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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.
| 51. |
Assertion (A) : The rate constant of a zero order reaction has same units as the rate of reaction. Reason (R ): Rate constant of a zero order reaction does not depend upon the units of concentration.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. |
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Answer» Correct Answer - C Correct (R ), `k=(Concentration)/(Time)= mol L^(-1) s^(-1)` i.e., it depends upon of concentration.Correct Answer - C Correct (R ), `k=(Concentration)/(Time)= mol L^(-1) s^(-1)` i.e., it depends upon of concentration. |
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| 52. |
For the reaction: `aA + bB rarr cC+dD` Rate `= (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt)` The rate of formation of `SO_(3)` in the following reaction `2SO_(2) + O_(2) rarr 2SO_(3)` is `100 g min^(-1)`. Hence the rate of disappearance of `O_(2)` isA. `2 g min^(-1)`B. `20 g min^(-1)`C. `200 g min^(-1)`D. `50 g min^(-1)` |
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Answer» Correct Answer - B For the given reaction `(-d[O_(2)])/(dt) = (1)/(2)[d[SO_(3)])/(dt)` `(d[SO_(3)])/(dt) = 100 g min^(-1) = (100)/(80) mol min^(-1)` `:. (-d[O_(2)])/(dt) = (1)/(2)((100)/(80)) mol min^(-1)` `= (1)/(2) xx (100)/(80) xx 32 g min^(-1) = 20 g min^(-1)` |
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| 53. |
Assertion (A) : For: `aA+bB rarr` Product. The order of reaction is equal to `(a+b)`. Reason (R ): Rate of reaction `=k[A]^(a)[B]^(b)`.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. |
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Answer» Correct Answer - B Rate of reaction `=k[A]^(a)[B]^(b)` `:.` Order of reaction `=a+b`Correct Answer - B Rate of reaction `=k[A]^(a)[B]^(b)` `:.` Order of reaction `=a+b` |
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| 54. |
The hydrolysis of ethyl acetate is a reaction of :` CH_(3)COOC_(2)H_(5)+H_(2)Ooverset(H^(+))rarr CH_(3)COOH+C_(2)H_(5)OH`A. zero orderB. pseudo first orderC. second orderD. third order |
| Answer» Correct Answer - B | |
| 55. |
Assertion (A) : The order of the reaction, `CH_(3)COOC_(2)H_(5) + H_(2)O hArr CH_(3)COOH + C_(2)H_(5)OH` is `2`. Reason (R ): The molecularity of this reaction is `2`.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. |
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Answer» Correct Answer - D Hydrolyiss of ester is pseudo first order reaction.Correct Answer - D Hydrolyiss of ester is pseudo first order reaction. |
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| 56. |
The hydrolysis of ethyl acetate is a reaction of :` CH_(3)COOC_(2)H_(5)+H_(2)Ooverset(H^(+))rarr CH_(3)COOH+C_(2)H_(5)OH`A. zero orderB. first orderC. second orderD. third order |
| Answer» Correct Answer - B | |
| 57. |
The hydrolysis of ethyl acetate is a reaction of :` CH_(3)COOC_(2)H_(5)+H_(2)Ooverset(H^(+))rarr CH_(3)COOH+C_(2)H_(5)OH`A. First orderB. Second orderC. third orderD. Zero order |
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Answer» Correct Answer - A Water is in excess for hydrolysis of ethyl acetate . |
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| 58. |
Half life of a first order reaction is 2 hours , what time is required for 90% of the reactant to be consumed ?A. 199 minutesB. 398 minutesC. 598 minutesD. 798 minutes |
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Answer» Correct Answer - B `t_(1//2) =120 min . K =(0.693)/(t_(1//2)) =(0.693)/(120 )mol / min .` `t=(2.303)/(k)log_(10)""(x_(0))/(X_(t)), X_(t)=100-90=10%` `=(2.303)/(0.693//120) log _(10)=(2.303)/(0.693)xx120=398 min .` |
