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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.
| 251. |
Catalyst decomposition of hydrogen peroxide is a….. Order reactionA. FirstB. SecondC. ThirdD. Zero |
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Answer» Correct Answer - A It is a first order reaction as is clear from the law expression, `r = k(H_(2)O_(2))` |
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| 252. |
In the reaction, `P+Q rarr R+S` the time taken for `75%` reaction of `P` is twice the time taken for `50%` reaction of `P`. The concentration of `Q` varies with reaction time as shown in the figure. The overall order of the reaction is A. `2`B. `3`C. `0`D. `1` |
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Answer» Correct Answer - D Overall order reaction can be decided by the data given `t_(75%) = 2t_(50%)` `:.` It is a first order reaction with respect to `P`. form graph `[Q]` is linearly decreaisng with time, i.e., order of reaction w.r.t. `Q` is zero and the rate expresison is `r = [P]^(1)[Q]^(0)`. Hence (d) is correct.Correct Answer - D Overall order reaction can be decided by the data given `t_(75%) = 2t_(50%)` `:.` It is a first order reaction with respect to `P`. form graph `[Q]` is linearly decreaisng with time, i.e., order of reaction w.r.t. `Q` is zero and the rate expresison is `r = [P]^(1)[Q]^(0)`. Hence (d) is correct. |
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| 253. |
`2N_(2)H_(4)(g)+N_(2)O_(4)(g)rarr3N_(2)(g)+4H_(2)O(g)` If `N_(2)H_(4)(g)` disappears at a rate of `0.12` mol L min, at what rate does `N_(2)(g)` appear?A. `0.080"mol.L"^(-1)."min"^(-1)`B. `0.12"mol.L"^(-1)."min"^(-1)`C. `0.18"mol.L"^(-1)."min"^(-1)`D. `0.30"mol.L"^(-1)."min"^(-1)` |
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Answer» Correct Answer - C |
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| 254. |
The rate of decomposition of hydrogen peroxide is first order in `H_(2)O_(2)`.AT `[H_(2)O_(2)]=0.150` M , the decomposition rate was measured to be `4.83xx10^(-6)M.s^(-1)` . What is the rate constant for the reaction?A. `2.15xx10^(-4)s^(-1)`B. `3.22xx10^(-5)s^(-1)`C. `4.83xx10^(-6)s^(-1)`D. `7.25xx10^(-7)s^(-1)` |
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Answer» Correct Answer - B |
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| 255. |
For the non-stoichiometric reaction `2A+BrarrC+D` The following kinetic data were obtained in theee separate experiment, all at `98 K` `|{:("Initial concentration (A)","Initial concentration (B)","Initial rate of formation of C" (molL^(-1) s^(-1))),(0.01 M,0.1 M,1.2 xx 10^(-3)),(0.1 M,0.2 M,1.2 xx 10^(-3)),(0.2 M,0.1 M,2.4 xx 10^(-3)):}|` The rate law for the formation of `C` is:A. `(dc)/(dt)=K[A][B]^(2)`B. `(dc)/(dt)=k[A]`C. `(dc)/(dt)=k[A][B]`D. `(dc)/(dt)=k[A]^(2)[B]` |
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Answer» Correct Answer - B (I) `R = k[A]^(x)[B]^(y)` `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` …(i) `1.2 xx 10^(-3) = k[0.1]^(x)[0.2]^(y)` …(ii) Divide Eq. (ii) by Eq. (i) `1 = (2)^(y)` `:. (2)^(0) = (2)^(y) :. y = 0` (II) `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` `= k[0.1]^(x)` , `(because y = 0)` ...(iii) `2.4 xx 10^(-3) = k[0.2]^(x)[0.1]^(0)` ...(iv) Divide Eq. (iv) by Eq. (iii) `(2)^(1) = (2)^(x) :. x = 1`. `R = k[A]` Hence answer (b) is correct.Correct Answer - B (I) `R = k[A]^(x)[B]^(y)` `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` …(i) `1.2 xx 10^(-3) = k[0.1]^(x)[0.2]^(y)` …(ii) Divide Eq. (ii) by Eq. (i) `1 = (2)^(y)` `:. (2)^(0) = (2)^(y) :. y = 0` (II) `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` `= k[0.1]^(x)` , `(because y = 0)` ...(iii) `2.4 xx 10^(-3) = k[0.2]^(x)[0.1]^(0)` ...(iv) Divide Eq. (iv) by Eq. (iii) `(2)^(1) = (2)^(x) :. x = 1`. `R = k[A]` Hence answer (b) is correct. |
