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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1351. |
For the reaction `A+B rarr C+D` The variation of the concentration of the products is given by the curve A. xB. yC. zD. w |
| Answer» Correct Answer - B | |
| 1352. |
For a rate law of the form, Rate = `k[A]^(m)[B]^(n)`, the exponents m and n are obtained from :A. changes in rate with changing temperature.B. the concentrations of A and B in a single experiment.C. the concentrations of A and B in a single experiment.D. changes in the reaction rates for different concentrations of A and B. |
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Answer» Correct Answer - D |
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| 1353. |
A + 2B `rarr C, the rate equation for this reaction is given as Rate = k[A] [B]. If the concentration of A is kept the same but that of B is doubled what will happen to the rate itelf?A. QuadrupledB. DoubledC. HalvedD. The same |
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Answer» Correct Answer - B |
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| 1354. |
For the reaction `2H_(2)(g)+2NO(g)rarrN_(2)(g)+2H_(2)O(g)` Rate = `k[H_(2)][NO]^(2)`. At a given temperature, what is the effect on the reaction rate if the concentration of `H_(2)` is doubled and the concentration of NO is halved?A. The reaction rate is halved.B. The reaction rate is unchanged.C. The reaction rate is doubled.D. The reaction rate increases eightfold. |
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Answer» Correct Answer - A |
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| 1355. |
The activation energy of a certain reaction is 87 kJ `"mol"^(-1)`. What is the ratio of the rate constants for this reaction when the temperature is decreases from `37^(@)C"to"15^(@)C`?A. `5//1`B. `8.3//1`C. `13//1`D. `24//1` |
| Answer» Correct Answer - C | |
| 1356. |
Propane reacts with iodine in acid medium accroding to the following equation: `CH_(3)-overset(O)overset(||)C-CH_(2)+I_(2)overset(H^(+))rarrCH_(3)-overset(O)overset(||)C-CH_(2)I+HI` These data were obtained when the reaction was studied : What is the rate equation for the reaction?A. Rate = `k[CH_(3)-overset(O)overset(||)C-CH_(3)][I_(2)]`B. Rate `k[CH_(3)-overset(O)overset(||)C-CH_(3)]^(2)`C. Rate = `k[CH_(3)-overset(O)overset(||)C-CH_(3)][I_(2)][H^(+)]`D. Rate= `k[CH_(3)-overset(O)overset(||)C-CH_(3)][H^(+)]` |
| Answer» Correct Answer - D | |
| 1357. |
Match the plots in column I with their order in column II and mark the appropriate choice. A. (A) `to`, (B) `to` (ii), (C ) `to` (i), (D) `to` (iv)B. (A) `to` (i), (B) `to` (ii) ,(C ) `to` (iii), (D) `to` (iv)C. (A) `to` (iv), (B) `to` (iii), (C ) `to` (ii), (D) `to` (i)D. (A) `to` (ii), (B) `to` (i), (C ) `to` (iii), (D) `to` (iv) |
| Answer» Correct Answer - B | |
| 1358. |
Match the graphs given in colum I with the order given in column II and mark the appropriate choice. A. (A) `to` (i), (B) `to` (ii), (C ) `to` (iii), (D) `to` (iv)B. (A) `to` (iii), (B) `to` (ii), (C ) `to` (iii), (D) `to` (iv)C. (A) `to` (ii), (B) `to` (i), (C ) `to` (iii), (D) `to` (iv)D. (A) `to` (iv), (B) `to` (iii), (C ) `to` (i), (D) `to` (ii) |
| Answer» Correct Answer - B | |
| 1359. |
For a reaction `X + Y to Z`, rate `prop [X]`. What is (i) molecularity and (ii) order of reaction ?A. (i) 2, (ii) 1B. (i) 2, (ii) 2C. (i) 1, (ii) 1D. (i) 1, (ii) 2 |
| Answer» Correct Answer - A | |
| 1360. |
Half life of a first order reaction in 10 min. What `%` of reaction will be completed in 100 min ? `A. 0.25B. 0.5C. 0.999D. 0.75 |
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Answer» Correct Answer - C `(0.693)/(t_(1//2)) = (2.303)/(t) "log" (a)/(a - x)` `(0.693)/(10) = (2.303)/(100) "log" (100)/(100 - x) implies x = 99.9%` |
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| 1361. |
Energy of activation of a reactant is reduced byA. Increased temperatureB. Reduced temperatureC. Reduced pressureD. Increased pressure |
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Answer» Correct Answer - a Energy of activation reduced by increasing temperature . |
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| 1362. |
Activation energy isA. The amount of energy to be added to the actual energy of a molecule so that the threshold energy is reachedB. The amount of energy the molecule must contains so that it reactsC. The energy which a molecule should have in order to enter into an effective collisionD. The average kinetic energy of the molecule |
