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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.
| 2001. |
The reaction `Ararr B` follows first order kinetics. The time taken for `0.8 mol` of `A` to produce `0.6 mol` of `B` is `1 hr`. What is the time taken for the conversion of `9.0 mol` of `A` to Product `0.675 mol` of `B` ?A. `0.25h`B. `2h`C. `1H`D. `0.5h` |
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Answer» Correct Answer - C Rate constant of first of this order of first order reaction `K=(2.303)/(t) Log""((A)_(0))/((A)_(t))` `or k=(2.303)/(t)xxlog_(10)"" (0.8)/(0.2)` because 0.6 mole of B is formed suppose `t_(1) ` hour are reguireed for changing the concentration of A from 0.9 mole to 0.675 mole of B remaining mole of A =0.9-0.675=0.225 `therefore k=(2.303)/(t_(1))log_(10)(0.9)/(0.225)` From Eqs. (i) and(ii) `(2.303)/(1) log_(10)(0.8)/(0.2)=(2.303)/(t_(1)) log_(10)(0.9)/(0.225)` `2.303 log _(10 ) 4=(2.303)/(t_(1))log _(10)4` `t_(1) =1h` |
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| 2002. |
The reaction `Ararr B` follows first order kinetics. The time taken for `0.8 mol` of `A` to profuce `0.6 mol` of `B` is `1 hr`. What is the time taken for the converison of `9.0 mol` of `A` to Product `0.675 mol` of `B` ?A. `1 hr`B. `0.5 hr`C. `0.25 hr`D. `2 hr` |
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Answer» Correct Answer - A Time for the converison of `0.8 mol` of `A` to `0.6 mol` of `B = (0.6)/(0.8) = 0.75 =t_(75%)` ismilarly, time for the converison of `0.9 mol` of `A` to `0.675 mol` of `B = (0.675)/(0.9) = 0.75 = t_(75%)` Hence, `t_(75%)` in both case `= 1 hr` |
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| 2003. |
For the reaction : `2N_(2)O_(5)rarr4NO_(g)+O_(2)(g)` if the concentration of `NO_(2)` increases by `5.2xx10^(-3)M` in 100 sec, then the rate of reaction is :A. `1.3xx10^(-5)Ms^(-1)`B. `5xx10^(-4)Ms^(-1)`C. `7.6xx10^(-4)Ms^(-1)`D. `2xx10^(-3)Ms^(-1)` |
| Answer» Correct Answer - A | |
| 2004. |
A decompositio has following mecfhanism`2N_(2)O_(5)rarr4NO_(2)+" "` (overall) `N_(2)O_(5)hArrNO_(2)+ " "` (fast decomposition) `NO_(2)NO_(3)rarrNO+NO_(2)+O_(2) " "` (slow) `NO+NO_(3)rarr2NO_(2) " "` (fast) Determine rate law, is the mechanism correspond. first order reacrtion |
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Answer» Correct Answer - `(-d[N_(2)O_(5)])/(dt)=(2k_(1)k_(3))/(k_(1)+k_(3))[N_(2)O_(5)]` |
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| 2005. |
Consider a reaction, `2A + B rarr` Products When concentration of `B` alone was doubled, the half-life did not change. When the concentration of `A` alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is :A. (a) `L mol^(-1) s^(-1)`B. (b) No unitC. (c ) `mol L^(-1) s^(-1)`D. (d) `s^(-1)` |
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Answer» Correct Answer - a `2A+B rarr Product` When conc. Of `B` is doubled, the half-life did not change, hence reaction is of first w.r.t. B. When concentration of A is doubled, reaction rate is doubled, hence reaction is of first order w.r.t. A. Hence, overall of reaction is `1+1=2` So, unit of rate constant is `mol^(-1)` litre `s^(-1)` |
