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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.
| 351. |
Assertion (A) : For a first order, the concentration of a reaction decreases exponentially with time. Reason (R ): The rate of reaction at any time depends upon the concentration of the reactant at that time.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explnation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is noth the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is incorrect, but `(R )` is correct. |
| Answer» Correct Answer - B | |
| 352. |
Consider the reaction `A to B`. The concentration of both the reactants and the products varies exponentially with time. Which of the following figure correctly describes the change in concentration of reactants and products with time ?A. B. C. D. |
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Answer» Correct Answer - B In a reaction `A to B` concentrations of reactant decreases as concentration of product increases during the course of a reaction. |
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| 353. |
Assertion: For a first order reaction, the concentration of the reactant decreases exponentially with time. Reason: Rate of reaction at any time depend upon the concentration of reactant at that time.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B Correct explanation. `[A] = [A]_(0)e^(-kt)` |
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| 354. |
Pick the appropriate choice about collision theory of reaction ratesA. It explains the effect of temperature on rate of reactionB. It assumes that the reactants must be in correct orientation to reactC. It says rate depends upon the frequency at which reactants collideD. The collision having energy higher than the threshold value give successful reaction |
| Answer» Correct Answer - d | |
| 355. |
Which of the following statement is not true according to collision theory of reaction rates ?A. Collision of molecules is percondition for any reaction to occurB. All collisions result in the formation of the productsC. Only activated collisions result in the formation of the productsD. Molecules which have acquired the energy of activation can collide effectively |
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Answer» Correct Answer - B All collisions are not effective and does result in the formation of the products |
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| 356. |
Find the two-thirds life `(t_(2//3))` of a first order reaction in which `k=5.48xx10^(-1) "sec"^(-1) ` ( log 3 = 0.4771 , log 2 = 0.3010) |
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Answer» Correct Answer - 2.0005 s `(t_(2//3))=(2.303)/(k)log_(10).(a)/(a-(2)/(3)a)` |
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| 357. |
How much time is requred for two - third completion of a first order reaction having, `K=5.48xx10^(-14)S^(-1)`?A. `2.01xx10^(11)s`B. `2.01xx10^(13)s`C. `8.08xx10^(13)s`D. `16.04xx10^(11)s` |
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Answer» Correct Answer - B Time required for two-third completion of a first order reaction, `[A]_(0)=a,[A]=a-(2)/(3)a=(a)/(3)` `t_(t//2)=(2.303)/(k)"log"([A]_(0))/([A])=(2.303)/(K)"log"(a)/(a/3)` `=(2.303)/(K) "log"3` `t(1//2)=(2.303xx0.4771)/(5.48xx10^(-14))=2.01xx10^(13)s` |
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| 358. |
Bicyclohexane was found to undergo two parallel first order rearrangements. At `730 K`, the first order rate constant for the formation of cyclohexene was measured as `1.26xx10^(-4)s^(-1)` and for the formation of methyl cyclopentene the rate constant was `3.8 xx 10^(-5) s^(-1)`. What is the percentage distribution of the rearrangement Products ? |
| Answer» Correct Answer - `0.768, 0.232;` | |
| 359. |
