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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.
| 601. |
What names apply to chemical species corresponding to locations 1 and 2 on this reaction coordiante diagram ? A. `{:(,"Location I","Location II"),((a),"Activated complex","Activated complex"):}`B. `{:(,"Location I","Location II"),((b),"Reaction intermediate","Activated complex"):}`C. `{:(,"Location I","Location II"),((c),"Activated complex","Intermediat"):}`D. `{:(,"Location I","Location II"),((d),"Reaction intemediate","Intermediat"):}` |
| Answer» Correct Answer - B | |
| 602. |
In general, with every `10^(@)C` rise in temperature, the rate of reaction becomes appproximately..........A. ten timesB. doubleC. halfD. one tenth |
| Answer» Correct Answer - B | |
| 603. |
For the reaction `2N_(2)O_(5) rarr 4NO_(2)+O_(2)` rate of reaction and rate constant are `1.02 xx 10^(-4)` and `3.4 xx 10^(-5) sec^(-1)` respectively. The concentration of `N_(2)O_(5)` at that time will beA. `1.732`B. `3`C. `1.02 xx 10^(-4)`D. `3.4 xx 10^(5)` |
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Answer» Correct Answer - B `R = K[A], 1.02 xx 10^(-4) = 3.4 xx 10^(-5), [N_(2)O_(5)]` or `(N_(2)O_(5)) = (1.02 xx 10^(-4))/(3.4 xx 10^(-5)), K = 3` |
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| 604. |
rate constant of a reaction at 290 K was found to be `3.2 xx 10^(-3)`. At 300 K it will beA. `1.6xx10^(-3)`B. `6.4xx10^(-3)`C. `3.2xx10^(-4)`D. `3.2xx10^(-2)` |
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Answer» Correct Answer - B `6.4xx10^(-3)` 10 K rise , the velosity constant becomes nearly double . |
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| 605. |
Which of the following is the correct expression for Arrhenius equation ?A. `l n (k_(2))/(k_(1))=(E_(a))/(R ) ((1)/(T_(1))-(1)/(T_(2)))`B. `l n k = l n A - (E_(a))/(RT ) `C. `k=A*e^((E_(a))/(RT))`D. All the above |
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Answer» Correct Answer - D Different forms of Arrhenius equation . |
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| 606. |
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)` againstA. TB. `log T `C. `(1)/(T ) `D. `log ""(1)/(T)` |
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Answer» Correct Answer - C Arrhenius equation . |
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| 607. |
Following is the graph between `(a-x)` and time `t` for second order reaction `theta=tan^(-1)(0.5)OA=2L mol^(-)` Hence, the rate at the start of the reaction isA. `1.25 mol L^(-1) mi n^(-1)`B. `0.5 mol L^(-1)mi n^(-1)`C. `0.125 mol L^(-1) mi ^(-1)`D. `12.5 mol L^(-1) min^(-1)` |
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Answer» Correct Answer - C `kt=((1)/(a-x)-(1)/(a))` `(1)/(a-x)=kt+(1)/(a)` Graph between `(a-x)^(-1)` and time `t` is a straight line, hence, `k= tan theta= 0.5` `OA=(1)/(a)=2` `:. A=(1)/(2)=0.5` `:. R=k[A]^(2)` `= 0.5xx(0.5)^(2)` `=0.125Lmol^(-1)min^(-1)` |
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| 608. |
The half-life periof of the reaction in the above question isA. `1.386y` minB. `4` minC. `16` minD. `2` min |
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Answer» Correct Answer - B `t_(1//2)=(1)/(ka)=(1)/(0.5xx0.5)=4 min` |
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| 609. |
In the above question, the order w.r.t. B is:A. 3B. 2C. 1D. zero |
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Answer» Correct Answer - D d) In the second experiment, Rate `propto [A]^(x)[B]^(y)` `8 propto [2]^(3)[2]^(y)` y=0. order w.r.t. B is zero. |
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| 610. |
Why does the use of pressure cooker reduce cooking time ? |
| Answer» The pressure inside the cooker is independent of the atmospheric pressure. As the boiling point of water in the pressure cooker gets raised, the rate of cooking process therefore, increases or the cooking is done in less time. | |