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| 2006. |
The forward reaction rate for the nitric oxide-oxygen reaction `2NO+O_(2) rarr 2NO_(2)` has the rate law as: `Rate = k[NO]^(2)[O_(2)]`. If the mechanism is assumed to be: `2NO+O overset(k_(eq))hArr` , (rapid equilibration) `N_(2)O_(2) + O_(2) overset(k_(2))rarr 2NO_(2)` (slow step), then which of the following is (are) correct? (I) Rate constant `= k_(eq)k_(2)` , (II) `[N_(2)O_(2)] = k_(eq)[NO]^(2)` (III) `[N_(2)O_(2)] = k_(eq)[NO]` , (IV) Rate constant `= k_(2)` The correct option isA. I, IIB. III, IVC. I, IIID. None of these |
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Answer» Correct Answer - D `k_(eq) = ([N_(2)O_(2)])/([NO]^(2)) rArr [N_(2)O_(2)] = k_(eq)[NO]^(2)` and `RDS : Rate = k_(2)[N_(2)O_(2)][O_(2)]` `= k_(2)k_(eq)[NO]^(2)[O_(2)] = k [NO]^(2)[O_(2)]` |
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| 2007. |
The air oxidation of nitric oxide is one of the reactions that contributes to the formation of acid rain: `2NO(g)+O_(2)(g)rarr 2NO_(2)(g)` Some initial rate data are collected Two overall order of the reaction isA. twoB. threeC. oneD. zero |
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Answer» Correct Answer - B Note that pairs of experiments are designed to invertigate the effect on the intial rate of a change in the intial concentration of a single reactant. In the first two experments, the concentration of `NO` is doubled from `0.015M` to `0.03M` while the concentration of `O_(2)` is held constant. The intial rate increases by a factor of `4` , from `0.048Ms^(-1)` to `0.192Ms^(-1)` , indicating that the rate depends on the concentration of `NO` squared, `[NO]^(-2)` . When `[NO]` is held constant and `[O_(2)]` is doubled (experiments `1` and `3`), the intial rate doubles from `0.048Ms^(-1)` to `0.096Ms^(-1)` , indicating that the rate depends on the rate concentration of `O_(2)` to the first power, `[O_(2)]^(1)` . Therefore, the rate law for formation of `NO_(2)` is Rate `=(d[NO_(2)])/(dt)=K[NO]^(2)[O_(2)]` The overall reaction order s three. In accord with this rate law, which is second order in `NO` , first order in `O_(2)` , and third order overall, the intial rate increases by a factor of `8` when the concentratiobn of both `NO` and `O_(2)` are doubled (experiments `1` and `4`). |
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| 2008. |
In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change form 0.1 M to 0.025 M is :A. 30 minB. 15 minC. `7.5` minD. 60 min |
| Answer» Correct Answer - A | |
| 2009. |
The reaction `Ararr B` follows first order kinetics. The time taken for `0.8 mol` of `A` to produce `0.6 mol` of `B` is `1 hr`. What is the time taken for the conversion of `9.0 mol` of `A` to Product `0.675 mol` of `B` ?A. 0.5 hB. 0.25 hC. 1 hD. 2 h |
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Answer» Correct Answer - C Time required for the conversion of 0.8 mol of A into 0. 6mol of `B=(0.6)/(0.8)=0.75rArrt_(75%)` Similarly, the time required for the conversion of 0.9 mol of A to 0.675 mol of B is `(0.675)/(0.9)=0.75rArrt_(75%)` Hence, both cases requires 1 h. |
