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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.
| 751. |
For zero order reaction which is (are) true?A. The rate constant is dimenisonless.B. Amount of the reactant remains the same theoughout.C. `t_(1//2) prop` initial concentration of the reactant.D. A plot of concentration of reactant vs time is a straight. line with slope equal to `-k`. |
| Answer» Correct Answer - C::D | |
| 752. |
The activation energy of a reaction is zero. The rate constant of this reactionA. Increases with increase of temperatureB. Decreases with an increase of temperatureC. Decreases with decrease of temperatureD. Is nearly independent of temperature |
| Answer» Correct Answer - D | |
| 753. |
Which of the following is/are examples of pseudo unimolecular reactions?A. `CH_(3)CO_(2)C_(2)H_(5)+H_(2)Ooverset(H^(o+))rarrCH_(3)CO_(2)H + C_(2)H_(5)OH`B. `C_(12)H_(22)O_(11)+H_(2)Ooverset(H^(o+))rarrunderset(("Glucose"))(C_(6)H_(12)O_(6))+underset(("Fructose"))(C_(6)H_(12)O_(6))`C. `CH_(3)COCl+H_(2)OrarrCH_(3)CO_(2)H+HCl`D. `CH_(3)CO_(2)C_(2)H_(5)+H_(2)Ooverset(overset(ɵ)OH)rarrCH_(3)CO_(2)H+C_(2)H_(5)OH` |
| Answer» Correct Answer - A::B::C | |
| 754. |
The value of log `(k_(2))/(k_(1))` is equal toA. `(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`B. `(E_(a))/(2.303R)[(T_(1)T_(2))/(T_(2)-T_(1))]`C. `(E_(a))/(2.303R)[(T_(2)+T_(1))/(T_(1)T_(2))]`D. `(E_(a))/(2.303R)[(T_(1)T_(2))/(T_(1)+T_(2))]` |
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Answer» Correct Answer - A The value of `log.(k_(2))/(k_(1))` is equal to `(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` |
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| 755. |
Two first order reaction have half-life in the ratio `3:2`. Calculate the ratio of time intervals `t_(1):t_(2)`. The time `t_(1)` and `t_(2)` are the time period for `25%` and `75%` completion for the first and second reaction respectively:A. `0.311:1`B. `0.420:1`C. `0.273:1`D. `0.119:1` |
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Answer» Correct Answer - a `t_(1)=(2.303xxt_(1//2))/(0.693)"log"(100)/(100-25)` (For I) `t_(2)=(2.303xxt_(1//2))/(0.693) "log" (100)/(100-75)`, (For II) `:. t_(1)/t_(2)=3/2xx("log"(100)/(75))/("log"(100)/(25))=3/2xx0.125/0.602` `= 0.311:1` |
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| 756. |
In which of the following ways does an activated complex differ form an ordinary molecule?A. It is quite unstable and has no independent existence.B. `Delta_(f)H^(ɵ)` is probably poistive.C. The system has no vibrational character.D. The system has no vibrational character. |
| Answer» Correct Answer - A::C | |
| 757. |
In accordance to Arrhenius equation, the plot of log k against `(1)/(T)` is a straight line. The slope of the line is equal toA. `-(E_(a))/(R )`B. `-(E _(a))/(RT)`C. `-(E _(a))/(2.303 R)`D. `-2.303 E_(a)R` |
| Answer» Correct Answer - A | |
| 758. |
In which of the following ways does an activated complex differ from an ordinary complex?A. `DeltaH_(f)^(@)` is probably positiveB. The formation of activated complex need the presence of catalystC. The activated complex has lower energy level than the ordinary complexD. All of these |
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Answer» Correct Answer - d These are facts about activated complex. |
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| 759. |
The difference of energy between activated complex and that of the reactants is called ……………… . |
| Answer» Correct Answer - activation energy | |
| 760. |
The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by `10^(@)C` isA. The value of threshold energy increasesB. Collision frequency increasesC. The fraction of the molecules having energy equal to threshold energy or more increasesD. Activation energy decreases |
