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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.
| 1101. |
By the overall order of a reaction , we meanA. The number of concentration terms in the equation for the reactionB. The sum of powers to which the concentration terms are raised in the velocity equationC. The least number of molecules of the reactants needed for the reactionD. The number of reactants which take part in the reaction |
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Answer» Correct Answer - b The overall order of a reaction is sum of powers raised on concentration terms in order to write rate expression. |
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| 1102. |
The rates of a certain reaction (dc/dt) at different times are as follows The reaction isA. Zero orderB. First orderC. Second orderD. Third order |
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Answer» Correct Answer - a The concentration of reactant does not change with time for zero order reaction (unit of k suggests zero order ) since reactant is in excess. |
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| 1103. |
The reaction `H_(2)O_(2)to H_2O +[O]` is aA. first order reactionB. Second order reactionC. zero order reactionD. third order reaction |
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Answer» Correct Answer - A Rate `=K [H_(2) O_(2)] , n=1` |
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| 1104. |
The reaction `3A to 2B+C` is carried in a closed vessel. The rate of disappearance of `A[(-d[A])/(dt)]` is 0.01 `molL^(-1)s^(-1)`. Calculate `(d[B])/(dt)` and `(d[C])/(dt)`. |
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Answer» For the reaction, `3A to 2B + C` Rate = Rate = `1/3 xx (-d[A])/(dt) = 1/2 xx(d[B])/(dt)=(d[C])/(dt)` `(d[B])/(dt) = 2/3 xx (d[A])/(dt) = 2/3 xx 0.01 = 0.0067 mol L^(-1)s^(-1)` `(d[C])/(dt) = 1/3 xx (d[A])/(dt) = 1/3 xx 0.01 = 0.0033 mol L^(-1)s^(-1)` |
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| 1105. |
The term `(-(dc)/(dt))` in a rate equation refers to theA. Concentration of the reactantB. Decrease in concentration of the reactant with timeC. Increase in concentration of the reactant with timeD. Velocity constant of the reaction |
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Answer» Correct Answer - b `-(dc)/(dt)` refers as decrease in concentration of the reactant with time . |
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| 1106. |
The unit of the rate constant for first order reaction isA. `Mol ^(-1)`B. `Sec^(-1)`C. `Sec^(-1) mol_(-1) dm^(3)`D. `Mol Sec^(-1) dm ^(3)` |
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Answer» Correct Answer - B `(Mol Lit ^(-1) )^(1-n)S^(-1) ` when `n=1` unit is ` S^(-1)` |
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| 1107. |
What is the order of a reaction if the rate constant has the units `"L.mol:"^(-1).s^(-1)`?A. zeroB. FirstC. SecondD. Third |
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Answer» Correct Answer - C |
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| 1108. |
Correct order for a first order reaction isA. `t_(1//2)prop C^(-1)`B. `t_(1//2)prop C `C. `t_(1//2)prop C^(0)`D. `t_(1//2)prop C^(1/2)` |
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Answer» Correct Answer - C `t_(1//2) prop C^(@)` for 1 st order reaction . |
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| 1109. |
The initial rate , `-(d [A])/(dt)` at t = 0 was found to be `2 . 6 xx 10^(-2)` mol `L^(-1) s^(-1)` for the reaction `A + 2B to ` Products The initial rate , `- (d[B])/(dt)` at t = 0 isA. `0.01` mol `L^(-1) s^(-1)`B. `2.6 xx 10^(-2)` mol `L^(-1) s^(-1)`C. `5.2 xx 10^(-2)` mol `L^(-1) s^(-1)`D. `6.5 xx 10^(-3)` mol `L^(-1) s^(-1)` |
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Answer» Correct Answer - c `A + 2 B implies ` products `-(d[A])/(dt) = (-d[B])/(2dt)` `2.6 xx 10^(-2)` mol `L^(-1) s^(-1) = (-d[B])/(2dt)` `(-d[B])/(dt) = 2 xx 2.6 xx 10^(-2)` `= 5.2 xx 10^(-2)` mol `L^(-1) s^(-1)` |
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| 1110. |
The unit of the rate constant of nth order isA. `"mol"^(1-n)L^(n-1)s^(-1)`B. `"mol"^(n-1)L^(1-n)s^(-1)`C. `"mol"^(n-1)L^(n-1)s`D. `"mol"^(n)L^(1-n)s^(-1)` |
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Answer» Correct Answer - A The units of rate constant of nth order of `mol^(1-n)L^(n-1)s^(-1)` |
