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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.
| 1201. |
Select correct statement(s).A. The emission of gamma radiation involves transition between energy levels within the nucleus.B. `(4)/(2)` He is formed due to emission of beta particle from tritium `(3)/(1)` H.C. When positron`(overset(0)(+1)e)` is emitted, `(n)/(P)` ration increases.D. None of above |
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Answer» Correct Answer - A::C |
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| 1202. |
For a zero order reaction of the type `A rarr` products, the integrated rate equation may be expressed asA. `k = ([A]_(0)-[A])/(2).t`B. `k = ([A]_(0)-[A])/(2t)`C. `k = ([A]-[A]_(0))/(t)`D. `k = ([A]_(0)-[A])/(t)` |
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Answer» Correct Answer - D Consider the reaction `Ararrprocts` Rate `=-(d[A])/(dt)=K[A]^(@)` As any equantity raised to power zero is unity Rate `=-(d[A])/(dt)=Kxx1` or `d[A]=-Kdt` Intergrating both sides `[A]=-KtI` where `I` is the constant of integration. At `t=o` , the concentration of the reaction `A=[A]_(0)` , where `[A]_(0)` is intial concentration of the reaction. Substitution in previous equation, we get `[A]_(0)=-K(0)+I` or `[A]_(0)=I` Substituting the value of `I` in the equation `(1)` , we get `[A]=-Kt+[A]_(0)` Further simplifying equation `(2)` , we get the rate constant, `K` as `K=([A]_(0)-[A])/(t)` |
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| 1203. |
In the reaction `A + B rarr C+D`, the concentration of `A` and `B` are equal and the rate of the reaction is rate `= k[A][B]`. The integrated rate equation for this reaction isA. `k = (x)/(t(a-x))`B. `k = (xa)/((a-x))`C. `k = (1)/(t).(x)/(a(a-x))`D. `k = (1)/(t).(x)/(a(x-a))` |
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Answer» Correct Answer - C It is a second order reaction. |
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| 1204. |
In the Wilhelmey equation of a first order reaction `c_(t) = c_(0)e^(-kt)`. If the initial concentration `c_(0)` is increased `m` times, thenA. The value of `k` will increase `m` timesB. The value of `k` will decrease `m` timesC. The value of `k` will remain unchangedD. None of these |
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Answer» Correct Answer - C First order reaction is independent of `c_(0)`. Hence, `k` is constant. |
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| 1205. |
The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea. |
| Answer» `Ea =52.897 KJ mol^(-1)` | |
| 1206. |
The activation energy for the reaction `:`. `2Hl(g) rarr H_(2)(g)+I_(2)(g)` is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ? |
| Answer» Correct Answer - `1.471xx10^(-19)` | |
| 1207. |
Which of the following statement is not correct about order of a reaction ?A. The order of a reaction can be a fractional numberB. Order of a reaction is experimentally determined quantityC. The order of a reaction is always equal to the sumof the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.D. The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression. |
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Answer» Correct Answer - C |
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| 1208. |
Which of the following statements is correct?A. The rate of fa reaction decreases with passage of time as the concentration of reactants decreases.B. The rate of a reaction is same at any time during the reaction.C. The rate of a reaction is independent of temperature change.D. The rate of a reaction decreases with increases in concentration of reactant (s) |
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Answer» Correct Answer - A According to rate law expression, Rate `prop [R]` Thus, rate of a reaction decreases with passage of tiem as the concentration of reactants decreases. |
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| 1209. |
