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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.
| 101. |
The inversion of cane sugar proceeds with half life of 50 minutes pH = 5 for any concentration of sugar. However if pH = 6, the half life changed to 500 minutes. The law expression of sugar inversion can be written as:A. `r=K["sugar"]^(2)[H^(+)]^(0)`B. `r=K["sugar"]^(1)[H^(+)]^(0)`C. `r=K["sugar"]^(1)[H^(+)]^(1)`D. `r=K["sugar"]^(0)[H^(+)]^(0)` |
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Answer» Correct Answer - C |
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| 102. |
Inversion of cane sugar is a `cdotsAcdots` order reaction `underset("Cane sugar")(C_(12))H_(22)O_(11)+H_(2)Ooverset(H^(+))to C_(6)undersetB(H_(12))O_(6)+ underset("Fructose")C` Rate =D Here A,B,C and D respectively areA. `A="first,B=fructose",C=C_(6)H_(12)O_(6),D=K[H_2O]`B. `A="second,B=fructose",C=C_(7)H_(14)O_(7),D=K[H_2O]`C. `A="pseudo first, B=glucose,"C=C_(6)H_(12)O_(6),` `D=K[C_(12)H_(22)O_(11)]`D. `A="pseudo first, B=fructose,"C=C_(6)H_(12)O_(6),` `D=K[C_(12)H_(22)O_(11)][H_(2)O]` |
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Answer» Correct Answer - C Inversion of cane sugar is a pseudo first order reaction . `underset("Cane sugar")(C_(12))H_(22)O_(11)+H_(2)Ooverset(H+)to underset"Glucose"(C_(6))H_(12)O_(16)+ underset("Fructose")(C_(6)H_(12)O_(6)` Rate `=K[C_(12)H_(22)O_(11)]` |
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| 103. |
If initial concentration is doubled, the time for half-reaction is also doubled, the order of reaction isA. zeroB. firstC. secondD. third |
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Answer» Correct Answer - A For nth order reaction, `t_(1//2)propa^(1-n)` Here, `t_(1//2)propa` Hence, 1 - n or n = 0 (i.e. zero order reaction) |
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| 104. |
In an exothermic reaction `X rarr Y`, the activation energy is `100 kJ mol^(-1)` of `X`. The enthalphy of the reaction is `-140 kJ mol^(-1)`. The activation energy of the reverse reaction `Y rarr X` is A. `40 kJ mol^(-1)`B. `340 kJ mol^(-1)`C. `240 kJ mol^(-1)`D. `100 kJ mol^(-1)` |
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Answer» Correct Answer - C For `X rarr Y:(E_(a))_(f) = 100 kJ mol^(-1)` `Delta H = -140 kJ mol^(-1) =| 140 kJ mol^(-1)|` Form the figure above `(E_(a))_(b) = (E_(a))_(f) + Delta H` `= 100 + 140 + 240 = kJmol^(-1)` |
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| 105. |
Which of the following theory is not related to the chemical kinetics?A. Colliisons theoryB. Absolutely theoryC. Absolute reaction rateD. VSEPR theory |
| Answer» Correct Answer - D | |
| 106. |
Phosphrous undergoes slow combustion and glows in dark. The precess is calledA. Photochemical changeB. ChemiluminescenceC. FlouresceneD. Phosphorescene |
| Answer» Correct Answer - D | |
| 107. |
4g of hydrogen and 128 g of hydrogen iodide are present in a 2litre flask. What are their active masses? |
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Answer» Mass of hydrogen =4g Mol . Mass of hydrogen =2 Volume of th flask =2 liter Active mass of hydrogen `=(4)/(2xx2)=1"mol"L^(-1)` Mass of HI =128 Mol. Mass HI =128 g Volume of the flash =2 litre Active mass of hydrogen iodide=`=(128)/(128xx2)=0.5"mol"L^(-1)` |
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| 108. |
A radioactive substance has 0.1gm at a particuular instant aand has an average life of 1 day. The mass of the substance which always during the 4th day is given by:A. 6.25mgB. 12.5mgC. 3.15mgD. 1.25mg |
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Answer» Correct Answer - C |
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| 109. |
What will be the energy change in the following nuclear reaction, `X^(40) + ._(0)n^(1) rarr Y^(30)+Z^(11)` if binding energy per nucleon of X,Y and Z is 9, 7 and 6 MeV respectively.A. Energy released 84 MeVB. Energy absorved 84 MeVC. Energy released 4 MeVD. Energy absorbed 4 MeV |
