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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.
| 501. |
The plot of In `(c_(0))/(c_(0)-x)` against `t` is a straight line, showing the reaction to be aA. Zero order reactionB. First order reactionC. Second order reactionD. Half order reaction |
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Answer» Correct Answer - B The plot of In `(c_(0))/(c_(0)-x)` or In `([A]_(0))/([A]_(0)-x) vs t` is a straight line for first order reaction. |
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| 502. |
The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `58.5 kJ "mol"^(-1)`B. `60.5 kJ"mol"^(-1)`C. `53.6 kJ"mol"^(-1)`D. `48.6 kJ"mol"^(-1)` |
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Answer» Correct Answer - C `log_(10)((k_(2))/(k_(1)))=(E_(a))/(2.303R)xx((1)/(T_(1))-(1)/(T_(2)))` `log2=(E_(a))/(2.303xx8.314)xx((1)/(300)-(1)/(310))` `E_(a)=(0.301xx2.303xx8.314xx300xx310)/(10)` `=53598 J "mol"^(-1)` `~~53.6kJ "mol"^(-1)` |
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| 503. |
The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `53.6 kJ mol ^(-1)`B. `48.6 kJ mol ^(-1)`C. `58.5kJ mol ^(-1)`D. `60.5 kJ mol ^(-1)` |
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Answer» Correct Answer - a |
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| 504. |
The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `53.6kJ" mol"^(-1)`B. `48.6kJ" mol"^(-1)`C. `58.5kJ" mol"^(-1)`D. `60.5kJ " mol"^(-1)` |
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Answer» Correct Answer - A From Arrhenius equation, `"log"(k_(2))/(k_(1))=-E_(a)/(2.303R)((1)/(T_(2))-(1)/(T_(1)))` Given, `k_(2)/k_(1)=2,T_(2)=310k,T_(1)=300k` On substituting the values, `"log"2=(-E_(a))/(2.303xx8.314)((1)/(310)-(1)/(300))` `orE_(a)=(log2xx2.303xx8.314xx310xx300)/(10)` `E_(a)=53598.6J//"mol"=53.6kJ//"mol"` |
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| 505. |
Calculate the overall order of a reaction which has the rate expresison. (a) Rate`= k[A]^((1)/(2))[B]^((3)/(2))` , (b) Rate `= k[A]^((3)/(2))[B]^(-1)` |
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Answer» (a) Rate `= k[A]^(x)[B]^(y)` "order" `= x+y` "So order" `= 1//2+3//2 =2`, i.e., "secound order" "(b)So order" `= 3//2+3//(-1) =1//2`, i.e., "helf order." |
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| 506. |
which of the following are correct about rate of reaction?A. Average rate and instantaneous rate can never be equal.B. Rate of reaction increases with increases in temperature.C. Concentration of cataylst affect rate of reaction.D. Small quantity of enzyme is sufficient to increases the rate of biological reaction. |
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Answer» Correct Answer - B::C::D |
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| 507. |
The rate at which a substance reacts, depends on its:A. atomic massB. equivalent massC. molecular massD. active mass |
| Answer» Correct Answer - C | |
| 508. |
Unit of rate of a reaction isA. `L mol ^(-1 ) t^(-1)`B. `mol dm^(-3)t^(-1)`C. `Ms`D. `M^(-1) S^(-1)` |
| Answer» Correct Answer - B | |
| 509. |
The unit of rate constant for first order reaction isA. `min""^(-2)`B. `s`C. `s^(-1)`D. `min` |
| Answer» Correct Answer - C | |
| 510. |
Give the units of the rate constant for second order reaction. |
| Answer» Correct Answer - `mol^(-1)Ls^(-1)` | |
| 511. |