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| 2010. |
Consider the following reaction, `2N_(2)O_(5)rarr4NO_(2)+O_(2),(d[NO_(2)])/(dt)=k_(2)[N_(2)O_(5)]`, `(d[O_(2)])/(dt)=k_(3)[N_(2)O_(5)]" and "(d)/(dt)[N_(2)O_(5)]=k_(1)` The relation between `k_(1), k_(2)` and `k_(3)` isA. `k_(1)=k_(2)=k_(3)`B. `2k_(1)=k_(2)=4k_(3)`C. `2k_(1)=4k_(2)=k_(3)`D. None of these |
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Answer» Correct Answer - B `2N_(2)O_(5)rarr4NO_(2)+O_(2)` `Rate=-(1)/(2)(d[N_(2)O_(5)])/(dt)=(1)/(4) (d[NO_(2)])/(dt)=(d[O_(2)])/(dt)` `rArr" "2k_(1)=k_(2)=4k_(3)` |
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| 2011. |
Consider the following reaction, `2A + B + C rarr` Products How will the rate of reaction changes when the concentration of A is doubled and that of B is triplet while C is taken in excess ?A. The rate reduces 8 times of its original valueB. The rate reduces 12 times of its original valueC. The rate increases 8 times of its original valueD. The rate increases 12 times of its original value |
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Answer» Correct Answer - D `2A+B+Crarr` Products `r_(1)=k[A]^(2)[B]` [`because` C is taken in excess, rate does not depend on C] `r_(2)=k[2A]^(2)[3B]` `(r_(2))/(r_(1))=(12k[A]^(2)[B])/(k[A]^(2)[B])implies r_(2)=12r_(1)` |
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| 2012. |
If the rate of reaction, `2SO_(2)(g)+O_(2)(g)overset(Pt)rarr2SO_(3)(g)` is given by: Rate`=K([SO_(2)])/([SO_(3)]^(1//2))` which statements are correct?A. The overall order of reaction is `-(1)/(2)`B. the overall order of reaction is `+(1)/(2)`C. The reaction slows down as the product `SO_(3)` is build upD. The rate of reactions does not depend upon concentraion of `SO_(3)` formed |
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Answer» Correct Answer - B::C |
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| 2013. |
Nitrogen dioxide `(NO_(2))` dissociates into nitric oxide (NO) and oxygen `(O_(2))` as follows `NO_(2) to 2NO + O_(2)` If the rate of decrease of concentration of `NO_(2)` is `6.0 xx 1-^(12)` mol `L^(-1) s^(-1)`. What will be the rate of increase of concentration of `O_(2)`?A. `3 xx 10^(-12) mol L^(-1) s^(1)`B. `6 xx 10^(-12) mol L^(-1) s^(-1)`C. `1 xx 10^(-12 mol L^(-1) s^(1)`D. `1.5 xx 10^(-12) mol L^(-1) s^(1)` |
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Answer» Correct Answer - A For the reaction, `2NO_(2) to 2NO + O_(2)` `-(1)/(2) (d[NO_(2)])/(dt) = (1)/(2) (d[NO])/(dt) = (d[O_(2)])/(dt)` `- (d[NO_(2)])/(dt) = 6 xx 10^(-12) mol L^(-1) s^(-1)` `(d[O_(2)])/(dt) = 3 xx 10^(-12) mol L^(-1) s^(-1)` |
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| 2014. |
In a reaction `2X to Y`, the concentration of X decreases from 3.0 mole/litre to 2.0 moles/litre in 5 minutes. The rate of reaction isA. 0.1 mol `L^(-1) "min"^(-1)`B. 5 mol `L^(-1) "min"^(-1)`C. 1 mol `L^(-1) "min"^(-1)`D. 0.5 mol `L^(-1) "min"^(-1)` |
