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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.
| 651. |
On adding ` AgNO_(3) ` to ` NaCl ` white ppt , occursA. InstananeosulyB. with a measurable speedC. SlowlyD. Depending on condition |
| Answer» Correct Answer - A | |
| 652. |
Minimum energy required for molecules to react is called :A. Potential energyB. concentration of productC. timeD. activation energy |
| Answer» Correct Answer - D | |
| 653. |
Half life period of a first order reaction `A to ` product is 6.93 hour . What is the value of rate constant ?A. `1.596h^(1)`B. `0.1h^(-1)`C. `4.802h^(-1)`D. `10h^(-1)` |
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Answer» Correct Answer - B We have the formulae for half-life of first order reaction, `t_(1//2)=0.693/k`where k = rate constant `therefore" "6.93h=0.693/kRightarrowk=(0.693)/(6.93h)` `k=0.1h^(-1)` |
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| 654. |
If a is the initial concentration and K is the rate constant of a zero order reaction , the time for the reaction to go to completion wil beA. `(k)/(a)`B. `(a)/(k)`C. `(a)/(2K)`D. `(K)/(2a)` |
| Answer» Correct Answer - B | |
| 655. |
Minimum energy required for molecules to react is called :A. Kinetic energyB. Potential energyC. Heat energyD. Activation energy |
| Answer» Correct Answer - D | |
| 656. |
The number of molecules of the reactants taking part in a single step of the reaction tells about. :A. molecularity of the reactionB. mechanism of the reactionC. order of reactionD. all of the above |
| Answer» Correct Answer - A | |
| 657. |
The number of molecules of the reactants taking part in a single step of the reaction tells about. :A. Molecularity of the reactionB. Mechanism of the reactionC. Order of reactionD. Mole fraction |
| Answer» Correct Answer - A | |
| 658. |
The relationship between rate constant and half life period of zero order reaction is given by _______.A. `t_(1//2)=[A]_(O)2k`B. `t_(1//2)=(0.693)/(k)`C. `t_(1//2)=([A]_(O))/(k)`D. `t_(1//2)=([A]_(O))/(2k)` |
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Answer» Correct Answer - C The relationship between rate constant (k) and half-life period of zero order reaction is written as, `k=([A]_(0)-[A])/(t_(1//2))` `or " " k=([A]_(0)-1//2[A]_(0))/(t_(1//2))" "[because at" "t_(1//2),[A]=1/2[A]_(0)]` `or" "t_(1//2)=[A]_(0)/(2k)` |
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| 659. |
For the reaction `O_(3(g)) +O_((g)) to 2O_(2(g)) ` , if the rate law expression is , rate `=k [O_3][O]` the molecularity and order of the reaction are respectively ______.A. 2 and 2B. 2 and 1.33C. 2 and 1D. 1 and 2 |
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Answer» Correct Answer - A (a) Given reaction is `O_(3)(g)+O(g)rarr2O_(2)(g)` Also, rate law expression is given as rate `=k[O_(3)][O]`From the reaction, it is clear that two molecules(one `O_(3)` and one O) colloide to form `O_(2)(g)`hence the molecularity is 2. From the rate law expression, sum of power of concentration of reactants `1+1=2`. |
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| 660. |
Match the rate law given in column I with the dimensions of rate constants given in column II and mark the appropriate choice. `{:("Column I","Column II"),((A) "Rate" = k[NH_(3)]^(0),(i) mol L^(-1) s^(-1)),((B)"Rate" = k[H_(2)O_(2)][I^(-)],(ii) L mol^(-1) s^(-1)),((C )"Rate" = k[CH_(3)CHO]^(3//2),(iii) S^(-1)),((D) "Rate" = k [C_(5)H_(5)CI],(iv)L^(1//2) mol^(-1//2) s^(-1)):}`A. (A) `to` (iv), (B) `to` (iii), (C ) `to` (ii), (D) `to` (i)B. (A) `to` (i), (B) `to` (ii) (C ) `to` (iii) ,(D) `to` (iv)C. (A) `to` (iii), (B) `to` (ii) (C ) `to` (i) ,(D) `to` (iv)D. (A) `to` (ii), (B) `to` (i), (C ) `to` (vi), (D) `to` (iii) |