Write the units of the rate constant for zero order reaction. |
| Answer» Correct Answer - `MolL^(-1)s^(-1)` | |
| 512. |
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`B. `1.2`C. `0.04`D. `0.8` |
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Answer» Correct Answer - D Rate `= k[N_(2)O_(5)]` `2.4 xx 10^(-5) mol L^(-1) s^(-1) = (3.0 xx 10^(-5) s^(-1)) [N_(2)O_(5)]` `[N_(2)O_(5)] = (2.4 xx 10^(-5) mol L^(-1) s^(-1))/(3.0 xx 10^(-5) s^(-1)) = 0.8 mol L^(-1)` Since the unit of `k` is `s^(-1)`, hence the decompoistion of `N_(2)O_(5)` is first order reaction. |
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| 513. |
Molecularity of an elementary reactionA. may be zeroB. is always integralC. may be semi - integralD. may be integral fractional or zero |
| Answer» Correct Answer - B | |
| 514. |
A reaction of first order with respect to reactant A and second order with respect to reactant B. the rate law for the reaction is given byA. `"Rate"=k[A][B]^(2)`B. Rate `=[A][B]^(2)`C. Rate `=[A]^(2)[B]`D. Rate `=k [A]^(0)[B]^(2)` |
| Answer» Correct Answer - A | |
| 515. |
The reaction between `Cr_(2)O_(7)^(2-)` and `HNO_(2)` in an acidic medium is `Cr_(2)O_(7)^(2-) + 5H^(o+) + 3HNO_(2) rarr 2Cr^(3+) + 3NO_(3)^(ɵ) + 4H_(2)O`. The rate of disappearance of `Cr_(2)O_(7)^(2-)` is found to be `2.4 xx 10^(-4) mol L^(-1) s^(-1)` during measured time interval. What will be the rate of disappearance of `HNO_(2)` during the same time interval?A. `2.4xx10^(-4)`B. `7.2xx10^(-4)`C. `4.8xx10^(-4)`D. `0.8xx10^(-4)` |
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Answer» Correct Answer - B Rate `= (-d[Cr_(2)O_(7)^(2-)])/(dt) = (1)/(3) (-d[HNO_(2)])/(dt)` or `(d[HNO_(2)])/(dt) = (3d[Cr_(2)O_(7)^(2-)])/(dt)` `:. (d[HNO_(2)])/(dt) = 3 xx 2.4 xx 10^(-4) = 7.2 xx 10^(-4)` |
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| 516. |
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. `(1)/(4)A + (1)/(2)(C ) rarr B+(1)/(3)D`B. `4A + 2C rarr B + 3D`C. `B + 3D rarr 4A + 2C`D. (d) `B + (1)/(3)D rarr (1)/(4)A + (1)/(2)C` |
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Answer» Correct Answer - C Rate `= (1)/(2)(d[C])/(d t) = (1)/(3) (-d[D])/(d t) = (1)/(4)(d[A])/(d t) = (-d[B])/(d t)` `:.` Stoichiometric coefficient of `B = 1` Stoichiometric coefficient of `D = 3` Stoichiometric coefficient of `A = 4` Stoichiometric coefficient of `C = 2` Therefore, the reaction is : `B + 3D rarr 4A +2C` |
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| 517. |
`A + 2 B to C + D . ` if `- (d [A])/(dt ) = 5 xx 10^(-4)` mol `l^(-1) s^(-1)` , then `- (d[B])/(dt)` isA. `2.5 xx 10^(-4)` mol `l^(-1) s^(-1)`B. `5.0 xx 10^(-4)` mol `l^(-1) s^(-1)`C. `2.5 xx 10^(-3)` mol `l^(-1) s^(-1)`D. `1.0 xx 10^(-3)` mol `l^(-1) s^(-1)` |
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Answer» Correct Answer - d ` A + 2 B to C + D` `(-d[A])/(dt) = 5 xx 10^(-4) implies (-1)/(2) (d[B])/(dt) = 5 xx 10^(-4)` `(d[B])/(dt) = 10 xx 10^(-4) = 10^(-3)` |
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| 518. |
Which of the following will react at the highest rate ?A. `1 mol` of `A` and `1 mol B` in a `1-L` vesselB. `2 mol` of `A` and `2 mol B` in a `2-L` vesselC. `3 mol` of `A` and `3 mol B` in a `3-L` vesselD. All would react at the same rate |
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Answer» Correct Answer - D Since all have same concentration of reactants all would react at same rate. |