| 611. |
If half life of a reaction is inersely proportional to initial concentration of the reactant, what is the order of reaction? |
| Answer» The reaction is of second order because for a second order reaction, `t_(1//2)=1/(ka)` | |
| 612. |
Rate if a reaction can be expressed by Arrhenius equations as : `k = Ae^(-E//RT)` In this equation , E representsA. The energy above 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 , TD. The fraction of molecules with energy greater than the activation energy of the reaction |
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Answer» Correct Answer - b None of the given options is perfectly correct . The closest choice is option (b) . |
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| 613. |
Which of the following is incorrect regarding catalysis?A. Change in catalyst of reaction may change the reaction product.B. Catalyst changes the pathway of reaction.C. In case of negative catalysis, each step of mechanism occur at higher threshould energy than uncatalysed reaction.D. Enzyme catalysis is homogeneous catalysis. |
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Answer» Correct Answer - C |
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| 614. |
According to the adsorption theory of catalysis , the speed of the reaction increases becauseA. Adsorption produces heat which increases the speed of the reactionB. Adsorption lowers the activation energy of the reactionC. The concentration of reactant molecules at the active centres of the catalyst becomes high due to adsorptionD. In the process of adsorption , the activation energy of the molecules become large |
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Answer» Correct Answer - b A catalyst lowers the activation energy of the reaction following adsorption mechanism . `because` Rate of reaction `prop (1)/("Activation energy")` |
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| 615. |
Which one of the following statements regarding catalysis in not true ?A. A small amount of the catalyst can catalyse a large amount of reactants.B. A catalyst does not alter Gibbs energy, `Delta_(r)G^(@)` of a reactionC. It catalyses the spontaneous but does not catalyse non-spontaneous reactions.D. A catalyst changes the equilibrium constant of a reversible reaction. |
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Answer» Correct Answer - D Catalyst does not change the equilibrium constant of a reverible reaction, rather it helps in attaining th equilibrium faster because it catalyes the forward as well as the backward reactions to the same exent so that the equilibrium state remains the same extent so but is attained erlier. A Small amount of the catalyst is needed because a catalyst does not take part in the overall reaction, but all of it is reformed in later steps. Thus, a catalyst does not appear in the balanced equation for the reaction. |
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| 616. |
For a first order reaction the units of A in Arrhenius equation will beA. `s^(-1)`B. `mol L^(-1)s^(-1)`C. `Jk^(-1)s^(-1)`D. `Jk^(-1)L^(-1)s^(-1)` |
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Answer» Correct Answer - A A is a constant having the same units as the rate constant. Thus, A may be considered as frequency `(s^(-1))` for first order reaction only. |
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| 617. |
Which of the following graphs describes the typical dependence of the rate constant of a reaction on temperature ?A. B. C. D. |
| Answer» When we increases temperature `(T)` , the rate constant in most reaction increases exponetially in accordance with the Arrhenius equation, `k=Ae^(-E_(a)//RT)` | |
| 618. |
