. 2D. 6 |
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Answer» Correct Answer - A |
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| 477. |
Which of the following isomerization reactions is/are of the first order?A. `"Cylopropane" rarr "Propane"`B. cis-But-`2`-ene`rarr`Trans-but-`2`-eneC. Vinyl allyl ether `rarr` Pent-`4`-enalD. `CH_(3)NCrarrCH_(3)CN` |
| Answer» Correct Answer - A::B::C::D | |
| 478. |
If half life period for a first order reaction in A is 2 minutes, how long will it take [A] to reach (i) `25%` of its initial concentration ii) `10%` of its initial concentration? |
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Answer» `k=0.693/t_(1//2) = 0.693/(2min) = 0.3465 min^(-1)` Ist case, `a=100%,(a-x)=25%`, `t=2.303/k loga/(a-x) = (2.303)/(0.3465 min^(-1)) xx log 100/25` `=2.303/(0.3465 min^(-1)) log 4 = (2.303 xx 0.602)/(0.3465 min^(-1)) = 4`min IInd case, `a=100%, (a-x)=10%`, `t=2.303/(0.3465 min^(-1)) log 10 = 6.65 min` |
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| 479. |
A first order reaction is `75%` complete in 60 min. Find the half life period of the reaction? |
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Answer» Step I: Calculation of rate constant, `a=100% (a-x) = 25%, t=60` min for first order reaction, `k=(2.303)/t log a/(a-x) = (2.303)/(a-x) log 100/25 = (2.303)/(60"min") xx log4 = (2.303 xx 0.6020)/(60 "min") = 0.0231 min^(-1)` Step-II: Calculation of half life period `(t_(1//2))` `t_(1//2) = 0.693/k = 0.693/(0.0231 min^(-1)) = 30` min. |
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| 480. |
For a zero order reaction. Which of the following statement is false:A. the rate is independent of the temperature of the reaction.B. the rate is independent of the concentration of the reactants.C. the half life is depends upon the concentration of the reactants.D. the rate constant has the unit `"mole" L^(-1) sec^(-1)`. |
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Answer» Correct Answer - A Rate `= k("conc.")` order here `k` depends on temperature. |
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| 481. |
The maximum value of activation energy is equal to:A. zeroB. heat of the reactionC. threshold energyD. none of these. |
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Answer» Correct Answer - D Activation energy may be greater than heat of reaction or lesser than threshold energy. |
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| 482. |
The gas phase decomposition of dimethyl ether follows first order kinetics. `CH_(3)-O-CH_(3)(g)rarrCH_(4)(g)+H_(2)(g)+CO(g)` The reaction is carried out in a constant volume container at `500^(@)C` and has a half life of `14.5 min`. Initially, only dimethyl ether is present at a pressure `0.40 atm`. What is the total pressure of the system after `12min`? (Assume ideal gas behaviour) |
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Answer» `{:(,CH_(3).OCH_(3(g)),rarr,CH_(4(g)),+,H_(2(g)),+,CO_((g))),("Initial P","0.40 atm",,,,,,),("Final P (at 12 min)",(0.40-P),,P,,P,,P):}` For ideal gas behaviour, mole `prop` Pressure, (at consatnt V and T) `:. a prop 0.40, (a-x) prop (0.40-P)` `:. K=K=2.303/t"log"_(10)a/(a-x)` `0.693/14.5=2.303/12"log"_(10)0.40/((0.40-P))` `:. P=0.175 atm` `:.` Pressure of ether decomposed= `0.175` Now, total pressure `=0.40-P+P+P+P` `=0.40+2P=0.40+2xx0.175` `=0.75 atm` |
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| 483. |
Which equation indicates the presence of a catalyst in the reaction?A. `A + B rarr D +B`B. `A+B rarr C+D`C. `A+A rarr D`D. `A rarr B+C` |
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Answer» Correct Answer - A |
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| 484. |
The rate constant, activation energy, and Arrphenius parameter of a chemical reaction are `3.0xx10^(-4)s^(-1), 104.4KJ mol^(-1)`, and `6.0xx10^(14)s^(-1)`, respectively. The value of rate constant as `Trarroo` isA. (a) `2.0xx10^(18) sec^(-1)`B. (b) `6.0xx10^(14) sec^(-1)`C. (c ) InifinityD. (d) `3.6xx10^(30) sec^(-1)` |
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Answer» Correct Answer - b `K=A.e^(-E_(a)//RT) [if Trarr oo, e^(-E_(a)//RT)]` or `K=A` |
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| 485. |
For which of the following reactions `k_(310)//k_(300)` would be maxiumum?A. `A+BrarrC , E_(a)=50 kJ`B. `X+YrarrZ , E_(a)=40 kJ`C. `P+QrarrR , E_(a)=60 kJ`D. `E+FrarrG , E_(a)=100 kJ` |