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Answer» Correct Answer - A Rate `= - (1)/(2) (Delta[X])/(Delta t)` `= - (1)/(2) ((3 - 2))/(5) = - 0.1 mol L^(-1) "min"^(-1)` Negative sign signifies the decrease in concentration. |
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| 2015. |
The rate of formation of `SO_(3)` in the following reaction: `2SO_(2) + O_(2) rarr 2SO_(3)` is `10 g sec^(-1)`. The rate of disappearance of `O_(2)` will beA. `5 g sec^(-1)`B. `100g sec^(-1)`C. `20 g sec`D. `2 g sec^(-1)` |
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Answer» Correct Answer - D `2SO_(2) + O_(2) rarr 2SO_(3)` `r_(SO_(3))=10 g//sec =10/80 "mol"//sec` Also, `r_(O_(2))/1=r_(So_(3))/2 implies r_(O_(2))=10/80 xx1/2 "mol"//sec` `=(10/160)xx32 g//sec=2 g//sec` |
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| 2016. |
The rate of formation of `SO_(3)` in the following reaction is `100 g " min"^(-1).2SO_(2)+O_(2)rarr2SO_(3)` The rate of disappearance of `O_(2)` isA. 29 g `"min"^(-1)`B. 20 g `"min"^(-1)`C. 200 g `"min"^(-1)`D. 50 g `"min"^(-1)` |
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Answer» Correct Answer - B `2SO_(2)+O_(2)rarr2SO_(3)` `Rate=-(d[O_(2)])/(dt)=(1)/(2)(d[SO_(3)])/(dt)` The rate of formation of S)_(3) is `(d[SO_(3)])/(dt)=100g min^(-1)=(100)/(80) mol min^(-1)` `[because "molar mass" of SO_(3)=80 g " mol"^(-1)]` `(d[O_(2)])/(dt)=(1)/(2)((100)/(80))mol min^(-1)` `(d[O_(2)])/(dt)=(1)/(2)((100)/(80))xx32gmin^(-1)` `[because "molar mass of "O_(2)=32g" mol"^(-1)]` `=20g" min"^(-1)` |
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| 2017. |
The rate of formation of `SO_(3)` in the following reaction is `100 g " min"^(-1).2SO_(2)+O_(2)rarr2SO_(3)` The rate of disappearance of `O_(2)` isA. 29 g `"min"^(-1)`B. 20 g `"min"^(-1)`C. 50 g `"min"^(-1)`D. 200 g `"min"^(-1)` |
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Answer» Correct Answer - B `2SO_(2)+O_(2)rarr2SO_(3)` Rate `=-(d[O_(2)])/(dt)=(1)/(2)(d[SO_(3)])/(dt)` `-(d[O_(2)])/(dt)=(1)/(2)((100)/(80))" mol min"^(-1)=(1)/(2)((100)/(80))xx32 g " min"^(-1)` `[because" Molar mass of "O_(2)=32 g//"mol"]` `=20g " min"^(-1)` |
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| 2018. |
The rate of disappearance of `SO_(2)` in the reaction `2SO_(2) + O_(2) rarr 2SO_(3)` is `1.28 xx 10^(-3) g//sec` then the rate of formation of `SO_(3)` isA. `0.64 xx 10^(-5) M s^(-1)`B. `0.32 xx 10^(-5) M s^(-1)`C. `2.56 xx 10^(-5) M s^(-1)`D. `1.28 xx 10^(-5) M s^(-1)` |
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Answer» Correct Answer - D The rate of disappearance of `SO_(2)` and the rate of formation of `SO_(3)` are same. |
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| 2019. |
The rate of disappearance of `SO_(2)` in the reaction `2SO_(2) + O_(2) rarr 2SO_(3)` is `1.28 xx 10^(-3) g//sec` then the rate of formation of `SO_(3)` isA. `0.64 xx 10^(-3) g//sec`B. `0.80 xx 10^(-3) g//sec`C. `1.28 xx 10^(-3) g//sec`D. `1.60 xx 10^(-3) g//sec` |