Decomposition of `H_(2)O_(2)` is a first-order reaction. A solution of `H_(2)O_(2)` labelled as `20` volumes was left open. Due to this some `H_(2)O_(2)` decomposed. To determine the new volume strength after `6` hours, `10 mL` of this solution was diluted to `100 mL`. `10 mL` of this diluted solution was titrated against `25 mL `of `0.025 m KMnO_(4)` acidified solution. Calculate the rate constant for decomposition of `H_(2)O_(2)`. |
| Answer» Correct Answer - `0.022 hr^(-1);` | |
| 360. |
Ethylene is produced by, `underset("Cyclobutane")(C_(4)H_(8))overset(Delta)(rarr) 2C_(2)H_(4)` The rate constant is `2.48xx10^(-4) sec^(-1)`. In what time will the molar ratio of the ethylene to cyclobutane in reaction mixture attain the value `1` ? |
| Answer» Correct Answer - (a) `27.25 minute, (b) 264.2 minute;` | |
| 361. |
Statement: The rate of reaction whether exothermic or endothermic usually, increases with temperature. Explanation: The rate of reaction `=K["reactant"]^(n)` and `K` increases with temperature.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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| 362. |
For the reaction `2N_(2)O_(5)rarr 4NO_(2)+O_(2)`, if rate formation of `O_(2) is 16 g//hr`, then rate of decomposition of `N_(2)O_(5)` and rate of formation of `NO_(2)` respectively is:A. `54 g//hr and 46 g//hr`B. `32 g//hr and 64 g//hr`C. `108 g//hr and 92 g//hr`D. None of these |
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Answer» Correct Answer - c `(d[O_(2)])/(dt)=16 g//hr=16/32 mol//hr` `(d[NO_(2)])/(dt)=(4d[O_(2)])/(dt)=4xx16/32=2 mol//hr` `=2xx46=92 g//hr` `(-d[N_(2)O_(5)])/(dt)=(2d[O_(2)])/(dt)=2xx16/32=1 mol//hr` `=1xx108=108 g//hr` |
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| 363. |
For the reaction `4NH_(3) + 5O_(2) to 4NO + 6H_(2) O`, if the rate of disappearance of `NH_(3)` is `3.6 xx 10^(-3)` mol `L^(-1) s^(-1)`, what is the rate of formation of `H_(2) O`A. zero `mol L^(-1)S^(-1)`B. First `mol L^(-1) S^(-1)`C. first `S^(-1)`D. Zero `L mol ^(-1)S^(-1)` |
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Answer» Correct Answer - A For a zero order reaction rate `=K =(dx)/(dt)` unit of `k= mol L^(-1)S^(-1)` |
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| 364. |
For the reaction `4NH_(3) + 5O_(2) to 4NO + 6H_(2) O`, if the rate of disappearance of `NH_(3)` is `3.6 xx 10^(-3)` mol `L^(-1) s^(-1)`, what is the rate of formation of `H_(2) O`A. `5.4xx10^(-3) mol L^(-1)s^(-1)`B. `3.6xx10^(-3)mol L^(-1)s^(-1) `C.D. |
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Answer» Correct Answer - A `-(1)/(4) (d[NH_(3)])/(dt) =+(1)/(6) (d[H_(2)O])/(dt) ` ` (d[H_(2)O])/(dt) =(6)/(4)xx3.6xx10^(-3) =5.4xx10^(-3) mol L^(-1)S^(-1)` |
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| 365. |
The commercial poduction of ammonia is represented by the equation, `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`. If the rate of disappearance of `H_(2)(g)` is `1.2xx10^(-3)` mol/min, what is the rate of appearance of `NH_(3)(g)`?A. `2.4xx10^(-3)"mol"//"min"`B. `1.8xx10^(-3)"mol"//"min"`C. `1.2xx10^(-3)"mol"//"min"`D. `8.0xx10^(-4)"mol"//"min"` |
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Answer» Correct Answer - D |
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| 366. |
Ammonia reacts with oxygen according to the equation : `4NH_(3)(g)+5O_(2)(g)rarr4NO(g)+6H_(2)O(l)` In an experiment in which the rate of change of nitric oxide is found to be `1.10"M=-min"^(-1)`, what is the rate of change of oxygen gas?A. `-1.38 "M.min"^(-1)`B. `-0.880 "M.min"^(-1)`C. `-0.275 "M.min"^(-1)`D. `-0.220 "M.min"^(-1)` |
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Answer» Correct Answer - A |
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| 367. |
