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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.
| 1601. |
If 50second are required to decay 2//3rd fraction of a radionactive sample, what fraction will be left after 25sec?A. `(1)/(2)`B. `(1)/(sqrt3)`C. `(1)/(3)`D. `sqrt((2)/(3))` |
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Answer» Correct Answer - B |
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| 1602. |
A first order reaction `A_(2)B_(2)(g) to 2A(g) + 2B(g)` at the temperature `400^(@)`C has the rate constant `k=2.0 xx 10^(-4)s^(-1)`. What percentage of `A_(2)B_(2)` is decomposed on heating for 900 seconds? |
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Answer» For first order reaction, k`=2.303/t log a/(a-x), 2.0 xx 10^(-4)s^(-1)= 2.303/(900s)loga/(a-x)` `loga/(a-x) = (2.0 xx 10^(-4)s^(-1)) xx (900s)/(2.303) = 0.0781` `a/(a-x)= "Antilog" 0.0781 = 1.197, a=1.197a-1.197, x=0.197/1.197a = 0.1645a` this means that `16.45%` of the initial concentration has changed into the products. |
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| 1603. |
For, `underset(("excess"))(2A) + B+C rarr` Products, calculate the: (a) rate expression. (b) units of rate and rate constant. ( c) effect on rate if concentration of `A` is doubled and of `B` is tripled. |
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Answer» (a) Rate `=K[A]^(0)[B]^(1)[C]^(1)` (b) Unit of rate = mol `litre^(-1) time^(-1)` Unit of rate constant = `litre mol^(-1) time^(-1)` (c ) Let initial conc. Of A, B and C be a, b and c, respectively. `:. r_(1)=K(a)^(0)(b)^(1)(c )^(1) …(1)` Now `[A]=2a,[B]=3b` `:. r_(2)=K(2a)^(0)(3b)^(1) (c )^(1) ...(2)` By eqs. (1) and (2), `r_(1)/r_(2)=1/3` `:. r_(2)=3 r_(1)` |
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| 1604. |
For the reaction, `2A + B to A_(2)B`, the reaction rate = `k[A][B]^(2)` with k =`2.0 xx 10^(-6) mol^(-2) L^(2)s^(-1)`. Calculate the initial rate of the reaction when [A] = 0.1 mol `L^(-1)`, [B]= 0.2 mol`L^(-1)`. Also calculate the reaction rate when [a] is reduced to 0.06 mol `L^(-1)`. |
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Answer» First case: Rate = `k[A][B]^(2)` =`2.0 xx 10^(-6)mol^(-2)L^(2)s^(-1) xx (0.1 xx mol L^(-1)) xx (0.2 mol L^(-1))^(2)` `8 xx 10^(-9) mol L^(-1)s^(-1) = 8 xx 10^(-9)Ms^(-1)`. Second case: The concentration of A after taking part in the reaction = 0.06 mol `L^(-1)` Amount of B reacted `= (0.1 - 0.06) = 0.04 mol L^(-1)` Amount of B reacted `= 1/2 xx 0.04 mol L^(-1) = 0.02 mol L^(-1)` The concentration of B after taking part in the reaction = `(0.2-0.02)=0.18 mol L^(-1)` Rate = `k[A][B]^(2)` `=2.0 xx 10^(-6) mol^(-2)L^(2)s^(-1) xx (0.06 mol L^(-1)) xx (0.18 mol L^(-1))^(2)` `=3.89 xx 10^(-9)mol L^(-1)s^(-1) = 3.89 xx 10^(-9)Ms^(-1)` |
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| 1605. |
When initial concentration of a reactant is doubled in a reaction , its half-life period is not effected . The order of the reaction isA. FirstB. SecondC. More than zero but less than firstD. Zero |
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Answer» Correct Answer - a `t_(1//2) = (0.6932)/(K)` Half life of first order reaction is independent on initial conc., of reactant |
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| 1606. |
The reaction `2A + B + C rarr D + 2E` is found to be first order in `A`, second order in `B` and zero order in `C`. (a) Write the rate expresison. (b) What is the effect on rate on increaisng the concentration of `A, B`, and `C` two times ? |
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Answer» (a) Rate `=K[A]^(1)[B]^(2)[C]^(0)` (b) Let initial conc. Of `A, B` and `C` be a, b, and c mol `litre^(-1)` respectively Then, Rate `r_(1) =K a^(1) b^(2) c^(0) …(1)` Now if conc. Of `A, B` and `C` are double, i.e., `2a, 2b` and `2c` respectively, then, `r_(2)=K(2a)^(1) (2a)^(2) (2c)^(0) ...(2)` By eqs. (1) and (2), `r_(1)/r_(2)=1/8` `r_(2)=8r_(1)` |
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| 1607. |
A reactions is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half?A. The rate of the reaction is proportional to [X]B. The rate of the reaction is proportional to `[X]^(2)`C. Two molecules of X are present in the stoichiometric equationD. The reaction occurs in two steps |