| Answer» Correct Answer - D | |
| 661. |
The number of molecules of the reactants taking part in a single step of the reaction tells about. :A. order of a reactionB. molecularity of a reactionC. fast step of the mechanism of a reactionD. half-life of the reaction |
| Answer» Correct Answer - B | |
| 662. |
The activation energy of a reaction can be determined byA. Changing concentration of reactantsB. knowing rate constant at two different temperatureC. knowing rate at 298 KD. knowing concentration of reactants at 298 K |
| Answer» Correct Answer - B | |
| 663. |
In Arrhenius equation , `K=Ae^(-E_a//RT)` . The A isA. fraction of collisionB. frequency factorC. collision frequencyD. Collision factor |
| Answer» Correct Answer - B | |
| 664. |
Chemical kinetics, a branch of physical chemistry, deals with :A. heat changes in a reactionB. physical changes in a reactionC. rates of reactionsD. structure of a molecules |
| Answer» Correct Answer - C | |
| 665. |
The Unit of rate constant and rate of reaction are same forA. zero orderB. first orderC. second orderD. none of these |
| Answer» Correct Answer - A | |
| 666. |
Two reacants `A` and `B` are present such that `[A_(0)] = 4[B_(0)]` and `t_(1//2)` of `A` and `B` are `5` and `15` mintute respectively. If both decay folliwing `I` order, how much time later will concentrations of both of them would be equal?A. `15 min`B. `10 min`C. `5 min`D. `12 min` |
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Answer» Correct Answer - A Amount of `A` left in `n_(1)` halves`= ((1)/(2))^(n_(1)) [A_(0)]` Amount of `B` left in `n_(2)` halves `= ((1)/(2))^(n_(2))[B_(0)]` At the end, according to the question `([A_(0)])/(2^(n_(1))) ([B_(0)])/(2^(n_(2))) rArr (4)/(2^(n_(1))) = (1)/(2^(n_(2))), ([A_(0)] = 4[B_(0)])` `(2^(n_(1)))/(2^(n_(2))) = 4 rArr 2^(n_(1)-n_(2)) = (2)^(2) rArr n_(1) - n_(2) = 2` `:. n_(2) = (n_(1) - 2)` ....(i) Also `t = n_(1) t_(1//2(A)), t = n_(2)t_(1//2(B))` (Let concentration of both become equal after time `t`) `:. (n_(1) xx t_(1//2(A)))/(n_(2) xx t_(1//2(B))) = 1 rArr (n_(1) xx 5)/(n_(2) xx 15) = 1 rArr (n_(1))/(n_(2)) = 3` ...(ii) form Eqs. (i) and (ii), we get `n_(1) = 3, n_(2) = 1` `t = 3 xx 5 = 15 min` |
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| 667. |
What specific name can be given to the following sequence of steps: `Hg+hv rarr Hg^(**)` `Hg^(**) + H_(2) rarr H_(2)^(**)+Hg`A. FluorescenceB. PhosphoresecenceC. PhotosenistizationD. Chemiluminescence |
| Answer» Correct Answer - C | |
| 668. |
If the order of the reaction `x + y overset(hv) (to) ` xy is zero , it means that the rate ofA. Reaction is independent of temperatureB. Formation of activated complex is zeroC. Reaction is independent of the concentration of reacting speciesD. Decomposition of activated complex is zero |
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Answer» Correct Answer - c In photochemical reactions the rate of reaction is independent of the concentration of reacting species . |
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| 669. |
In pseudo-unimolecular reactions :A. one of the reactants is present in large excessB. both the reactants have same concentrationC. both the reactants are present in low concentrationD. one of the reactants is less reactive |
| Answer» Correct Answer - A | |
| 670. |