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Answer» Correct Answer - a The definition of activation energy . |
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| 1363. |
Activation energy of any reaction depends onA. TemperatureB. Nature of reactantsC. Number of collisions per unit timeD. Concentration of reactants |
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Answer» Correct Answer - b The value of activation energy for a chemical reaction is primarily dependent on the nature of reacting species . |
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| 1364. |
Assertion : If the activation energy of a reaction is zero , temperature will have no effect on the rate constant . Reason : Lower the activation energy , faster is the reaction .A. If both assertion and reason are true and the reason is the correct explanation of the assertion .B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false .D. If the assertion and reason both are false . |
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Answer» Correct Answer - b According to Arrhenius equation , `k = Ae^(-E_(a)//RT)` when `E_(a) = 0` , k = A . |
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| 1365. |
Activation energy of a chemical reaction can be determined byA. Changing concentration of reactantsB. Evaluating rate constant at standard temperatureC. Evaluating rate constants at two different temperaturesD. Evaluating velocities of reaction at two different temperatures |
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Answer» Correct Answer - c log `(k_(2))/(k_(1)) = (E_(a))/(2.303R) [(T_(2) - T_(1))/(T_(1)T_(2))]` |
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| 1366. |
The rate constant k , for the reaction `N_(2)O_(5) to 2NO_(2) (g) + (1)/(2) O_(2)(g)` correspond to concentration of `N_(2)O_(5)` initially and at time t .A. `[N_(2)O_(5)]_(t) [N_(2)O_(5)]_(0) + kt`B. `[N_(2)O_(5)]_(0) = [N_(2)O_(5)]_(t) e^(kt)`C. `log_(10) [N_(2)O_(5)]_(t) = log_(10)[N_(2)O_(5)]_(0) - kt`D. ln `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))= kt` |
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Answer» Correct Answer - d Rate constant = `2.3 xx 10^(-2) sec^(-1)` It means it is a first order reaction (because unit of rate constant is `sec^(-1)`) For first order reaction k = `(1)/(t)` ln `(a)/(a-x)` Kt = ln `(a)/(a-x)` = ln `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))`. |
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| 1367. |
The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation , `K = Ae^(-E^(**) //RT)` . Activation energy `(E^(**))` of the reaction can be calculated by plottingA. log k vs `(1)/(logT)`B. k vs TC. k vs `(1)/(log T)`D. log k vs `(1)/(T)` |
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Answer» Correct Answer - d k = `Ae^(-E^(0) //RT) underset(y) ("log")k = underset(c) ("log") A - underset("mx")(E^(0))//RT therefore` log k vs `(1)/(T)`. |
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| 1368. |
The Activation energy for a chemical reaction mainly depends uponA. TemperatureB. Nature of reacting speciesC. Concentration of the reacting speciesD. Collision frequency |
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Answer» Correct Answer - B `E_(a)` depends on bond energy of reactant |
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| 1369. |
The Activation energy for a chemical reaction mainly depends uponA. the difference in energies of reactants and productsB. the sum of energies of reactants and productsC. the difference in energy of intermediate complex with the average energy of reactantsand productsD. the difference in energy of intermediate complex and the average energy of reactants. |
| Answer» Correct Answer - D | |
| 1370. |
The Activation energy for a chemical reaction mainly depends uponA. temperatureB. nature of reacting speciesC. concentration of the reacting speciesD. collision frequecny |
| Answer» Correct Answer - B | |
| 1371. |
The Activation energy for a chemical reaction mainly depends uponA. temperatureB. nature of reacting speciesC. collision frequencyD. concentration of reactants |
| Answer» Correct Answer - B | |
| 1372. |