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Answer» Correct Answer - b `r prop [X]^(2)` or `r = k [X]^(2)`. |
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| 1608. |
Which of the following expresison can be used to describe the instantaneous rate of the reaction ? `2A+B rarr A_(2)B`A. `-(dA)/(2dt)`B. `-(dA)/(dt)`C. `(d(A_(2)B))/(2dt)`D. `-1/2(dA)/(dt)(dB)/(dt)` |
| Answer» Correct Answer - A | |
| 1609. |
The average rate and instantaneous rate are the same. |
| Answer» Correct Answer - T | |
| 1610. |
The instantaneous rate of reaction `2A+B to C +3D ` is given byA. `(dA)/(dt)`B. `(1)/(2) (d[A])/(dt)`C. `(d[B])/(dt)`D. `(1)/(3)(d[D])/(dt)` |
| Answer» Correct Answer - D | |
| 1611. |
In the reaction `A+3B to 2C,` the rate of formation of C isA. the same as rate of concentration of AB. the same as the rate consumption of BC. twice the rate of consumption of AD. 3/2 times the rate of consumption of B |
| Answer» Correct Answer - C | |
| 1612. |
For the reaction `A+3B to 2C` `rate = k[A]^(1//2)[B]^(1//2)`. Find the order of reaction. |
| Answer» Order = `1/2 + 1/2=1` | |
| 1613. |
In the reaction `A+3B to 2C,` the rate of formation of C isA. the same as rate of concentration of A B. the same as the rate consumption of BC. twice the rate of consumption of AD. 3/2 times the rate of consumption of B |
| Answer» Correct Answer - C | |
| 1614. |
For a chemical reaction the energy of activation is 85 kJ `"mol"^(-1)` If the frequency factor is `4.0xx10^(9) L "mol"^(-1)s^(-1)`, what is the rate constant at 400 K?A.B.C.D. |
| Answer» Correct Answer - `k=3.19xx10^(-2) L "mol"^(-1)s^(-1)` | |
| 1615. |
From the concentrations of `C_(4)H_(9)Cl` (butyl chloride) at different times given below, calculate the average rate of the reaction: `C_(4)H_(9)Cl +H_(2)Orarr C_(4)H_(9)OH+HCl` during different intervals of time. `{:(t//s,0,50,100,150,200,300,400,700,800),([C_(4)H_(9)Cl]//"mol"^(-1),0.100,0.0905,0.0820,0.0741,0.0671,0.0549,0.0439,0.0210,0.017):}` |
| Answer» We can determine the difference in concentration over different intervals of time and thus determine the average rate by dividing ∆[R] by ∆t (Table 4.1). | |
| 1616. |
For a gaseous reaction, `A+B to 2C,` the rate law is given as `R=K[A]^(-1)[B]^(2)`. Select the statements which are not correct.A. If initially concentration of A is double of B then half life of reaction will remian same throughtout the reaction.B. If initial A is taken in very large amount then half life of reaction will keep on increasing as reaction proceeds.C. If initially B is taken in very large amount then half life of reaction will remian same throughout the reaction.D. If A and B are taken with equal concentration, then both will be reduced to half of original in same time interval. |
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Answer» Correct Answer - A::C |
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| 1617. |
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.B.C.D. |
| Answer» Correct Answer - `53.59 kJ "mol"^(-1)` | |
| 1618. |
Statement-1: Th plot of atomic number (y-axis) versus number of neutrons (x-axis) for stable nuclei shows a curvature towards x-axis from the line of `45^(@)C` slope as the atomic number is increased. Statement-2: Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides.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 - a |
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| 1619. |
For two reactions, `I: A to B " " "Rate constant"=k_(1)sec^(-1)` `II: C to D " " "Rate constant"=k_(2) M^(-1)sec^(-1)` Starting with initial concentration of 1M each, time taken to reach to 0.5M is same, then, identify the correct options.A. `k_(1)gtk_(2)`B. `k_(1)lt k_(2)`C. Rate of first reaction at 0.5M concentration gt rate of second reaction at 0.5M concentration.D. Initial rate of first reaction lt initia rate of second reaction. |
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Answer» Correct Answer - B::C::D |
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| 1620. |
Statement: A catayst does not affect the heat of reaction. Explanation: It increases the rate of reaction.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 - D Both statement and explanation are correct. |