The gas phase dicomposition of dinitrogen pentoxide is represented by this equation, `2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)` What is the rate of formation of oxygen gas (in mol.`L.s^(-1)`) in an experiment where `0.080` mol of `N_(2)O_(5)` is consumed in a `4.0` L container every `0.20` second?A. `0.020`B. `0.050`C. `0.10`D. `0.20` |
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Answer» Correct Answer - B |
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| 368. |
The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i) Half life of `nth` order reaction depends on the initial concentration according to the following relation: `t_(1//2) prop (1)/(a^(n-1))` ...(ii) The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii) The differential rate law for `nth` order reaction may be given as: `(dx)/(dt) = k[A]^(n)` ...(iv) where `A` denotes the reactant. In a chemical reaction `A rarr B`, it is found that the rate of the reaction doubles when the concentration of `A` is increased four times. The order of the reaction with respect to `A` is:A. `0`B. `(1)/(2)`C. `1`D. `2` |
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Answer» Correct Answer - B `r_(1) = k[A]^(n)` …(i) `r_(2) = 2r_(1) = k[4A]^(n)` …(ii) `(2r_(1))/(r_(1)) = (4)^(n)` `rArr (2)^(1) = (2)^(2n)` `:. 2n = 1 rArr n = (1)/(2)` |
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| 369. |
The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i) Half life of `nth` order reaction depends on the initial concentration according to the following relation: `t_(1//2) prop (1)/(a^(n-1))` ...(ii) The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii) The differential rate law for `nth` order reaction may be given as: `(dX)/(dt) = k[A]^(n)` ...(iv) where `A` denotes the reactant. The rate constant for zero order reaction is where `c_(0)` and `c_(t)` are concentration of reactants at respective times.A. `k=(C_(0))/(2t)`B. `k=(C_(0)-C_(t))/(t)`C. `k=ln.(C_(0)-C_(t))/(2t)`D. `k=(C_(0))/(C_(t))` |
| Answer» Correct Answer - B | |
| 370. |
The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i) Half life of `nth` order reaction depends on the initial concentration according to the following relation: `t_(1//2) prop (1)/(a^(n-1))` ...(ii) The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii) The differential rate law for `nth` order reaction may be given as: `(dx)/(dt) = k[A]^(n)` ...(iv) where `A` denotes the reactant. For a reaction: `I^(ɵ) + OCl^(ɵ) rarr IO^(ɵ) + Cl^(ɵ)` in an aqueous medium, the rate of the reaction is given by `(d[IO^(ɵ)])/(dt) = k([I^(ɵ)][OCl^(ɵ)])/([overset(ɵ)(OH)])` The overall order of the reaction isA. `-1`B. 1C. zeroD. 2 |
| Answer» Correct Answer - B | |
| 371. |
The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i) Half life of `nth` order reaction depends on the initial concentration according to the following relation: `t_(1//2) prop (1)/(a^(n-1))` ...(ii) The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii) The differential rate law for `nth` order reaction may be given as: `(dX)/(dt) = k[A]^(n)` ...(iv) where `A` denotes the reactant. The half life for a zero order reaction equalsA. `(1)/(2)(k)/(a^(2))`B. `(a^(2))/(2k)`C. `(2k)/(a)`D. `(a)/(2k)` |
| Answer» Correct Answer - D | |
| 372. |
The reaction `A+B to ` C has zero order. Write its rate equatons. |
| Answer» Rate = `k[A]^(0)[B]^(0)`. | |
| 373. |
Time required to decompose `SO_(2)Cl_(2)` to half of its intial amount is `60 mi n.` If the decomposition is a first order reaction, calculate the rate constant of the reaction. |
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Answer» For first order reaction, Rate constant (k) = `(0.693)/(60 "min")=0.01155 min^(-1)` |
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| 374. |