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| 256. |
For the reaction, `2A+BrarrC` which relationship is correct?A. `Delta[A]=Delta[C]`B. `-Delta[A]=Delta[C]`C. `-2Delta[A]=Delta[C]`D. `=Delta[A]=2Delta[C]` |
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Answer» Correct Answer - D |
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| 257. |
At T (K) , if the rate constant of a first order reaction is `4.606xx10^(-3)s^(-1)`, the time to reduce the initial concentration of the reactant to `(1)/(10)` in seconds is :A. 500B. 1000C. 100D. 50 |
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Answer» For first order reaction : `k=(2.303)/(t)log.([R_(0)])/([R_(t)])` `4.606xx10^(-3)=(2.303)/(t)log.((1)/(1//10))` `t=(2.303)/(4.606xx10^(-3))=500 sec` |
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| 258. |
The rate of reaction can be expressed by Arrhenius equation `R=Ae^(-K//RT)` . In this equation. E representsA. the energy below which all the colliding molecules will reactB. the energy below which colliding molecules will not reactC. The total energy of the reacting molecules at a temperature T.D. The fraction of the molecules with energy energy greater than the activation energy of the reactants. |
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Answer» Correct Answer - B b) It is the correct answer. |
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| 259. |
What will be the half-life of the first order reaction for which the value of rate constant is `200 s^(-1)`?A. `3.46 xx 10^(-2) x`B. `3.46 xx 10^(-3) s`C. `4.26 xx 10^(-2) s`D. `4.26 xx 10^(-3) s` |
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Answer» Correct Answer - B `t_(1//2) = (0.693)/(k) = (0.693)/(200) = 3.46 xx 10^(-3) s` |
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| 260. |
The half life period of a first reaction is 1 min 40 seconds. Calulate its constantA. `6.93xx10^(-3) "min"^(-1)`B. `6.93xx10^(-3) "sec"^(-1)`C. `6.93xx10^(-3) sec`D. `6.93xx10^(-3) sec` |
| Answer» `k=(0.693)/(t_(1//2))=(0.693)/(100)=6.93xx10^(-3) sec^(-1)` | |
| 261. |
The rate equation for the reaction `2A+B rarr C` is found to be: `rate = k[A][B]`. The correct statement in relation of this reaction is thatA. unit of k must be `sec^(-1)`B. `t_(1//2)` is constantC. rate of formation of C is twice the rate of disappearance of AD. value of k is independent of initial concentrations of A and B |
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Answer» Correct Answer - C |
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| 262. |
The rate constant of first order reaction is `10^(-2)"min"^(-1)`. The half-life period of reaction isA. 69.3 minB. 34.65 minC. 17.37 minD. 3.46 min |
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Answer» Correct Answer - B `t_(1//2) = (0.693)/(k) = (0.693)/(2 xx 10^(-2)) = 34.65` min |
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| 263. |
For reaction, A `rarr` B, the rate law is rate = k[A]. If the reaction is `40.0%` complete after `50.0` minutes, what is the value of the rate constant, k?A. `8.00xx10^(-3)"min"^(-1)`B. `1.02xx10^(-2)"min"^(-1)`C. `1.39xx10^(-2)"min"^(-1)`D. `1.83xx10^(-2)"min"^(-1)` |
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Answer» Correct Answer - B |
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| 264. |
A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will:A. remain unchangedB. be tripledC. increase by a factor of 4D. be doubled |
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Answer» Correct Answer - C |
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| 265. |
A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will:A. DoubleB. Remain unchahgedC. TripleD. Increase by a factor of 4 |