| Answer» Increase in rate constant is maximum for the reaction having maximum activation energy | |
| 486. |
The half-life for a reaction is ………. of temperature:A. IndependentB. Increased with increaseC. Decreased with increaseD. Dependent |
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Answer» Correct Answer - c Half-life depends upon rate constant and rate constant `(K)` varies with temperature. Since `K=A.e^(-E_(a)//RT)`, thus `K` increases with temperature. Also `t_(1//2) prop 1/K` |
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| 487. |
Combustion of carbon is exothermic, but coal stored in coal depots does not durn automatically because of:A. High threshold energy barrierB. Kinetic stability of coalC. Higher energy of ectivation needed for burningD. All of these |
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Answer» Correct Answer - d The higher threshold energy barrier prevents coal to burn spontaneously and provides kinetic stability to fuel. |
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| 488. |
The rate constant, activation energy, and Arrphenius parameter of a chemical reaction are `3.0xx10^(-4)s^(-1), 104.4KJ mol^(-1)`, and `6.0xx10^(14)s^(-1)`, respectively. The value of rate constant as `Trarroo` isA. `2.0xx10^(18)s^(-1)`B. `6.0xx10^(14)s^(-1)`C. `3.6xx10^(30)s^(-1)`D. None of these |
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Answer» Correct Answer - B Given `k= 3.0xx10^(-4)s^(-1), A= 6.0xx10^(14)s^(-1)` `E_(a)=104.4KJ mol^(-1)` Arrhenius equation is `k=Ae^(-E_(a)//RT)` as `Trarroo` `k=Ae^(-E_(a)//Rxxoo)oare^(0) [:.(-E_(a))/(Rxxoo)=0]` or `k=A` `:. k = 60 xx10^(14)s^(-1)` as `Trarroo` |
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| 489. |
For which of the following reactions `k_(310)//k_(300)` would be maximum?A. `A+Brarr, E_(a)= 500kJ`B. `X+YrarrZ, E_(a)= 40kJ`C. `P+QrarrR, E_(a)= 600kJ`D. `E+FrarrG, E_(a)= 100kJ` |
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Answer» Correct Answer - D Increase in the rate constant is maximum for the reaction having maximum activation energy. |
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| 490. |
Two reactions `A rarr` products and `B rarr` products have rate constants `k_(a)` and `k_(b)` respectively at temperature `T` and activation energies are `E_(a)` and `E_(b)` respectively. If `k_(a) gt k_(b)` and `E_(a) lt E_(b)` and assuming the freuency factor `A` in both the reactions are same thenA. on increasing temperature `k_` will be greater than `k_(b)`B. At lower temperature `k_(a)` and `k_` will differ moreC. As temperature rises `k_(a)` and `k_(b)` will be close to each other in magnitudeD. All of the above are correct |
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Answer» Correct Answer - D ln `k_(a) = A - E_(a)//RT`, ln `k_(b) = A - E_(b)//RT` As temperature rises `E//RT` become negligible. |
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| 491. |
Conisder a reaction `A rarr B + C` . The initial concentration of `A` was reduced form `2 M `to `1 M` in `1 h` and form `1M` to `0.25 M` in `2h`, the order of the reaction isA. `1`B. `0`C. `2`D. `3` |
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Answer» Correct Answer - A (a) Half life of the given reaction is independent of initial concentration, hence it is a first order reaction. `[t_(1//2)prop(1)/(a^(n-1))]` |
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| 492. |
Two reaction : `X rarr` Products and `Y rarr` Products have rate constants `k_(1)` and `k_(2)` at temperature `T` and activation energies `E_(1)` and `E_(2)`, respectively. If `k_(1) gt k_(2)` and `E_(1) lt E_(2)` (assuming that the Arrhenius factor is same for both the Products), then (I) On increaisng the temperature, increase in `k_(2)` will be greater than increaisng in `k_(1)`. (II) On increaisng the temperature, increase in `k_(1)` will be greater than increase in `k_(2)`. (III) At higher temperature, `k_(1)` will be closer to `k_(2)`. (IV) At lower temperature, `k_(1) lt k_(2)`A. At higher temperature `K_(A)` will be greater than `K_(B)`B. At lower temperature `K_(A)` and `K_(B)` will differ more and `K_(A) gt K_(B)`C. As temperature rises `K_(A)` and `K_(B)` will be close to each other in magnitudeD. All of these |