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| 1111. |
Which of the following is a first order reaction ?A. `NH_(4)NO_(2) to N_(2)+2H_(2)O`B. `2HI hArr H_(2) +I_(2)`C. `2NO_(2)hArr 2NO+O_(2)`D. `2NO_(2) hArr 2NO +O_(2)` |
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Answer» Correct Answer - A Rate `k[A][B]` `"volume" prop =(1)/("presesure or concentration ")` Volume reduces by `1//4^(th) ` concentration increases by 4 times Rate of reaction increases by `(4)^(2)` times |
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| 1112. |
For the reaction, `A(g)+2B(g)rarrC(g)+D(g)" "((dx)/(dt))=k[A][B]^(2)` Initial pressure of A and B are respectively `0.60` and `0.80` atm. At a time when pressure of C is `0.20` atm rate of the reaction, relative to the initial value is :A. `(1)/(6)`B. `(1)/(48)`C. `(1)/(4)`D. `(1)/(24)` |
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Answer» Correct Answer - A |
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| 1113. |
In the presence of acid, the initial concentration of cane sugar was reduced from 0.2 M to 0.1 M in 5 h and to 0.05 M in 10 h. The reaction must be ofA. Zero orderB. First orderC. Second orderD. Fractional order |
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Answer» Correct Answer - B `0.2 M underset(t_(1//2) = 5 hr)rarr 0.1 M underset(t_(1//2) = 5 hr)rarr0.05 M` form `0.2 M underset(t = 10 hr)rarr 0.05 M` So `t_(1//2)` is constant which is characteristic of first order reaction. Hence, `t_(1//2) prop (a)^(0)`.Correct Answer - B `0.2 M underset(t_(1//2) = 5 hr)rarr 0.1 M underset(t_(1//2) = 5 hr)rarr0.05 M` form `0.2 M underset(t = 10 hr)rarr 0.05 M` So `t_(1//2)` is constant which is characteristic of first order reaction. Hence, `t_(1//2) prop (a)^(0)`. |
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| 1114. |
The rate constant of a reaction is `2.5xx10^(-2) "minutes"^(-1)` The order of the reaction isA. oneB. zeroC. twoD. three |
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Answer» Correct Answer - A `k=("Rate ")/(["concentration "])` for first order k is independent of concentraion . |
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| 1115. |
For a reaction between gaseous compounds, `2A + B rarrC + D` , the reaction rate law is rate k[A][B]. If the volume of the container is made `1//4^(th)` of the initial, then what will be the rate of reaction as compared to the initial rate?A. `(1)/(16)`B. `(1)/(8)`C. `16` timesD. 8 times |
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Answer» Correct Answer - C Concentration will increase by 4 times of each reactant `therefore ` Rate will become ` (4)^(2)` I .e 16 times |
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| 1116. |
Gasesous cyclobutane isomerizes to butadiene following first order process which has half life of `150.5` minute at certain temperature. How long will take for the process to occur to the extent of `40%` at the same temperature ?A. 103 minutesB. 121 minutesC. 111 minutesD. None of these |
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Answer» Correct Answer - 3 `k=(0.693)/(t_(1//2))=(0.693)/(150.5)mi n^(-1),t=(2.303)/(k)log.(100)/(100-40)=(2.303xx150.5)/(0.693)log.(100)/(60)=111` minutes |
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| 1117. |
Half life of a first order reaction is `69.3` minutes. Time required to complete `99.9%` of the reaction will be `:`A. 693 minutesB. 999 minutesC. `99.9` minutesD. 691 minutes |
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Answer» Correct Answer - 4 `t=(2.303xxt_(1//2))/(0.693)log.(100)/(100-99.9)=(2.303xx69.3)/(0.693)log1000=691mi n` |
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| 1118. |
For the reaction: `[Cu(NH_(3))_(4)]^(2+) + H_(2)O hArr [Cu(NH_(3))_(3)H_(2)O]^(2+)+NH_(3)` The net rate of reaction at any time is given net by: rate `=2.0xx10^(-4) [[Cu(NH_(3))_(4)]^(2+)][H_(2)O]-3.0xx10^(5) [[Cu(NH_(3))_(3)H_(2)O]^(2+][NH_(3)]` Then correct statement is (are) :A. Rate constant for forward reaction `=2xx10^(-4)`B. Rate constant for backward reaction= `3xx10^(5)`C. Equilibrium constant for the reaction= `6.6xx10^(-10)`D. All of these |
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Answer» Correct Answer - d Net rate of reaction = rate of forward recation - rate of backward reaction `=K_(f) ["reactant"]-K_(b)`[product] Also `K_(c )=K_(f)/K_(b)` (at equilibrium) |