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Answer» Correct Answer - c When the temperature is increased , heat energy is supplied which increases the kinetic energy of the reacting molecules . This will increase the number of collisions and ultimately the rate of reaction will be enhanced . |
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| 761. |
Arrhenius equation isA. `A=ke ^(-E_(a)//RT)`B. `K=Ae^(Ea//RT)`C. `(A )/(K)=e^(E_(a)//RT )`D. `K=Ae^(-RT//E_(a)` |
| Answer» Correct Answer - C | |
| 762. |
The slope of a line in the graph of log k versus `1/T` for a reaction is `-5841`K. Calculate energy of activation for the reaction. |
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Answer» Slope = `-E_(a)/(2.303R)` or `E_(a) = -2.303 R xx "slope"` Slope =`-5841 K, R=8.314 JK^(-1)mol^(-1)` `therefore E_(a) = -2.303 xx (8.314 JK^(-1) mol^(-1)) xx (-5841 K) = 1.118 xx 10^(5)J mol^(-1)` |
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| 763. |
A graph plotted between `log k` versus `1//T` for calculating activation energy is shown byA. B. C. D. |
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Answer» Correct Answer - B According to Arrhenius equation, `k prop T` and `k = Ae^(-Ea//RT)` `log k = log A - (E_(a))/(2.303 xx R) xx(1)/(T)` (`y = c + mx`, equation of straight line) The value of `log k` should increase uniformly with `T` or decrease with `1//T`. Hence, the answer is (b). |
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| 764. |
The collision frequency isA. inversely proportional to the concentration of the treacting moleculesB. Proportional to the concentration of the reacting moleculesC. equal to the concentration of reactantsD. equal to the concentration of products |
| Answer» Correct Answer - B | |
| 765. |
Arrhenius equation isA. `A=ke ^(-E_(a)//RT)`B. `(A )/(k)=e^(-E_(a)//RT)`C. `k=Ae ^(-E_(a )//RT)`D. `k=Ae^(-RT//R_(a))` |
| Answer» Correct Answer - C | |
| 766. |
The activation energy of reaction is equal toA. Threshold energy + Energy of the prosuctsB. Threshold energy - energy of the reactantsC. Threshold energy + Energy of the reactantsD. Threshold energy - Energy of the products |
| Answer» Correct Answer - B | |
| 767. |
The reactions of higher order are rare becauseA. many body collisions involve very high activation energyB. many body collisons have a low energetically favouredC. many body collisions are not energetically favouredD. many body collisions can take place only in the gaseous phase |
| Answer» Correct Answer - B | |
| 768. |
The rate of reaction of spontaneous reaction is generally very slow . This is due to the fact thatA. the equolibrium energy of the reaction is largeB. the activation energy of the reaction if largeC. The reaction are exothermicD. the reaction are endothermic |
| Answer» Correct Answer - B | |
| 769. |
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 M`B. `3 M`C. `1.02 xx 10^(-4) M`D. `3.5 xx 10^(5) M` |
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Answer» Correct Answer - B `r = K[N_(2)O_(5)]` `:. [N_(2)O_(5)] = (r )/(J) = (1.02 xx 10^(-4))/(3.4 xx 10^(-5)) = 3 M` |
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| 770. |
For a zero order reaction, will the molecularity be equal to zero? Explain. |
| Answer» No, the molecularity for a reaction can neither be zero nor fractional. | |
| 771. |
If the rate of the reaction id equal to the rate constant, the order of the reaction isA. `3`B. `0`C. `1`D. `2` |
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Answer» Correct Answer - B For zero-order reaction `r = k`. |
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| 772. |
Rate of a reactionA. increases with increases in temperatureB. decreases with increases in temperatureC. does not depend on temperatureD. Does not depend on concentration |
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Answer» Correct Answer - A Increase in termperature favours favours collision of molecules |
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| 773. |
Molecularity is always a whole…………… while the order of reaction may be……………, ……………, ………………., or even zero. |