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Answer» Correct Answer - D `r_(0) = K[CO]^(2)` `r_(1) = K[2CO]^(2)` `:. r = 4r_(0)` |
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| 266. |
The rate constant of a first order reaction is doubled when the temperature is increased from `20^(@)C` to `25^(@)C` . How many times the rate constant will increase if the temperature is raised from `20 ^(@)C ` to `40^(@)C`A. 4B. 8C. 16D. 32 |
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Answer» Correct Answer - c `5^(@)C` rise causes increase of 2 time . `20^(@)C` will cause `(2)^(4) = 16` times . |
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| 267. |
In a first order reaction, The concentration of reactant is reduced to 1/8th of the initial concentration in 75 minutea at 298 K. What is the half period of the reaction in minutes?A. 50 minB. 15 minC. 30 minD. 25 min |
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Answer» Correct Answer - D Let a = 1, a - x = 1/8 th, t == 75 min `k = (2.303)/(t) "log" (a)/(a - x) = (2.303)/(75) "log" (1)/(1//8) = (2.303 xx 0.903)/(75) "min"^(-1)` for first order reaction, `t_(1//2) = (0.693)/(k) = (0.693 xx 75)/(2.303 xx 0.903) = 25` min |
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| 268. |
In the following reaction , the initial concentration of the reactant and initial rate at 298 K are given `2 A to C + D ` `{:([A]_(0) "mol" L^(-1) ,, "Initial rate in mol" L^(-1) s^(-1)), (0.01 ,, 5. 0 xx 10^(-5)), (0.02 ,, 2xx 10^(-4)):}` The value of rate constant of this reaction at 298 K isA. `0.01 s^(-1)`B. `5 xx 10^(-3) mol L^(-1) s^(-1)`C. `2.0 xx 10^(-2) mol^(-1) L s^(-1)`D. `5 xx 10^(-1) mol^(-1) L s^(-1)` |
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Answer» Correct Answer - d Let the order with respect to A is `alpha` . Rate = k `[A]^(alpha)` `5 xx 10^(-5) = k[0.01]^(alpha) " " … (i)` `2 xx 10^(-4) = k[0.02]^(alpha) " "… (ii)` By dividing eqn. (ii) by (i) , `(2xx 10^(-4))/(5 xx 10^(-5)) = (k[0.02]^(alpha))/(k[0.01]^(alpha)) implies 4 (2)^(alpha)` i.e., `alpha = 2` Substituting the values of `alpha` in eq. (i) , `5 xx 10^(-5) = k[0.1]^(2) implies k = (5 xx 10^(-5))/(10^(-4)) = 5 xx 10^(-1) mol^(-1) L s^(-1)` |
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| 269. |
Which one of the following formula represents a first-order reaction?A. `K = (x)/(t)`B. `K = (1)/(2t)[(1)/((a - x)^(2)) - (1)/(a^(2))]`C. `K = (2.303)/(t)"log"_(10)(a)/((a - x))`D. `K = (1)/(t) (x)/(a(a - x))` |
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Answer» Correct Answer - C For Ist-order reaction `K = (2.303)/(t)"log"_(10) (a)/((a-x))` |
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| 270. |
A reaction was found to be of second order with respect to concentration of carbon monoxide. If the concentration of carbon monoxide is doubled with everything else keep the same, the rate of reaction willA. remain unchangedB. become tripleC. increase by a factor 4D. become double |
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Answer» Correct Answer - C c) rate = `k[CO]^(2)` By doubling the concentration of carbon monoxide (CO),the rate of reaction will become four times. |
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| 271. |
In a I order reaction `A rarr` products, the concentration of the reactant decrease to `6.25%` of its initial value in `80` minutes. What is (i) the rate constant and (ii) the rate of the reaction, `100` minutes after the start, if the initial concentration is `0.2 "mole"//"litre"` ?A. `2.17xx10^(-2)"min"^(-1),3.47xx10^(-4)"mol.litre"^(-1)"min"^(-1)`.B. `3.465xx10^(-2)"min"^(-1),2.166xx10^(-4) "mol.litre"^(-1) "min"^(-1)`.C. `3.465xx10^(-3)"min"^(-1),2.17xx10^(-3)"mol.litre"^(-1)"min"^(-1)`.D. `2.166xx10^(-3)"min"^(-1),2.667xx10^(-4)"mol.litre"^(-1)"min"^(-1)`. |