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Answer» Correct Answer - B |
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| 110. |
A periodic table has 18 groups numbered from 1 to 18. What will be the group number of the final daughter nucleus formed , if `._(63)Eu^(150)` shows sequential decay emitting `1 alpha and 1 beta` particles.A. 3B. 2C. 4D. 5 |
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Answer» Correct Answer - A |
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| 111. |
Two radioactive nuclides A and B have half-lives 50 min and 10 min respectively . A fresh sample contains the nuclides of B to be eight times that of A. How much time should elapse so that the mumber of nuclides of A becomes double of B ?A. 30B. 40C. 50D. 100 |
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Answer» Correct Answer - C |
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| 112. |
Hydrogeneration of oils (manufacture of vegetable ghee) is carried out in the presence of………..(name of catalyst). |
| Answer» Correct Answer - finely divided nickel | |
| 113. |
The temeprature cofficient of a reaction is :A. ratio of rate conctants at two temperatures differing by `1^(@)C`B. ratio of rate conctants at two temperatures `35^(@)C and 25^(@)C`C. ratio of rate conctants at two temperatures `30^(@)C and 25^(@)C `D. specific reaction rate at `25^(@)C` |
| Answer» Correct Answer - B | |
| 114. |
For the reaction, `2A+2Bto"Product"` the rate law is Rate` =k[A][B]^(2)` .which mechanism is constant with this information ?A. `B+BhArrC`` C+Ato"Product"("slow")`B. `A+BtoC ("slow")` `C+Dto "poduct"`C. `A+AtoC` `B+BtoC` `C+Bto "Product"`D. `A+BhArrC` `B+CtoD`("slow")` `D+Ato"Product"` |
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Answer» Correct Answer - D |
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| 115. |
Which of the following statements is incorrect about the collison theory of chemical reaction?A. It considers reacting molecules or atoms to be hard spheres and ignores their structural features.B. Number of effective collisions determines the rate of reaction.C. Collision of atoms or molecules possessing suffieient threshold energy results into the product formation.D. Molecules should collide with sufficient threshold energy and proper orientation for the collsision to be effective. |
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Answer» Correct Answer - C Not only sufficient threshold energy of colliding atoms or molecules but also the proper orientation for the collision is required for the formation of products. |
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| 116. |
For the reaction, `N_(2)(g) + 3H_(2)(g) to 2NH_(3)`, how are the rate of reaction expressin inter- related? |
| Answer» Rate `= (-dN_(2))/(dt) =-1/3dH_(2)[Br_(2)]^(1//2), "order" = 11/2` | |
| 117. |
The chemical reaction `2O_(3) overset(k_(1))rarr 3O_(2)` proceeds as follows: `O_(3) overset(k_(eq))hArr O_(2)+O` (fast) `O + O_(3) overset(k)rarr 2O_(2)` (slow) What should be the rate law expresison ?A. `r = k[O_(3)]^(2)`B. `r = k[O_(3)]^(2)[O_(2)]^(-1)`C. `r = k[O_(3)][O_(2)]`D. Unpredictable |
| Answer» Correct Answer - B | |
| 118. |
For a reversible reaction `A underset(k_(2)=2 xx 10^(2)s^(-1))overset(k_(1)=4xx10^(2)s^(-1))hArr B` initial concentration of A is 21 mol `L^(-1)` . Select the correct statement:A. Equilibrium concentration of A is 14 mol `L^(-1)`B. Equilibrium conentration of B is 14 mol `L^(-1)`.C. The concentration of A reduce by 50% of equilibrium concentration after 11.55 sec .D. The concentration of A reduce to 50% after 23.1 sec. |
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Answer» Correct Answer - B::C::D |
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| 119. |