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| 59. |
Assertion (A) : The order of the reaction, `CH_(3)COOC_(2)H_(5) + H_(2)O hArr CH_(3)COOH + C_(2)H_(5)OH` is `2`. Reason (R ): The molecularity of this reaction is `2`.A. If both (A) and (R) are correct but (R) is the correct explanation of (A) .B. If both (A) and (R) are correct but (R) is not the correct exaplanation of (A)C. If (A) is correct but (R) is incorrectD. If (A) is incorrect but (R) is correct |
| Answer» Correct Answer - D | |
| 60. |
The order and molecularity of the chain reaction, `H_(2)(g)Cl_(2)(g)overset(kv)rarr2HCl(g)`are :A. 2,0B. 0,2C. 1,1D. 3,0 |
| Answer» Correct Answer - B | |
| 61. |
Which of the following values of the molecularity are not possible ? |
| Answer» Correct Answer - A::D | |
| 62. |
(i) Distinguish between order and molecularity of a reaction. (ii) when would order and molecularity of a reaction be the same? |
| Answer» for reactions taking place in single step. | |
| 63. |
In a first reaction, `10%` of the reactant is consumed in 25 minutes. Calcualte. i) Half life period `(t_(1//2))` ii) Time taken to complete `87.5%` of the reaction. |
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Answer» For the first order reaction `k= 2.303/tloga/(a-x)` `=2.303/(25 mm) log 100/(100-10)= 0.0042145 mm^(-1)` Half life period `(t_(1//2))` = `0.693/k = 0.693/(0.0042145 mm^(-1))=164.43 mm` ii) Again for the reaction: `t= 2.303/t loga/(a-x)` or `t=2.303/tlog a/(a-x)` `t= 2.303/(0.004215min^(-1)) log 100/(100-87.5) = 2.303/(0.0042145 mm^(-1)) xx 0.9030 = 493.45 mm` |
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| 64. |
Half life period for a reaction `A to` Products at 298 K is 3.33 hours. Calcualte the rate constant for the reaction. If the reaction is started from one mole of A, what amount of A would remain unreacted at the end of 9 hours? |
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Answer» Step i. Calculation of the rate constant. Rate constant(k)`=(0.693)/(t_(1//20))= (0.693)/(3.33 "hours")=0.2081 hr^(-1)` Step II. Calculation of amount of A left unreacted `k=0.2081 hr^(-1)`, a=1mol, t=9 hours. `k=2.303/t log a/(a-x)` `log a/(a-x) = (kxxt)/(2.303) = (0.2081 hr^(-1) xx 9hr)/(2.303) = 0.8133` `a/(a-x) = "Antilog" 0.8133=6.506` `(a-x) = a/6.506 = 1/(6.506)=0.1538` mol |
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| 65. |
Calcualte the time required for the initial concentration of `2.0 mol//dm^(3)` to get reduced to `1.2 mol//dm^(3)`. Given specific rate constant(k) = `0.009 min^(-1)` |
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Answer» For a first order reaction, `t=(2.303)/(k) log(2.0 mol dm^(-3))/(1.2 mol dm^(-3))` `(2.303)/(0.09 min^(-1)) log 1.66=(2.303)/(0.009min^(-1)) xx 0.2218=56.8 min` |
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| 66. |
In order to react, a molecule at the time of collision, must posses a certain amount of energy known as :A. free enegyB. kinetic energyC. threshold energyD. internal enegy |
| Answer» Correct Answer - C | |
| 67. |
The rate of chemical reactionA. Increases as the reaction proceedsB. Decreases with as the reaction proceedsC. May increase or decrease during the reactionD. Remains constant as the reaction proceeds |
| Answer» Correct Answer - B | |
| 68. |
The rate constant for the reaction `2N_(2)O_(5)rarr 4NO_(3)+O_(2)` is `3.0 xx10^(-4)s^(-1)`. If start made with `1.0 mol L^(-1)` of `N_(2)O_(5)`. Calculate the rate of formation of `NO_(2)` at the moment of the reaction when concentration of `O_(2)` is `0.1 mol L^(-1)`A. `2.7 xx 10^(-4)mol L^(-1)s^(-1)`B. `2.4xx10^(-4)mol L^(-1)s^(-1)`C. `4.8 xx 10^(-4) mol L^(-1)s^(-1)`D. `9.6xx10^(-1)mol L^(-1)s^(-1)` |