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Answer» Correct Answer - B |
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| 272. |
In a first order reaction , the concentration of the reactant is reduced to 1/8 of the initial concentration in 75 minutes at 298 K . What is the half-life period of the reaction in minutesA. 50B. 15C. 45D. 25 |
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Answer» Correct Answer - d Let a = 1 , a-x = 1/8 , t = 75 min For first order reaction `k = (2.303)/(t)` log `(a)/(a-x)` `k = (2.303)/(75)` log `(1)/(1//8) = (2.303 xx 0.903)/(75) "min"^(-1)` For first order reaction `t_(1//2) = (0.693)/(k) = (0.693 xx 75)/(2.303 xx 0.903) = 25` min . |
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| 273. |
Rate constant for a reaction is `10^(-3) S^(-1)`. How much time is required to reduce the initial concentration of reactant to 25%A. 693 sB. 1386 sC. 6930 sD. 2029 s |
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Answer» Correct Answer - B `t=(2.303)/(K)"log"(a)/(a-x)` where," "k=rate constant=`10^(-3)s^(-1)` a=I"initial amount "=100 a-x= amount left after time, t =25 t=time required to reduce the initial concentration of reactant to 25% reaction `therefore " " t=(2.303)/(10^(-3))"log"(100)/(25)=(2.303)/(10^(-3))`"log" 4 =` (2.303xx0.6020)/(10^(-3))=1386s` |
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| 274. |
Rate of a reaction can be expressed by Arrhenlus equation, `k=Ae^(-E_(a)//RT)`. Here, E isA. the total energy of the reacting molecules at a temperature TB. the energy above which all the colliding molecules will reactC. the energy below which colliding molecules will not reactD. the fraction of molecules with energy greater than the activation energy of the reaction |
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Answer» Correct Answer - C In the Arrhenius equation for the rate of reaction, E is the energy below which colliding molecules will not react. |
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| 275. |
In a I order reaction `A rarr` products, the concentration of the reactant decrease to `6.25%` of its initial value in `80` minutes. What is (i) the rate constant and (ii) the rate of the reaction, `100` minutes after the start, if the initial concentration is `0.2 "mole"//"litre"` ?A. `2.17 xx 10^(-2) "min"^(-1), 3.47 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)`B. `3.465 xx 10^(-2) "min"^(-1), 2.166 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)`C. `3.465 xx 10^(-3) "min"^(-1), 2.17 xx 10^(-3) "mol.litre"^(-1) "min"^(-1)`D. ( d) `2.166 xx 10^(-3) "min"^(-1), 2.667 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)` |
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Answer» Correct Answer - B `K = (2.303)/(80)"log"((100)/(6.25)) = 3.465 xx 10^(-2)mm^(-1)` Then `3.465 xx 10^(-2) = (2.303)/(100)"log"((0.2)/(a_(t)))` `a_(t) = 0.00625` Rate `= K xx [a_(t)]` `= 0.00625 xx 3.465 xx 10^(-2)` `= 2.166 xx 10^(-4) sec^(-1)` |
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| 276. |
For a first-order reaction `A rarr B` the reaction rate at reactant concentration of `0.10 M` is found to be `2.0 xx 10^(-5) "mol" L^(-1) s^(-1)`. The half-life period of the reaction isA. 30 sB. 220sC. 300sD. 347 s |
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Answer» Correct Answer - D d) `A to B` rate of reaction = `2 xx 10^(-5) mol L^(-1)s^(-1)` Order of reaction (n) =1 rate = `k[A]^(n) = k[A]` `k=("rate")/[A]=(2 xx 10^(-5) mol L^(-1)s^(-1))/(0.01 mol L^(-1)) = 2 xx 10^(3) s^(-1)` `t_(1//2) = 0.693/k = 0.693/(2 xx 10^(-2)s^(-1)) = 347s` |
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| 277. |
For a first-order reaction `A rarr B` the reaction rate at reactant concentration of `0.10 M` is found to be `2.0 xx 10^(-5) "mol" L^(-1) s^(-1)`. The half-life period of the reaction isA. `220 s`B. `30 s`C. `300 s`D. `347 s` |