| Answer» Correct Answer - number, poistive, negative | |
| 774. |
During the course of a chemical reaction the rate of reactionA. cannot be predictedB. decreases as the reaction proceedsC. Increase as the proceedsD. remains constant throughout the reaction |
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Answer» Correct Answer - B Rate `=-(dx )/(dt) ` hence decrease as reaction proceeds |
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| 775. |
In the elementary reaction `A + B to AB ` , if the concentration A and B is doubled , the rate of reaction willA. Be doubledB. Be halvedC. Increase bu 6 timesD. Incrase by 4 |
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Answer» Correct Answer - D Rate `=[A][B]` ,If concentration of A and B is doubled , the rate will increase by ` 2xx2=4` times |
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| 776. |
Which of the following statement (s) `is//are true?A. For zero order reaction, `t_(1//2) prop[R]_(0)`B. For first order reaction, `t_(1//2) "is independent of" [R]_(0)`C. Both (a) and (b)D. None of the above |
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Answer» Correct Answer - C For zero order reaction `t_(1//2) prop [R]_(0)` First order reaction `t_(1//2)"is independent of" [R]_(0)`. |
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| 777. |
`cdotsAcdots` dependence of rate is caled defferential rate equation. Choose the suitable word to replace A.A. ConcentrationB. VolumeC. OrderD. Pressure |
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Answer» Correct Answer - A Concentration dependence of rate is called defferential rate equation. |
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| 778. |
What is the instantaneous rate of a reaction. |
| Answer» It is the reaction rate at a particular time. | |
| 779. |
What are the units of rate constant for third order reaction? |
| Answer» `k=mol^(-2)L^(2)s^(-1)` | |
| 780. |
`._(z)^(M)A (g) to ._(z-4)^(M-8)B(g)+(alpha-"particles")` `(alpha-"participles are helium nuclei, so will form helium gas by trapping electrons")` The radioactive disintegration follows first-rder kinetics. Starting with 1 mol of A in a 11-litre closed flask at `27^(@)C` pressure developed after two half-lives is approximately:A. 25 atmB. 12 atmC. 61.5 atmD. 40 atm |
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Answer» Correct Answer - C |
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| 781. |
The reaction , `X+2Y+Z rarr N` occurs by the following mechanism: (P) `X + Y hArr M` (rapid equilibrium) (Q) `M+Z rarr P` (slow) (R) `O+Y+P rarr N` (very fast) What is the rate law for this reaction?A. Rate =`k`[Z]B. Rate =`k [X][Y]^(2)[Z]`C. Rate =`k` [N]D. Rate =`k`[X][Y][Z] |
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Answer» Correct Answer - D |
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| 782. |
The rate of constant of a reaction depends onA. TemperatureB. MassC. WeightD. Time |
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Answer» Correct Answer - a k = `Ae^(-E_(a)//RT)` by this equation it is clear that rate constant of a reaction depends on temperature . |
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| 783. |
The half-life of a first order reaction having rate constant K = `1.7 xx 10^(-5) s^(-1)` isA. 12.1 hB. 9.7 hC. 11.3 hD. 1.8 h |
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Answer» Correct Answer - c k = `1.7 xx 10^(-5) s^(-1)` `t_(1//2) = (0.693)/(k) = (0.693)/(1.7) xx 10^(5) = 11.32` h. |
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| 784. |
If in a chemical reaction, `A+B to `Products, the rate law is given the expression : rate = `k[A](1/2)[B]^(3/2)`. What is the order of the reaction? |
| Answer» Order of reaction is `:1//2 + 3//2=2` | |
| 785. |
For a first order reaction half life is 14 sec. The time required for the initial concentration to reduce 1/8 of the value isA. `(14)^(3)s`B. 28sC. 42D. `(14)^(2)s` |