Assertion (A) : The rate constant of a pseudo unimolecular reaction has the units of a second order reaction. Reason (R ): A pseudo unimolecular reaction is a reaction of second order in whiich one of the reactant is present in large excess.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 | |
| 671. |
The mechanism for the reaction is given below `2 P + Q to S + T` ` P + Q to R + S ` (slow ) `P + R to T` (fast) The rate law expression for the reaction isA. `r = k [P]^(2) [Q]`B. `r = k [P][Q]`C. r = k[A] [R]D. `r = k [P]^(2)` |
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Answer» Correct Answer - b The rate law expression for the reaction is r = k[P][Q] . |
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| 672. |
Keen observation of the rates of reaction about the series of steps leading to the formation of products is called the reaction mechanism . The reaction between `H_(2)` and `I_(2)` to form hydrogen iodide was originally postulated as a simple one step reaction `H_(2) + I_(2) to 2 HI` Rate = `k [H_(2)][I_(2)]` but the formation of HI has been explained on the basis of the following mechanism `({:( I_(2) , to , 2 I "(fast)" , ... (i) ), ( H_(2) + I , to , H_(2) I "(fast)" , ... (ii)) , (H_(2) I + I , to , 2HI "(slow)" , ... (iii) ):})/("overall" H_(2) + I_(2) to 2 HI` For the reaction `2NO_(2) + F_(2) to 2 NO_(2) F` following mechanism has been provided `NO_(2) + F_(2) overset("slow")(to) NO_(2)F + F` `NO_(2) + F overset("fast")(to) NO_(2)F` Thus rate expression of the above reaction can be written asA. `r = k[NO_(2)]^(2) [F_(2)]`B. `r = k[NO_(2)]`C. `r = k [NO_(2)][F_(2)]`D. `r = k [F_(2)]` |
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Answer» Correct Answer - c Rate is given by the slowest step of the reaction so rate is r = `k[NO_(2)][F_(2)]`. |
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| 673. |
Which of the following is correct ?A. molecularity of a raction is same as the same as the order of a reactionB. in some cases order of a reaction may be same as the molecularity of the reactionC. both (a) and (b) are correctD. Molecularity is different from order of reaction |
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Answer» Correct Answer - B For elementary reaction order and molecularity is same . |
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| 674. |
Ethyl acetate is hydrolysed in acidic medium Its order of reaction and molecularity are respectively :A. 1 and 1B. 1 and 2C. 2 and 1D. 2 and 2 |
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Answer» Correct Answer - B `CH_(3) COOC_(2)H_(2) +H_(2)O to CH_(3) COOH+C_(2)COOH+C_(2)H_(5)OH` |
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| 675. |
The concentration of R in the reaction `R to P` was measure as a function of time and the following data is obtained . The order reaction is |
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Answer» `K = (C_(0) - C)/(t) = (1-0.75)/(0.05) = (0.25)/(0.05) = 5` `K = (0.75 - 0.40)/(0.07) = (0.35)/(0.07) = 5` So ,reaction must be of zero order . |
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| 676. |
An organic compound undergoes first-order decomposition . The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are `t_(1//8)` and `t_(1//10)` respectively . What is the value of `([t_(1//8)])/([t_(1//10)]) xx 10` (take `log_(10) 2 = 0.3`) |
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Answer» Correct Answer - 9 `Kt_(1//8) =` ln `{(C_(O))/(C_(O)//8)} = `ln 8 `K t_(1//10)` = ln `{(C_(O))/(C_(O)//10)}` = ln 10 then `(t_(1//8))/(t_(1//10)) xx 10 = ("ln"8)/("ln"10) xx 10 = (3"log"2)/("log"10) xx 10 = 9` |