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Answer» Correct Answer - D `R = K[A]` `2 xx 10^(-5) = underline(K) xx 10^(-2)` `K = 2 xx 10^(-3) sec^(-1)` `t_(1//2) = (0.693)/(K) = (.693)/(2 xx 10^(-3)) = 347 sec` |
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| 278. |
Compounds A and B react according to the following chemical equation : `A(g) + 2B(g) rarr2C(g)` Concentration of either A or B were changed keeping the concentrations of one of the reactant constants and rates were measured as a function of initial concentration, Following results were obtained. Choose the correct option for the rate equations for this reaction.A. Rate = `k[A]^(2)[B]`B. Rate = `k[A][B]^(2)`C. Rate = `k[A][B]`D. Rate = `k[A]^(2)[B]^(0)` |
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Answer» Correct Answer - B Let Rate law, `r=k[A]^(x)[B]^(y)" "....(i)` On putting values, we get `0.10 = K [0.30]^(x)[0.30]^(y) " "....(ii)` `0.40 = K[0.30]^(x)[0.60]^(y)" "....(iii)` `0.20=K[0.60]^(x)[0.30]^(y)" "....(iv)` From Eqs. (ii) and (iii), we get `(0.1)/(0.4)=[(0.30)/(0.30)]^(x)[(0.30)/(0.60)]^(y)` `(1)/(4)=[(1)/(2)]^(y)implies y = 2` From Eqs. (iii) and (iv), we get `(0.40)/(0.20)=((0.30)/(0.60))^(x)((0.60)/(0.30))^(y)` `2=((1)/(2))^(x)(2)^(y)` On putting the values of y, we get `2=((1)/(2))^(x)(2)^(2)` `implies " "x=1` On putting the values of x and y in Eq. (i), we get `r=k[A]^(1)[B]^(2)=k[A][B]^(2)` |
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| 279. |
The rate constant for the reaction `A rarr B` is `2 xx 10^(-4) mol^(-1)`. The concentration of `A` at which rate of the reaction is `(1//2) xx 10^(-5) M sec^(-1)` is -A. `0.25 M`B. `(1//20) sqrt(5//3) M`C. `0.5 M`D. None of these |
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Answer» Correct Answer - B Order of reaction is `2` (from unit of `K`) `:. rate = K C_(A)^(2)` `C_(A) = sqrt((1//12 xx 10^(-5))/(2 xx 10^(-4))) = sqrt((1)/(24) xx 10^(-1)) = sqrt((1)/(240))` `= sqrt((5)/(1200)) = (1)/(20) sqrt((5)/(3))` |
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| 280. |
Select the law that corresponds to data shown for the following reaction `A+B rarr` Products `{:(Exp,[A],[B],"Initial rate",),(1,0.012,0.035,0.1,),(2,0.024,0.070,0.8,),(3,0.024,0.035,0.1,),(4,0.012,0.070,0.8,):}`A. Rate = `k[B]^(3)`B. Rate = `k[B]^(4)`C. Rate = `k[A][B]^(3)`D. Rate = `k[A]^(3)[B]` |
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Answer» Correct Answer - A It is seen that, in experiments (3) and (2), [A] is constant and [B] is doubled and rates becomes 8 times, so order with respect to [B] = 3. In experiments (1) and (3), [B] is constant and [A] is doubled, but rate does not change, so order with respect to [A] = 0 Thus, rate = `k[B]^(3)` |
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| 281. |
For the reaction X-Y, the concentration of X are `1.2` M, `0.6`M, order of reaction is:A. ZeroB. HalfC. oneD. two |
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Answer» Correct Answer - C c) `underset("Initialtime")1.2M to underset("1 hr")0.6M to underset(2 hr)0.3M to underset(3 hr)0.15 M` `t_(1//2) = 1hr`. For the nth order reaction. `(t_(1//2))_(1)/(t_(1//2))_(2) = ([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1))` `1=(6/3)^(n-1)` or `(2)^(0) = (2)^(n-1)` n-1=0 or n=1 Order of reaction is one. |
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| 282. |
`t_(1//4)` can be taken as the time taken for concentration of reactant to drop to `.^(3)//_(4)` of its initial value. If the rate constant for a first order reaction is `K`, then `t_(1//4)` can be written as:A. `0.10//K`B. `0.29//K`C. `0.69//K`D. `0.75//K` |