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| 1119. |
Mark the correct statement about given graph: A. `X` is threshold energy levelB. `Y` and `Z` are energy of activation for forward and backward reaction respectivelyC. `Q` is heat of reaction and reaction is exothermicD. All of these |
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Answer» Correct Answer - d All these are standard facts for given graph. |
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| 1120. |
A molecule of gas is struck by another molecule of the same gas, the first molecule shows:A. An exchange of potential energyB. An exchange of kinetic energyC. An exchange of chemical energyD. No exchange of energy |
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Answer» Correct Answer - b No doubt total kinetic energy of gas remains same but exchange of kinetic energy takes place between molecules. |
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| 1121. |
The maximum value of activation energy is equal to:A. ZeroB. Heat of the reactionC. Threshold energyD. None of these |
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Answer» Correct Answer - d Activation energy may be greater than heat of reaction or lesser than threshold energy. |
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| 1122. |
At the point of intersection of the two curves shown, the conc. of `B` is given by ……… for, `A rarr nB`: A. `(nA_(0))/2`B. `A_(0)/(n-1)`C. `(nA_(0))/(n+1)`D. `((n-1)/(n+1))A_(0)` |
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Answer» Correct Answer - c From fig., `[A]_("left")=[B]_("formed")=nxx[A]_("decayed")` `:. [A_(0)]e^(-lambdat)=nxxA_(0)[1-e^(-lambdat)]` `:. e^(-lambdat)=n/(n+1)` `:. [B]_("formed")=nxxA_(0)xx[1-n/(n+1)]` `=(nA_(0))/(n+1)` |
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| 1123. |
At the point of intersection of the two curves shown, the conc. of `B` is given by ……… for, `A rarr nB`: A. `(nA_(0))/(2)`B. `(A_(0))/(n-1)`C. `(nA_(0))/(n+1)`D. `((n-1)/(n+1))A_(0)` |
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Answer» Correct Answer - C From figure `[A]_(l e ft)=[B]_("formed")=nxx[A]_("decayed")` `:." "[A]_(b)e^(lambdat)=nxxA_(0)[1-e^(-lambdat)]` `:." "[B]_("formed")=nxxA_(0)xx[1-(n)/(n+1)]` `=(nA_(0))/(n+1)` |
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| 1124. |
The half-lives for two samples are 0.1 and 0.8 s whose concentrations are 400 and 50 mol `L^(-1)` respectively. The order of the reaction is |
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Answer» Correct Answer - C The ratio of two half-lives and their concentrations are related as `((t_(1//2))_(1))/((t_(1//2))_(2))=[(a_(2))/(a_(1))]^(n-1)` [Here, n = order] Therefore, `(0.1)/(0.8)=[(50]/(400)]^(n-1)` On taking log of both sides `log .(0.1)/(0.8)=(n-1)log. (50)/(400)` `1 = n-1implies n=2` |
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| 1125. |
The rate constant for the first order reaction is `60 s^(-1)`. How much time will it take to reduce the concentration of the reactant to `1//16th` value ?A. `4.6 xx 10^(4) s`B. `4.6 xx 10^(-4) s`C. `4.6 xx 10^(-2) s`D. `4.6 xx 10^(2) s` |
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Answer» Correct Answer - C For the first order reaction, `k=(2.303)/(t)log.([A_(0)])/([A])` Then, `k=(2.303)/(t)log. ((a)/(a//16))=0.0462=4.62xx10^(-2)s` |
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| 1126. |
Statement -1: In `A overset(k_(1))to B overset(k_(2))to C ` If half life of A is very less as compared to B, then, net reaction is A to C with rate constant `(k_(1)xxk_(2))` Statement-2: Slowest step is the rate determining step si B to C is rate determining step.A. Statement-1 is True, statement-2 is True, Statement-2 is a correct explantion for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - d |
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| 1127. |
Statement-1: For a reaction `A(g) to B(g)` `-r_(A)=2.5P_(A) " " at 400K` `-r_(A)=2.5P_(A) " "at 600K ` Activation energy is 4135 J//mol//k, `(Take: R=8.314J//mol//k, " " log 2=0.3, log 3=0.48) ` Statement-2: Since for any reaction , values of rate constant at two different temperatures are same therefore activation energy of the reaction is zero.A. Statement-1 is True, statement-2 is True, Statement-2 is a correct explantion for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - b |
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| 1128. |