Statement: In order for molecules to interact, they must approach each-other so closely to collide with each other. Explanation: Rearrangement of chemical bonds occur during collision.A. (a) `S` is correct but `E` is wrongB. (b) `S` is wrong but `E` is correctC. (c ) Both `S` and `E` are correct and `E` is correct explanation of `S`D. (d) Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C `H_(2)+Cl_(2)rarr 2H-Cl` `H-H+Cl-Clrarr [[H,...,Cl],[equiv,+,equiv],[H,...,Cl]]rarr 2H-Cl` They must collide each other and rearrangement of chemical bonds occur during collision. |
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| 1373. |
Statement: Fluresence is the emission of light by sucstances after absorbing light. Explanation: It may continue for appreciable time after the exciting light is switched off.A. (a) `S` is correct but `E` is wrongB. (b) `S` is wrong but `E` is correctC. (c ) Both `S` and `E` are correct and `E` is correct explanation of `S`D. (d) Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - A It is a fact |
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| 1374. |
Assertion: Rate of constant increases with the increase in temperature. Reason: Number of collisions increase with the increase in temperatures.A. If both assertion and reason are correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - B Correct explanation: Rate of reaction increases with increases in temperature because the number of effective collisions increase. |
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| 1375. |
Assertion (A) : Formation of HI is a bimolecular reaction. Reason (R ): Two molecules of reactants are involved in this reaction.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explnation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is noth the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is incorrect, but `(R )` is correct. |
| Answer» Correct Answer - A | |
| 1376. |
Assertion (A) : If the activation energy of a reaction is zero, temperature will have no effect on the rate constant. Reason (R ): Lower the activation energy, faster is the reaction.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explnation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is noth the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is incorrect, but `(R )` is correct. |
| Answer» Correct Answer - B | |
| 1377. |
Assertion (A) : If the activation energy of a reaction is zero, temperature will have no effect on the rate constant. Reason (R ): Lower the activation energy, faster is the reaction.A. If both assertion and reason are correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - B Correct explanation: According to Arrhenius equation, `k=Ae^(-Ea//RT)`. When `E_(a) = 0, k=A`. |
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| 1378. |
Assertion: For a reaction `A(g) rarr B(g)` `-r_(A) = 2.5 P_(A)` at `400 K` `r_(A) = 2.5 P_(A)` at `600 K` activation energy is `4135 J//"mol"`. Reason: Since for any reaction, values of rate constant at two different temp is same therefore activation energy of the reaction is zero.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B `-r_(A) = [2.5 R(400)]C_(A)` (at `400 K`) `-r_(A) = [2.5 R(600)]C_(A)` (at `600 K`) `log((6)/(4)) = (E_(a))/(2.303 R) [(1)/(400) - (1)/(600)]` `E_(a) = 4135` So Assertion and Reason both are correct and reason is not the correct explanation. |
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| 1379. |
It is observed that when ozone undergoes decomposition to form `O_(2)`(g) a two step mechanism is observed. (a) `O_(3)(g)underset(k_(2))overset(k_(1))(hArr) O_(2)(g)+O(g)` (b)`O_(3)(g)+O(g) overset(k_(3))rightarrow2O_(2)` Also it is known that `k_(1) gt gt k_(3) and k_(2) gt gt k_(3)` If `DeltaH` of first reaction is 20 kj and `E_(a)` of second step is 50 kJ then net activation energy will be:A. 50 kJB. 20 kJC. 30 kJD. 70 kJ |
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Answer» Correct Answer - D |
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| 1380. |
Decomposition of ozone follows the given mechanism . From the mechanism select the option which is not correct. Step (i) : `O_(3) underset(K_(2))overset(K_(1))hArr O_(2)(g) +O(g)" "` [Fast] Step (ii): `O_(3)(g) +O(g) overset(K_(3))rarr 2O_(2)(g)" "` [Slow]A. Overall rate will be dependent on rate of (ii) stepB. order of the reaction is equal to 2.C. Molecularity of step (ii) will be 2.D. Rate of formation of ozone will decrease with increase in concentration of oxygen gas. |