The following data were obtained during the first order thermal decomposition of `SO_(2)Cl_(2)` at a constant volume `SO_(2)Cl_(2)(g)to SO_(2)(g)+Cl_(2)(g)` Calculate the rate of the reaction when total pressure is 0.65 atmA. `0.35 atmS^(-1)`B. `2.235 xx10^(-3)atmS^(-1)`C. `7.8xx10^(-4) atm s^(-1)`D. `1.55 xx10^(-4) atm S^(-1)` |
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Answer» Correct Answer - C ` SO_(2) Cl_(2 ) to SO_(2) +Cl_(2)` initial pressure ` P_(0) ,0,0` Pressure pressure `P_(0)-p " " P" " p` Let initial pressure `p_(0) prop R_(0)` pressure at time t, `P_(t) =p_(0)-p+p+p=P_(0) +p` Pressure of reactants at time t ` p_(0)-P=2P_(0)-p_(1)prop R` `k=(2.303)/(t) log ""(p_(0))/(2p_(0)-p_(t))` ` =(2.303)/(100) log""(0.5)/(2xx0.5-0.6)=(2.303)/(100) log 1.25 ` `=2.318 xx10^(-3)S^(-1)` pressure of `SO_(2)Cl_(2)` at time `t (P_(SO_(2)Cl_(2))) ` `=2p_(0)-P_(t) =2xx0.50-0.65 atm + 0.35 atm ` rate at that time `=k xxp_(SO_(2)Cl_(2))` `=(2.318xx10^(-3))xx(0.35)=7.8 xx10^(-4) atmS^(-1)` |
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| 375. |
A graph plotted between log `t_(50%)` vs log concentration is a straight line. What conclusion can you draw from this graph? A. `n=1: t_(1//2)=1/(x xxa)`B. `n=2: t_(1//2)=1/a`C. `n=1: t_(1//2)=0.693/k`D. None of these |
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Answer» Correct Answer - c `t_(1//2) prop (a)^((1-n))` or `t_(1//2)=z. (a)^((1-n))` `:. Log_(1//2)=log z+(1-n)(log a)` or `y=c+mx` |
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| 376. |
After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is `1.155xx10^(-3)s^(-1)`?A. 600B. 100C. 60D. 10 |
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Answer» Correct Answer - A Rate constant `k=1.155xx10^(-3)s^(-1)` `k=(2.303)/(t)"log"(a)/((a-x))` `because" "a=a,(a-x)=(a)/(2)` `t_(1//2)=(2.303)/(k)"log"(a)/(a//2)=(2.303)/(1.155xx10^(-3))"log "2` `or" "t_(1//2)=600s` |
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| 377. |
Consider figure and mark the correct option. A. Activation energy of forward reaction is `E_(1) + E_(2)` and product is less stable than reactant.B. Activation energy of forward reaction is `E_(1) + E_(2)` and product is more stable than reactant.C. Activation energy of both forward and backward reaction is `E_(1) + E_(2)` and reactant is more stable than product.D. Activation energy of backward reaction is `E_(1)` and product is more stable than reactant. |
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Answer» Correct Answer - A `E_(a("Forward reaction")) = E_(1) + E_(2)` and product is less stable as it has higher energy |
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| 378. |
In the presence of a catalyst, the heat evolved or absorbed during the reaction:A. increasesB. decreasesC. remains unchangedD. may increase or decrease |
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Answer» Correct Answer - C Catalyst alters the activation energy of both forward and backward reactions eqully hence heat of reaction remains unchanged. |
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| 379. |
Activation energy of a chemical reaction can be determined byA. evaluting rate constant at standard temperatureB. evaluting velocities of reaction at two different temperaturesC. evaluating rate constants at two different temperaturesD. changing concentration of reactants |
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Answer» Correct Answer - C According to Arrhenius equation `k=Ae^(-E_(a)//RT)` Taking natural logarithm of both sides, we get `Ink=InA-(E_(a))/(RT)` At temperature `T_(1))` , equation becomes `Ink_(1)=InA-(E_(a))/(RT_(1)` At temperature `T_(2))` , equation becomes `Ink_(1)=InA-(E_(a))/(RT_(1)` (Since `A` is constant for a given reaction). `K_(1)` and `k_(2)` are the value of rate constant at temperature `T_(1)` and `T_(2)` respectively. Subtracting equation `(1)` from equation `(2)` , we obtain `Ink_(2)-Ink_(1)=(-(E_(a))/(RT_(2)))-(-(E_(a))/(RT_(1)))` `In(k_(2))/(k_(1))=-(E_(a))/(R)((1)/(T_(2))-(1)/(T_(1)))` or log`(k_(2))/(k_(1))=-(E_(a))/(2.303R)((1)/(T_(2))-(1)/(T_(1)))` or log`(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` Thus, we can evaluate the activation energy of a chemical reaction by knowing rate constants `(k_(1) and k_(2))` at two differences temperatures `(T_(1) and T_(2))` . |