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Answer» Correct Answer - C The rate of formation of `SO_(3)` is `1.28 xx 10^(-3) g//sec`. `(-d[SO_(2)])/(d t) = 1.28 xx 10^(-3)` `r = -(1)/(2) (d[SO_(2)])/(d t) = (1)/(2)(d[SO_(3)])/(d t) = -(d[O_(2)])/(d t)` `rArr -(1)/(2) (d[SO_(2)])/(d t) = -(1)/(2) (d[SO_(3)])/(d t)` `rArr (d[SO_(3)])/(d t) = 1.28 xx 10^(-3)` |
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| 2020. |
The rate of formation of `SO_(3)` in the reaction `2SO_(2)+O_(2) rarr 2SO_(3)` is 100 g `"min"^(-1)` Hence rate of disappearance of `O_(2)` isA. 50 `"g min"^(-1)`B. 40 `"g min"^(-1)`C. 200 `"g min"^(-1)`D. 20 `"g min"^(-1)` |
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Answer» Correct Answer - D |
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| 2021. |
a)Express the relationship between the rate of production of water and the rate of disappearance of oxygen in the following reaction: `2H_(2) + O_(2) to 2H_(2)O` b) For the chemical reaction `X_(2)(g) + 2Y_(2)(g) to 2XY_(2)(g)` Write the rate equation in terms of disappearance of `Y_(2)`. |
| Answer» Self explanatory | |
| 2022. |
A reaction rate constant is given by `k = 1.2 xx 10^(14) e^(-(25000//RT)) sec^(-1)` . It meansA. log k versus log T will give a straight line with slope as `- 25000`B. log k versus T will give a straight line with slope as `- 25000`C. log k versus log 1/T will give a straight line with slope `-25000`D. log k versus 1/T will give a straight line |
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Answer» Correct Answer - d According to the Arrhenius equation a straight line is to be obtained by plotting the logarithm of the rate constant of a chemical reaction (log k) against 1/T. |
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| 2023. |
Arrehnius equation for a reaction is given by `k=24xx10^(14)e^(-25000//RT)` Which of the following statement is correct about this expression ?A. log k versus log T will be a plot of straight line with a slope of 25000B. log k versus log T will be a plot of straight line with a slope of -25000C. log k versus T will be a plot of straight line with a slope of -25000D. log k versus `(1)/(T)` will gie a straight line. |
| Answer» Correct Answer - D | |
| 2024. |
Show reactions requires less activation energy as compared to fast reactions. Do you agree with the statement? |
| Answer» Correct Answer - No | |
| 2025. |
The rate of a first order reaction is 0.04 mol `"litre"^(-1)s^(-1)` after 10 minutes and 0.03 mol `"litre"^(-1)s^(-1)` after 20 minutes. Find the half life period of the reaction. |
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Answer» Step. I. Calculation of rate constant (k). For first order reaction, rate( r) `propto C` or kC Rate `(r_(1)) = kC_(1)` and Rate `(r_(2)) =kC_(2)` `therefore` Rate `(r_(1))= kC_(1)` and Rate`(r_(2) = kC_(2)` `therefore` `(r_(1)("at 10 min"))/(r_(2)("at 20 min")) = C_(1)/C_(2)= (0.04 mol L^(-1))/(0.03 mol L^(-1)) =0.04/0.03` Now, for first order reaction, `t=(2.303)/k log(C_(1))/C_(2), k = 2.303/t log C_(1)/C_(2) = 2.303/(10 min) log (0.04)/(0.03)` `=2.303/(10 min) xx log(4/3) xx log(4/3) = 2.303/(10 min) xx 0.1249 = 0.0287 min^(-1)` Step II. Calculation of half life period `(t_(1//2))` Half life period `t_(1//2)= 0.693/k = (0.693)/(0.0287 min^(-1)) = 24.15` min |