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Answer» Correct Answer - B |
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| 1381. |
The rate of reaction of `A+B rarr C` is found to be same when the reaction is carried out at `300^(@)` C or `200^(@)` C. Which of the following mechanism will become improbable?A. `A+B hArr D` (fast) `D rarr C` (slow)B. `A+B hArr D` (slow) `D rarr C` (fast)C. `A+B hArr D` (fast) `D hArr E` (fast) `E rarr C` (slow)D. `A overset(hv)hArr A` (fast) `A+B rarr C` (slow) |
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Answer» Correct Answer - B |
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| 1382. |
Determining the order of reaction from the rate law: Bromide ion is oxidized by bromate ion in acidic solution. `5Br^(-)(aq.)+BrO_(3)^(-)(aq.)+6H^(+)(aq.)rarrBr_(2)(aq.)+3H_(2)O(l)` The expermentally determined rate law is `"Rate" = k[Br^(-)][BrO_(3)^(-)][H^(+)]^(2)` What is the order of reaction with respect to each of the reactants and what is teh overall reaction order ? Strategy : To find the reaction order with respect to each reactant, look at the exponents in the rate law, not the coefficients in the balanced chemical equation, and then sum the exponents to obtain the overall reaction order. |
| Answer» Because the exponewnt on `[Br^(-)]` (understood) is 1, the reaction is first order with respect to `Br^(-)` (an order with respect to a species equals the exponent of its concentration). Similary, the reaction is first order with respect to `BrO_(3)^(-)` qand is second order with repect to `H^(+)`. The reaction is fourth order overall because the sum of the exponents `(1+1+2)` is 4. | |
| 1383. |
Determining the law from a mechanism with an initial fast, equilibrium step: The oxidation of iodide ion by hydrochlorite ion : `CIO^(-)(aq.)+I^(-)(aq.)rarrCI^(-)(aq.)+IO^(-)(aq.)` has been postulated to occur by the two step mechanism : 1. `CIO^(-)(aq.)+H_(2)O(l)hArrHCIO(aq.)+OH^(-)(aq.)` fast, equilibrium 2. `I^(-)(aq.)+HCI(aq.)overset(k_(2))rarr(HIO(aq.)+CI^(-)(aq.)` slow What rate law is predicted by this mechanism ? Strategy : Write the rate equation for the rate determining (slow) step. However, in this case the equation contains a species that does not appear in the overall eqwuation for the reaction. We need to eliminate it from the final from of the rate law. For this purpose we use the fast and equilibrium step. |
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Answer» The rate - determining step is step 2. Its law is `Rate = k_(2)[I^(-)][HCIO]` This rate law includes the concentration of HCIO, which is not part of the overall reaction as it is not one of the orginial reactants. To convert the rate law to a more usful form, we must find the relationship between the concentration of HCIO and the concentration of substances originally present in the solution. Step 1, which is fast and reaches equilibrium, controls the [HCIO]. To eliminate [HCIO] from the rate law, we use the fact than step 1 is described by the equilibium constant `k_(eq.)= ([HCIO][OH^(-)])/([CIO^(-)])` Note that `[H_(2)O]` is included in the value of `k_(eq.)` as it is constant on according of the fact that `H_(2)O` (sovent) is present in excess. Solving for `[HCIO]` gives `[HCIO] = k_(eq.) ([CIO^(-)])/([OH^(-)])` Substituting this expression into the rate law gives `Rate = k_(2)k_(eq.) ([I^(-)][CIO^(-)])/([OH^(-)])` This is the overall rate law for the reaction. The product `k_(2)k_(eq.)` is aconstant. The reaction is first order in iodide ion, i.e., as the concentration of `OH^(-)` increases, the rate of the reaction decrases. This decrease in rate can be attributed to the decrease in the concentration of HCIO as the solution becomes more alkaline. |
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| 1384. |
The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i) Half life of `nth` order reaction depends on the initial concentration according to the following relation: `t_(1//2) prop (1)/(a^(n-1))` ...(ii) The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii) The differential rate law for `nth` order reaction may be given as: `(dX)/(dt) = k[A]^(n)` ...(iv) where `A` denotes the reactant. The half life for a zero order reaction equalsA. `(1)/(2) (k)/(a^(2))`B. `(a^(2))/(2k)`C. `(2k)/(a)`D. `(a)/(2k)` |