For a reversible reaction `C hArr D`, heat of reaction at constant volume is `-33.0 kJ mol^(-1)`, calculate: (i) The equilibrium constant at `300 K`. (ii) If `E_(f)` and `E_(b)` are energy of activation for forward and backward reactions respectively, calculate `E_(f)` and `E_(b)` at `300 K`. Given that `E_(f) : E_(b)=20:31` Assumw pre exponential factor same for forward and backward reaction. |
| Answer» Correct Answer - `5.57xx10^(5), 60 kJ, 93 kJ;` | |
| 120. |
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 same rate of converison of hydrogen under the same condition is................`kg h^(-1)`. |
| Answer» Correct Answer - `N_(2)=8.12xx10^(-4) kg hr^(-1), H_(2)=1.76xx10^(-4) kg hr^(-1);` | |
| 121. |
For a reversible first-order reaction, `A underset(K_(2))overset(K_(1))(hArr)B, K_(f)=10^(-2) s^(-1) and B_(eq.)/A_(eq.)=4,` If `A_(0)=0.01 ML^(-1)` and `B_(0)=0`, what will be concentration of `B` after `30 sec`? |
| Answer» Correct Answer - `2.50xx10^(-3) M;` | |
| 122. |
A mixture of reactants may be thermodynamically unstable but kinetically stable. |
| Answer» Correct Answer - T | |
| 123. |
If the half-life of a reaction increases as the initial concentration of substance increases, the order of the reaction is :A. zeroB. FirstC. secondD. third |
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Answer» Correct Answer - A |
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| 124. |
The dissociation of HI molcules as shown below, occurs at a temperature os 639 K. The rate constant `k=3.02xx10^(-5)M^(-1)s^(-1)` `HI(g)rarr(1)/(2)H_(2)(g)+(1)/(2)I_(2)(g)` What is order of reaction? |
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Answer» Correct Answer - C |
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| 125. |
Activation energy `(E_(a))` and rate constants (`K_(1)` and `K_(2)`) of a chemical reaction at two different temperatures (`T_(1)` and `T_(2)` ) are related byA. ln `(K_(2))/(K_(1)) = - (E_(a))/ ( R) ((1)/(T_(1)) - (1)/(T_(2)))`B. ln `(K_(2))/(K_(1)) = (E_(a))/ ( R) ((1)/(T_(1)) - (1)/(T_(2)))`C. ln `(K_(2))/(K_(1)) = - (E_(a))/ ( R) ((1)/(T_(1)) + (2)/(T_(1)))`D. ln `(K_(2))/(K_(1)) = - (E_(a))/ ( R) ((1)/(T_(2)) - (1)/(T_(1)))` |
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Answer» Correct Answer - bd Arrhenius equation ln `(K_(2))/(K_(1)) = (E_(a))/(R) ((1)/(T_(1)) - (1)/(T_(2)))` ln `(K_(2))/(K_(1)) = - (E_(a))/(R) ((1)/(T_(2)) - (1)/(T_(1)))` |
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| 126. |
For a chemical reaction at `27^(@)C` , the activation energy is 600 R . The ratio of the rate constants at `327^(@)C` to that of at `27^(@)C` will beA. 2B. 40C. eD. `e^(2)` |
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Answer» Correct Answer - c ln `(K_(2))/(K_(1)) = (E_(a))/(R) ((1)/(T_(1)) - (1)/(T_(2)))` or , ln `(K_(2))/(K_(1)) = (600R)/(R) ((1)/(300) - (1)/(600))` or , ln `(K_(2))/(K_(1)) = (600R)/(R) ((2-1)/(600)) = 1 ` ln `(K_(2))/(K_(1))` = ln e `" " (K_(2))/(K_(1)) = e` |
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| 127. |
The reaction `C_(2)H_(5)I to C_(2)H_(4) + HI` is of first order and its rate constants are `3.20 xx 10^(-4) s^(-1)` at 600 K and `1.60 xx 10^(-2)s^(-1)` at 1200 K. Calculate the energy of activation for the reaction. (Given `R= 8.314 JK^(-1)mol^(-1))` |
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Answer» `(log) k_(2)/k_(1) = E_(a)/(2.303R) [1/T_(1)-1/T_(2)]` `k_(1) = 3.20 xx 10^(-4)s^(-1), k^(2) = 1.60 xx 10^(-2)s^(-1), T_(1) = 600 K, T_(2)=1200K` `k_(1) = 3.20 xx 10^(-4)s^(-1), k_(2) = 1.60 xx 10^(-2)s^(-1), T_(1)= 600K, T_(2)=1200K` `(log) (1.60 xx 106(-2)s^(-1))/(3.20 xx 10^(-4)s^(-1)) = (E_(a))/(2.303 xx (8.314JK^(-1) mol^(-1))) xx [(1200K - 600K)/(600K) xx 1200K]` `log50 = (E_(a))/(2.303 xx (8.314 JK^(-1)mol^(-1))) xx (600)/(600 xx 1200K)` `1.6900 = (E_(a))/(2.303 xx 8.314 xx 1200(J mol^(-1)))` `E_(a) = 1.6990 xx 2.303 xx 8.314 xx 1200(Jmol^(-1))=39037 J mol^(-1)` `=39.037 kJ mol^(-1)` |