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| 2026. |
For a chemicla reaction A`to`B, the change in concentration of A in 40s is `-0.04` mol/litre. What is the rate of chemical reaction? |
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Answer» `Delta[A] = -0.004 mol L^(-1), Deltat=40s`. `therefore` Average rate of reaction = `-(Delta[A])/(Deltat)= -(-0.004 molL^(-1))/(40s) = 1.0 xx 10^(-4)mol L^(-1)s^(-1)`. |
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| 2027. |
Under the same reactions conditions , initial concentration of 1.386 mol `dm^(-3)` of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics , respectively . Ratio `(k_(1))/(k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reactions isA. `0.5 mol^(-1) dm^(-3)`B. `1.0 mol dm^(-3)`C. `1.5 mol dm^(-3)`D. `2.0 mol^(-1) dm^(-3)` |
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Answer» Correct Answer - a For first order `(t_(1//2))_(1) = (0.693)/(k_(1)) implies K_(1) = (0.693)/((t_(1//2))_(1))` for zero `(t_(1//2))_(0) = (A_(0))/(2k_(0)) implies k_(0) = (A_(0))/(2(t_(1//2))_(0))` `(K_(1))/(K_(0)) = (0.693 xx 2(t_(1//2))_(0))/((t_(1//2))_(1)A_(0)) implies (K_(1))/(K_(0)) = (0.693)/(1.386) = 0.5 mol^(-1) dm^(3)`. |
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| 2028. |
Under the same reaction condition, initial concentration of 1.386 mol `dm^(-3)` of a substance becomes half in 40s and 20s through first order and zero order kinetics respectively. Find out the `k_(1)/k_(0)` ratio for first order `(k_(1))` and zero order `(k_(0))` of the reaction.A. `0.5 mol^(-1)dm^(3)`B. `1.0 "mol" dm^(-3)`C. `1.5 "mol" dm^(-3)`D. `2.0 mol^(-1)dm^(3)` |
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Answer» Correct Answer - A a) For first order reaction, `k_(1) = 0.693/t_(1//2) = 0.693/(40)` For zero order reaction `k_(0) = ([A]_(0))/(2t_(1//2)) = (1.386 dm^(-3))/(2 xx20)` `k_(1)/k_(0) = 0.693/(40) xx 40/(1.386 "mol"dm^(-3))` `0.5 mol^(-1)dm^(3)` |
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| 2029. |
Under the same reaction condition, initial concentration of 1.386 mol `dm^(-3)` of a substance becomes half in 40s and 20s through first order and zero order kinetics respectively. Find out the `k_(1)/k_(0)` ratio for first order `(k_(1))` and zero order `(k_(0))` of the reaction. |
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Answer» For the first order kinetics, `k_(1)= 0.693/t_(1//2) = 0.693/(40s)` For zero order kinetics, `k_(0) = [A]_(0)/2t_(1//2)= (1.386 mol dm^(-3))/(2 xx 20s)` `k_(1)/k_(0) = 0.693/(40s) xx (40s)/(1.386 mol dm^(-3))= 0.693/(1.386 mol dm^(-3)) = 0.5 mol ^(-1)dm^(3)` |
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| 2030. |
Half-life period of a first-order reaction is `1386` seconds. The specific rate constant of the reaction isA. `0.5xx10^(-2)s^(-1)`B. `0.5x10^(-3)s^(-1)`C. `5.0xx10^(-2)s^(-1)`D. `5.0xx10^(-3)s^(-1)` |