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| 380. |
Activation energy of a chemical reaction can be determined byA. determining the rate constant at standard temperatureB. determining the rate constants at two temperaturesC. determining probaiility of collisionD. using catalyst |
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Answer» Correct Answer - B `"log" (K_(2))/(k_(1)) = (E_(a))/(2.303) [(T_(2) - T_(1)])/(T_(1)T_(2)]` |
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| 381. |
Activation energy of a chemical reaction can be determined byA. changing concentration of reactantsB. evaluating rate constant at standard temperatureC. evaluating rate constant at two different temperturesD. evaluating velocities of reaction at two different temperatures |
| Answer» Correct Answer - C | |
| 382. |
Activation energy of a chemical reaction can be determined by:A. evaluating rate constant at standard temeprature.B. evaluating velocities of reaction at two different temperaturesC. evaluating rate constant at two different temperatureD. cahnging concentration of reactants |
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Answer» Correct Answer - B::C |
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| 383. |
The experimental data for the reaction `2A+B_(2)rarr2AB,` is : The rate equation for the above data is :A. rate = `k[B]^(2)`B. rate = `k[B]`C. rate =`k[A]^(2)[B]^(2)`D. rate =`k[A]^(2)[B]` |
| Answer» Correct Answer - B | |
| 384. |
In hypothetical reaction, `A_(2)+B_(2)rarr2AB`, follows the follows the mechnism as given below : `{:(A_(2)hArrA+A,("fast reaction")),(A+B_(2)rarrAB+B,("slow reaction")),(A+BrarrAB,("fast reaction")):}` Give the rate law and order of reactionA. zeroB. 1C. 2D. `3//2` |
| Answer» Correct Answer - D | |
| 385. |
The experimental data for the reaction `2A + B_(2) to 2AB` is The rate equation for the above data isA. Rate = `k [B_(2)]`B. Rate = `k [B_(2)]^(2)`C. Rate = `k [A]^(2) [B]^(2)`D. Rate = `k [A]^(2) [B]` |
| Answer» Correct Answer - a | |
| 386. |
In hypothetical reaction, `A_(2)+B_(2)rarr2AB`, follows the follows the mechnism as given below : `{:(A_(2)hArrA+A,("fast reaction")),(A+B_(2)rarrAB+B,("slow reaction")),(A+BrarrAB,("fast reaction")):}` Give the rate law and order of reaction |
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Answer» Slowest step is rate eleminated. `K_(c)=([A][A])/([A_(2)])=([A]^(2))/([A_(2)])` `[A]=K_(c)^(1//2)[A_(2)]^(1//2)` From eq. (i), Rate `=kK_(c)^(1//2)[A_(2)]^(1//2)[B_(2)]` `=K[A_(2)]^(1//2)=[B_(2)],[K=*K_(c)^(1//2)]` Order =1+1/2=3/2 |
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| 387. |
Activation energy of a chemical reaction can be determined byA. evaluating rate constant at standard temperatureB. evaluating velocities of reaction at two different temperaturesC. evaluating rate constants at two different temperaturesD. changing concentration of reactants |
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Answer» Correct Answer - C Activation energy can be calculated by using arrhenius equation the Arrhenius equation is `Log""(K_(2))/(K_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` where `K_(1) and k_(2)` =rate constants at two different temperature i.e., `T_(1) and T_(2)` respectively . `E_(a)`= Activation energy R= Gas constant So, Activation energy of a chemical reaction can be detetmined by evevaluating rate constants at two different temperatures. |