Identify the statement which is correct from the followingA. Rate of reaction increases substantially with increases in temperature since frequency of collision increases.B. Rate of reaction decreases on increasing temperature if the reaction is exothermic.C. Rate of reaction remains unchanged if temperature changes for reactions having no activation energy.D. Rate of reactions decreases on decreasing temperature only if the reaction is endothermic |
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Answer» Correct Answer - C |
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| 1129. |
The energy required to form the intermediate, called activated complex `(C )` is known as activation energy `(E_(a))`. The diagram is obtained by plotting potential energy vs. reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the heat of the reaction depends upon the nature of reactants and products. If a reaction `A+Brarr C`, is exothermic to the extent of `30 kJ mol^(-1)` and the forward reaction has an activation energy of `249 kJ mol^(-1)`, the activation energy for reverse reaction in `kJ mol^(-1)` is:A. (a) `324`B. (b) `279`C. (c ) `40`D. (d) `100` |
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Answer» Correct Answer - b `DeltaH=E_(f)-E_(b)` `-30=249-E_(b)` `:. E_(b)=279 kJ mol^(-1)` |
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| 1130. |
For a simple reaction `A to B` it is observed that % of activated molecules at a temperature of 400 K is equal % `100/(e^(10)) %`. Activation energy of the reaction is given by :A. 33.256 k JB. 3.3256 JC. 33.256 JD. 3.3256 kJ |
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Answer» Correct Answer - A |
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| 1131. |
The energy required to form the intermediate, called activated complex `(C )` is known as activation energy `(E_(a))`. The diagram is obtained by plotting potential energy vs. reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the heat of the reaction depends upon the nature of reactants and products. If a reaction `A+Brarr C`, is exothermic to the extent of `30 kJ mol^(-1)` and the forward reaction has an activation energy of `249 kJ mol^(-1)`, the activation energy for reverse reaction in `kJ mol^(-1)` is:A. (a) `E_(1) gt E_(2)`B. (b) `E_(1) lt E_(2)`C. (c ) `E_(2) = 2E_(1)`D. (d) `sqrt(E_(1)E_(2)^(2))=1` |
| Answer» Correct Answer - a | |
| 1132. |
The energy required to form the intermediate, called activated complex `(C )` is known as activation energy `(E_(a))`. The diagram is obtained by plotting potential energy vs. reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the heat of the reaction depends upon the nature of reactants and products. If a reaction `A+Brarr C`, is exothermic to the extent of `30 kJ mol^(-1)` and the forward reaction has an activation energy of `249 kJ mol^(-1)`, the activation energy for reverse reaction in `kJ mol^(-1)` is:A. (a) `E_(a_(1)) gt E_(a_(2))`B. (b) `E_(a_(1)) = E_(a_(2))`C. (c ) `E_(a_(1)) lt E_(a_(2))`D. (d) `E_(a_(1)) ge E_(a_(2))` |
| Answer» Correct Answer - c | |
| 1133. |
Rate of reaction is given by the equation Rate = `k[A]^(2)[B]` What are the units for the rate and rate constant for the section? |
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Answer» Units of rate ( R) : Moles `L^(-1)s^(-1)` Units of `k:L^(2)mol^(-2)s^(-1)` |
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| 1134. |
Select the correct statement (s).A. The value of rate constatnt cannot exced the value of Arrhenius factor.B. Molecularity of multiple step reaction can be obtained from mechanism.C. Half lifeof a foruth order reaction is linearly dependent on intial concentration of reactantD. reactions with order `ge1` cannot get completed in finite time interval. |
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Answer» Correct Answer - A::D |
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| 1135. |
A certain reaction `A rarr B` follows the given concentration (Molarity)-time graph. Which of the following statements is/are true? A. The reaction is second order with respect to `A`.B. The rate for this reaction at `20 s` will be `7 xx 10^(-3) M s^(-1)`C. The rate for this reaction at `80 s` will be `1.75 xx 10^(-3) M s^(-1)`D. The `[B]` will be `0.35 M` at `t = 60 s`. |