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Answer» Correct Answer - 4 `Mol L^(-1)` of `N_(2)O_(5)` reacted `=2xx0.1=0.2,[N_(2)O_(5)]`left`=1.0-0.2=0.8mol L^(-1)` Rate of reaction `=kxx[N_(2)O_(5)]=3.0xx10^(-4)xx0.8=2.4xx10^(-4)mol L^(-1)s^(-1),` Rate of formation of `NO_(2)=4xx2.4xx10^(-4)=9.6xx10^(-4)mol L^(-1)s^(-1),` |
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| 69. |
The efficiently of an enzyme in catalyzing a reaction is due to its capacityA. To form a strong enzyme-substrate complex.B. To decrease the bond energy of all substrate molecules.C. To change the shape of the substrate molecule.D. To lower the activation energy of the reaction. |
| Answer» Correct Answer - D | |
| 70. |
The reaction `2N_(2)O_(5)(g) rarr 4NO_(2)(g)` is first order w.r.t. `N_(2)O_(5)`. Which of the following graphs would yield a straight line?A. `log p_(N_(2)O_(5))` vs time with `-ve` slopeB. `(p_(N_(2)O_(5)))^(-1)` vs timeC. `p_(N_(2)O_(5)` vs timeD. `log p_(N_(2)O_(5)` vs time with `+ve` slope |
| Answer» Correct Answer - A | |
| 71. |
When `KClO_(3)` is heated, it decomposes into `KCl` and `O_(2)`. If some `MnO_(2)` is added, the reaction goes much faster becauseA. `MnO_(2)` decomposes to give `O_(2)`.B. `MnO_(2)` provides heat by reacting.C. Better contact is provided by `MnO_(2)`.D. `MnO_(2)` acts as a catatlyst. |
| Answer» Correct Answer - D | |
| 72. |
Which statement about eh behavior of a catalyst is correct?A. A catalyst reacts with the product and shifts the equilibrium to the right speeding up the reaction.B. A catalyst lowers the activation energy of the original reaction pathway.C. A catalyst provides additional energy to a reactant so it can achieve the necessary activation energy.D. A catalyst provides an alternative reaction pathway with a lower activation energy. |
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Answer» Correct Answer - D |
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| 73. |
`C(a) + CO_(2)(g) rarr 2CO(g) " " DeltaH=172 KJ "mol"^(-1)` In a suitable reaction vessel. Pieces of graphite are mixed with carbon dioxide gas at 1.00 atm and 1000 K . Which of the following changes will result in an increase in reaction rate?A. Decrease in size of the graphite piecesB. Decrease in temperatureC. Decrease in partial pressure of CO(g)D. Decrease in partial pressure of `CO_(2)(g)` |
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Answer» Correct Answer - A |
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| 74. |
For a reaction `2NH_(3)rarrN_(2)+3H_(2)` , it is observed that `(-d(NH_(3)))/(dt)=k_(1)(NH_(3)),(d(N_(2)))/(dt)=k_(2)(NH_(3)),(d(H_(2)))/(dt)=k_(3)(NH_(3))` What is the reletion between `k_(1),k_(2) and k_(3)`?A. `k_(1) = k_(2) = k_(3)`B. `k_(1) = 3k_(2) = 3k_(3)`C. `1.5k_(1) = 3k_(2) = k_(3)`D. `2k_(1) = k_(2) = 3k_(3)` |
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Answer» Correct Answer - C `2NH_(3) to N_(2) + 3H_(2)` Rate `= - (1)/(2) (d[NH_(3)])/(dt) = (d[N_(2)])/(dt) = (1)/(3) (d[h_(2)])/(dt)` `(1)/(2) k_(1) [NH_(3)] = k_(2) [H_(3)] = (1)/(3) k_(3) [NH_(3)]` `1. k_(1) = 3K_(2) = k_(3)` |
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| 75. |
Select to correct statements.A. Catalyst can change the spontaneity of reactionB. `If"(n)/(p)` is higher than `((n)/(p))("stabel")` then `._(-1)^(0)beta` partical is emittedC. Binding energy per atom first incjreases then decreases with atomic massD. Rate of radiocative disintegration is endependent of temperature |
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Answer» Correct Answer - B::D |
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| 76. |