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Answer» Correct Answer - B `K = (2.303)/(t)"log"(a)/(a-x)` `= (2.303)/(t_(1//4))"log"(a)/(3a) = (2.303)/(t_(1//4))"log"(4)/(3)` `K = (2.303 xx 0.125)/(t_(1//4)) = (0.29)/(t_(1//4))` |
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| 283. |
The reaction `[Co (NH_(3))_(5)Br]^(2) + (H_(2)O) rarr [Co(NH_(3))_(5)(H_(2)O)]^(3+) + Br^(-)` is followed by measuring a property of the solution known as the optical density of the solution which may be taken to be linearly related to the concentration of the reactant. The values of optical density are `0.80, 0.35` and `0.20` at the end 0f `20` minutes, `40` minutes and infinite time after the start of the reaction which is first order. Calculate the rate constant.A. `6.93 xx 10^(-3) "min"^(-1)`B. `3.51 xx 10^(-3) "min"^(-1)`C. `6.93 xx 10^(-2) "min"^(-1)`D. `3.51 xx 10^(-3) "min"^(-1)` |
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Answer» Correct Answer - C `(2.303)/((40 - 20))"log"((0.80 - 020))/((0.35 - 0.20)) = 6.93 xx 10^(-2)` |
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| 284. |
`60%` of a first order reaction was completed in `60 min`. The time taken for reactants to decompose to half of their original amount will beA. `45` minB. `60` minC. `40` minD. `50` min |
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Answer» Correct Answer - A `60 = (2.303)/(K) "log" (100)/(40)` `t = (2.303)/(K)"log"(100)/(50)` `:. (60)/(t) = (0.40)/(0.310)` `:. t = 45` minutes |
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| 285. |
In a second order reaction, when the concentration of both the reactants are equal, the reaction is completed `20%` in 500 s. How long would it take for the reaction to go to `60%` completion?A. 3000sB. 5000sC. 1000 sD. 2000s |
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Answer» Correct Answer - A a) For the second order reaction, `k=1/(t xx a). (X/(a-x))` According to available data, in the first case t=500s , a=100 a-x `=80%` of 100 = 80 `k=(1 xx 20)/(500s xx 100 xx 80)= 5 xx 10^(-6)s^(-1)` In the second case, a=100 `(a-x) = 40%` of 100 = 40 `t=(1 xx 60)/((5 xx 10^(-6)s^(-1)) xx (100 xx 40))=3000s`. |
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| 286. |
……………………… reaction obeys the expresison `t_(1//2) = 1//ka` in chemical kinetics.A. firstB. secondC. thirdD. fourth |
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Answer» Correct Answer - B `T_(1//2)` for nth order reaction `= (2^(n-1) - 1)/((n-1)k(a)^(n-1))` On subsitituting `n = 2`, we obtain `T_(1//2) = (2-1)/(1.k.a) = (1)/(k a)` Therefore, the order of the reaction is `2`. |
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| 287. |
The initial concentration of both the reactants of a second order reaction are equal and `60%` of the reaction gets completed in `30s`. How much time will be taken in `20%` completion of the reaction? |
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Answer» Correct Answer - 5 For second order: `k_(2) = (1)/(t).(x)/(a(-x))` , [Let `a=1`] `= (1)/(30 s) xx (0.6)/(1(1-0.6)) = (1)/(30) xx (0.6)/(0.4)` Now for `20%` completion, `k_(2) = (1)/(t).(x)/(a(1-x))` (Since `k_(2)` is constant) `(1)/(30) xx (0.6)/(0.6) = (1)/(t) xx (1)/(4)` `t = (30)/(0.6) xx (0.4)/(4) = 5s`Correct Answer - 5 For second order: `k_(2) = (1)/(t).(x)/(a(-x))` , [Let `a=1`] `= (1)/(30 s) xx (0.6)/(1(1-0.6)) = (1)/(30) xx (0.6)/(0.4)` Now for `20%` completion, `k_(2) = (1)/(t).(x)/(a(1-x))` (Since `k_(2)` is constant) `(1)/(30) xx (0.6)/(0.6) = (1)/(t) xx (1)/(4)` `t = (30)/(0.6) xx (0.4)/(4) = 5s` |
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| 288. |
……………………… reaction obeys the expresison `t_(1//2) = 1//ka` in chemical kinetics.A. `1`B. `0`C. `3`D. `2` |