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| 677. |
An organic compound undergoes first order decomposition. The time taken for the decomposition of `1//8` and `1//10` of its initial concentration are `1//8` and `1//10` respectively. What is the value of `[t_(1//8)]/([t_(1//10)]) xx 10` (take `log_(10)2=0.3)` |
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Answer» `k=2.303/t log ([A]_(0))/([A])` `[A] = 1//8 [A]_(0)` at `t_(1//8)` `t_(1//8) = 2.303/k log ([A]_(0))/([A]_(0//8))` `=2.303/k log8` `[A] = 1/10 [A]_(0)` at `t_(1//10)`, `t_(1//10) = 2.303/k log ([A]_(0))/([A]_(0//10))` `=2.303/k log 10`...........(ii) `([t_(1//8)])/([t_(1//10)]) xx 10 = (log 8)/(log 10) xx 10 = (3 log 2)/(log 10) xx 10` `= 3 xx (0.3010 xx 10) =9` |
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| 678. |
The rate constant (K) for the reaction ` 2A + B to ` Product was found to be `2.5 xx 10^(-5)` litre `mol^(-1) sec^(-1)` after 15 sec , `2.60 xx 10^(-5)` litre `mol^(-1) sec^(-1)` after 30 sec and `2.55 xx 10^(-5)` litre `mol^(-1) sec^(-1)` after 50 sec . The order of reaction is |
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Answer» Correct Answer - 2 By observing the unit of rate constant `mol^(1-n) "litre"^(n-1) "sec"^(-1) = mol^(-1) "litre" sec^(-1)` 1 - n= `-1 implies n = 2` |
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| 679. |
Following is the graph between log `t_(1//2)` and log a (a initial concentration ) for a given reaction at `27^(@)C` Find the order of reaction |
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Answer» `t_(1//2) prop (1)/(a_(0)^(n-1)) implies t_(1//2) k = (1)/(a_(0)^(n-1))` log `t_(1//2)` = log k - (n-1) log `a_(0)` tan `45^(@) = -(n-1) = -(0-1)` tan `45^(@) = 1 " " therefore n = 0` |
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| 680. |
Following is the graph between `log T_(50)` and `log a` (`a =` initial concentration) for a given reaction at `27^(@)C`. Hence order is |
| Answer» Correct Answer - A | |
| 681. |
A first order reaction is found to have a rate constant, `k=7.39 xx 10^(-5)s^(-1)`. Find the half life of the reaction. |
| Answer» Half life period `(t_(1//2))=0.693/k =(0.693)/(7.39 xx 10^(-5)s^(-1)) = 9.3775 xx 10^(3)s` | |
| 682. |
The hydroliss of ethyl acetate in………… medium is a…………..order reaction. |
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Answer» Correct Answer - acidic, first order `OR` alkaline, second order The hydrolyiss of ethyl acetate in acidic medium is first order reaction. The hydrolyiss of ethyl acetate in alkaline medium is a second order reaction.Correct Answer - acidic, first order `OR` alkaline, second order The hydrolyiss of ethyl acetate in acidic medium is first order reaction. The hydrolyiss of ethyl acetate in alkaline medium is a second order reaction. |
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| 683. |
The rate first order reaction is `2.4 xx 10^(-3) mol L^(-1) s^(-1)`, when the initial concentration is `0.2 mol L^(-1)`. The rate constant is…………. . |
| Answer» Correct Answer - `1.2 xx 10^(-3) s^(-1)` | |
| 684. |
For the first order reaction, the rate constant is `4.62 xx 10^(-2)s^(-1)`. What will be the time required for the initial concentration `1.5` mol of the reactant to be be reduced to 0.75 mol? |
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Answer» According to available data: `[A]_(0) =1.5 mol, [A]=0.75 mol, k=4.62 xx 10^(-2)s^(-1)`. `t=(2.303)/(k) log ([A]_(0))/([A]) = (2.303)/(4.62 xx 10^(-2)s^(-1)) = 15s` |