Which one of the following statements regarding catalysis is true ?A. A catalyst alters the position of equilibrium.B. A catalyst speeds up the reaction by providing an altrnative path of lower activation energyC. A catalyst speeds up the reaction by decreasing the frequency factorD. A catalyst speed up the reaction by providing an altranative path of higher activation energy |
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Answer» Correct Answer - B Catalyst does not change the equilibrium constant of a reverible reaction, rather it helps in attaining th equilibrium faster because it catalyes the forward as well as the backward reactions to the same exent so that the equilibrium state remains the same but is attaned earlier. A Small amount of hte catayst is needd because a catalst does not take part in the overall reaction, but all of it is reformed in later steps. Thus, a catalyst does not appear in the balanced equation for the reaction. |
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| 619. |
Which of the following relations represents the temperature coefficient of a reaction ?A. `r_(t+10)/r_(t)`B. `(t_(k))/(t_(t)+10)`C. `k_(t+10)/k_(t)`D. `(E_(a)-K_(t+10))/(E_(a)+K_(t))` |
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Answer» Correct Answer - C The temperature coefficient of a reaction gives the number of times the rate of a reaction increases by a rise in temperature of `10^(@)C` It is expressed as ther ratio of the rate constants of a reaction at two different temperatureas (usually `35^(@)C` and `25^(@)C`) differing by `10k` . Thus Temperature coefficient `=(k_(35^(@))/(k_(25^(@)` The temperature coefficient of most of the reactions is positive and lies between `2` and `3` . A reaction whose temperature coefficient is negative is `2NO(g)+O_(2)(g)rarr2NO_(2)(g)` Thus, the rate of this reaction decreases with increases of temperature. It can be explained following: Reaction Mechanism `2NO(g)rarrN_(2)O_(2)(g)` `N_(2)O_(2)(g)rarr2NO_(2)(g)` At higher temperature more `N_(2)O_(2)` molecules are dissociated. The lower concentration of `N_(2)O_(2)` molecules at higher temperature result in decrease of reaction rate. |
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| 620. |
Which of the following relations are correct if `Delta H` represents only magnitude ?A. a. Exothermic reactions: `E_(a(f)) + Delta H = E_(a(b))`B. b. Endothermic reactions: `E_(a(f)) = E_(a(b)) + Delta H`C. c. Exothermic reactions: `Delta H gt E_(a)`D. d. Exothermic reactions: `Delta H lt E_(a)` |
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Answer» Correct Answer - A::B::D For any reaction, `Delta H =E_(a(f)) - E_(a(b))` Also, for exothermic reaction `Delta H ge E_(a)` |
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| 621. |
Which of the following statements is correct?A. Rate of reaction `prop(1)/(E_(a))`B. At lower temp. increase in temp. causes more change in the value of kC. Both (a) and (b) are correctD. None is correct |
| Answer» Correct Answer - C | |
| 622. |
Which of the following graphs represents zero order if `A rarr P` At `t = 0 rArr [A]_(0)` At `t = t rArr [A]_(t)`A. B. C. D. |
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Answer» Correct Answer - A::B::C::D `t_(1//2) prop (a)^(1-n)`, `[A]_(t) = [A]_(0) - kt` and `X = kt` are integrated form of zero order. |
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| 623. |
For a reaction `A+3BrarrP`, Rate = `(-d[A])/(dt)` ,the expression for the rate of reaction in terms of change in the concentration of `B,(-d[B])/(dt)` will be :A. `k[A]^(2)[B]`B. `k[A]^(2)[3B]`C. `3k[A]^(2)[B]`D. `(1//3)k=[A]^(2)[B]` |
| Answer» Correct Answer - C | |
| 624. |
The specific rate constant of a first order reaction depends on theA. timeB. concentration of the reactantC. temperatureD. concentration of the product |
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Answer» Correct Answer - C The rrate contant `(k)` is a proportionality constant in the relationship between rate and concentration. For every reaction (irrespective or reaction order), it has a fixed value at any given temperature, but it varties with temperature. |