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| 519. |
The term `-dx//dt` in the rate expresison refers to theA. Concentration of the reactantsB. Increase in concentration of the reactantsC. Instantaneous rate of reactionD. (d) Average rate of reaction |
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Answer» Correct Answer - C It is an expresison for instantaneous rate. |
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| 520. |
Which of the following expression can be used to describe the instantaneous rate of the reaction ? `2A+B rarr A_(2)B`A. `(1)/(2)(-d[A])/(d t)`B. `(-d[A])/(d t)`C. `(1)/(2) (d[A_(2)B])/(d t)`D. `(1)/(2) (-d[A])/(d t).(d[B])/(d t)` |
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Answer» Correct Answer - A The instantaneous rate of the reaction can be expressed by any of the following expresisons. `(1)/(2)(-d[A])/(d t)` or `(-d[B])/(d t)` or `(d[A_(2)B])/(d t)` |
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| 521. |
In a reaction `2A + B rarr A_(2)B`, the reactant `A` will disappear atA. half the rarte will B will decreaseB. twice the rarte will B will decreaseC. the same rate B will decreaseD. the same rate`A_(2)B` will decrease |
| Answer» Correct Answer - B | |
| 522. |
If a reaction involves gaseous reactants and Products, the units of its rate areA. atmB. atm-`s`C. atm-`s^(-1)`D. `atm^(2) s^(2)` |
| Answer» Correct Answer - C | |
| 523. |
Order of a reaction can have ______ values . |
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Answer» Correct Answer - d Order of reaction can be integer , fraction and zero. |
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| 524. |
Which of the following reaction ends in finite time ?A. 0 orderB. 1st orderC. 2nd orderD. 3rd order |
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Answer» Correct Answer - a In case of zero order reaction , the concentration of reactant decrease linearly with time , as its rate is independent of the concentration of the reactants. |
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| 525. |
The decomposition of `NH_(3)` on platinum surface is zero order reaction the rate of production of `H_(2)` is `(k=2.5xx10^(-4) M s^(-1))`A. `3.35xx10^(-4) and 1.25xx10^(-4),` respectivelyB. `1.25xx10^(-4) and 3.75xx10^(-4),` respectivelyC. `3.75xx10^(-3) and 2.45xx10^(-3),` respectivelyD. `1.25xx10^(-3) and 3.25xx10^(-3),` respectively |
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Answer» Given,`k=2.5xx10^(-4) molL^(-1) S^(-1)` `2H_(3)(g)toN_(2)(g)+3H_(2)(g)` On diving the Eq.by 2, `NH_(3)(g)to(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)` Rate=`(d[NH_(3)])/(dt)=+(2d[N_(2)])/(dt)=(2)/(3)(d[H_(2)])/(dt)` For zero order reaction, rate of reaction =k so,`(-d[NH_(3)])/(dt)=(2d[N_(2)])/(dt)=(2)/(3)(d[H_(2)])/(dt)=2.5xx10^(-4)" ""mol"L^(-1)S^(-1)` `therefore "Rate of production of" N_(2)=(d[N_(2)])/(dt)=(k)/(2)` `=2.5xx10^(-4)" ""mol"L^(-1)S^(-1)` `therefore "Rate of production of" H_(2)=(d[H_(2)])/(dt)=(3)/(2)xx(2.5xx10^(-4)molL^(-1)S^(-1))` `=3.75xx10^(-4)molL^(-1)S^(-1)` |
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| 526. |
A reaction is of second order with respect to a reactant . How is its effected if the concentration of the reactant is (i)doubled (ii) reduced to half ?A. Becomes 2 times and `(1)/(2)` times respectivelyB. Becomes` (1)/(2) "timses and" (1)/(2)` times respectivelyC. Becomes 2 times and 4 times, respectivelyD. Becomes 4 time and 2 times, respectively |