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| 685. |
The rate constant of a first order reaction is `2.0 xx 10^(-5)s^(-1)` and the initial concentration is 0.10 mol `L^(-1)`. The initial rate isA. `2.0 xx 10^(-6)"mol " L^(-1)s^(-1)`B. `1.0 xx 10^(-6)"mol " L^(-1)s^(-1)`C. `1.5 xx 10^(-6)"mol " L^(-1)s^(-1)`D. `0.5 xx 10^(-6)"mol " L^(-1)s^(-1)` |
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Answer» Correct Answer - A Rate `((dx)/(dt))` = Rate constant (k) `xx` Concentration `=2.0xx10^(-5)s^(-1)xx0.10" mol "L^(-1)` `=2.0 xx 10^(-6)" mol "L^(-1)s^(-1)` |
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| 686. |
The concentration of `R` in the reaction `R rarr P` was measured as a function of time and the following data were obtained. What is the order of the reaction? `|{:([R] (mol),1.0,0.75,0.40,0.10),(T (min),0.0,0.05,0.12,0.18):}|` |
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Answer» Zero order `|{:([R] (mol),1.0,0.75,0.40,0.10),(T (min),0.0,0.05,0.12,0.18):}|` On careful observation, it can be conclude that the rate of the reaction is not decreaisng with decrease in the concentration of the reactant, so the order of the reaction must be zero.Zero order `|{:([R] (mol),1.0,0.75,0.40,0.10),(T (min),0.0,0.05,0.12,0.18):}|` On careful observation, it can be conclude that the rate of the reaction is not decreaisng with decrease in the concentration of the reactant, so the order of the reaction must be zero. |
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| 687. |
The rate of chemical change is directly proportional to …………… . |
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Answer» Correct Answer - the concentration of the reactant at that time The rate of chemical changes is directly proportional to the concentration of the reactant at that time.Correct Answer - the concentration of the reactant at that time The rate of chemical changes is directly proportional to the concentration of the reactant at that time. |
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| 688. |
An organic compound undergoes first decompoistion. The time taken for its decompoistion to `1//8` and `1//10` of its initial concentration are `t_(1//8)` and `t_(1//10)`, respectively. What is the value of `([t_(1//8)])/([t_(1//10)]) xx 10`? `(log_(10)2 = 0.3)` |
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Answer» `1^(st)` Methof : `Kt_(1//8) = ln{(C_(0))/(C_(0)//8)} = ln 8` `Kt_(1//10) = ln{(C_(0))/(C_(0)//10)} = ln 10` Then `(t_(1//8))/(t_(1//10)) xx 10 = (log 8)/(log 10) = (3 log 2)/(log 10) xx 10 = 9` `2^(nd)` Methof : `(t_(1//8))/(t_(1//10)) = (log((1)/(1//8)))/(log((1)/(1//10))) = (log 8)/(log 10) = 3 log 2 = 3 xx 0.3 = 0.9` `:. (t_(1//8))/(t_(1//10)) xx 10 = 0.9 xx 10 = 9``1^(st)` Methof : `Kt_(1//8) = ln{(C_(0))/(C_(0)//8)} = ln 8` `Kt_(1//10) = ln{(C_(0))/(C_(0)//10)} = ln 10` Then `(t_(1//8))/(t_(1//10)) xx 10 = (log 8)/(log 10) = (3 log 2)/(log 10) xx 10 = 9` `2^(nd)` Methof : `(t_(1//8))/(t_(1//10)) = (log((1)/(1//8)))/(log((1)/(1//10))) = (log 8)/(log 10) = 3 log 2 = 3 xx 0.3 = 0.9` `:. (t_(1//8))/(t_(1//10)) xx 10 = 0.9 xx 10 = 9` |
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| 689. |
The rate equation for the reactions `2A + B to C` is found to be: rate = `k[A][B]`. The correct statement in relation to this reaction is that theA. Rate of formation of C is twice the rate of disappearance of AB. `t_(1//2)` is a constantC. Unit of k must be `s^(-1)`D. Value of k is independent of the initial concentration of A and B |