A graph of volume of hydrogen released vs time for the reaction between zinc and dil. HCl is given in figure . On the basis of this markt the correct options. A. Average rate upto 40 seconds is `(V_(3)-V_(2))/(40)`B. Average rate upto 40 seconds is `(V_(3)-V_(2))/(40-30)`C. Average rate upto 40 seconds is `V_(3)/40`D. Average rate upto 40 seconds is `(V_(3)-V_(1))/(40-20)` |
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Answer» Correct Answer - C |
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| 1210. |
Consider the graph given in Q.9 Which of the following options does not show instantaneous rate of reaction at `40^(th)` second?A. `(V_(5) - V_(2))/(50 -30)`B. `(V_(4) - V_(2))/(50 - 30)`C. `(V_(3) - V_(2))/(40 - 30)`D. `(V_(3) - V_(1))/(40 - 20)` |
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Answer» Correct Answer - B Instantaneous rate can be determind graphically by drawing a tangent on the curve. Thus, volume change during 50-30 sec. interval = `V_(5) - V_(2)` and not `v_(4) - V_(2)` Hence, `(V_(4) - V_(2))/(50 - 30)` does not show Instantaneous rate of reaction at `40^(th)` second. |
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| 1211. |
A graph of volume of hydrogen released vs time for the reaction between zinc and dil. HCI is given in figure. On the basic of this mark the correct option. A. Average rate upto 40 seconds is `(V_(3) - V_(2))/(40)`B. Average rate upto 40 seconds is `(V_(3) - V_(2))/(40 - 30)`C. Average rate upto 40 seconds is `(v_(3))/(40)`D. Average rate upto 40 seconds is `(V_(3) - V_(1))/(40 - 20)` |
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Answer» Correct Answer - C Average rate upto 40 sec `= (V_(3) - V_(0))/(40 - 0) = (v_(3))/(40)` |
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| 1212. |
The thermal decomposition of `N_(2)O_(5)` occurs in the following steps: Step I: `N_(2)O_(5) overset("slow")(rarr) NO_(2)+NO_(3)` Step II: `N_(2)O_(5)+NO_(3)overset("fast")(rarr)3NO_(2)+O_(2)` Overall reaction, `2N_(2)O_(5)rarr 4NO_(2)+O_(2)` Suggest the rate expression. |
| Answer» Rate `=K[N_(2)O_(5)].` (from slow step) | |
| 1213. |
`X overset("Step"1)to Y overset("StepII")to Z` is a complex reaction. Total order of reaction is 2 and step II is a slow step. What is the molecularity of step II?A. 1B. 3C. 2D. 4 |
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Answer» Correct Answer - C c) Since step II is a slow step and order (2) is expressed for this step, this means that the molecularity of reaction is also 2. |
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| 1214. |
The rate constant of the reaction, `2N_(2)O_(5) to 4NO_(2) + O_(2)` at 300 K is `3 xx 10^(-5)s^(-1)`. If the rate of the reaction at the same temperaturre is `2.4 xx 10^(-5)` mol `dm^(-3)s^(-1)`, then the molar concentration of `N_(2)O_(5)` isA. 0.4 MB. 0.8 MC. `0.04` MD. `0.08` M |
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Answer» Correct Answer - B b) For first order reaction, `Rate =k[N_(2)O_(5)]` `[N_(2)O_(5)]=("Rate")/k=(2.4 xx 10^(-5))/(3 xx 10^(-5)) "mol" dm^(-3)` `=0.8 "mol" dm^(-3)` |
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| 1215. |
A first order reaction has `k=1.5 xx 10^(-6) s^(-1)` at `200^(@)`C. If the reaction is allowed to run for 10 hours, what percentage of initial concentration would have changed in products? |
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Answer» For the first order reaction, `t=2.303/k log a/(a-x)` `loga/(a-x) = (kxxt)/(2.303)=((1.5 xx 10^(-6)s^(-1)) xx (10 xx 60 xx 60s))/(2.303)=0.0235` `a/(a-x) ="Antilog" 0.0235=1.055` or `100=105.5 -1.055 x or x=(5.5)/(1.055) = 5.2=5.2%` |
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| 1216. |
Identify the order of reaction from each of the following rate constants? a) `k=2.3 xx 10^(-3) L mol^(-1)s^(-1)` b) `k=1.25 xx 10^(-2)s^(-1)` c) `k=2.9440 xx 10^(6) mol L^(-1)s^(-1)` d) `k=2.8 xx 10^(-8) atm^(-1)s^(-1)` |