At high pressure the following reaction is zero order `2NH_(3)(g)-underset("Platinum castalyst")overset(1130K)rarrN_(2)(g)+3H_(2)(g)` Which of the folowing option s are correct for this reaction ?A. Rate of reaction = Rate constantB. Rate of the reaction depends on concentration of ammonia.C. Rateof decompostion of ammonia will remain constant until ammonia disppears completely.D. Further increase in pressure will change the rate of reaction. |
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Answer» Correct Answer - A::C |
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| 77. |
At high pressure the following reaction is zero order. `" "2NH_(3)(g) underset("Platinum catalyst")overset(1130 K)to N_(2)(g) + 3H_(2)(g)` Which of the following options are correct for this reaction ?A. Rate of reaction = Rate constantB. Rate of the reaction depends on concentration of ammoniaC. Rate of decomposition of ammonia will remin constant until ammonia disappears completelyD. Further increases in pressure will change the ratio of reaction |
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Answer» Correct Answer - A::C::D Given, chemical reaction is `" "2NH_(3)(g) underset("Platinum catalyst")overset(1130 K)to N_(2)(g) + 3H_(2)(g)` At very high pressure reactions become independent of ammonia i.e., zero order reaction Hence, `" ""Rate"= k[p_(NH_(3))]^(0)` `" ""Rate" = K` (a) Rate of reaction = Rate constant (b) Rate of decomposition of ammonia will remain constant until ammonia disappears completly. (c) Since, formation of ammonia is reversible process furhter increases in pressure will change the rate of reaction. According to Le-Chatelier principle increases will favour in backward reaction. |
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| 78. |
For a reaction of second order , `t_(75%)=xt_(50%)`. The value of X is: |
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Answer» For the second order reaction: `t=1/k[1/[[A]]-1/[A]_(0)]` For `t_(1//2)` or `t_(50%)` `t_(50%) = 1/k[(1/[A]_(0))/2-1/[A]_(0)]=1/k[2/[A]_(0)-1/[A]_(0)]` `=1/k xx 1/[A]_(0)` For `t_(3//4)` or `t_(75%)` `t_(75%) = 1/k[(1/[A]_(0))/4-1/[A]_(0)]=1/k[4/[A]_(0)-1/[A]_(0)]=1/k xx 3/[A]_(0)` `t_(75%) = x t_(50%)` `t_(75%)/(t_(50%))=(1/k xx 3/[A]_(0))/(1/k xx 1/[A]_(0))=3` `therefore x=3` |
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| 79. |
Statement -1: Time taken for the completion of 75% of a 1st order reaction is double that `t_(1//2)`. Statement-2: Time taken for completion of any fraction of 1st order reaction is a constant value.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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| 80. |
`75%` of a first-order reaction was completed in `32` minutes. When was `50%` of the reaction completedA. `24` minB. `4` minC. `16` minD. `8` min |
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Answer» Correct Answer - C Let original amount, `N_(0) = 100` Since `75%` completed, so final amount `N = 100 - 75 = 25` As we know `(N_(0))/(N) = 2^(n)`, where `n =` no. of half lives or, `(100)/(25) = 2^(n)` or, `4 = 2^(n)` or, `2^(2)` or, `2^(2) = 2^(n)` `:. n=2` Since total time `= n xx t_(1//2)` `32` minutes `= 2 xx t_(1//2)` `:. t_(1//2) = 16` minutes |
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| 81. |
The rate of a reaction becomes 4 times when temperature is raised from 293 K to 313 K. The activation energy for such reaction would beA. 50.855 kJ `" mol"^(-1)`B. 52.849 kJ `" mol"^(-1)`C. 54.855 kJ `" mol"^(-1)`D. 56.855 kJ `" mol"^(-1)` |
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Answer» Correct Answer - B From the Arrhenius equation, `log.(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1)]/(T_(1).T_(2))]` Hence, `log4=(E_(a))/(2.303xx8.314)[(313-293)/(293xx313)]` `E_(a)=(11.521)/(0.000218)=52.849kJ" mol"^(-1)` |