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Answer» (For first order reaction ), rate =k [A] we have to find the rate expression when conecentration is doubled or reduced to half and compare it with the normal rate. E.g. for a reaction A `to` products Rate =k[A]=ka (i) When the concentration of A is doubled [A]=2a Rate =k (2a)=k2a Rate of reaction becomes 2 times (ii) When concentration of Ais reduced to `(1)/(2)` ` [A] =(1)/(2)a` Rate = `k ((a)/(2)) =(ka)/(2)` Rate of reaction becomes `(1)/(2)` times i.e reduced to twice |
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| 527. |
In a reaction, `n_(1)A+n_(2)B rarr m_(1)C+m_(2)D, 5mol litre^(-1)` of `A` are followed to react with `3 mol litre^(-1)` of `B`. After `5` second, the concentration of `A` was found to be `B`. After `5` second, the concentration of `A` was found to be `4M`. Calculate rate of reaction in terms of `A` and `D`. |
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Answer» Change in `[A]` in `5 sec =5M-4M =1M` `:.` Rate of reaction in terms of `A=-(d[A])/(dt)` `=1/5 =0.2 M sec^(-1)` Also, `=1/n_(1)(d[A])/(dt)=1/m_(2)(d[D])/(dt)` `:.` Rate of reaction in terms of `D` `=(d[D])/(dt)=m_(2)/n_(1)xx0.2 M sec^(-1)` |
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| 528. |
The rate of reaction increases with increase in activation energy. |
| Answer» Correct Answer - F | |
| 529. |
From the rate expression for the following reactions, determine their order of reaction and dimensions of the rate constants. `a. 3NO(g) rarr N_(2)O(g),` Rate`=k[NO]^(2)` `b. H_(2)O_(2)(aq)+3I^(-)(aq)+2H^(o+) rarr 2H_(2)O(l)+I_(3)^(-),` Rate`=k[H_(2)O_(2)][I^(-)]` `c. CH_(3)CHO(g)rarr CH_(4)(g)+CO(g),` Rate`=k[CH_(3)CHO]^(3//2)` `d. C_(2)H_(5)Cl(g) rarr C_(2)H_(4)(g)+HCl(g),` Rate `k[C_(2)H_(5)Cl]` |
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Answer» (i) Given rate = `k[NO]^(2)` Therefore , order of the reaction = 2 `k = ("Rate")/([NO]^(2))` Dimension f = `( "mol" L^(-1) s^(-1))/(( "mol" L^(-1))^(2))` = `("mol" L^(-1) s^(-1))/( mol^(2) L^(-2))` = `L mol^(-1) s^(-1)` (ii) Given that = `k [H_(2)O_(2)][l^(-)]` Therefore , order of the reaction = 2 `k = ("Rate")/([H_(2)O_(2)][I^(-)])` Dimension of = `( "mol" L^(-1) s^(-1))/(("mol" L^(-1)) ("mol" L^(-1)))` = `L mol^(-1) s^(-1)` Given rate = `k[CH_(3)CHO]^(3//2)` Therefore , order of reaction = `(3)/(2)` `k = ( "Rate")/([CH_(3)CHO]^(3/2))` Dimension of `( "mol" L^(-1) s^(-1))/(("mol" L^(-1))^((3)/(2)))` = `("mol" L^(-1) s^(-1))/( mol^((3)/(2)) L^((3)/(2)))` = `L^((1)/(2)) mol^((1)/(2)) s^(-1)` (iv) Given rate = k `[C_(2)H_(5)Cl]` Therefore , order of the reaction = 1 `k = ("Rate")/([C_(2)H_(5)Cl])` Dimension of `= ( "mol" L^(-1)s^(-1))/( "mol" L^(-1))` =` s^(-1)` |
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| 530. |
The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10^(-4) "mol litre"^(-1)s^(-1)`? |
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Answer» The decomposition of `NH_(3)` on platinum surface is represented by the following equation . `2 NH_(3 (g)) overset("Pt")(to) N_(2 (g)) + 3 H_(2 (g))` Therefore , Rate = `-(1)/(2) (d[NH_(3)])/("dt") = (d [N_(2)])/("dt") = (1)/(3) (d [H_(2)])/("dt") = k` `= 2.5 xx 10^(-4) "mol" L^(-1) s^(-1)` Therefore , the rate of production of `N_(2)` is `(d [N_(2)])/("dt") = 2.5 xx 10^(-4) "mol" L^(-1) s^(-1)` And , the rate of production of `H_(2)` is `(d [H_(2)])/("dt") = 3 xx 2.5 xx 10^(-4) "mol" L^(-1) s^(-1)` = `7.5 xx 10^(-4) "mol" L^(-1) s^(-1)` |