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Answer» Correct Answer - D Uisng Eq. (ii), when `t = t_(1//2)`, then `x = a//2`. Uisng Eq. (i), `t_(1//2) = (c_(0)-c_(t))/(k) = (a-(a//2))/(k) = (a)/(2k)` |
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| 1385. |
The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i) Half life of `nth` order reaction depends on the initial concentration according to the following relation: `t_(1//2) prop (1)/(a^(n-1))` ...(ii) The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii) The differential rate law for `nth` order reaction may be given as: `(dX)/(dt) = k[A]^(n)` ...(iv) where `A` denotes the reactant. The rate constant for zero order reaction is where `c_(0)` and `c_(t)` are concentration of reactants at respective times.A. `k = (c_(0))/(2t)`B. `k = (c_(0)-c_(t))/(t)`C. `k = ln.(c_(0)-c_(t))/(2t)`D. `k = (c_(0))/(c_(t))` |
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Answer» Correct Answer - B On uisng equation (i), for `n = 0`, we get `t xx k = ((1)/(0-1))((1)/((a-x)^(0-1))-(1)/(a^(0-1)))` `= -1[(a-x)-a]` or `k = (a-(a-x))/(t) or (c_(0)-c_(t))/(t)` |
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| 1386. |
The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i) Half life of `nth` order reaction depends on the initial concentration according to the following relation: `t_(1//2) prop (1)/(a^(n-1))` ...(ii) The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii) The differential rate law for `nth` order reaction may be given as: `(dX)/(dt) = k[A]^(n)` ...(iv) where `A` denotes the reactant. The unit of rate and rate constant are same forA. Zero order reactionB. First order reactionC. Second order reactionD. Half order reaction |
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Answer» Correct Answer - A For zero order reaction `(dx)/(dt) k[A]^(0) = k` `:.` Units of rate `= mol L^(-1) time^(-1)` Unit of rate constant `= mol L^(-1) time^(-1)` |
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| 1387. |
A colliison between reactant molecules must occur with a certain minimum energy before it is effective in yielding Product molecules. This minimum energy is called activation energy `E_(a)` Large the value of activation energy, smaller the value of rate constant `k`. Larger is the value of activation energy, greater is the effect of temperature rise on rate constant `k`. `E_(f) =` Activation energy of forward reaction `E_(b) =` Activation energy of backward reaction `Delta H = E_(f) - E_(b)` `E_(f) =` threshold energy The activation eneries for forward and backward reactions in a chemical reaction are `30.5` and `45.4 kJ mol^(-1)` respectively. The reaction isA. ExothermicB. EndothermicC. Neither exothermic nor endothermicD. Independent of temperature |
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Answer» Correct Answer - A `Delta H = E_(f)-E_(b) = 30.5-45.4 = -14.9 kJ mol^(-1)` i.e., reaction is expthermic because heat is evolved during the reaction. |
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| 1388. |
A colliison between reactant molecules must occur with a certain minimum energy before it is effective in yielding Product molecules. This minimum energy is called activation energy `E_(a)` Large the value of activation energy, smaller the value of rate constant `k`. Larger is the value of activation energy, greater is the effect of temperature rise on rate constant `k`. `E_(f) =` Activation energy of forward reaction `E_(b) =` Activation energy of backward reaction `Delta H = E_(f) - E_(b)` `E_(f) =` threshold energy The rate constant of a certain reaction is given by `k = Ae^(-E_(a)//RT)` (where `A =` Arrhenius constant). Which factor should be lowered so that the rate of reaction may increase?A. `T`B. `Z`C. `A`D. `E_(a)` |
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Answer» Correct Answer - D Lower the activation energy higher is the rate of reaction. |
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| 1389. |