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Answer» Correct Answer - B For the first order reaction `t_(1//2)=(0.693)/(k)` Thus, `k=(0.693)/(t_(1//2))` `=(0.693)/(1386s)=0.0005s^(-1)` `0.5xx10^(-3)s^(-1)` |
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| 2031. |
In the reaction `BrO^(-3)(aq) + 5Br^(-) (aq) + 6H^(+) rarr 3Br_(2)(1) + 3H_(2)O(1)` The rate of appearance of bromine `(Br_(2))` is related to rate of disapperance of bromide ions as folllwoing : |
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Answer» `BrO_(3)^(-)(aq)+5Br^(-)(aq) + 6H^(+) (aq) to 3Br_(2)(l) + 3H_(2)O(l)` `BrO_(3)^(-)(aq)+5Br^(-)(aq) + 6H^(+) (aq) to 3Br_(2)(l) + 3H_(2)O(l)` Rate of reaction in terms of `Br^(-)` ions and `Br_(2)` is: Rate (r) = `-1/5 (d[Br^(-)])/(dt) = 1/3 (d[Br_(2)])/(dt)` `(d[Br_(2)])/(dt) = -3/5(d[Br^(-)])/(dt)` |
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| 2032. |
In the reaction `BrO^(-3)(aq) + 5Br^(-) (aq) + 6H^(+) rarr 3Br_(2)(1) + 3H_(2)O(1)` The rate of appearance of bromine `(Br_(2))` is related to rate of disapperance of bromide ions as folllwoing :A. `(d[Br_(2)])/(dt) = -(5)/(3) (d[Br^(-)])/(dt)`B. `(d[Br_(2)])/(dt)= (5)/(3) (d[Br^(-)])/(dt)`C. `(d[Br_(2)])= (3)/(5) (d[Br^(-)])/(dt)`D. `(d[Br_(2)])/(dt) = -(3)/(5) (d[Br^(-)])/(dt)` |
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Answer» Correct Answer - D For the reaction `BrO_(3)^(-)(aq)+5Br^(-)(aq)+6H^(+)(aq)rarr3Br_(2)(l)+3H_(2)O(l)` We have Rate `=-(d[BrO_(3)^(-)])/(dt)=-(d[Br^(-)])/(5dt)=-(d[H^(+)])/(6dt)` `=(d[Br_(2)])/(3dt)=-(d[H_(2)O])/(3dt)` Thus `(d[Br_(2)])/(dt)`=-(3)/(5)(d[Br^(-)])/(dt)` |
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| 2033. |
In the reaction `BrO^(-3)(aq) + 5Br^(-) (aq) + 6H^(+) rarr 3Br_(2)(1) + 3H_(2)O(1)` The rate of appearance of bromine `(Br_(2))` is related to rate of disapperance of bromide ions as folllwoing :A. `(d[Br_(2)])/(d t) = (3)/(5)(d[Br^(-)])/(d t)`B. `(d[Br_(2)])/(d t) = -(3)/(5)(d[Br^(-)])/(d t)`C. `(d[Br_(2)])/(d t) = -(5)/(3)(d[Br^(-)])/(d t)`D. `(d[Br_(2)])/(d t) = (5)/(3)(d[Br^(-)])/(d t)` |
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Answer» Correct Answer - B `(1)/(3) (d[Br_(2)])/(d t) = -(1)/(5)(d[Br^(-)])/(d t) = R` |
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| 2034. |
For a chemical reaction `A rarr B`, the rate of reaction increases by a factor of `1.837` when the concentration of `A` is increased by `1.5` time. The order of reaction with respect to `A` is:A. `1`B. `1.5`C. `2`D. `-1` |
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Answer» Correct Answer - B (b) Rate `= k[A]^(n)` …(i) `1.837 xx Rate = k[1.5A]^(n)` …(ii) Dividing Eq. (ii) by Eq. (i), `1.837 = (1.5)^(n)` `:. n = 3//2` (Solving by logarithimic methof) |
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| 2035. |
Conisder the reaction represented by the equation: `CH_(3)Cl(g) + H_(2)O(g) rarr CH_(3)OH(g) + HCl(g)` These kinetics data were obtained for the given reaction concentrations: `{:("Initial conc (M)",,"Initial rate of disappearance"),([CH_(3)Cl],[H_(2)O],"of "CH_(3)Cl(Ms^(-1))),(0.2,0.2,1),(0.4,0.2,2),(0.4,0.4,8):}` The rate law for the reaction will beA. `r=k[CH_(3)Cl][H_(2)O]`B. `r=k[CH_(3)Cl]^(2)[H_(2)O]`C. `r=k[CH_(3)Cl][H_(2)O]^(2)`D. `r=k[CH_(3)Cl]^(2)[H_(2)O]^(4)` |