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| 1621. |
Statement-1: Temperature coefficient of a one step reaction may be negative. Statement-2: The rate of reaction having negative order with respect to a reactant decreases with the increase in concentration of the reactant.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 - d |
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| 1622. |
Statement-1: A catayst provides an alternative path to the reaction in which conversion of reactants into products take place quickly. Statement-2: The catalyst forms an activated comlex of lower potential energy, with the reactants by which more number of molecules are able to cross the barrier per unit of time.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 - a |
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| 1623. |
Statement-1: In a reversible endothemic reactiuon, `E_(act)` of forward reaction is higher than that of backward reaction. Statement-2: The threshold energy of forward reaction is more than that of backward reaction.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 - c |
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| 1624. |
The acid catalysed ionisation of `gamma`-hydroxy butyric acid proceeds as a reversible reaction. Which is I order in both the forward and backward steps: `underset("(Acid)")(A) underset(K_(1)) overset(K_(2))(hArr) underset("(Lactose)")(B)` The rate `-(d[A])/(d t)` is given by:A. `K_(1)[A]`B. `-K_(2)[B]`C. `K_(1)[A]-K_(2)[B]`D. `(K_(1)[A])/(K_(2)[B])` |
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Answer» Correct Answer - C rate `= K[A] - K_(2)[B]`, This is reversible reaction. |
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| 1625. |
The energy of activations for forward and backward change for an endothermic reaction, `X rarr Y` are `E_(f)` and `E_(b)` respectively. Which of the following is correct ?A. `E_(b) lt E_(f)`B. `E_(b) gt E_(f)`C. `E_(b) = E_(f)`D. No relation between them |
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Answer» Correct Answer - A `Delta H = E_(f)-E_(b)` `because Delta H = +ve` `:. E_(f) gt E_(b)` |
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| 1626. |
Assertion : `E_(a)` of the forward reaction is higher than that of backward reaction in a reversible endothermic reaction. Reason : Increasing the tempreature of the substance increases the fraction of molecules which collide with energies greater than `E_(a)`.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B Threshold energy of reactants is less than that of products in reversable endothermic reactions hence, `E_(a)` for forward reaction is more than `E_(a)` for backward reaction |
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| 1627. |
Assertion : The reaction `underset("Cane sugar")(C_(12)H_(22)O_(11)) + H_(2) O overset(H^(+))(t0) underset("Glucose")(C_(6)h_(12)O_(6)) + underset("Fructose")(C_(6)H_(12)O_(6))` is a first order reaction. Reason : Change in concentration of `H_(2)O` is negligible.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A such reactions are called pseudo first order reactions. |
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| 1628. |
The time for half life period of a certain reaction `A to` Products A is `2.0 mol L^(-1)`, how much time does it take for its concentration to come from `0.50` to `0.25` mol `L^(-1)` if it is a zero order reaction?A. 1 hB. 4 hC. `0.5`hD. `0.25`h |
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Answer» Correct Answer - D d) For zero order of reaction. `t_(1//2) = ([A_(0)])/(2k)` or `k=([A_(0)])/(2t_(1//2))` `k=(2.0molL^(-1))/(2 xx 1hr)=1 molL^(-1)hr^(-1)` For zero order reaction, `t=([A]_(0)-[A])/(k)` `((0.5-0.2s)mol L^(-1))/(1 mol L^(-1) hr^(-1))` = 0.25 hr. |
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| 1629. |
A hydrogenation reaction is carried out at `500 K`. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is `400 K`. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by `20 kJ mol^(-1)`. |
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Answer» Correct Answer - `100kJmol^(-1)` |
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| 1630. |
Identify the reaction order from each of the following rate. i) `k=2.3 xx 10^(5)` L `mol^(-1)s^(-1)` ii) `k=2.3 xx 10^(5)Lmol^(-1)s^(-1)` ii) `k=3.1 xx 10^(-4)s^(-1)` |