A colliison between reactant molecules must occur with a certain minimum energy before it is effective in yielding Product molecules. This minimum energy is called activation energy `E_(a)` Large the value of activation energy, smaller the value of rate constant `k`. Larger is the value of activation energy, greater is the effect of temperature rise on rate constant `k`. `E_(f) =` Activation energy of forward reaction `E_(b) =` Activation energy of backward reaction `Delta H = E_(f) - E_(b)` `E_(f) =` threshold energy For two reactions, activation energies are `E_(a1)` and `E_(a2)`, rate constant are `k_(1)` and `k_(2)` at the same temperature. If `k_(1) gt k_(2)`, thenA. `E_(a1) gt E_(a2)`B. `E_(a1) = E_(a2)`C. `E_(a1) lt E_(a2)`D. `E_(a1)geE_(a2)` |
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Answer» Correct Answer - C Higher the rate constant lower is the activation energy. |
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| 1390. |
A colliison between reactant molecules must occur with a certain minimum energy before it is effective in yielding Product molecules. This minimum energy is called activation energy `E_(a)` Large the value of activation energy, smaller the value of rate constant `k`. Larger is the value of activation energy, greater is the effect of temperature rise on rate constant `k`. `E_(f) =` Activation energy of forward reaction `E_(b) =` Activation energy of backward reaction `Delta H = E_(f) - E_(b)` `E_(f) =` threshold energy In a hypothetical reaction `A rarr B`, the activation energies for the forward and backward reactions are `15` and `9 kJ mol^(-1)`, respectively. The potential energy of `A` is `10 kJ mol^(-1)`. Which of the following is wrong?A. The threshold energy of the reaction is `25 kJ`.B. The potential energy of `B` is `16 kJ`.C. The heat of reaction is `6 kJ`.D. The reaction is exothermic. |
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Answer» Correct Answer - D `Delta H = E_(f) -E_(b) = 15-9 = 6 kJ mol^(-1)`, i.e., endothermic. THE `=E_(a(f)) + PE` of `A = 15 + 10 = 25 kJ` `PE` of `B = THE-E_(a(b)) = 25-9 = 16 kJ` So statements (a), (b) and (c ) are correct and (d) is wrong. |
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| 1391. |
A colliison between reactant molecules must occur with a certain minimum energy before it is effective in yielding Product molecules. This minimum energy is called activation energy `E_(a)` Large the value of activation energy, smaller the value of rate constant `k`. Larger is the value of activation energy, greater is the effect of temperature rise on rate constant `k`. `E_(f) =` Activation energy of forward reaction `E_(b) =` Activation energy of backward reaction `Delta H = E_(f) - E_(b)` `E_(f) =` threshold energy For the following reaction at a particular temperature, according to the equations `2N_(2)O_(5) rarr 4NO_(2)+O_(2)` `2NO_(2) + (1)/(2)O_(2) rarr N_(2)O_(5)` The activation energies are `E_(1)` and `E_(2)`, respectively. ThenA. `E_(1) gt E_(2)`B. `E_(1) lt E_(2)`C. `E_(1) = 2E_(2)`D. `sqrt(E_(1)E_(2)^(2)) = 1` |
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Answer» Correct Answer - A The reactions: (i) `2N_(2)O_(5) rarr 4NO_(2) + O_(2)` is of `(C rarr A+B)` type. (ii) `2NO_(2) + (1)/(2)O_(2)rarrN_(2)O_(5)` is of `(A+B rarr C)` type. And for the reaction `(A + B rarr C)`, `E_(a(f))` is `249` and `E_(a(b))` is `279 kJ mol^(-1)`. |
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| 1392. |
A colliison between reactant molecules must occur with a certain minimum energy before it is effective in yielding Product molecules. This minimum energy is called activation energy `E_(a)` Large the value of activation energy, smaller the value of rate constant `k`. Larger is the value of activation energy, greater is the effect of temperature rise on rate constant `k`. `E_(f) =` Activation energy of forward reaction `E_(b) =` Activation energy of backward reaction `Delta H = E_(f) - E_(b)` `E_(f) =` threshold energy If a reaction `A + B rarr C` is exothermic to the extent `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)` isA. `324`B. `279`C. `40`D. `100` |
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Answer» Correct Answer - B `Delta H = E_(f)-E_(b)` `-30 = 249-E_(b)` `:. E_(b) = 249+30 = 279 kJ mol^(-1)` |
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| 1393. |
In the start of summer, a given sample of milk turns sour at room temperature `(27^(@)C)` in `48` hours. In a refrigerator at `2^(@)C`, milk can be stored there times longer before it sours. Calculate the rate constant at `310 K`, when rate constant at `300 K` is `1.6 xx 10^(5)`A. `2.363 xx 10^(5)`B. `2.4 xx 10^(5)`C. `2.450 xx 10^(5)`D. `3.123 xx 10^(5)` |