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| 128. |
Mass defect in the nuclear reactions may be expressed in terms of the atomic masses of the parent and daughter nuclides in place of their nuclear masses. The mass defect of the nuclear reaction: `._(5)B^(8) rarr ._(4)Be^(8)+e^(+)` isA. `Deltam=at` mass of `._(5)B^(8)-at` mass of `._(4)Be^(8)`B. `Deltam=at` mass of `._(5)B^(8)-at` mass of `._(4 )Be^(8)-` mass of one electronC. `Deltam=at` mass of `._(5)B^(8)-at` mass of `._(4)Be^(8)+` mass of one electronD. `Deltam=at` mass of `._(5)B^(8)-at` mass of `._(4)Be^(8)-` mass of two electrons |
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Answer» Correct Answer - D |
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| 129. |
Mass defect in the nuclear reactions may be expressed in terms of the atomic masses of the parent and daughter nuclides in place of their nuclear masses. The mass defect of nuclear reaction, `._(4)Be^(10) rarr ._(5)B^(10)+e^(-)` isA. `Deltam = at` mass of `._(4)Be^(10)`- at mass of `._(5)B^(10)`B. `Deltam=at` mass of `._(4)Be^(10)`-at mass of `._(5)B^(10)`- mass of one electronC. `Deltam=at` mass of `._(4)Be^(10)` at mass of `._(5)B^(10)+` mass of one electronD. `Deltam=at` mass of `._(4)Be^(10)` -at mass of `._(5)B^(10)-` mass of two electrons |
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Answer» Correct Answer - A |
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| 130. |
The rate of reaction is expressed as : `(1)/(2)(+d)/(d t)[C] = (1)/(3)(-d)/(d t)[D] = (1)/(4)(+d)/(d t)[A] = -(d)/(d t)[B]` The reaction is:A. `4 A+B rarr 2C + 3D`B. `B + 3D rarr 4A + 2C`C. `4 A + 2 B rarr 2C + 3D`D. `B + (1//2)D rarr 4 A + 3` |
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Answer» Correct Answer - B `-(1)/(3) (d[D])/(d t) = -(d[B])/(d t) = (1)/(2)(d[C ])/(d t) = (1)/(4)(d[D])/(d t)` `B + 3D rarr 2C + 4A` `B + 3D rarr 4A + 2C` |
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| 131. |
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 initial concentration of `N_(2)O_(5)` (in `mol L^(-1)`) isA. `1.4 (M)`B. `1.2 (M)`C. `0.04 (M)`D. `0.8 (M)` |
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Answer» Correct Answer - D Order of rection is 1st `:. [N_(2)O_(5)] = ("rate")/("rate constant") = (2.4 xx 10^(-5))/(3 xx 10^(-5))` `= 0.8 (M)` |
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| 132. |
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 initial concentration of `N_(2)O_(5)` (in `mol L^(-1)`) isA. 1.4B. 1.2C. 0.04D. 0.8 |
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Answer» Correct Answer - D d) The unit of rate constant (k) indicates reaction to be of first order. `therefore` rate = `k[N_(2)O_(5)]` `[N_(2)O_(5)] = ("Rate")/k = (2.4 xx 10^(-5)mol L^(-1)s^(-1))/(3.0 xx10^(-5)s)`=0.8 |
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| 133. |
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 initial concentration of `N_(2)O_(5)` (in `mol L^(-1)`) isA. 0.8B. 0.7C. 1.2D. 1 |
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Answer» Correct Answer - B Rate `= K [N_(2) O_(5)]` (first order as unit of rate constant is `s^(-1)`) `[N_(2)O_(5)] = ("rate")/(k) = (1.4 xx 10^(-5) mol L^(-1) s^(-1))/(2 xx 20^(-5) s^(-1)) = 0.7 mol L^(-1)` |
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| 134. |
An endothermic reaction, `Ararr B` have an activation energy `15 kcal//mol` and the heat of the reaction is `5 kcal//mol`. The activation energy of the reaction, `Brarr A` is:A. `20kcal//mol`B. `15kcal//mol`C. `10kcal//mol`D. zero |
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Answer» Correct Answer - C For endothermic reaction, `ArarrB` Activation energy =`15"kcal"//mol` Energy of reaction=`5"kcal"//,mol` Hence, the activation energy for `BrarrA ` is `15-5=10"kcal"//mol.` |