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| 531. |
For the reaction `:` `2A+B rarr A_(2)B` the rate `=k[A][B]^(2)` with `k=2.0xx10^(-6)mol^(-2)L^(2)s^(-1)`. Calculate the initial rate of the reaction when `[A]=0.1 mol L^(-),[B]=0.2 mol L^(-1)`. Calculate the rate of reaction after `[A]` is reduced to `0.06 mol L^(-1)`. |
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Answer» The initial rate of reaction is Rate = `k[A][B]^(2)` = `(2 .0 xx 10^(-6) mol^(-2) L^(2) s^(-1)) 0.1 "mol" L^(-1))^(2)` =`8.0 xx 10^(-9) mol^(-2) L^(2) s^(-1)` When [A] is reduced from `0.1` mol `L^(-1)` to `0.06 mol^(-1)` the concentration of A reacted = `(0.1 - 0.06) "mol" L^(-1) = 0.04 "mol" L^(-1)` Therefore , concentration of B reacted `= (1)/(2) xx 0.04 "mol" L^(-1) = 0.02 "mol" L^(-1)` Then , concentration of B available , [B] = `(0.2 - 0.02)` mol `L^(-1)` = `0.18 "mol" L^(-1)` After [A] is reduced to `0.06` mol `L^(-1)` , the rate of the reaction is given by . Rate = `k[A][B]^(2)` = `(2.0 xx 10^(-6) mol^(-2) L^(2) s^(-1)) (0.06 "mol" L^(-1)) (0.18 "mol" L^(-1))^(2)` = `3.89 "mol" L^(-1) s^(-1)` |
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| 532. |
For the reaction `:` `2A+B rarr A_(2)B` the rate `=k[A][B]^(2)` with `k=2.0xx10^(-6)mol^(-2)L^(2)s^(-1)`. Calculate the initial rate of the reaction when `[A]=0.1 mol L^(c-),[B]=0.2 mol L^(-1)`. Calculate the rate of reaction after `[A]` is reduced to `0.06 mol L^(-1)`. |
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Answer» Initial rate is obtained by substituting the values of rate constant (k) and the concentration of both the reactants (A and B) into the rate law expression: Rate `= k[A][B]^(2)` Initial rate `= (2.0xx10^(-6)mol^(2)L^(2)s^(-1)(0.1mol L^(-1))(0.2mol L^(-1))(2)` `= 0.008xx10^(-6)mol L(-1)s^(-1)` `= 8.0xx10^(-9)mol L^(-1)s^(-1)`. In the next calculation, we are given the concntration of the reactant, A. It is `0.06 mol L^(-1)` . But to calculate the reaction rate we also need the concentration of the other reactant B. According to the balanced equation, when 2 mol of A are consumed, only 1 mol of B disappears. Thus, when [A] is reduced from `0.1mol L^(-1) to 0.06 mol L^(-1), i.e., 0.04 mol L^(-1)` of A has reacted, concentration of B consumed will be half of the concentration of A reacted: `= (1)/(2) (0.04 mol L^(-1))` `= 0.02 mol L^(-1)` Thus, final `[B] = "Intial" [B]- "Reacted" [B]` `= (0.2mol L^(-1)) - (0.02 mol L^(-1))` `= 0.81mol L^(-1)` and `"Rate" = (2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.06mol L^(-1))(0.18mol L^(-1))^(2)` `= 0.00389xx10^(-6)mol^(-2)L^(2)s^(-1)` `= 3.89xx10^(-9)mol L^(-1)s^(-1)` |
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| 533. |
For the reaction, `2A + B+C rarr A_(2)B+C` The rate `= k[A][B]^(2)` with `K = 2.0 xx 10^(-6) M^(-2) s^(-1)`. Calculate the initial rate of the reaction when `[A] = 0.1M, [B] = 0.2M` and `[C] = 0.8M`. IF the rate of reverse reaction is negligible then calculate the rate of reaction after `[A]` is reduced to `0.06M`. |
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Answer» Rate `=K[A][B]^(2)` `[A]=0.1 M, [B]=0.2M, K=2.0xx10^(-6)` `:.` Initial rate `=2.0xx10^(-6)xx0.1xx(0.2)^(2)` `=8xx10^(-9) mol litre^(-1) time^(-1)` New rate when `A` is reduced to `0.06 M` and `B` is `0.18M`. See stoichiometry of reaction. Rate=`2.0xx10^(-6)xx(0.06)xx(0.18)^(2)` `=3.89xx10^(-9) mol litre^(-1) time^(-1)` |