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Answer» Correct Answer - A `(k_(310))/(k_(300)) = 1.477` `:. k_(310) = 1.477 xx k_(300) = 1.477 xx 1.6 xx 10^(5)` `= 2.363 xx 10^(5)` |
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| 1394. |
In the start of summer, a given sample of milk turns sour at room temperature `(27^(@)C)` in `48` hours. In a refrigerator at `2^(@)C`, milk can be stored there times longer before it sours. The time taken by the milk to sour at `37^(@)C`A. `35.2 hr`B. `32.5 hr`C. `35.3 hr`D. `32.3 hr` |
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Answer» Correct Answer - B Now `T_(1) = 300 K, T_(2) = 37^(@)C = 310 K` `E_(a) = 30146 J mol^(-1)` `log.(k_(310))/(k_(300)) = (30146 J mol^(-1))/(2.303 xx 8.314) xx (310 - 300)/(300 xx 310) = 0.1693` `:. (k_(310))/(k_(300)) = "antilog"(0.1693) = 1.477` Higher the rate constant, faster is the reaction, i.e., lesser is the time taken, hence `(t_(310))/(t_(300)) = (k_(300))/(k_(310)) = (1)/(1.477)` `:. t_(310) = t_(300) xx(1)/(1.477) = (48)/(1.477) = 32.5 hr` |
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| 1395. |
In the start of summer, a given sample of milk turns sour at room temperature `(27^(@)C)` in `48` hours. In a refrigerator at `2^(@)C`, milk can be stored there times longer before it sours. The activation energy of the souring of milk is `(kJ mol^(-1))`A. `30.210`B. `30.146`C. `30.0`D. `35.126` |
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Answer» Correct Answer - B At `2^(@)C (275 K)`, the reaction is theee times slower than at `27^(@)C(300 K)`. This implies that for souring of milk. `(k_(300))/(k_(275)) = 3, T_(1) = 275 K, T_(2) = 300 K` Applying Arrhenius equation `log.(k_(300))/(k_(275)) = (E_(a))/(2.303 R) ((T_(2)-T_(1))/(T_(2)T_(1)))` `log 3 = (E_(a))/(2.303 xx 8.314) ((300-275)/(275 xx 300))` or `E_(a) = 30.146 kJ mol^(-1)` |
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| 1396. |
In Arrhenius equation K `=Ae^(-E_(a)//RT)`, as `underset(Trarroo)"lim" "log"_(10)` K equals to:A. In AB. AC. `"log"_(10)A`D. none of these |
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Answer» Correct Answer - C |
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| 1397. |
A given sample of milk turns sour at room temperature `(27^(@)C)` in five hours. In a refrigerator at `-3^(@)C`, it can be stored 10 times longer. The energy of acrivation for the souring of milk isA. `2.303xx5RkJ" mol"^(-1)`B. `2.303xx3RkJ" mol"^(-1)`C. `2.303xx2.7RkJ" mol"^(-1)`D. `2.303xx10RkJ" mol"^(-1)` |
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Answer» Correct Answer - C Energy of activation is given by `"log"(k_(2))/(k_(1))=E_(a)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `Here, (k_(2))/(k_(1))=(1)/(10),T_(1)=27^@C=300k,T_(2)=-3^@C=270k` `therefore" log"(1)/(10)=E_(a)/(2.303R)[(270-300)/(270xx300)]` `-1=-E_(a)/(2.303R)((30)/(270xx300))` `E_(a)=+2.303xx2700xxRJ" mol"^(-1)` `=+2.303xx2.7xxRkJ" mol"^(-1)` |
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| 1398. |
Rate constant `k = 1.2 xx 10^(3) mol^(-1) L s^(-1)` and `E_(a) = 2.0 xx 10^(2) kJ mol^(-1)`. When `T rarr oo`:A. `A = 2.0 xx 10^(2) kJ mol^(-1)`B. `A = 1.2 xx 10^(3) mol^(-1) L s^(-1)`C. `A = 1.2 xx 10^(3) mol L^(-1) s^(-1)`D. `A = 2.4 xx 10^(3) kJ mol^(-1) s^(-1)` |
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Answer» Correct Answer - B `k=Ae^(-Ea//RT)` if `Trarroo, k=A` |
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| 1399. |
A catalyst decreases `E_(a)` form `100 KJ mol^(-1)` to `80 KJ mol^(-1)` At what temperature the rate of reaction in the absence of catalyst at `500 K` will be equal to rate reaction in the presence of catalyst?A. 400 KB. 200 KC. `"log"_(10)`AD. none of these |
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Answer» Correct Answer - A |
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| 1400. |
Rate constant `k = 1.2 xx 10^(3) mol^(-1) L s^(-1)` and `E_(a) = 2.0 xx 10^(2) kJ mol^(-1)`. When `T rarr oo`:A. `A=2.0xx10^(2)kJ"mol"^(-1)`B. `A=1.2xx10^(3)" mol "L^(-1) S^(-1)`C. `A=1.2xx10^(3)" mol"^(-1)LS^(-1)`D. `A=2.4xx10^(3)kJ" mol"^(-1)S^(-1)` |
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Answer» Correct Answer - C `k=A_e^(-E_a//RT)` when, `Trarroo,k=Ae^(o)=A` `A=k=1.2xx10^(3)"mol "Ls^-1` |
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