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Answer» Correct Answer - D `t_(1//2) = (1)/(Ka)` for second order reactions. `t_(1//2) = (1)/(K(a)) rArr t_(1//2) prop (1)/([a]^(1))` `rArr t_(1//2) prop [a]^(-1)` Also, `t_(1//2) prop [a]^(1-n)` `[a]^(n-1) = [a]^(-1) rArr 1 - n = -1` `n = 2` |
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| 289. |
Following is the graph between `log T_(50)` and `log a` (`a =` initial concentration) for a given reaction at `27^(@)C`. Hence order is A. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - D `t_(1//2)prop((1)/(a))^(n-1)" or "t_(1//2)=k(a)^(1-n)` `logt_(1//2)=logk+(1-n)loga` (It represents straight line equation, y = c + mx) Slope `=(1-n)=tan45^(@)=1` `(1-n)=1implies n=0` |
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| 290. |
Following is the graph between `log T_(50)` and `log a` (`a =` initial concentration) for a given reaction at `27^(@)C`. Hence order is A. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - A `t_(1//2)prop((1)/(a))^(n-1) or t_(1//2)=k(a)^(1-n)` `log.t_(1//2)=logk+(1-n)log a (y=c+mx)` Slope `=(1-n)=tan 45=1` `:. (1-n)=1impliesn=0` |
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| 291. |
An organic compound undergoes first decompoistion. The time taken for its decompoistion to `1//8` and `1//10` of its initial concentration are `t_(1//8)` and `t_(1//10)`, respectively. What is the value of `([t_(1//8)])/([t_(1//10)]) xx 10`? `(log_(10)2 = 0.3)`A. 2B. 3C. 3D. 9 |
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Answer» Correct Answer - D For a first order process, `kt="In[A]_(0)/([A])` `rArrkt_(1//8)=In[A]_(0)/([A]_(0)//8)="In "8` `and" "kt_(1//10)="In"[A]_(0)/([A]_(0)//10)="In "10` …..(ii) Therefore, " `(t_(1//8))/(t_(1//8))= ("In 8")/("In 10")=log8 = 3 log 2 = 3 xx 0.3= 0.9 ` ` rArr " "t_(1//8)/t_(1//10)xx 10 xx 0.9 xx 10 = 9.0 ` |
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| 292. |
Two substances `A (t_(1//2) = 5 min)` and `B(t_(1//2) = 15 min)` follow first order kinetics and are taken in such a way that initially `[A] = 4[B]`. The time after which the concentration of both the substance will be equal is `5x min`. Find the value of `x`. |
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Answer» Correct Answer - 3 `|{:([A],[B],"Time (min)",),(4a,a,0,),(2a,-,5 min,),(a,-,10 min,),(a//2,a//2,15 min,):}|` `:. 5x = 15 min` `x = 3 min`Correct Answer - 3 `|{:([A],[B],"Time (min)",),(4a,a,0,),(2a,-,5 min,),(a,-,10 min,),(a//2,a//2,15 min,):}|` `:. 5x = 15 min` `x = 3 min` |
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| 293. |
A first order reaction has `t_(1//2) = 6.93 min`. The rate constant is……….. . |
| Answer» Correct Answer - `0.1 min^(-1)` | |
| 294. |
Conisder the following statement for a second order reaction and score of each statement. `2A rarr P` `|{:(,,"Score"),(a.,[A]=([A_(0)])/(1+kt),3),(b.,"A plot of" 1//[A]^(2) "vs time will be straight line",2),(c.,"Half life is long when the concentration is low",1):}|` Find the total score of the correct statements. |
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Answer» Correct Answer - 4 Statement (a) and ( c) are correct. So total score is `3+1 = 4`.Correct Answer - 4 Statement (a) and ( c) are correct. So total score is `3+1 = 4`. |
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| 295. |
Following are two first order reactions with their half times given at `25^(@)C` `A overset(t_(1//2) = 30 "min")(to)` Products `B overset(t_(1//2) = 40 "min") (to) ` Products The temperature coefficients of their reaction rates 3 and 2 , respectively , between `25^(@)C` and `35^(@)C` . if the above two reactions are carried out taking `0.4`M of each reactant but at different temperatures `25^(@)C` for the first reaction and `35^(@)C` for the second reaction , find the ratio of the concentration of A and B after an hour |