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| 82. |
The integrated rate expression for a first order reaction can be written asA. (a-x)=a exp `(-k_(1)t)`B. x=a exp `(-k_(1)t)`C. (a-x)=t exp `(-k_(1)t)`D. (a-x)=a exp `(-k_(1)//t)` |
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Answer» Correct Answer - A Let the reaction be `{:(a" "0" ""initial concentration"),(A" "to" "B" "),((a-x)" "x" ""concentration at time"t):}` `In(a-x)/(a)=-K_(1)t)` or `(a-x)/(a)=e^(-K_(1)t)` In the above expression `(a-x)` will be zero only at infinite time. |
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| 83. |
Two radioactive material `A_(1) and A_(2)` and decay constant of `10lambda_(0) and lambda_(0)`. If initially they have same number of nuclei, then after time `(1)/(9lambda_(0))` the ratio of number of their undercayed nuclei will be:A. `(1)/(e)`B. `(1)/(e^(2))`C. `(1)/(e^(3))`D. `(sqrt3)/(1)` |
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Answer» Correct Answer - A |
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| 84. |
Which of the following processes represents a gamma-decay only.A. `.^(A)X_(Z)+gammararr.^(A)X_(Z-1)+a+b`B. `.^(A)X_(Z)+.^(1)n_(0)rarr.^(A-3)X_(Z-2)+C`C. `.^(A)X_(Z)rarr .^(A)X_(Z)+f`D. `.^(A)X_(Z)+e_(-1) rarr .^(A)X_(Z-1)+g` |
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Answer» Correct Answer - c |
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| 85. |
Consider the following nuclear reactions : `._(92)^(238)M to _(Y)^(X)N+2_(2)^(4)"He " ,_(Y)^(X)N to _(B)^(A)L+2beta^(+)` The number of neutrons in the element L is :A. 142B. 144C. 140D. 146 |
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Answer» Correct Answer - b |
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| 86. |
For the reaction `H_(2)(g) + Br_(2)(g) rarr 2HBr (g)` the experimental data suggested that `r = k[H_(2)][Br_(2)]^(1//2)` The molecularity and order of the reaction are respectively:A. `2, 3//2`B. `3//2, 3//2`C. Not defined, `3//2`D. `1, 1//2` |
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Answer» Correct Answer - C `H_(2)(g) + Br_(2) (g) rarr 2HBr(g)` `r = K[H_(2)][Br_(2)]^(1//2)` Molecularity rarr not defined, (`because` complete reaction) order `= 1 + (1)/(2) =(3)/(2)` |
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| 87. |
Statement 1 : The molecularity of the reaction `H_(2) + Br_(2) to 2 HBr ` is 2 . Statement 2 : The order of the reaction is 3/2 . |
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Answer» Correct Answer - b Molecularity is 2 because two molecules of the reactants are involved in the given elemantary reactions. |
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| 88. |
Statement 1 : The rate of reaction increases generally by 2 to 3 times for every `10^(@)C` rise in temperature . Statement 2 : An increase in temperature increases the collision frequency . |
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Answer» Correct Answer - b With increase of temperature by `10^(@)` , the fraction of molecules having effective collisions becomes 2 or 3 times. |
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| 89. |
Assertion (A): The rate of reaction increases generally by `2` to `3` times for every `10^(@)C` rise in temperature. Reason (R ): An increase intemperature increases the colliison frequency.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. |
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Answer» Correct Answer - B Correct reason: With increase of temperature by `10^(@)C`, the fraction of molecules having effective colliisons becomes `2` or `3` times.Correct Answer - B Correct reason: With increase of temperature by `10^(@)C`, the fraction of molecules having effective colliisons becomes `2` or `3` times. |