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Answer» Correct Answer - C `N=N_(0)xx((1)/(2))^(n)` `(1)/(8)N_(0)=N_(0)xx((1)/(2))^(n)` n=3 `T=nxxt_(t//2)=3xx14=42s` |
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| 786. |
For a first order reaction half life is 14 sec. The time required for the initial concentration to reduce 1/8 of the value isA. `(14)^(3) sec`B. 28 secC. 42 secD. `(14)^(2) sec ` |
| Answer» Correct Answer - C | |
| 787. |
The slope of a graph l n `[A_(t)]` versus t for a first order reaction is ` -2.5xx10^(-3) s^(-1)` . The rate constant for the reaction will beA. `5.76 xx10^(-3) S^(-1) `B. `1.086xx10^(-3)S^(-1) `C. `-2.5xx10^(-3) S^(-1)`D. `2.5 xx10^(-3) S^(-1)` |
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Answer» Correct Answer - A Slope `=(-k)/(2.303)` `k=2.5xx10^(-3)xx2.303=5.76xx10^(-3)S^(-1)` |
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| 788. |
For the single step reaction of the type `A+2B to E +2F ` , the rate law isA. `"Rate " =k [A][B]`B. `"rate"=(k[E][F]^(2))/([A][B]^(2))`C. `"Rate"=k [A][2B]`D. `"rate" =k[A][B]^(2)` |
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Answer» Correct Answer - D Elementary reaction . |
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| 789. |
For the first order reaction with half life is 150 seconds , the time for the concentration of the reactant to fall from m/ 10 to m/100 will be approximatelyA. 600 sB. 900sC. 500sD. 1500s |
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Answer» Correct Answer - C `(M )/(10)overset(t_(1//2))to (M )/(20)overset(t_(1//2))to (M )/(40 )overset(t_(1//2))to (M )/(80 )overset(t_(1//2)to (M)/(180)` ` therefore T=~3xxt_(1//2)=~450 "to" 600s` |
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| 790. |
Observe the following reaction: `A(g) + 3B(g) rarr 2C(g)` The rate of this reaction `{(-d[A])/(d t)}` is `3 xx 10^(-3) "mol litre"^(-1) "min"^(-1)`. What is the value of `(-d[B])/(d t)` in mol `"litre"^(-1) "min"^(-1)` ?A. `3xx10^(-3)`B. `9xx10^(-3)`C. `10^(-3)`D. `1.5xx10^(-3)` |
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Answer» `(-d[A])/(dt)=-(1)/(3)(d[B])/(dt)` `:. (-d[B])/(dt)=3{(-d[A])/(dt)}=3xx3xx10^(-3)` `9xx10^(-3) "mol litre"^(-1) "mon"^(-1)` |
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| 791. |
Which of the following statement is wrong?A. Law of mass action and rate law expression are same for single step reactionsB. Both order of reaction and molecularity have normally a miximum value of `3`C. Order of reaction and molecularity for elementary reaction are sameD. Molecularity of a complex reaction `A+2Brarr C` is `3` |
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Answer» Correct Answer - d Molecularity of complex reaction is not to be given by simply looking the net chemical change. |
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| 792. |
The rate law for the decomposition of `N_(2)O_(5)` is given as: Rate =`k[N_(2)O_(5)]`. What is the significance of k in the equation? |
| Answer» k represents the rate constant for the reaction. It becomes equal to the reaction rate when the molar concentration of `N_(2)O_(5)` is 1 mol `L^(-1)`. | |
| 793. |
What is the effect of temperature on the rate constant of reaction? How can this temperature effect on the rate constant be represented quantitatively? |
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Answer» The rate constant (k) for a reaction increase with rise in temperature and becomes nearly double with about every `10^(@)` rise in temperature. The effect is expressed with Arrhenius equation. `k =(Ae^(-Ea))/(RT)` |
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| 794. |
The `DeltaH` value of the reaction `H_(2) + Cl_(2) hArr 2 HCl` is `-44.12` kcal . If `E_(1)` is the activation energy of the products , then for the above reactionA. `E_(1) gt E_(2)`B. `E_(1) lt E_(2)`C. `E_(1) = E_(2)`D. `DeltaH` is not related to `E_(1)` and `E_(2)` |