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Answer» Correct Answer - d `K_(A)=Ae^(-E_(A)//RT)` and `K_(B)=Ae^(-E_(B)//RT)` Also `K_(A) gt K_(B)` `E_(A) lt E_(B)` Now notice that all the given facts are satisfied. |
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| 493. |
Two reaction : `X rarr` Products and `Y rarr` Products have rate constants `k_(1)` and `k_(2)` at temperature `T` and activation energies `E_(1)` and `E_(2)`, respectively. If `k_(1) gt k_(2)` and `E_(1) lt E_(2)` (assuming that the Arrhenius factor is same for both the Products), then (I) On increaisng the temperature, increase in `k_(2)` will be greater than increaisng in `k_(1)`. (II) On increaisng the temperature, increase in `k_(1)` will be greater than increase in `k_(2)`. (III) At higher temperature, `k_(1)` will be closer to `k_(2)`. (IV) At lower temperature, `k_(1) lt k_(2)`A. IB. IIC. I, IIID. I, III, IV |
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Answer» Correct Answer - B Since `k_(2) lt k_(2)` and `E_(2) gt E_(1)` form Arrehenius equation `(k = Ae^(-E_(a)//RT))`, if `k_(1)` is high then `E_(1)` is low. Therefore, with increase in temperature increase in `k_(1)` will be greater than increase in `k_(2)`. |
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| 494. |
The rate constants `k_(1)` and `k_(2)` of two reactions are in the ratio `2:1`. The corresponding energies of acativation of the two reaction will be related byA. `E_(1) gt E_(2)`B. `E_(1) gt E_(2)`C. `E_(1)=E_(2)`D. `E_(1)=2E_(2)` |
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Answer» Correct Answer - B `k_(1)//k_(2)= 2:1 or k_(1)=2k_(1), i.e., k_(1) gt k_(2)`. Uisng Arrhenius eqaution, `k=Ae^(-E_(a)//RT)` Therefore, if `k` is high `E_(a)` is low. Hence, `E_(2) gt E_(1)` |
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| 495. |
For a reaction, `E_(a)= 0 and k= 3.2xx10^(4)s^(-1)` at `300 K`. The value of `k` at `310 K` would beA. `6.4 xx 10^(4) s^(-1)`B. `3.2 xx 10^(4) s^(-1)`C. `3.2 xx 10^(8) s^(-1)`D. `3.2 xx 10^(5) s^(-1)` |
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Answer» Correct Answer - B `k = Ae^(-E_(a)//RT)` When `E_(a) = 0, k = A =` constant `:. k_(310) = k_(300) = 3.2 xx 10^(4) s^(-1)`. |
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| 496. |
For a reaction, `E_(a)= 0 and k= 3.2xx10^(4)s^(-1)` at `300 K`. The value of `k` at `310 K` would beA. `6.4xx10^(4)s^(-1)`B. `3.2xx10^(8)s^(-1)`C. `3.2xx10^(4)s^(-1)`D. `3.2xx10^(5)s^(-1)` |
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Answer» Correct Answer - C `k=Ae^(-E_(a)//RT)` When `E_(a)=0, k=A("Constant")` `k_(310)=k_(300)= 3.2xx10^(4)s^(-1)` |
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| 497. |
When the temperature of a reaction is raised from 300 K to 310 K, the reaction rate doubles. Determine the activation energy , `E_(a)` associated with the reaction.A. 6.45 KJ `"mol"^(-1)`B. 23.3 KJ `"mol"^(-1)`C. 53.58 KJ `"mol"^(-1)`D. 178 KJ `"mol"^(-1)` |
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Answer» Correct Answer - C |
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| 498. |
Statement: The elementary reaction is single step reaction and does not possess mechanism. Explanation: An elementry reaction has order of reaction and molecularity same.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 - C Explanation is correct reason for statement. |
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| 499. |
For a certain reaction, a plot of `([C_(0)-C])/(C )` against time `t`, yields a straight line. `C_(0)` and C are concentrations of reaction at `t=0` and `t=t` respectively. The order of reaction is:A. `3`B. ZeroC. `1`D. `2` |
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Answer» Correct Answer - d For II order, `(-dC_(A))/(dt)=K[C_(A)]^(2)` On interating `1/C_(A)=Kt+ constt`., `(at t=0, C_(A)= C_(0))` `1/C=Kt+ 1/C_(0)` `((C_(0)-C))/(C_(0)-C)=Kt` or `([C_(0)-C])/C=K.C_(0).t, (y=mx+0)` |
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| 500. |
Statement-1: For the reaction `2A +B to C`, the rate of disappearance of A is twice the rate of disappearance of A is twice the rate of disappearance of B. Statement-2: For the reaction : `2A +B to C` Rate of reactions is `(d[C])/(dt)`A. Statement-1 is True, statement-2 is True, Statement-2 is a correct explantion for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - b |
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