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Answer» Correct Answer - d For ` 2 A + B to C ` Rate = k[A] [B] Value of rate constant `K = Ae^(-Ea//RT)` here k is independent of the initial concentration of A and B . |
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| 690. |
The rate equation for the reactions `2A + B to C` is found to be: rate = `k[A][B]`. The correct statement in relation to this reaction is that theA. rate of formation of C is twice the rate of dissappearing of AB. `t_(1//2)` is a constant.C. unit of k must be `s^(-1)`D. value of k is independent of the initial concentration of A and B. |
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Answer» Correct Answer - D d) is the correct answer. |
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| 691. |
Select the law that correponds to data shown for the following reaction `A+B rarr` Products `{:(Exp,[A],[B],"Initial rate",),(1,0.012,0.035,0.1,),(2,0.024,0.070,0.8,),(3,0.024,0.035,0.1,),(4,0.012,0.070,0.8,):}`A. Rate `= k[B]^(3)`B. Rate `= k[B]^(4)`C. Rate `= k[A][B]^(3)`D. Rate `= k[A]^(2)[B]^(2)` |
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Answer» Correct Answer - A form experiments `(3)` and `(2)`, `[A]` is constant and `[B]` is doubled and rates becomes `8` times, so order w.r.t. `[B] = 3`. form experiments `(1)` and `(3)`, `[B]` is constant and `[A]` is doubled, but rate does not change, so order w.r.t `[A] = 0`. Thus, rate `= k[B]^(3)` |
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| 692. |
For the reaction `A + B to` products, what will be the order of reaction with respect to A and B? `{:("Exp",[A](mol L^(-1)),[B](mol L^(-1)),"Initial rate (mol L^(-1) s^(-1))),(1,2.5 xx 10^(-4), 3 xx 10^(-5), 5 xx 10^(-4)),(2,5 xx 10^(-4),6 xx 10^(-5), 4 xx 10^(-3)),(3, 1xx 10^(-3), 6 xx 10^(-5), 1.6 xx 10^(-2)):}`A. 1 with respect to A and 2 with respect to BB. 2 with respect to A and 1 with respect to BC. 1 with respect to A and 1 with respect to BD. 2 with respect to A and 2 with respect to B |
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Answer» Correct Answer - B Rate `= k [A]^(x) [B]^(y)` From exp.(1), `5 xx 10^(-4) = k (2.5 xx 10^(-4))^(x (3 xx 10^(-5))^(y)` Frox exp. (2), `4 xx 10^(-3) = k (5 xx 10^(-4))^(x) (6 xx 10^(-5))^(y)`….(ii) Dividing (ii) by (i), `(4 xx 10^(-3))/(5 xx 10^(-4))` From exp. (3) `1.6 xx 10^(-2) = k (1 xx 10^(-3))^(x) (6 xx 10^(-5)^(-6))^(y)`...(iii) Dividing (iii) by (ii), `(1.6 xx 10^(-2))/(4 xx 10^(-3)) = 2^(x) = 4` or x = 2, y = 1 Hence order with respect to A is 2 and with respect to B is 1. |
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| 693. |
A first order reaction is 60% complete in 20 min. How long will the reaction take to be 84% complete?A. 68 minB. 40 minC. 76 minD. 54 min |
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Answer» Correct Answer - B for a first order reaction, `k=2.303/tlog""(a)/(a-x)`case i `k=2.303/20log""(100)/(100-60)=0.045`case ii`k=2.303/20log""(100)/(100-60)=0.045` `t=2.303/0.045log""100/16=40.73`min |
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| 694. |
A first order reaction is `10%` complete in 20 min. the time taken for `19%` completion is :A. 30 minB. 40 minC. 50 minD. 38 min |
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Answer» Correct Answer - B 10% of the reaction is completed in 20 min. Next 10% of the reaction wil be completed in next 20 min.Hence, 10%+9%=10% of the reaction will be completed in approx=40min. |