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Answer» Correct Answer - a Because reaction is exothermic . |
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| 795. |
Which of the following is the fastest reactionA. `C + (1)/(2) O_(2)overset(250^(@)C)(to) CO`B. ` C + (1)/(2) O_(2) overset(500^(@)C)(to) CO`C. `C + (1)/(2) O_(2) overset(750^(@)C)(to) CO`D. `C + (1)/(2) O_(2) overset(1000^(@)C)(to) CO` |
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Answer» Correct Answer - d Combination is an exothermic process , which is expected to be favoured by low temperature , but it is not true . Combustion include burning of particles which takes place at higher temperature . Thus at high temperature due to combustion of more particles , the reaction proceeds at higher rate . |
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| 796. |
A Substance undergoes first order decomposition . The decomposition follows two parallel first order reactions as The percentage distribution of B and C areA. 75% B and 25% CB. 80% B and 20% CC. 60% B and 40% CD. 76.83% B and 23.17% C |
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Answer» Correct Answer - d % distribution of B = `(k_(1))/(k_(1) + k_(2)) xx 100` `= (1.26 xx 10^(-4))/(1.26 xx 10^(-4) + 3.8 xx 10^(-4)) xx 100` B% = 76.83 % %Distribution of C = `(k_(2))/(k_(1) + k_(2)) xx 100` =` (3.8 xx 10^(-4))/(1.26 xx 10^(-4) + 3.8 xx 10^(-4)) xx 100` C% = 23.17 %. |
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| 797. |
A substance undergoes first order decomposition. The decomposition follows two parallel first order reaction as: and `KP_(1)=1.26xx10^(-4)sec^(-1)` `K_(2)=3.80xx10^(-5)sec^(-1)` The percentage distribution of B and C are:A. `80%B and 20%C`B. `76.83% and 231.7%C`C. `90%B and 10%C`D. `60%B and 40%C` |
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Answer» Correct Answer - b For parallel path reaction, `K_("average")=K_(1)+K_(2)` `=1.26xx10^(-4)+3.8xx10^(-5)` `1.64xx10^(-4) sec^(-1)` Also fractional yield of `B=K_(B)/K_(av)` `=(1.26xx10^(-4))/(1.64xx10^(-4))=0.7683` Fraction yield of `A=K_(A)/K_(av)` `=(3.8xx10^(-5))/(1.64xx10^(-4))=0.2317` |
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| 798. |
For a first order reaction, which of the following relation is not correct?A. `K=K_(1)+K_(2)`B. `(1)/(tau)=(1)/(tau_(1))+(1)/(tau_(2))`C. `(1)/(t_(1//2))=((1)/t_(1//2))_(1)+((1)/(t_(1//2)))_(2)`D. `(1)/(K)=(1)/(K_(1))+(1)/(K_(2))` |
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Answer» Correct Answer - d For parallel path decay, `K_(av)=K_(1)+K_(2)` Also `K=1/T and K=0.693/t_(1//2)` |
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| 799. |
Half-life `(t_(1))` of the first order reaction and half-life `(t_(2))` of the second order reaction are equal. Hence ratio of the rate at the start of the start of the reaction:A. `1`B. `2`C. `0.693`D. `1.44` |
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Answer» Correct Answer - c For I order, `K_(1)=0.693/t_(1//2)`, For II order, `K_(2)=1/(t_(1//2)a)` `K_(1)=0.693/T_(1), K_(2)=1/(T_(2)a)` If `T_(1)=T_(2)`, then `K_(1)/K_(2)=0.693a` Initially `r_(1)=K_(1)[a]^(1), r_(2)=K_(2)[a]^(2)` `:. R_(1)/r_(2)=K_(1)/(K_(2)a)=(K_(2)xx0.693xxa)/(K_(2)xxa)=0.693` |
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| 800. |
In a second order reaction, first order in each reactant `A` and `B`, which one of the following reactant mixtures will provide the highest initial rate?A. `0.1` mole of `A` of 0.1 mole of `B` in `0.1`litre solution.B. `0.2` mole of `A` of 0.2 mole of `B` in `0.1` litre solution.C. `0.1` mole of `A` and `0.1` mole of `B` in `1` litre solution.D. `0.1` mole of `A` and `0.1` mole of `B` in `0.2`litre solution. |
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Answer» Correct Answer - b Rate=`K[A]^(1)[B]^(1)=K[([0.2])/0.1][([2.0])/0.1]` is maximum. |
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