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| 534. |
If the half life period of a first order reaction is `2.31 xx 10^(3)` minutes, how long will it take for `1//5`th of the reactant to be left behind? |
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Answer» `k=2.303/k log a/(a-x)` `t=(2.303)/(3.0 xx 10^(-4) mm^(-1)) log a/(a//5) = (2.303 xx 0.699)/(3.0 xx 10^(-4) min^(-1))` `=0.536 xx 10^(4) mm=5.36 xx 10^(3)` min |
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| 535. |
The rate of a first order reaction is `0.04 mol l^(-1) s^(-1)` at 10 seconds and `0.03` mol `l^(-1) s^(-1)` at 20 seconds after initiation of the reaction . The half-life period of the reaction isA. 24.1 sB. 34.1 sC. 44.1 sD. 54.1 s |
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Answer» Correct Answer - a `K = (2.303)/((t_(2) - t_(1))) "log" ((a-x_(1)))/((a-x_(2))) K = (2.303)/((20-10)) "log" ((0.04)/(0.03))` `K = (2.303 xx 0.1249 )/(10) (2.303 xx "log" 2)/(t_(1//2) = (2.303 xx "log" 2)/(10)` `t_(1//2) = (0.3010 xx 10)/(0.1249) = 24.1` sec |
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| 536. |
The rate constant of the reaction ` A to B is 0.6 xx 10^(-3)` mole per second . If the concentration of A is 5 M then concentration of B after 20 minutes isA. 1.08 MB. 3.60 MC. 0.36 MD. 0.72 M |
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Answer» Correct Answer - d It is zero order reaction `therefore 6 xx 10^(-4) = ("con")/(20 xx 60)` `therefore` Conc., of B = 0.72 M |
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| 537. |
When initial concentration of the reactant is doubled, the half-life period of a zero order reactionA. is halvedB. is doubledC. is tripledD. remains unchanged |
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Answer» Correct Answer - B `(t_(1//2))_(zero) = ([A]_(0))/(2K)` for zero order `t_(1//2) prop [A_(0)]^(1)` `:.` If `[A]_(0) =` doubled, `t_(1//2) =` doubled. |
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| 538. |
The rate of a reaction increases four times when the temperature changes from 300K to 320 K. Calculate the energy of activation of the reaction. `(R=8.314JK^(-1)mol^(-1))` |
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Answer» According to the available data: `k_(2)//k_(1) = 4, T_(1)=300K , T_(2)=320K`. `logK_(2)/k_(1) = E_(a)/(2.303 xx (8.314JK^(-1)mol^(-1)))[(320K-300K)/(320Kxx 300K)]` `0.6020 = (E_(a) xx 20)/((2.303) xx (8.314Jmol^(-1)) xx (320) xx (300))` `E_(a) = (0.6020 xx 2.303 xx 8.314 J mol^(-1) xx 320 xx 300)/(20)` `=55327.6 Jmol^(-1) = 55.328 kJ mol^(-1)` |
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| 539. |
At 300K, a certain reaction is `50%` complete in 20 minutes. At 350K, the same reaction is `50%` complete in 5 minutes. Calcualte the activation energy for the reaction. |
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Answer» `k_(1) = 0.693/(20 mm) = 0.03465 mm^(-1), T_(1)=300K` `k_(2) = 0.693/(5 mm) = 0.1386 mm^(-1), T_(2)=350K` `log k_(2)/k_(1) = E_(a)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2)]` `log(0.1386 min^(-1))/(0.03465 min^(-1)) = (E_(a)xx[350K -300K])/(2.303 xx 8.314 JK^(-1)mol^(-1) xx 350K xx 300K)` `log4=(E_(a) xx 50)/(2.303 xx 8.314 xx 350 xx 300(Jmol^(-1))` `E_(a)=(0.6021 xx 2.303 xx 8.314 xx 350 xx 300)/(50) Jmol^(-1)` `24210 J mol^(-1)=24.21 kj mol^(-1)`. |