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| 135. |
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)` (in `mol L^(-1)`) isA. `1.4`B. `1.2`C. `0.04`D. `0.8` |
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Answer» Correct Answer - D |
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| 136. |
The rate law for the dimerisation of `NO_(2)` is : `-(d[NO_(2)])/(dt)=k[NO_(2)]^(2)` which of the following changes will change the value of specific rate constant, k :A. doubling the total pressure on the systemB. doubling the temperatureC. both (a) and (b)D. None of the above |
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Answer» Correct Answer - B |
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| 137. |
If energy of activation of the rection is 53.6kJ`mol^(-1)` and the temperature changes from `27%^(@)` to `37^(@)` C, then the value of `k_(37^(@)C)/(k_(27^(@)C)` isA. 2.5B. 1C. 2D. 1.5 |
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Answer» log `(k_(2)/k_(1)) = E_(a)/(2.303 R) (T_(2)-T_(1))/(T_(1)T_(2))` `E_(a) = 53.6 xx 10^(3) J mol^(-1), T_(1)=300K, T_(2)=310K` `log (k_(2)/k_(1)) = (53.6 xx 10J mol^(-1))/(2.303 xx (8.314 JK^(-1)mol^(-1))) xx (310-300)/(310 xx 300)` `=0.3010` `k_(2)/k_(1) = "Antilog"(0.3010)=2` |
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| 138. |
The above plot is for _______ order rection to calculate value of rate constant .A. SecondB. firstC. zeroD. First and zero |
| Answer» Correct Answer - C | |
| 139. |
For a first order reaction, the rate of the reaction doubles as the concentration of the rection (S)doubles. |
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Answer» Correct Answer - T |
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| 140. |
For a first order decomposition of `N_(2)O_(5)(g)` to give `NO_(2)(g)` and `O_(2)(g)`, what will be the rate constant if at initial instant, after 10 minutes and after a very long time, tolal pressure is 200 mm of Hg, 325 mm of Hg and 450 mm of Hg?A. 0.693min^(-1)`B. `6.93min^(-1)`C. `6.93xx10^(-2)min^(-1)`D. `(6.93)/(2)xx10^(-2)min^(-1)` |
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Answer» Correct Answer - C |
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| 141. |
The rate of a reaction quadruples when the temperature changes from `293K` to `313K`. Calculate the energy of activation of the reaction assuming that it does not change with temperature.A. `48.625kJ" mol"^(-1)`B. `654.35kJ" mol"^(-1)`C. `354.20kJ" mol"^(-1)`D. `52.854kJ" mol"^(-1)` |
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Answer» Correct Answer - D According to Arrhenius equation, `"log"k_(2)/(k_(1))=E_(a)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `"log"(4)/(1)=E_(a)/(2.303xx(8.314Jmol^(-1)k^(-1)))xx(20)/(293xx313)` `E_(a)=52.854kJmol` |
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| 142. |
The decomposition of `A` into product has value of `k` as `4.5xx10^(3)s^(-1)` at `10^(@)C` and energy of activation of `60kJmol^(-1)`. At what temperature would `k` be `1.5xx10^(4)s^(-1)?`A. 273.15 kB. `24.01^@C`C. 280.39 KD. `45.29^@C` |
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Answer» Correct Answer - B According to Arrhenius equation, `"log"(k_(2))/(k_(1))=E_(a)/(2.303R)xx(T_(2)-T_(1))/(T_(1)T_(2))` `"log"(1.5xx10^(4))/(4.5xx10^(3))=(60000J"mol"^(-1))/(2.303xx(8.314J"mol"^(-1)))((T_(2)-283)/(283T_(2)))` `log3.333=3.1333.62((T_(2)-283)/(283T_(2)))` `T_(2)=297.01k=(297.01-273.0)^@C=24.01^@C` Temperature =`24.01^@C` |
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| 143. |
For the rection 3A `rarr` Products the value of `k=1xx10^(-3)` L/(mol-min) the value of `-(d[A])/(dt)` in mol/L-sec when [A] = 2 M is :A. `6.67xx10^(-3)`B. `1.2xx10^(-2)`C. `2xx10^(-4)`D. `4xx10^(-3)` |
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Answer» Correct Answer - C |