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Answer» Correct Answer - 2 0.4M of `A overset(30 "min")(to) 0.2 M overset(30 "min")(to) 0.1 M ` (for A) 0.4 M of `B overset(20 "min")(to) 0.2M overset(20 "min") (to) 0.1 M overset(20 "min")(to) 0.05` (for B) `"("(-d[B])/(dt)` will be doubled and hence `t_(1//2)` will be halved) `therefore ([A])/([B]) = 2` |
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| 296. |
Following are two first order reaction with their half times given at `25^(@)C`. `A overset(t_(1//2) = 30 min)rarr` Products `B overset(t_(1//2)=40 min)rarr` Products The temperature coefficients of their reactions rates are `3` and `2`, respectively, beween `25^(@)C` and `35^(@)C`. IF the above two resctions are carried out taking `0.4 M` of each reactant but at different temperatures: `25^(@)C` for the first order reaction and `35^(@)C` for the second order reaction, find the ratio of the concentrations of `A` and `B` after an hour. |
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Answer» Correct Answer - 2 `0.4M of Aoverset(30 min)rarr0.2Moverset(30 min)rarr0.1M ("for"A)` `0.4M of B overset(20 min)rarr0.2 M overset(20 min)rarr 0.1 M overset(20 min)rarr0.05 M("for"B)` `((-d[B])/(dt) "will be doubled and hence" t_(1//2) "will be halved")` `:. ([A])/([B]) = 2`Correct Answer - 2 `0.4M of Aoverset(30 min)rarr0.2Moverset(30 min)rarr0.1M ("for"A)` `0.4M of B overset(20 min)rarr0.2 M overset(20 min)rarr 0.1 M overset(20 min)rarr0.05 M("for"B)` `((-d[B])/(dt) "will be doubled and hence" t_(1//2) "will be halved")` `:. ([A])/([B]) = 2` |
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| 297. |
In pseudo-unimolecular reactions :A. both the reactants are present in low concentraionB. both the reactants are present in same concentrationC. one of the reactant is present in excessD. one of the reactant is non-reactive. |
| Answer» Correct Answer - C | |
| 298. |
the velocity of a reaction with molar concentration of each of the reactants is unity is calledA. Rate constantB. specific reaction rateC. Rate of reactionD. Both (a) and (b) |
| Answer» Correct Answer - D | |
| 299. |
The value of rate of a pseudo first order reaction depends uponA. the concentration of both the reactants present in the reactionB. the concentration of the reactant present small amountC. the concentration of the reactant present in excessD. the value of `Delta H` of the reaction. |
| Answer» Correct Answer - B | |
| 300. |
A first order reaction: `A rarr` Products and a second order reaction: `2R rarr` Products both have half time of `20 min` when they are carried out taking `4 mol L^(-1)` of their respective reactants. The number of mole per litre of `A` and `R` remaining unreacted after`60 min` form the start of the reaction, respectively, will beA. `1` and `0.5 M`B. `0.5 M` and negligibleC. `0.5` and `1 M`D. `1` and `0.25 M` |
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Answer» Correct Answer - C In the case of first order reaction `t_(1//2)` will remain constant independent of initial concentration. So, `4 mol L^(-1) overset(20 min)rarr 2 mol L^(-1) overset(20 min)rarr 1 mol L^(-1) overset(20 min)rarr 0.5 mol L^(-1)`. That is, after `60 min 0.5 mol L^(-1)` of `A` will be laft unreacted. In the case of second order reaction, `t_(1//2)` is inversely proportional to the initial concentration of reactant, i.e., `t_(1//2)` will go on doubling as the concentration of reactant will go on getting half, i.e., `t_(1//2) a` will be constant. So `4mol L^(-1) overset(20 min)rarr 2mol L^(-1) overset(40 min)rarr 1 mol L^(-1)`. That is, after `60 min`, the concentration of `R` remaining unreacted will be `1 mol L^(-1)`. Note: `t_(1//2) a = 20 min xx 4 M = 40 min xx 2M = 80 mol L^(-1) min^(-1), a` constant. |
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