| Answer» Conceptual problems. | |
| 1217. |
For a decomposition, the values of rate constants at two different temperature are given below: `k_(1) = 2.15 xx 10^(-8)L mol^(-1)s^(-1)` at 650 K `k_(2) = 2.39 xx 10^(-7) L mol^(-1)s^(-1)` at 700 K Calculate the value of activation energy for the reaction (R=`8.314 JK^(-1) mol^(-1))` |
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Answer» According of Arrhenius equation: `(log)k_(2)/k_(1) = E_(a)/(2.303R)[1/T_(1)-1/T_(2)]` `k_(1) = 2.25 xx 10^(-8) L mol^(-1)s^(-1), k_(2)=2.39xx10^(-7)Lmol^(-1)s^(-1)` `T_(1) = 650K,T_(2)= 700K, R=8.314 JK^(-1)mol^(-1)` `log(2.39 xx 10^(-7))/(2.15 xx 10^(-8))= E_(a)/(2.303 xx (8.314 JK^(-1)mol^(-1)))[(320K-300K)/((320K)xx(300K))]` `0.6020 = (E_(a) xx 20)/((2.303) xx (8.314 J mol^(-1)) xx (320) xx (300))` `E_(a) = (0.6020 xx (2.303) xx (8.314J mol^(-1)) xx 320 xx 300)/(20)` `=55327.6 J mol^(-1) = 55.328 kJmol^(-1)` |
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| 1218. |
What percentage of the initial concentration will react in 5 hours in a first order reaction whose rate constants is `5.78 xx 10^(-5)s^(-1)?`. |
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Answer» For the first order reaction: `k=(2.303)/t log a/(a-x)` `k=5.78 xx 10^(-5)s^(-1), t=5hr=5 xx 60 xx 60=18000s` `log a/(a-x) = (kxxt)/(2.303)=(5.78 xx 10^(-5)s^(-1) xx 18000s)/(2.303)=0.4518` `(a/(a-x)) = "Antilog" 0.4518=2.83` Let `a=100%`, Therefore, `(100)/(100-x) =2.83` or `100 =283-2.83x` Let `a=100%`, Therefore, `(100)/(100-x)=2.83` or `100 = 283-2.83x` or `x=183/(2.83)=64.660 =64.66%` |
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| 1219. |
The plote between concentration versus time for a zero order reaction is represented by :A. B. C. D. |
| Answer» Correct Answer - D | |
| 1220. |
The activation energy of a reaction can be lowered by :A. using a positive catalystB. increasing temperatureC. decreasing temperatureD. increasing concentration of the reactants |
| Answer» Correct Answer - A | |
| 1221. |
The rate constant of a reaction depends onA. initial concentration of the reactantsB. time of reactionC. temperatureD. extent of reaction |
| Answer» Correct Answer - C | |
| 1222. |
The activation energy of a reaction is zero. The rate constant of the reactionA. increases with increase in temperatureB. decreases with decrease of temperatureC. decreases with increase of temperatureD. is nearly independent of temperature |
| Answer» Correct Answer - D | |
| 1223. |
The radioisotpe, N-13,has a half-life of 10.0 minutes is used to image organs in the body. If an injected sample has an activity of 40 microcuries `(40,muCi)`, what is its activity after 25 minutes in the body?A. `0.75 muCi`B. `3.5 muCi`C. `7.1 muCi`D. `12 muCi` |
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Answer» Correct Answer - C |
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| 1224. |
The following data were observed for the following reaction at `25^(@)C, CH_(3)OH + (C_(6)H_(5))_(3)C Clrarr(C_(6)H_(5))_(3)C.OCH_(3)+HCl` Rate constant of the above experiment is (in `L^(2) M^(-2) min^(-1)`) :A. `2.6 xx 10^(-2)`B. `2.6 xx 10^(-1)`C. `2.6 xx 10^(-4)`D. `1.3 xx 10^(-2)` |
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Answer» Correct Answer - B `((dx)/(dt)) = k[A]^(2)[B]` `:. k = ((dx//dt)/([A]^(2)[B])) = 0.26` |
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| 1225. |
The reduction of peroxydisulphate ion by `I^(-)` ion is expressed by `S_(2)O_(8)^(2-) + 3 I^(-) rarr 2SO_(4)^(2-) + I_(3)^(-)` If rate of disappearance of `I^(-)` is `9//2xx10^(-3)` mol `L^(-1) S^(-1)`, what is the rate of formation of `SO_(4)^(2-)` during same time ?A. `10^(-3) "mol"//L^(-1)s^(-1)`B. `2 xx 10^(-3) "mol"//L^(-1)s^(-1)`C. `3 xx 10^(-3) "mol"//L^(-1)s^(-1)`D. `4 xx 10^(-3) "mol"//L^(-1)s^(-1)` |