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| 388. |
Consider the following reaction, `N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)+22kcal" mol"^(-1)` The activation energy for the forward reaction is 50 kcal. What is the activation energy for the backward reaction?A. `28kcal" mol"^(-1)`B. `+28kcal"mol"^(-1)`C. `72kcal"mol"^(-1)`D. `+72kcal" mol"^(-1)` |
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Answer» Correct Answer - D The activation energy for forward reaction `=50kcal" mol"^(-1)` Energy released during the reaction =`22"kcal mol"^(-1)` Therefore, the acativation energy for the backward reaction `=50+22"kcal"=72"kcal mol"^(-1)` |
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| 389. |
A reaction having equal activation energies for forward and reverse reaction thatA. `Delta H = 0`B. `Delta S = 0`C. Zero orderD. None of these |
| Answer» Correct Answer - a | |
| 390. |
The experimental data for the reaction `2A+B_(2)rarr2AB,` is : The rate equation for the above data is :A. `rate =k [B_(2)]`B. `rate =k [B_(2)]_(2)`C. `rate =k[A]^(2)[B]^(2)`D. `rate=k [A]^(2)[B]` |
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Answer» Correct Answer - A Consider the following rate law equation . `(dx)/(dt)=K[A]^(m)[B_(2)]^(n)` `1.6xx10^(-4)=K[0.50]^(m)[0.50]^(n)` . ..(i) `3.2xx10^(-4)=K[0.50]^(m)[10]^(n)` . . .(ii) `3.2xx10^(-4)=K[100]^(m)[1.0]^(n)`. . .(iii) by dividing Eq.(iii) by (ii) we get , `(3.2xx10^(-4))/(3.2xx10^(-4))=(k[1.00]^(m)[1.0]^(n))/(k[0.50]^(m)[1.0]^(n))` `1=2 ^(m) or 2^(0)=2^(m)` ` therefore m=0` By dividing Eq(ii) by (i) `(3.2xx10^(-4))/(1.6xx10^(-4))=([0.50]^(m)[1.0]^(n))/([0.50]^(m)[0.50]^(n))` `1=2^(m) or 2^(0)=2^(n)` ` therefore m=0` by dividing Eq.s (ii) by (i) `(3.2xx10^(-4))/(1.6xx10^(-4))=([0.50]^(m)[1.0]^(n))/([0.50]^(m)[0.50]^(n))` `2=2^(n)` `or 2^(1)=2^(n)` `therefore n=1 ` hence rate . `((dx)/(dt))=K[A]^(0)[B_(2)]^(1)` `=K[b_(2)]` |
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| 391. |
in a reversible reaction the energy of activation of the forward reaction is 50 kcal . the energy of activation for the reverse reaction will beA. lt 50 kcalB. 50 kcalC. either greater tan or less than 50 kcalD. gt 50 kcal |
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Answer» Correct Answer - C the activation energy of a reaverse reaction decide whetaher the given reaction is exothermic or endothmic so the energy of activation of reverse reaction is either greater or less than 50 kcal in case of exothermic reaction the activation energy for reverse reaction is more than activation energy of tah forward reaction and in case of endothermic reaction the activation energy for reverse reaction is less than activation energy of the forward reaction . |
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| 392. |
What is the effect of adding a catalyst on the rate of a reversible reaction in the forwards and the reverse direction?A. It has no effect on the rate in either directionB. Both rated increase by the sme factor.C. The rate in the forward direction increases by a greater factor than the rate in the reverse directionD. The rate in the reverse direction increases by a greater factor than the rate in the forward direction. |
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Answer» Correct Answer - B |
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| 393. |
Select the incorrect statement.A. Rate of exothermic reactions (irreversible ) increasee with increase in temperature.B. Rate of endothermic reactins (irreversible 0 increases with increases in temperatureC. For `N_(2)+3H_(2) hArr 2NH_(3)` , If rate of formation of `NH_(3)` is 0.001 `Kghr^(-1)` , then rate of consumption of `H_(2)` is 0.0015 kg/hrD. In Arrhenius equation `K=Ae^(-Ea//RT)` if `T rarr oo implies K=A` |