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| 144. |
For a reaction , `AhArrB,(d[A])/(dt)=5xx10^(-4)[B]-4xx10^(-3)[A]M "min"^(-1)` Starting with only A at 0.1 M concentration calculate concentration of B after time t=9200sec. [Given:In 2=0.69]A. `(0.8)/(9)M`B. `(0.4)/(9)M`C. `(0.1)/(9)M`D. `(0.2)/(9)M` |
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Answer» Correct Answer - B |
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| 145. |
For the first order reaction `2 N_(2)O_(5)(g) to 4 NO_(2) (g) + O_(2)(g)`A. The concentration of the reaction decreases exponentially with timeB. The half-life of the reaction decreases with increasing temperatureC. The half-life of the reaction depends on the initial concentration of the reactantD. The reaction proceeds to 99.6 % completion in eight half-life duration |
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Answer» Correct Answer - abd `C_(t) = C_(0) e^(-kt)` `t_(1//2) prop (1)/(K) , " " K uarr` on increasing T . After eight half-lives , `C = (C_(0))/(2^(8))` `implies` % completion = `(C_(0) - (C_(0))/(2^(8)))/(C_(0)) xx 100 = 99.6%`. |
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| 146. |
The rate law for the reaction `RCl + NaOH(aq) rarr ROH + NaCl` is given by Rate `= k[RCl]`. The rate of the reaction will beA. is doubled by doubling the concentration of NaOHB. is halved by reducing the concentration of RCI by one halfC. is increased by increasing the temperature of the reactionD. is unaffected by change in temperature |
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Answer» Correct Answer - B `RCI+NaOHrarrROH+NaCI` Rate = `k[RCI]` For this reaction, rate of reaction depends upon the concentration of RCI. It means, the rate of reaction is halved by reducing the concentration of RCI by one half. |
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| 147. |
Consider a reaction `3A(g)rarr4B(g)+C(g)`, starting with pure A having pressure `(3)/(5)` atm, the pressure after 10 min. reaches to 1 atm. Calculate the value of rate of disappearance of A at initial instant.A. `0.06 "atm min"^(-1)`B. `0.6 "mole lit"^(-1) "min"^(-1)`C. `(3)/(50)"In"(3)/(5) "atm min"^(-1)`D. `0.36 "atm min"^(-1)` |
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Answer» Correct Answer - A |
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| 148. |
The rate law for the reaction `RCl + NaOH (aq) to ROH + NaCl` is given by , Rate = `k_(1)`[RCl] . The rate of the reaction will beA. Doubled on doubling the concentration of sodium hydroxideB. Halved on reducing the concentration of alkyl halide to one halfC. Increased on increasing the temperature of the reactionD. Unaffected by increasing the temperature of the reaction. |
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Answer» Correct Answer - bc As rate k = [RCl] , on decreasing the concentration of RCl to half , the rate will also be halved . Rate will also increase with temperature . |
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| 149. |
The rate law for the reaction `RCl + NaOH(aq) rarr ROH + NaCl` is given by Rate `= k[RCl]`. The rate of the reaction will beA. Doubled by doubling the concentration of NaOHB. Halved by reducing the concentration of RCI by one halfC. increased by increasing the temperature of the reactionD. unaffected by change in temperature. |
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Answer» Correct Answer - B::C (b,c) are correct options. |
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| 150. |
The rate law for the reaction `RCl + NaOH(aq) rarr ROH + NaCl` is given by Rate `= k[RCl]`. The rate of the reaction will beA. unaffected by increaing temperature of the reactionB. doubled on doubling the concentration of NaOHC. Halved on reducing the concentration of NAOH to one halfD. halved on reducing the concentration of RCI to one half |
| Answer» Correct Answer - D | |