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Answer» Correct Answer - C `(1)/(2)(d[SO_(4)^(2-)])/(d t) = (1)/(3) (-(d[I^(-)])/(d t))` `(1)/(2) xx (d[SO_(4)^(2-)])/(d t) = (1)/(3) xx (9)/(2)` `:. (d[SO_(4)^(2-)])/(d t) = 3 mol L^(-1)s^(-1)`. |
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| 1226. |
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):}` Unit of rate constant will beA. `"sec"^(-1)`B. `"litre"^(2)"mole"^(2)"sec"^(-1)`C. `"litre mole"^(2)"sec"^(-1)`D. `"mol litre"^(-1)"sec"^(-1)` |
| Answer» Correct Answer - B | |
| 1227. |
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):}` Order with respect to `[CH_(3)Cl]` will beA. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - B `OR` w.r.t. `[CH_(3)Cl] = 1` |
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| 1228. |
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):}` If `H_(2)O` is taken in large excess, the order of the reaction will beA. 1B. 0C. 3D. 2 |
| Answer» Correct Answer - A | |
| 1229. |
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)` |
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Answer» Correct Answer - C Comparing the rate law, the order w.r.t. `CH_(3)Cl = 1` and w.r.t. `H_(2)O = 2`. `:. r = [CH_(3)Cl][H_(2)O]^(2)` |
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| 1230. |
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):}` Overall order of the reaction will be |
| Answer» Correct Answer - D | |
| 1231. |
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):}` Order with respect to `[CH_(3)Cl]` will be |
| Answer» Correct Answer - B | |
| 1232. |
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):}` Overall order of the reaction will beA. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - D Overall order `= 1+2 = 3` |
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| 1233. |
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):}` Unit of rate constant will beA. `s^(-1)`B. `L^(2) mol^(-2) s^(-1)`C. `L mol^(-1) s^(-1)`D. `mol L^(-1) s^(-1)` |
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Answer» Correct Answer - B Units: `L^(2) mol^(-1) s^(-1)` |
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| 1234. |
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):}` If `H_(2)O` is taken in large excess, the order of the reaction will beA. `1`B. `0`C. `3`D. `2` |
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Answer» Correct Answer - A If `H_(2)O` is in excess, the order of the reaction will be `1`. |
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| 1235. |
A second order reaction requires `70 min` to change the concentration of reactants form `0.08 M` to `0.01 M`. The time required to become `0.04 M = 2x min`. Find the value of `x`. |
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Answer» Correct Answer - 5 For second order reaction: `[R]_("initial") = 0.08, M, [R]_("initial") = 0.01M` `x = 0.08 - 0.01 = 0.07M` `:. (a-x) = 0.08 - 0.07 = 0.01 M` `k_(2) = (1)/(t).(x)/(a(a-x))` `= (1)/(70 min) xx (0.07 M)/(0.08M xx 0.01 M)` …(i) Now, time required to become concentration `= 0.04 M`. i.e., `x = 0.04 M` `k_(2) = (1)/(t) xx (0.04 M)/(0.08 M xx (0.08-0.04)M)` ...(ii) form Eqs. (i) and (ii) `(0.07)/(70 xx 0.08 xx 0.01) = (0.04)/(t xx 0.08 xx 0.04)` `t = 10 min = 2x min` `:. x = 5 min`Correct Answer - 5 For second order reaction: `[R]_("initial") = 0.08, M, [R]_("initial") = 0.01M` `x = 0.08 - 0.01 = 0.07M` `:. (a-x) = 0.08 - 0.07 = 0.01 M` `k_(2) = (1)/(t).(x)/(a(a-x))` `= (1)/(70 min) xx (0.07 M)/(0.08M xx 0.01 M)` …(i) Now, time required to become concentration `= 0.04 M`. i.e., `x = 0.04 M` `k_(2) = (1)/(t) xx (0.04 M)/(0.08 M xx (0.08-0.04)M)` ...(ii) form Eqs. (i) and (ii) `(0.07)/(70 xx 0.08 xx 0.01) = (0.04)/(t xx 0.08 xx 0.04)` `t = 10 min = 2x min` `:. x = 5 min` |
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| 1236. |
Rate constant of two reactions are given below. Identifying their order of reaction `k = 5.3 xx 10^(-2) L mol^(-1) s^(-1)` `k = 3.8 xx 10^(-4) s^(-1)`A. (i) second order, (ii) first orderB. (i) first order, second orderC. (i) zero order, (ii) first orderD. (i) second order, (ii) zero order |