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Answer» Correct Answer - C |
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| 394. |
For the equilibrium, `A(g) rarr B(g) , DeltaH` is -40 KJ/mol. If the ratio of the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reactions is `(2)/(3)` then:A. `E_(f)` =60 KJ/mol, `E_(b)`=100 KJ/molB. `E_(f)` =30 KJ/mol, `E_(b)`=70 KJ/molC. `E_(f)` =80 KJ/mol, `E_(b)`=120 KJ/molD. `E_(f)` =70 KJ/mol, `E_(b)`=30 KJ/mol |
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Answer» Correct Answer - C |
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| 395. |
In a multistep reaction such as `A + B rarr Q rarr C`. The potential energy diagram is shown below. What is `E_(a)` for the reaction `Q rarr C` ? A. `3 kcal mol^(-1)`B. `5 kcal mol^(-1)`C. `8 kcal mol^(-1)`D. `11 kcal mol^(-1)` |
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Answer» Correct Answer - A For `Q rarr C` `E_(a) = 23-20=3 kcal mol^(-1)` |
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| 396. |
What is `Delta H` for the reaction `A + B rarr C` where the mechanism involves several kinetics steps. A. `11 "kcal mol"^(-1)`B. `4 "kcal mol"^(-1)`C. `5 "kcal mol"^(-1)`D. `22 "kcal mol"^(-1)` |
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Answer» Correct Answer - B `Delta H = 10-6 = 4 kcal mol^(-1)` |
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| 397. |
The half life of decompoistion of `N_(2)O_(5)` is a first order reaction represented by `N_(2)O_(5)rarrN_(2)O_(4)+1//2O_(2)` After `15`min the volume of `O_(2)` profuced is `9ml` and at the end of the reaction `35ml`. The rate constant is equal toA. `(1)/(15)log_(e).(35)/(26)`B. `(1)/(15)log_(e).(44)/(26)`C. `(1)/(15) log_(e).(35)/(36)`D. None of these |
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Answer» Correct Answer - A `k = (2.303)/(t)log.(V_(oo))/(V_(oo)-V_(t))` `= (1)/(t) log_(e).(V_(oo))/(V_(oo)-V_(t))` `= (1)/(15)log_(e).(35mL)/((35-9)mL) = (1)/(15)log_(e).(35)/(26)` |
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| 398. |
The reaction : `N_(2)O_(5)( "in" (C)Cl_(4) "solution") to2NO_(2)("same solution") +1//2O_(2)(g)` is of first order in `N_(2)O_(5)` with rate constant = `6.2 xx 10^(-4)s^(-1)`. What is the value of the rate of reaction when `[N_(2)O_(5)]` with rate constant = `6.2 xx 10^(-4)s^(-1)`. What is the rate of reaction when `N_(2)O_(5)=12.5 mol L^(-1)`?A. `5.15 xx 10^(-5) mol L^(-1)s^(-1)`B. `6.35 xx 10^(-3) mol L^(-1)s^(-1)`C. `7.75 xx 10^(-4)molL^(-1)s^(-1)`D. `3.85 xx 10^(-4) mol L^(-1)s^(-1)` |
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Answer» Correct Answer - C c) rate `=k xx [N_(2)O_(5)]` `(6.2 xx 10^(-4) s^(-1)) xx (1.25 mol L^(1))` `7.75 xx 10^(-4) mol L^(-1)s^(-1)` |
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| 399. |
Consider the following reaction , `2NO(g) + 2H_(2)(g) to N_(2)(g) + 2H_(2)O(g)` The rate law for the reaction is first order with respect to `H_(2)` and second order with respect to NO.Write the rate law for the reaction. |
| Answer» Self explanatory | |
| 400. |
The compoistion of `N_(2)O_(5)` is a first order reaction represented by: `N_(2)O_(5) rarr N_(2)O_(4) + 1//2O_(2)`. After `20 min` the volume of `O_(2)` profuced is `10 mL` and at the end of the reaction `40 mL`. The rate constant is equal toA. `(1)/(20)ln.(30)/(50)`B. `(1)/(20)ln.(50)/(30)`C. `(1)/(20)ln.(50)/(40)`D. `(1)/(20)ln.(40)/(30)` |
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Answer» Correct Answer - D `k = (2.303)/(t)log.(V_(oo))/(V_(oo)-V_(t))` or `k = (1)/(t)ln.(V_(oo))/(V_(oo)-V_(t))` `k = (1)/(20) ln.(40)/((40-10))` `= (1)/(20) ln.(40)/(30)` |
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