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| 540. |
The rate of a certain reaction increases by `2.3` times when the temperature is raised form `300K` to `310 K`. If `k` is the rate constant at `300K,` then the rate constant at `310 K` will be equal toA. `2k`B. `k`C. `2.3 k`D. `3 k^(2)` |
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Answer» Correct Answer - C Since `k prop T`. Hence, `k_(310K)=2.3k_(300 K)`Correct Answer - C Since `k prop T`. Hence, `k_(310K)=2.3k_(300 K)` |
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| 541. |
Find the order of reaction for the rate expression rate `= K[A][B]^(2//3)`. Also suugest the units of rate and rate constant for this expression. |
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Answer» Rate `=K[A][B]^(2//3)` `:.` Order of reaction `=1+2/3=5/3=1.67` Unit of rate `((dx)/(dt))` is mol `litre^(-1) time^(-1)` Unit of rate constant: rate constant `(K)` `=(dx//dt)/(["reactant"]^(5//3)]=("mol litre"^(-1) t^(-1))/((("mol")/("litre"))^(5//3))` `=mol^(-2//3) litre^(+2//3) time^(-1) :. n=2` |
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| 542. |
For the reaction A `rarr` B which is first order in A, which of the following change, as the concentration of A changes? (P) Rate (Q) Rate constant (R) Half-lifeA. P onlyB. R onlyC. Q and R onlyD. P, Q and R |
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Answer» Correct Answer - A |
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| 543. |
For a given reaction, `Ararr"Product, rate is" 1xx10^(-4)"M s^(-1)` when [A] = `0.01` M and rate is `1.41xx10^(-4)"M s"^(-1)` when[A] = `0.02` M. Hence, rate law is :A. `-(d[A])/(dt)=k[A]^(2)`B. `-(d[A])/(dt)=k[A]`C. `-(d[A])/(dt)=(k)/(4)[A]`D. `-(d[A])/(dt)=k[A]^(1//2)` |
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Answer» Correct Answer - D |
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| 544. |
In a first order reaction A `to` B , if k is rate constant an initial concentration of the reactant A is 0.5 M then the half - life isA. `(0.693)/(0.5 K)`B. `(log 2 )/(K)`C. `(log 2 )/(K sqrt(0.5))`D. `("In"2)/(K)` |
| Answer» Correct Answer - b | |
| 545. |
If initial concentration is reduced to `1//4^(th)` in a zero order reaction , the time taken for half the reaction to completeA. Remains sameB. becomes 4 timesC. becomes one - fourthD. Doubles |
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Answer» Correct Answer - C `t_(1//2)prop [conc.]"for" 0^(th) ` order reaction . |
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| 546. |
If initial concentration is reduced to its 1/4 th in a zero order reaction , the time taken for half of the reaction to completeA. Remains sameB. Becomes 4 timesC. Becomes one-fourthD. Doubles |
| Answer» Correct Answer - c | |
| 547. |
The rate of the reaction ` A + B to 3C ` gets increased by 72 times when the concentration of A is tripled and that of B is doubled . The order of the reaction with respect to A and B are -- and --- respectivelyA. 1,2B. 2,3C. 3,2D. 2,2 |
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Answer» Correct Answer - bRate = k[A]^x [B]^y 3x+2y=72 The option satisfying the equation is (2, 3) Answer: (B) |
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| 548. |