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Answer» Correct Answer - A For second order, rate constant `= (mol L^(-1))/(s) x (1)/((mol L^(-1))^(2)) = mol^(-1) L s^(-1)` For first order, rate constant`= (mol L^(-1))/(s) xx (1)/(mol L^(-1)) = s^(-1)` |
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| 1237. |
Find the values of A,B and C in the following table for the reaction `X + Y to Z`. The reaction is of first order w.r.t X and zero order w.r.t. Y. `{:("Exp",[X](mol L^(-1)),[Y](mol L^(-1)),"Initial rate (mol L^(-1) s^(-1))),(1,0.1,0.1,2xx10^(-2)),(2,A,0.2,4 xx 10^(-2)),(3,0.4,0.4,B),(4,C,0.2,2 xx 10^(-2)):}`A. `A = 0.2 mol L^(-1), B = 8 xx 10^(-2) mol L^(-1) s^(-1), C = 0.1 mol L^(-1)`B. `A = 0.4 mol L^(-1), B = 4 xx 10^(-2) mol L^(-1) s^(-1), C = 0.02 mol L^(-1)`C. `A = 0.2 mol L^(-1), B = 2 xx 10^(-2) mol L^(-1) s^(-1), C = 0.4 mol L^(-1)`D. `A = 0.4 mol L^(-1), B = 2 xx 10^(2) mol L^(-1) s^(-1), C = 0.4 mol L^(-1)` |
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Answer» Correct Answer - A Rate `= k [X][Y]^(0)` Rate is independent of the conc. Of Y and it depends only on the conc.of X and it is the first order reaction. From exp. (1), `2 xx 10^(-2) = k (0.1)` From exp, (2) `4 xx 10^(-2) = k(A)` Dividing (ii) and (i) , `(4 xx 10^(-2))/(2 xx 10^(-2)) = (k(A))/(k(0.1)) = (A)/(0.1)` `implies 2 xx 0.1 = A` `implies A = 0.2 mol L^(-1)` From exp. (3), `B = k(0.4)` Dividing (iii) and (i), `(B)/(2 xx 10^(-2)) - (k(0.4))/(k(0.1)) = 4` `implies B = 4 xx 2 xx 10^(2) = 8 xx 10^(-2) mol L^(-1) s^(-1)` From exp. (4), `2 xx 10^(2) = k (C )` dividing (iv) and (i), `(2 xx 10^(-2))/(2 xx 10^(-2)) = (k(C ))/(k(0.1)) = (C )/(0.1)` `implies C = 0.1 mol L^(-1)` |
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| 1238. |
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. 1B. `1.5`C. 2D. `-1` |
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Answer» Rate `=k [A]^(n)" "....(i)` `1.837x x"Rate"=k[1.5A]^(n) " "....(ii)` Dividing eq . (ii) by eq. (i), `1.837=1.5^(n)` `:. N=3//2` (solving by longarithmic method) |
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| 1239. |
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. 1B. 1.5C. 2D. 2.5 |
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Answer» Correct Answer - B Initially rate `(r_(1)) = k [X]^(n)` let n be the order of reaction then,` r_(1) xx 1.837 = k (k(1.5 X)^(n)` dividing (ii) by (i) `1.837 = (1.5)^(n)` `:. N = 1.5` Thus, order of reaction is 1.5 |
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| 1240. |
The rate of a gaseous reaction is given by the expresison `k[A]^(2)[B]^(3)`. The volume of the reaction vessel is suddenly reduced to one-half of the initial volume. The reaction rate relative to the original rate will beA. (a) `1//24`B. (b) `1//32`C. ( c) `32`D. (d) `24` |
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Answer» Correct Answer - C (c ) Rate `= k[A]^(2)[B]^(3)` When the volume is halved, the concentration will become double. `:.` Rate `= k[2A]^(2)[2B]^(3)` `= 32k[A]^(2)[B]^(3)` `= 32 xx` Original rate form Eq. (i). |
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| 1241. |
Under a given set of experiemental condition, with increase in the concentration of the reactants, the reate of a chemical reactionA. decreasesB. increasesC. remains unlteredD. First decreases and then increases |
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Answer» Correct Answer - B Rate of reaction `prop` concentration of reactant . |
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| 1242. |
For a reaction, `A+B to` Products, the rate law is given by: `r=k[A]^(1//2)[B]^(2)`. What is the order of reaction: |
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Answer» Rate law ( r) `= k[A]^(1//2) [B]^(2)` Order of reaction `= 1/2 + 2= 21/2 or 2.5`. |