| Answer» i) second order , ii) first order | |
| 1631. |
Activation energy of forward and backward process of reaction are 60 kJ and 40 kJ `"mol"^(-1)` respectively . Which of the following are true the reaction ?A. It is endothermic reactionB. It is exothermic reactionC. Heat of reaction is +20kJ `"mol"^(-1)`D. Thershol enregy of reaction is 100 kJ `"mol"^(-1)` |
| Answer» Correct Answer - A::B::C::D | |
| 1632. |
The energy of activation for a reaction is `100 KJ mol^(-1)`. The peresence of a catalyst lowers the energy of activation by `75%`. What will be the effect on the rate of reaction at `20^(@)C`, other things being equal? |
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Answer» The arrhenius equation is , `k=Ae^(-E_(a)//R)` In absence of catalyast, `k_(1)=Ae^(-100//RT)` In presence of catalyst, `k_(2)=Ax^(-25//RT)` So, `(k_(2))/(k_(1))=^(75//RT)or2.303 "log" (k_(2))/(k_(1))=(75)/(RT)` `or 2.303 "log" (k_(2))/(k_(1))=(75)/(8.314xx10^(-3)xx293)` `or "log" (k_(2))/(k_(1))=(75)/(8.314xx10^(-3)xx293xx2.303)` or `(k_(2))/(k_(1))=2.314xx10^(13)` As the things beings equal in presence or absence of a catalyst, `(k_(2))/(k_(1))` must be =`("rate in presence of catalyst")/("rate in absence of catalyst")` `i.e., (r_(2))/(r_(1))=(k_(2))/(k_(1))=2.34xx10^(13)` |
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| 1633. |
A poistive catalyst increases the activation energy of the reaction. |
| Answer» Correct Answer - F | |
| 1634. |
Assertion (A): Poistive catalyst lowers the activation energy of the reaction whereas the heat of reaction remains same. Reason (R ): The heat of reaction is equal to the difference between activation energies for forward and backward reactions.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 - B | |
| 1635. |
what is the activation energy? How is rate constant related to activation energy? |
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Answer» For the definition of activation energy `(E_(a))` Rate constant(k) for a reaction is related to activation energy as: In k = In A -`E_(a)/RT` |
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| 1636. |
Which of the followoing expression is correct for first order reaction ? `(CO)` refers to initial concentration of reactant.A. `t_(1//2) prop CO`B. `t_(1//2) prop CO^(-1)`C. `t_(1//2) prop CO^(-2)`D. `t_(1//2) prop CO^(0)` |
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Answer» Correct Answer - D `t_(1//2) prop (CO)^(@)` i.e. half life for `I^(st)` order is independent of initial concentration. |
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| 1637. |
For a zero order reaction `A rarr B, t_(1//2)` is (k is rate constant)A. `([A]_(o))/(2k)`B. `( ln 2)/( k)`C. `(1)/(k[A]_(0))`D. `(ln2)/([A]_(0)k)` |
| Answer» Correct Answer - A | |
| 1638. |
The decomposition of HI on the surface of gold is :A. Pseudo first orderB. zero orderC. first orderD. second order |
| Answer» Correct Answer - B | |
| 1639. |
The rate constant for an isomerisation reaction `Ato B` is `4.5 xx 10^(-3)min^(-1)`. If the initial concentration of A is 1M, Calculate the rate of the reaction after 1 hour. |
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Answer» Step I. Calculating of concentration after 1 h (60 min). `k=2.303/t log ([A])_(0)/([A])_(t)` `(log)[A]_(0)/[A]_(t)=(kt)/(2.303) = ((4.5 xx 10^(-3) min^(-1))xx (60 min))/2.303 = 0.11729` `(log)[A]_(0)/[A]_(t)= Antilog0.1172//9 = 1.310, [A]_(t) = [A]_(0)/1.310 = 0.763M` Step II. Calculation of rate after 1 h(60 min). Rate after 60 min= `k[A]_(t)= 4.5 xx 10^(-3) min^(-1) xx 0.763 M= 3.4354 xx 10^(-3) M min^(-1)` |
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| 1640. |
Consider following two reactions `Ato"Product", (d[A])/(dt)=k_(1)[A]_(0)^(0)` `Bto"Product", (d[B])/(dt)=k_(1)[A]_(0)^(0)` `k_(a) and k_(2)` are expressed are expressed in term of molarity (mol `L^(-1))` and time `(s^(-1))` as :A. `s^(-1),M s^(-1)L^(-1)`B. `M s^(-1),Ms^(-1)`C. ` s^(-1),M^(-1)s^(-1)`D. ` M s^(-1), s^(-1)` |
| Answer» Correct Answer - D | |
| 1641. |
At room temperature, the reaction between NO and `O_(2)` to give `NO_(2)` is fast, while that between CO and `O_(2)` is slow. It is due to:A. CO is smaller in size that of NOB. CO is poisonousC. the activation energy for the reaction, `2NO+O_(2)rarr 2NO_(2)` is less than `2CO+O_(2)rarr 2CO_(2)`D. none of the above |