Which one of following formula represents the first order reaction ?A. `k= (2.303)/(t) log"" ([A])/([A_(0)])`B. `k= 2.303log"" (a-x)/(a)`C. `[A=[A_(0)]e^(-kt)`D. `k= (2.303)/(t) log "" ( a+x)/(a)` |
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Answer» Correct Answer - C Rate law equation . |
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| 549. |
The reaction `2AX(g)+2B_(2)(g)rarr A_(2)(g)+2B_(2)X(g)` has been studied kinetically and on the baiss of the rate law following mechanism has been proposed. I. `2A X hArr A_(2)X_(2) " " ("fast and reverse")` II. `A_(2)X_(2)+B_(2)rarrA_(2)X+B_(2)X` III. `A_(2)X+B_(2)rarrA_(2)+B_(2)X` where all the reaction intermediates are gases under ordinary condition. form the above mechanism in which the steps (elementary) differ conisderably in their rates, the rate law is derived uisng the principle that the slowest step is the rate-determining step (RDS) and the rate of any step varies as the Product of the molar concentrations of each reacting speacting species raised to the power equal to their respective stoichiometric coefficients (law of mass action). If a reacting species is solid or pure liquid, its active mass, i.e., molar concentration is taken to be unity, the standard state. In qrder to find out the final rate law of the reaction, the concentration of any intermediate appearing in the rate law of the RDS is substituted in terms of the concentration of the reactant(s) by means of the law of mass action applied on equilibrium step. What is the initial rate of formation of `A_(2)` when `[AX]=0.1M` and `[B_(2)]=1.0M`A. `1.2xx10^(-5)mol L^(-1)min^(-1)`B. `4.5xx10^(-6)molL^(-1)min^(-1)`C. `3xx10^(-7)molL^(-1)min^(-1)`D. `6xx10^(-7)molL^(-1)min^(-1)` |
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Answer» Correct Answer - D `(+d[A_(2)])/(dt)=k_(II).k_(c )[AX]^(2)[B^(2)]` `=3xx10^(-2)xx2xx10^(-3)(0.1)^(2)(1)` `=6xx10^(-7) mol L^(-1) min^(-1)` |
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| 550. |
The reaction `2AX(g)+2B_(2)(g)rarr A_(2)(g)+2B_(2)X(g)` has been studied kinetically and on the baiss of the rate law following mechanism has been proposed. I. `2A X hArr A_(2)X_(2) " " ("fast and reverse")` II. `A_(2)X_(2)+B_(2)rarrA_(2)X+B_(2)X` III. `A_(2)X+B_(2)rarrA_(2)+B_(2)X` where all the reaction intermediates are gases under ordinary condition. form the above mechanism in which the steps (elementary) differ conisderably in their rates, the rate law is derived uisng the principle that the slowest step is the rate-determining step (RDS) and the rate of any step varies as the Product of the molar concentrations of each reacting speacting species raised to the power equal to their respective stoichiometric coefficients (law of mass action). If a reacting species is solid or pure liquid, its active mass, i.e., molar concentration is taken to be unity, the standard state. In qrder to find out the final rate law of the reaction, the concentration of any intermediate appearing in the rate law of the RDS is substituted in terms of the concentration of the reactant(s) by means of the law of mass action applied on equilibrium step. How many times the rate of formation of `A_(2)` will increase if concentrations of `AX` is doubled and that of `B_(2)` is increased theee fold?A. `36`B. `12`C. `6`D. `8` |
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Answer» Correct Answer - B Rate `prop[AX]^(2)[B_(2)]` |
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