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| 625. |
For an elementary processA. the order and the molecularity are identicalB. the order is always fractionC. the order is lesser than the molecularityD. the order is greater than the molecularity |
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Answer» Correct Answer - A For the general overall reaction `aA+bBrarrcC+dD` the exerimentally determined rate law expression has the from Rate `=k[A]^(x)[B]^(y)` The values of `x` and `y` are related to the coefficients of the reactants in the slowest step, influenced in some cases by earlier steps. |
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| 626. |
The rate of chemical reactionA. keeps on increasing with timeB. remains constant with timeC. keeps on decreasing with timeD. shows irregular trend with time. |
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Answer» Correct Answer - C Rate of reaction `prop` conc. Of reactants. As the reaction proceeds, concentration of the reactants decreases hence the rate also keeps on decreasing with time. |
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| 627. |
Which of the followinf is/are correct?A. It is unimolecular nucleophilic substitution reaction `S_(N)1` if or `II` is formed.B. It is bimolecular nucleophilic substitution reaction `S_(N)2` is `I` or `II` is formed.C. It is `S_(N) 1` if `I` and is enantomer are formed so that the mixture is racemic.D. It is `S_(N)2` if `II` is formed. |
| Answer» Correct Answer - C::D | |
| 628. |
In any unimolecular reactionA. Only one reacting species is involved in the rate determining stepB. the order and the molecularity of slowest step are equal to one.C. the molecularity of the reaction is one and order is zero.D. both molecularity and order of the reaction are one. |
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Answer» Correct Answer - A::B |
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| 629. |
A first order reaction has a rate constant `1.15xx10^(-3)s^(-1)`. How long will `5g` of this reactant take to reduce to `3g`` ?A. 444sB. 400sC. 528sD. 669s |
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Answer» Correct Answer - A `t = (2.303)/(k) "log" ([R_(0)])/([R]) = (2.303)/(1.15 xx 10^(-3)) "log" (5)/(3)` `=2.00 xx 10^(3) log 1.667 = 2 xx 10^(3) xx 0.2219 = 444s` |
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| 630. |
When the concentration of a reactant in reaction `A rarr B` is increased by `8` times but rate increases only `2` times, the order of the reaction would beA. `2`B. `1//3`C. `4`D. `1//2` |
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Answer» Correct Answer - B `r_(1) = k[A]^(n)` `r_(2) = 2r_(1) = k[8A]^(n)` `:. (2r_(1))/(r_(1)) = (8)^(n)` `(2)^(1) = (2)^(3n)` `:. 3n = 1` `n = (1)/(3)`Correct Answer - B `r_(1) = k[A]^(n)` `r_(2) = 2r_(1) = k[8A]^(n)` `:. (2r_(1))/(r_(1)) = (8)^(n)` `(2)^(1) = (2)^(3n)` `:. 3n = 1` `n = (1)/(3)` |
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| 631. |
The rate constant of forward and backward reactions for certain hypothetical reaction are `1.1 xx 10^(-2)` and `1.5 xx 10^(-3)`, respectively. The equilibrium constant of the reaction isA. `7.33`B. `0.733`C. `73.3`D. `733` |
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Answer» Correct Answer - A `k_(eq) = (1.1 xx 10^(-2))/(1.5 xx 10^(-3)) = 7.33`Correct Answer - A `k_(eq) = (1.1 xx 10^(-2))/(1.5 xx 10^(-3)) = 7.33` |
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| 632. |
A reaction that is of the first order with respect to reactant A has a rate constant `6 "min"^(-1)`. If we start with `[A] = 0.5 "mol" 1^(-1)`, when would `[A]` reach the value `0.05 mol 1^(-1)`A. `0.384 min`B. `0.15 min`C. `3 min`D. `3.84 min` |