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Answer» Correct Answer - B::D Use initial rate law methof, let the reaction be first order, So, `k = (2.303)/(t) log.(a)/(a-x)` (b) At `t =20 s, k_(1) = (2.303)/(20)log.((0.4)/(0.2))` and at `t = 40 s, k_(1) = (2.303)/(40)s, k_(2) = (2.303)/(40) log. ((0.4)/(0.1)) = k_(1)` `rArr` Assumption is correct `(because k_(1) = k_(2))` Rate at `20 s = k[A] = (0.693)/(20) xx 0.2` `= 0.0063 ~~ 7 xx 10^(-3) M s^(-1)` Clearly, half life `t_(1//2) = 20 s` (d) In `60 s`, number of life `= (60)/(20) =3` `rArr [B]` at `60 s` `= 0.4 xx 0.4 ((1)/(2))^(3) = 0.35 M` |
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| 1136. |
Select the correct statements.A. The molecularity of an elementary reaction indicates how many reactant species take part in the step.B. The rate law of an elementary reaction can be predicted by simply seeing the stoichiometry of reaction.C. The slowest elementary step in sequence of the reafion governs the overall rate of formation of product.D. A rate law is often derived from a proposed mechanism by imposing the steady state approximation. Or assuming that there is a pre-equilibrium. |
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Answer» Correct Answer - A::B::C::D |
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| 1137. |
The rate at which a substance react depends upon its :A. increases as the reaction proceedsB. decreases as the reaction proceedsC. my increases or decreases during the reactionD. reamins constant as the reaction proceeds |
| Answer» Correct Answer - D | |
| 1138. |
For a first order reaction,A. The degree of dissociation is equal to `(1-e^(-kt))`.B. A plot of reciprocal concentration of the reactant `vs` time gives a straight line.C. The time taken for the completion of `75%` reaction is theice the `t_(1//2)` of reaction.D. The pre-exponential factor in the Arrhenius equations has the dimenison of time, `T^(-1)`. |
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Answer» Correct Answer - A::D For a first order reaction,`A rarr` Products, we have `ln.([A]_(t))/([A]_(0)) = -kt` or `[A]_(t) = [A]_(0) e^(-kt)` In terms of degree of dissociation, m `[A]_(t) = [A]_(0) (1-alpha)` Hence, `[A] (1- alpha) = [A]_(0) e^(kt)` or `alpha = 1-e^(kt)` The Arrhenius equation is `k = Ae^(-Ea//RT)` The dimenison of pre-exponential factor `A` in the above expresison is the same as that of rate constant `k`. For a first order reaction, the dimenison of `k` is `t^(-1)`. |
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| 1139. |
order of reaction isA. never zeroB. never fractionalC. always equal to number of molecules taking part in reactionD. an experimentally determined quantity |
| Answer» Correct Answer - D | |
| 1140. |
The rate of a chemical reaction can be expressed in terms ofA. rate of concentration of catalystB. rate of concentration of reactants onlyC. rate of consumption of reactants and formation of products bothD. rate of formation of products only |
| Answer» Correct Answer - C | |
| 1141. |
The rate at which a substance reacts depends on itsA. Atomic weightB. Equivalent weightC. Molecular weightD. Active mass |
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Answer» Correct Answer - d According to law of mass action . |
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| 1142. |
Which of the following statements are correct about halr-period ?A. It is proportional to intial concentration for zeroth orderB. Average life `=1.44xx` half-life for first order reaction.C. Tiem of `75%` comletion of reaction is thrice of halr-life (intial harl-life) period in second order reactionD. `99.9%` reaction takes place in `100` minutes for the case when reate constant is `0.693 min^(-1)` |
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Answer» Correct Answer - A::B::C::D |
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| 1143. |
Which of the following statement `(s)` is/are correctA. The rate constant for the reaction `2N_(2)O_(5) rarr 4NO_(2)+O_(2)`, is `3.0 xx 10^(-5) s^(-1)`. If the rate is `2.40 xx 10^(-5) mol L^(-1) s^(-1)`, then the concentration of `N_(2)O_(5) = 0.8 mol L^(-1)`.B. In the intensity equation, `k = A exp (-E//RT)`. A may be termed as the rate constant at very low temperature.C. If `I` is the intensity of absorbed light and `c` is the concentration of `AB` for the photochemical process `AB+hv rarr AB^(**)`, the rate of formation of `AB^(**)` is diretly proportional to `I^(2)`.D. The rate constant, the activation energy, and the Arrhenius parameter of a chemical reaction at `25^(@)C` are `3.0 xx 10^(-4) s^(-1)`, `104.4 kJ mol^(-1)`, and `6.0 xx 10^(14) s^(-1)`, respectively. The value of the rate constant as `T rarr oo` is `6.0 xx 10^(14) s^(-1)`. |