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| 90. |
In a zero-order reaction for every `10^(@)` rise of temperature, the rate is doubled. If the temperature is increased from `10^(@)C` to `100^(@)C`, the rate of the reaction will becomeA. `64` timesB. `512` timesC. `256` timesD. `128` times |
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Answer» Correct Answer - B `(r + (r +100))/(r_(t)) = 2` for each `10^(@)` rise in temperature `:. (r_(100))/(r_(10)) = (2)^(9) = 512` |
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| 91. |
In a zero order reaction, for every `10^(@)` rise of temperaure, reaction rate is doubled. If the temperature `10^(@)`C to `100^(@)`C, the reaction rate will become:A. 256 timesB. 512 timesC. 64 timesD. 128 times |
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Answer» Correct Answer - B b) For every `10^(@)` rise in temperature, reaction rate increase by 2. `(r_(100^(@))C)/(r_(10^(@))C) = 2^((100-10))/(10) = 2^(9)=512` times. |
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| 92. |
The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every `10^(@)C` rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every `10^(@)C` change in temperature. `"Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C)` Arrhenius gave an equation which describes aret constant `k` as a function of temperature `k = Ae^(-E_(a)//RT)` where `k` is the rate constant, `A` is the frequency factor or pre-exponential factor, `E_(a)` is the activation energy, `T` is the temperature in kelvin, `R` is the universal gas constant. Equation when expressed in logarithmic form becomes `log k = log A - (E_(a))/(2.303 RT)` For the given reactions, following data is given `{:(PrarrQ,,,,k_(1) =10^(15)exp((-2000)/(T))),(CrarrD,,,,k_(2) = 10^(14)exp((-1000)/(T))):}` Temperature at which `k_(1) = k_(2)` isA. `434.22 K`B. `1000 K`C. `2000 K`D. `868.44 K` |
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Answer» Correct Answer - A `k_(1) = 10^(15)exp((-2000)/(T)), k_(2) = 10^(14)exp((-1000)/(T))` When `k_(1) = k_(2), 10^(15) exp ((-2000)/(T)) = 10^(14)exp ((-1000)/(T))` or `T = 434.22 K` |
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| 93. |
The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every `10^(@)C` rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every `10^(@)C` change in temperature. `"Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C)` Arrhenius gave an equation which describes aret constant `k` as a function of temperature `k = Ae^(-E_(a)//RT)` where `k` is the rate constant, `A` is the frequency factor or pre-exponential factor, `E_(a)` is the activation energy, `T` is the temperature in kelvin, `R` is the universal gas constant. Equation when expressed in logarithmic form becomes `log k = log A - (E_(a))/(2.303 RT)` For which of the following reactions `k_(310)//k_(300)` would be maximum?A. `P + Q rarr R, E_(a) = 10 kJ`B. `E + F rarr D, E_(a) = 21 kJ`C. `A + B rarr C, E_(a) = 10.5 kJ`D. `L+M rarr N, E_(a) = 5 kJ` |
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Answer» Correct Answer - B `k_(310) gt k_(300)`, and `k prop (1)/(E_(a))` and also `E_(a) prop (1)/(T)` But `E_(a)` at `310 K lt E_(a)` at `300 K` `:. k_(310)//k_(300)` will be maximum for the reaction having high `E_(a)`. |
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| 94. |
In a zero-order reaction for every `10^(@)` rise of temperature, the rate is doubled. If the temperature is increased from `10^(@)C` to `100^(@)C`, the rate of the reaction will becomeA. 256 timesB. 512 timesC. 64 timesD. 128 times |