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Answer» Correct Answer - A `t = (2.303)/(k)"log"(a)/(a-x)` Or `t = (2.303)/(6)"log"(0.5)/(0.05) = (2.303)/(6)log10` `= (2.303)/(6) = 0.384 min`. |
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| 633. |
Which one is not correctA. Rate of zero order reaction depends upon initial concentration of reactantB. Rate of zero order reaction does not depend upon initial concentration of reactantC. `t_(1//2)` of first order reaction is independent of initial concentration of reactantD. `t_(1//2)` of zero order reaction is dependent of initial concentration of reactant |
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Answer» Correct Answer - a The rate of zero order reaction is independent of the concentration of the reactants or the concentration of the reactant do not change with time . Thus , the rate of reaction remains constant. |
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| 634. |
Calculating time requireed to complete a definite fraction of first order reaction : A first order reaction is `20%` complete in 10 minutes. Calculate the time taken for the reaction to go to `75%` completion. Stratagy : First calculate the rate constant (k) using frist data and then calculate the required time to complete `75%` of the reaction by using the value of rate constant : |
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Answer» According to Eq. `(4.22)` we have `t_(f) = (2.303)/(k) log ((1)/(1-f))` or `k= (2.303)/(t_(f)) log( (1)/(1-f))` Substituting the values of `t_(f)(=10 min) and f(-20//100 or 0.2)`, we have `k = (2.303)/(10 min) log ((1)/(1-0.2))` `= (2.303)/(10 min) log ((1)/(0.8))` `= 0.0223 min^(-1)` Since rate constant dose not change as the reaction progresses, we can use its value to calculate the time takes for `75%` completion of the reaction. Let it be `t_(0.75)`, `t_(0.75) = (2.303)/(0.0223 minj^(-1)) log ((1)/(1-0.75))` `= 62.18` minutes . |
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| 635. |
A chemical reaction `2A rarr 4B+C`, in gaseous phase shows an increase in concentration of `B` by `5xx10^(-3) M` in `10` second. Calculate: (a) rate of appearance of `B`, (b) rate of the reaction, (c ) rate of disappearance of `A`. |
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Answer» (a) Rate of appearance of `B=("increase in[B]")/("time")=(5xx10^(-3))/10` `=5xx10^(-4) M s^(-1)` (`M` is mol `litre^(-1))` (b) Also, rate of reaction `=-1/2(d[A])/(dt)=1/4(d[B])/(dt)` `:. -1/2(d[A])/(dt)=1/4xx5xx10^(-4)` `:. -1/2(d[A])/(dt)=1.25xx10^(-4) M sec^(-1)` (c ) Rate of disappearance of `A=1.25xx2xx10^(-4)` `= 2.5xx10^(-4) M sec^(-1)` |
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| 636. |
The decompoistion of `N_(2)O_(5)` in `C CI_(4)` solution at `318 K` has been studied by monitoring the concentration of `N_(2)O_(5)` in the solution. Initially, the concentration of `N_(2)O` is `2.33 M` and after `184 min`, it is reduced to `2.08 M`. The reaction takes place according to the equation: `2N_(2)O_(5) rarr 4NO_(2) + O_(2)` Calculate the average rate of this reaction in terms of hours, minutes, and seconds. What is the rate of Production of `NO_(2)` during this period? |
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Answer» Average reaction rate `= -(1)/(2) ((Delta[N_(2)O_(5)])/(Delta t))` `= -(1)/(2) (([2.08-2.33]mol L^(-1))/(184 min))` `= -(1)/(2) ((-0.25 mol L^(-1))/(184 min))` `= 6.79xx10^(-4)mol L^(-1) min^(-1)` Average reaction rate in seconds `= 6.79xx10^(-4) (mol L^(-1))/(min)xx(1min)/(60s)` `= 0.113xx10^(-4) mol L^(-1)s^(-1)` `= 1.13xx10^(-5)mol L^(-1)s^(-1)` Average reaction rate in hours `= 6.79xx10^(-4)(mol L^(-1))/(min)xx(60min)/(1h)` `= 407.4xx10^(-4)mol L^(-1)h^(-1)` `= 4.07xx10^(-2) mol L(-1)h^(-1)` |
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| 637. |