| Answer» Correct Answer - C | |
| 2036. |
The reaction `S_(2)O_(8)^(2-) + 3I^(ɵ) rarr 2SO_(4)^(2-) + I_(3)^(ɵ)` is of first order both with respect to persulphate and iofide ions. Taking the initial concentration as `a` and `b`, respectively, and taking `x` as the concentration of the triofide at time `t`, a differential rate equation can be written. Two suggested mechanism for the reaction are: I. `S_(2)O_(8)^(2-)+I^(ɵ) hArr SO_(4)I^(ɵ)+SO_(4)^(2-) ("fast")` `I^(ɵ)+SO_(4)I^(ɵ) overset(k_(1))rarrI_(2) + SO_(4)^(2-)` (show) `I^(ɵ) + I_(2) overset(k_(2))rarr I_(3)^(ɵ) ("fast")` II. `S_(2)O_(8)^(2-) + I^(ɵ) overset(k_(1))rarr S_(2)O_(8) I^(2-) (slow)` `S_(2)O_(8)I^(3-) overset(k_(2))rarr2SO_(4)^(2-)+I^(o+) ("fast")` `I^(o+) + I^(ɵ) overset(k_(3)) rarr I_(2) ("fast")` `I_(2) + I^(o+) overset(k_(4))rarr I_(3)^(ɵ) ("fast")` For the reaction `I_(2)+2S_(2)O_(3)^(2-) rarr S_(4)O_(6)^(2-) + 2I^(ɵ)` I. `(-d[I_(2)])/(dt) = -(1)/(2) (d[S_(2)O_(3)^(2-)])/(dt)` II. `(-d[I_(2)])/(dt) = -2 (d[S_(2)O_(3)^(2-)])/(dt)` III. `(-d[I_(2)])/(dt) = -2 (d[I^(ɵ)])/(dt) xx (d[S_(2)O_(3)^(2-)])/(dt)` IV. `(d[S_(4)O_(6)^(2-)])/(dt) = (1)/(2)(d[I^(ɵ)])/(dt)` The correct option isA. Only IB. I and IVC. II and IVD. Only III |
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Answer» Correct Answer - B `I_(2)+2S_(2)O_(3)^(2-) rarr S_(4)O_(6)^(2-) + 2I^(ɵ)` Rate `= (-d[I_(2)])/(dt) = (-d[S_(2)O_(3)^(2-)])/(2dt)` `= (+d[S_(4)O_(6)^(2-)])/(dt) = = (+d[I^(ɵ)])/(2dt)` Hence, correct options are I and IV. |
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| 2037. |
The reaction `S_(2)O_(8)^(2-) + 3I^(ɵ) rarr 2SO_(4)^(2-) + I_(3)^(ɵ)` is of first order both with respect to persulphate and iofide ions. Taking the initial concentration as `a` and `b`, respectively, and taking `x` as the concentration of the triofide at time `t`, a differential rate equation can be written. Two suggested mechanism for the reaction are: I. `S_(2)O_(8)^(2-)+I^(ɵ) hArr SO_(4)I^(ɵ)+SO_(4)^(2-) ("fast")` `I^(ɵ)+SO_(4)I^(ɵ) overset(k_(1))rarrI_(2) + SO_(4)^(2-)` (show) `I^(ɵ) + I_(2) overset(k_(2))rarr I_(3)^(ɵ) ("fast")` II. `S_(2)O_(8)^(2-) + I^(ɵ) overset(k_(1))rarr S_(2)O_(8) I^(2-) (slow)` `S_(2)O_(8)I^(3-) overset(k_(2))rarr2SO_(4)^(2-)+I^(o+) ("fast")` `I^(o+) + I^(ɵ) overset(k_(3)) rarr I_(2) ("fast")` `I_(2) + I^(o+) overset(k_(4))rarr I_(3)^(ɵ) ("fast")` The general difference equation for the above reaction isA. `(dx)/(dt) = k[a-x][b-3x] (kgt0)`B. `(dx)/(dt) = -k[a-x][b-3x] (kgt0)`C. `(dx)/(dt) = k[a-x][b-x] (kgt0)`D. `(dx)/(dt) = -k[a-x][b-x] (kgt0)` |
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Answer» Correct Answer - A `S_(2)O_(8)^(2-) + 3I^(ɵ) underset(k)rarr 2SO_(4)^(2-) + I_(3)^(ɵ)` [Net reaction] `{:(a,b,0,0,),(a-x,b-3x,2x,x,):}` Rate `= k[S_(2)O_(8)^(2-)][I^(ɵ)]` [from slowest step] `= k(a-x)(b-3x) = (-d(a-x))/(dt) = (dx)/(dt)` |
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