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| 695. |
For a reaction ` A + B to C + 2D` , experimental results were collected for three trials and the data obtained are given below The correct rate law of the reaction isA. Rate = k`[A]^(0) [B]^(2)`B. Rate = k`[A][B]^(2)`C. Rate = k`[A][B]`D. Rate = k`[A] [B]^(0)` |
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Answer» Correct Answer - a Let the rate according to the rate law = `k [A]^(x) [B]^(y)` Then from the experiments `5.5 xx 10^(-4) = k [0.4]^(x) [0.2]^(y) ….. (i)` `5.5 xx 10^(-4) = k[0.8]^(x) [0.2]^(y) …. (ii)` `2.2 xx 10^(-3) = k [0.4]^(x) [0.4]^(y) …. (iii)` From equation (i) and (ii) `(5.5 xx 10^(-4))/(5.5 xx 10^(-4)) = (k[0.4]^(x) [0.2]^(y))/(k[0.8]^(x) [0.4]^(y))` `(1)/(4) = ((0.2)/(0.4))^(y) = ((1)/(2))^(y) implies y = 2` Thus , rate law Rate = `k[A]^(0) [B]^(2)` |
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| 696. |
What is the first-order rate constant for the reaction that is `36.5%` complete in `0.0200` seconds?A. `50.4s^(-1)`B. `27.7s^(-1)`C. `22.7s^(-1)`D. `9.86s^(-1)` |
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Answer» Correct Answer - C |
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| 697. |
The reaction of hydrogen and iodine monochloride is given as `H_(2 (g)) + 2I Cl_((g)) + I_(2 (g))` This reaction is of first order with respect to `H_(2 (g))` and `I Cl_((g))` following mechanisms were proposed Mechanism A : `H_(2(g)) + 2I Cl_((g)) to HCl_((g)) + I_(2(g))` Mechanism B : `{:(H_(2 (g)) + ICl_((g)) to HCl_((g)) + HI_((g)) "Slow") , (HI_((g)) + ICl_((g)) to HCl_((g)) + I_(2(g)) "Fast"):}` Which of the above mechanism(s) can be consists with the given information about the reactionA. B onlyB. A and B onlyC. Neither A nor BD. A only |
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Answer» Correct Answer - a Rate of reaction determined by rate determining step or slowest step . Because the mechanism B involves one molecule of I Cl in slowest step so mechanism B is approximate . |
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| 698. |
Concentration of the reactant in first-order is reduced to `(1)/(e^(2))` after : (Natural life `=(1)/(k)`)A. one natural life-timeB. two natural life-timeC. three natural life-timeD. four natural life-time |
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Answer» Correct Answer - B |
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| 699. |
The increase in rate constant of a chemical reaction with increasing temperature is(are) due to the fact(s) thatA. the number of collisons among the reactant molecules increases with increasing temperature.B. the activation energy of the reactions decreases with increasing temperature.C. the corresponding of the reactant molecules increases with increasing temperature.D. the number of rectant molecules acquiring the activation energy increases with increasing temperature. |
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Answer» Correct Answer - A::D (a,d) Both the statements are correct. |
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| 700. |
for the reaction, `2A + B rarr 3C + D`, which of the following does not express the reaction rateA. `(d[D])/(d t)`B. `-(d[A])/(2d t)`C. `-(d[C])/(3d t)`D. `-(d[B])/(d t)` |
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Answer» Correct Answer - C The rate of reaction is `= -(1)/(2) (d[A])/(d t) = -(d[B])/(d t) = (1)/(3)(d[C])/(d t) = (d[D])/(d t)` |
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