The decomposition of `N_(2)O_(5)` in `(C)Cl_(4)` at 318 K is studied by monitoring the concentration of `N_(2)O_(5)` in the solution. Initially the concentration of `N_(2)O_(5)` is 2.4 `mol L^(-1)`. What is the rate of production of `NO_(2)` during this period in `mol L^(-1)min^(-1)` ?A. `4 xx 10^(-3)`B. `2 xx 10^(-3)`C. `1 xx 10^(-3)`D. `2 xx 10^(-4)` |
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Answer» Correct Answer - A a) `2N_(2)O_(5) to 4NO_(2) + O_(2)` `1//2 xx "Rate of disappearance of" N_(2)O_(5)` `=1/4 xx "Rate of production of" NO_(2)` Rate of production of `NO_(2)` = `4/2` Rate of disappearance of `N_(2)O_(5)` `(2.4 -2.0)/(200) = 0.4/100 = 4 xx 10^(-3) mol L^(-1)min^(-1)` |
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| 638. |
Can activation energy for reactions be zero? |
| Answer» No it is not possible. Thisis explained with the help of Arrhenius equation according to which k = `Ae(E_(a)//RT)`. If `E_(a)` is zero, then `k=Ae^(0)`= A. This means the rate constant is equal to collision frequency and every collision among the reacting species will lead to the products. Since this does not happen. It is not possible for a chemical reaction to have activation energy equal to zero. | |
| 639. |
which of the following is a first order reaction ?A. `2H_(2)O _(2(aq))overset(pt)to 2H_(2)O+O_(2(g))`B. `2HI to H_(2) +I_(2)`C. `2NO_(2) to 2NO+O_(2)`D. `2NO+O_(2) to 2NO_(2)` |
| Answer» Correct Answer - A | |
| 640. |
Given one exmaple each of zero order and first order reactions: |
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Answer» `2NH_(3)(g) underset("pt-catalyst")overset(1130K/"pressure")to N_(2)(g) + 3H_(2)(g)` (Zero order) `H_(2)O_(2) to H_(2)O(g) + 1//2 O_(2)(g)` (First Order) |
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| 641. |
Explain the terms i) Order of a reaction ii) Molecularity of a reaction. |
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Answer» i) Order of a reaction ii) Molecularity of a reaction |
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| 642. |
Order of a reaction can beA. FractionalB. zeroC. NegativeD. both (a) and (b) |
| Answer» Correct Answer - D | |
| 643. |
A biological Catalyst isA. An amino acidB. a carbohydrateC. A nitrogen moleculeD. an enzyme |
| Answer» Correct Answer - D | |
| 644. |
The number of atoms or molecules whose concentration changes during a chemical change is itsA. Order of reactionB. MolecularityC. changes in reactionD. dynamics |
| Answer» Correct Answer - A | |
| 645. |
A catalystA. Increases the average kinetic energy of reacting moleculeB. Increases the activation energyC. Alters the reaction mechanismD. increases the frequency of collision of reacting species |
| Answer» Correct Answer - C | |
| 646. |
Number of moles of a substance present in one litre volume is known asA. ActivityB. Molar concentrationC. Mole fractionD. Molality |
| Answer» Correct Answer - B | |
| 647. |
The rate constant of `n^(th)` order has units :A. `"liter "^(1-n) Mol ^(1-n) sec^(-1)`B. `Mol "litre"^(1-n)Sec `C. `Mol ^(1-n^(2))"litre"^(n^(2))sec^(-1)`D. `"Mole"^(1-n) :"litre "^(n-1) sec^(-1) ` |
| Answer» Correct Answer - D | |
| 648. |
The graph between `log k` versus `1//T` is a straight line. |
| Answer» Correct Answer - T | |
| 649. |
Rate law for the reaction `A + B to ` product is rate `=K [A]^(2) [B].` What is the rate of reaction at a given temperature is `0.22Ms^(-1)` , when [A]=1 M and [B]=0.25 M?A. `3.52M^(-2)s^(-1)`B. `0.88M^(-2)s^(-1)`C. `1.136M^(-2)s^(-1)`D. `0.05M^(-2)s^(-1)` |
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Answer» Correct Answer - B For reaction, `A+BrarrProduct` `dx/dt=k[A]^(2)[B]Rightarrow0.22=k(1)^(2)0.25` `therefore" "k=0.22/0.22=0.88" "M^(-2)s^(-1)` |
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| 650. |
The difference between the energy maximum along a reaction path and the energy of the reactant is called activation energy. |
| Answer» Correct Answer - T | |