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Answer» Correct Answer - A::D (a) `r = k[N_(2)O_(5)]` `2.40 xx 10^(-5) M s^(-1) = (3.0 xx 10^(-5) s^(-1))[N_(2)O_(5)]` `:. [N_(2)O_(5)] = 0.80 M` (b) At very high temperature or zero energy of activation. (c ) Directly proportional to `I`. (d) According to Arrhenius equation, `k = Ae^(-Ea//RT)` As `T rarr oo, :. K = A` Hence, `A = 6.0 xx 10^(14) s^(-1)` |
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| 1144. |
A gaseous hypothetical chemical equations , is carried out in a closed vessel . The concentration of B is found to increase by `5 xx 10^(-3) mol l^(-1)` in 10 second . The rate of appearance of B isA. `5 xx 10^(-4) mol l^(-1) sec^(-1)`B. `5 xx 10^(-5) mol l^(-1) sec^(-1)`C. `6 xx 10^(-5) mol l^(-1) sec^(-1)`D. `4 xx 10^(-4) mol l^(-1) sec^(-1)` |
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Answer» Correct Answer - a Increase in concentration of `B = 5 xx 10^(-3)` mol `l^(-1)` Time = 50 sec Rate of apperance of B = `("Increase of conc. B")/("Time taken")` `= (5xx 10^(-3) mol l^(-1))/(10"sec") = 5 xx 10^(-4) mol l^(-1) sec^(-1)`. |
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| 1145. |
A first `-` order reaction is `20%` complete in 10 minutes. Calculate the rate constant of the reaction.A. `0.223 mi n^(-1)`B. `0.0322mi n^(-1)`C. `1.023mi n^(-1)`D. `0.123 mi n^(-1)` |
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Answer» Correct Answer - 2 `(i)` For the first`-` order reaction, we have `k=(2.303)/(t)log.([A])/([A]_(0))=-(2.303)/((10mi n))log.(80)/(100)=0.0223mi n^(-1)` |
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| 1146. |
Which of the following statements is/are correct?A. For the reaction `N_(2)(g) + 3H_(2)(g)rarr 2NH_(3)(g)` under certain conditions of temperature and partial pressure of the reactants, the rate of formation of `NH_(3)` is `0.001 kg h^(-1)`. The rate of converisons of `H_(2)` under the same conditions is `0.0015 kg hr^(-1)`.B. The rate law for the reaction `RCl + NaOH (aq) rarr ROH + NaCl` is given by, rate `= k[RCl]`. The rate of the reaction will be halved on reducing the concentration of alkyl halide to one half.C. The rate of the reaction in part (b) increased on decreasing the temperature of the reaction.D. The rate of the chemical change is inversely proportional to the concentration at that instant. |
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Answer» Correct Answer - B (a) `(-d[H_(2)])/(dt) = (3)/(2)(d[NH_(3)])/(dt) xx(Mw of H_(2))/(Mw of NH_(3))` `= (3)/(2) xx 0.001 xx (2)/(17)` `= 1.765 xx 10^(-4) kg hr^(-1)` (Hence, wrong) (b) Correct statement. (c ) Directly proportional (Wrong statement). (d) Increaisng the temperature (Wrong statement). |
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| 1147. |
A catalyst increases the rate of a chemical reaction byA. Increasing the activation energyB. Decreasing the activation energyC. Reacting with reactantsD. Reacting with products |
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Answer» Correct Answer - b Catalyst increases the rate by decreasing the activation energy. |
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| 1148. |
The rate of a chemical reactionA. Increases as the reaction proceedsB. Decreases as the reaction proceedsC. May increase or decrease during the reactionD. Remains constant as the reaction proceeds |
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Answer» Correct Answer - b Rate of reaction continuously decrease with time . |
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| 1149. |
Which of the following is/are examples of unimolecular reactions?A. `O_(3) rarr O_(2)+O`B. C. `NO+O_(3) rarr NO_(2)+O_(2)`D. `O+NO+N_(2) rarr NO_(2)+N_(2)` |
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Answer» Correct Answer - A::B In (a) and (b), number of reacting molecules are only one. |
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| 1150. |
Temerature dependent equation can be written as :A. ln k=ln `A-e(E_(a)//RT)`B. ln k=ln `A+e(E_(a)//RT)`C. ln k=ln `A+e(RT//E_(a))`D. all of these |
| Answer» Correct Answer - A | |