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Answer» Correct Answer - B For `10^(@)` rise in temperature, n=1 So, rate `=2^(n)=2^(1)=2` When temperature is increased from `10^(@)C" to "100^(@)C,` Change in temperature = 100-10=`90^(@)C` `i.e," "n=9` `rArrSo," "rate=2^(9)=512"times"` |
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| 95. |
The reaction `N_(2)O_(5)` (in `C Cl_(4)` solution ) `to 2 NO_(2)` (solution) `+ (1)/(2) O_(2)` (gas ) is of first order in `N_(2)O_(5)` with rate constant `6.2 xx 10^(-1) s^(-1)` . What is the value of rate of reaction when `[N_(2)O_(5)] = 1.25 "mole" l^(-1)`A. `7.75 xx 10^(-1) "mole" l^(-1) s^(-1)`B. `6.35 xx 10^(-3) "mole" l^(-1) s^(-1)`C. `5.15 xx 10^(-5) "mole" l^(-1) s^(-1)`D. `3.85 xx 10^(-1) "mole" l^(-1) s^(-1)` |
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Answer» Correct Answer - a Rate = `k (N_(2)O_(5)) = 6.2 xx 10^(-1) xx 1.25` = `7.75 xx 10^(-1)` mol `l^(-1) s^(-1)`. |
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| 96. |
If the concentration is expressed in moles per litre , the unit of the rate constant for a first order reaction isA. mole `"litre"^(-1) sec^(-1)`B. mol `"litre"^(-1)`C. `sec^(-1)`D. `"mole"^(-1) "litre"^(-1) sec^(-1)` |
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Answer» Correct Answer - c Unit of k for first order reaction = `sec^(-1)` . |
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| 97. |
The inversion of cane sugar is first order in [sugar] and proceeds with half-life of 600 min at pH=4 for a given concentration of sugar. However , if pH=5, the half-life changes to 60 min. The rate law expression for the sugar inversion can be written asA. `rate=k[sugar]^(1)[H^+]^(2)`B. `rate=k[sugar][H^(+)]^(1)`C. `rate=k[sugar][H^(+)]^(4)`D. `rate=k[sugar][H^(+)]^(0)` |
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Answer» Correct Answer - D `"Rate"=k[A]^(a)[B]^(b)` Overall order of reaction = a+b Inversion of cane sugar is first order reaction (given). Hence, `"rate"=k[sugar]^(a)[H^(+)]^(b)anda+b=1` In the given options, only (d) has value of a+b=1 |
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| 98. |
The inversion of cane sugar proceeds with half life of `500 min` at `pH 5` for any concentration of sugar. However, if `pH = 6`, if the half life changes to `50 min`. The rate law expression for the sugar inversion can be written as |
| Answer» Correct Answer - `K[Sugar]^(1) [H^(+)]^(0);` | |
| 99. |
The inversion of cane sugar proceeds with half life of `500 min` at `pH 5` for any concentration of sugar. However, if `pH = 6`, if the half life changes to `50 min`. The rate law expression for the sugar inversion can be written asA. `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^(+)]^(1)` |
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Answer» Correct Answer - b At `pH=5`, the half-life is `500 min` for all concentrations of sugar, i.e, `t_(1//2) prop [sugar]^(0)`. Thus, the reaction is I order w.r.t. sugar Now rate`=K[sugar]^(1)[H^(+)]^(m)` Also for `[H^(+)], t_(1//2) prop [H^(+)]^(1-m)` `:. 500 prop [10^(-5)]^(1-m)` `:. 50 prop[10^(-5)]^(1-m)` `:. 10=(10)^(1-m)` or `(1-m)=1 :. m=0` Therefore, rate=`K[sugar]^(1) [H^(+)]^(0)` |
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| 100. |
The inversion of cane sugar is represented by `C_(12) H_(22) O_(11) + H_(2)O to C_(6) H_(12)O_(6) + C_(6)H_(12) O_(6)` It is a reaction ofA. Second orderB. UnimolecularC. Pseudo unimolecularD. None of the three |
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Answer» Correct Answer - c Inversion of cane sugar is a Pseudo unimolecular reaction. |
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