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| 1243. |
For a first order reaction , to obtain a positive slope , we need to plot {where [A] is the concentration of reactant A}A. `- "log"_(10) [A]` vs tB. `- "log"_(e) [A]` vs tC. `log _(10) [A] vs ` log tD. [A] vs t |
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Answer» Correct Answer - b For a first order reaction to obtain a positive slope , a graph is plotted between `-"log"_(e)`[A] vs t. |
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| 1244. |
Orienation factor is direclty proportional toA. Callision frequencyB. Freaction of moleculeC. Rate of reactionD. Threshold energy |
| Answer» Correct Answer - C | |
| 1245. |
A reaction takes place in three steps with individual rate constant and activation energy, `{:(,"Rate constant","Activation energy"),("Step 1",k_(1),E_(a_(1))=180kJ //"mol"),("Step2", k_(2),E_(a_(2))=180kJ //"mol"),("Step3",k_(3),E_(a_(3))=180kJ //"mol"):}` oveall rate constant, `k=((k_(1)k_(2))/(k_(3)))^(2//3)` overall activation energy of the reaction will be :A. 140 kJ/molB. 150 kJ/molC. 130 kJ/molD. 120 kJ/mol |
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Answer» Correct Answer - A `Ae^(-E_(a)//RT)=[(Ae^(-E_(a_(1))//RT)xxAe^(-E_(a_(2))//RT))/(Ae^(-E_(a_(3))//RT))]^(2//3)` `=[Ae^((-E_(a_(1))-E_(a_(2))+E_(a_(3)))//RT)]^(2//3)` `E_(a)=(2)/(3)[E_(a_(1))+E_(a_(2))-E_(a_(3))]` `=(2)/(3)[180+80-50]=140 ` kJ/mol |
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| 1246. |
For a reaction taking place in three steps, the rate consatnt are `k_(1),k_(2) and k_(3)`. The oveall constant `k=(k_(1)k_(2))/(k_(3)` . If the energy of activation values of for the first, second and third stage are 40,50 and 60 kJ `"mol"^(-1)` is :A. 30B. 40C. 60D. 50 |
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Answer» Correct Answer - A `k=(k_(1)k_(2))/(k_(3))` `Ae^(-E_(a)//RT)=(Ae^(-E_(a_(1))//RT)xxAe^(-E_(a_(2))//RT))/(Ae^(-E_(a_(3))//RT))` `e^(-E_(a)//RT)=e^((E_(a_(1))-E_(a_(2))+E_(a_(1)))//RT)` `E_(a)=E_(a_(1))+E_(a_(2))-E_(a_(3))` `=40+50-60=30 kJ "mol"^(-1)` |
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| 1247. |
The activation energy of a reaction is `9.0 kcal//mol`. The increase in the rate consatnt when its temperature is increased from `298 K` to `308 K` isA. `10%`B. `100%`C. `50%`D. `63%` |
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Answer» Correct Answer - d `2.303 "log" K_(2)/K_(1)=E_(a)/R[(T_(2)-T_(1))/(T_(1)T_(2))]` `:. 2.303 "log" K_(2)/K_(1)=9/(2xx10^(-3))[(10)/(298xx308)]` `:. K_(2)/K_(1)=1.63, i.e., 63%` increse. |
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| 1248. |
For a given reaction, presence of catalyst reduces the energy of activation by `2` kcal at 27^(@)C`. Thus rate of reaction will be increased by:A. `20` timesB. `14` timesC. `28` timesD. `2` times |
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Answer» Correct Answer - c `K_(1)/K_(2)=e^(-[E_(a1)-E_(a2)]//RT)=e^(-[(2xx10^(3))//(2xx300)])` `"log"_(e) K_(2)/K_(1)=(2xx10^(3))/(600)` `K_(2)/K_(1)=28=r_(2)/r_(1)` `:. R_(2)=28r_(1)` `=28` |
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| 1249. |
For a reation: `A rarr` Product, rate law is `-(d[A])/(dt)=K[A]_(0)`. The concentration of `A` left after time `t` when `t=1/K` is:A. `([A]_(0))/(e)`B. `[A]_(0).e`C. `([A]_(0))/e^(2)`D. `1/([A]_(0))` |
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Answer» Correct Answer - a `N/N_(0)=e^(-lambdat)=e^(-lambda(1//lambda))=e^(-1)` `:. N=N_(0)/e` |
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| 1250. |
Expression for the half-life of zero order reaction is given asA. `t_(1//2)=(2[R]_0)/(K)`B. `t_(1//2)=(2[R]_0)/(2K)`C. `t_(1//2)=(0.693)/(K)`D. `t_(1//2)=(0.301)/(K)` |
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Answer» Correct Answer - B `t_(1//2)=([R]_(0))/(2k)` for a zero order reaction. |
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