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Answer» Correct Answer - C |
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| 1642. |
For a first order reaction `ArarrP` , the temperature (T) dependent rate constant `(K)` was found to follow the equation log `k=-(2000)(1)/(T)+6.0` . The pre- exponential factor A and the activation energy `E_(a)` , respectively, are :A. `1.0xx10^(6)s^(-1)` and `9.2 kJ mol^(-1)`B. `6.0 s^(-1)` and `16.6 kJ mol^(-1)`C. `1.0xx10^(6)s^(-1)` and `16.6 kJ mol^(-1)`D. `1.0xx10^(6) s^(-1)` and `38.3 kJ mol^(-1)` |
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Answer» Correct Answer - D |
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| 1643. |
In a first order reaction, `ArarrP,` the ratio of `a//(a-x)` was found to be 8 after 60 m in. If the concentration is 0.1 M then the rate of reaction isA. `2.226xx10^(3)" mol "L^(-1) " min"^(-1)`B. `4.455xx10^(-3)" mol "L^(-1)" min"^(-1)`C. `3.466xx10^(-3)" mol "L^(-1)" min"^(-1)`D. `5.532xx10^(-3)" mol "L^(-1)" min"^(-1)` |
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Answer» Correct Answer - C `k=(2.303)/(t)"log"((a)/(a-x))=(2.303)/(60)"log"B=0.03466"min"^(-1)` `((dx)/(dt))=k[A]=0.03466xx0.1` `=3.466xx10^(-3)" mol "L^(-1)"min"^(-1)` |
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| 1644. |
For a hypothetical reaction, `ArarrP` the rate constant is 0.12 `s^(-1).` which of the following relation is correct, if `[R]_(0)` is the initial concentration?A. `t_(1//2)=([R]_(0))/(0.12xx12)`B. `t_(1//2)=(3)/(2(0.12)[R]_(0)^(2))`C. `t_(1//2)=(0.693)/(0.12)`D. `t_(1//2)=(0.693)/(0.12xx3)` |
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Answer» Correct Answer - C Since, the unit of rate constant is `sec^(-1)`. Thus, the order of the reaction is 1. Hence , `t_(1//2)=(0.693)/(0.12).` |
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| 1645. |
The following data are for the reaction `A+B rarr` Products: `{:(Conc. of A(M),,Conc. of B(M),,Initial rate),(,,,,(mol L^(-1) s^(-1))),(0.1,,0.1,,4.0xx10^(-4)),(0.2,,0.2,,1.6xx10^(-3)),(0.5,,0.1,,1.0xx10^(-2)),(0.5,,0.5,,5.0xx10^(-2)):}` (a) What is the order with respect to `A` and `B` for the reaction? (b) Calculate the rate constant Determine the reaction rate when the concentrations of `A` and `B` are `0.2 M` and `0.35 M` respectvely. |
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Answer» Proceed as Problem `15`. (a) Rate `=K[A]^(2)[B]^(0)` (b) `K=4xx10^(-2) litre mol^(-1) sec^(-1)`, (c ) `Rate =16xx10^(-4) mol litre^(-1) sec^(-1)` |
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| 1646. |
For the hypothetical reaction `2A + B rarr` Products following data obtained: `|{:("Experiment number","Initial conc of (A)" (mol L^(-1)),"Initial conc of (B)" (mol L^(-1)),"Initial rate mol" L^(-1) s^(-1),),(1,0.10,0.20,3 xx 10^(2),),(2,0.30,0.40,3.6 xx 10^(10^(3)),),(3,0.30,0.80,1.44 xx 10^(4),),(4,0.10,0.40,...,),(5,0.20,0.60,...,),(6,0.30,1.20,...,):}|` Find out how the rate of the reaction depends upon the concentration of `A` and `B` and fill in the blanks given in the table. |
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Answer» From expts. (2) and (3) , it is clear that when concentration of A is kept constant and that of B is doubled, the rate increasefour times. This shows that the reaction is of second order with respect to B. Similarly, from expts. (1) and (2). It is observed that when concentration of A increased three times and that of B two times, the rate becomes twelve times. Hence. the reaction is first order with respect to A. |
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| 1647. |
For a chemical reactin `ArarrE,` it is found that rate of reaction is doubled when the concentration of A is increased four times. The order of the reaction is :A. 1B. 2C. `1//2`D. zero |
| Answer» Correct Answer - D | |
| 1648. |
The half of second order reactin is :A. inversely proportional to the square of the initial concentration of the reactantsB. inversely proportional to the intial concentration of reactantsC. proportional to the initial concentration of reactanstD. independent of the initial concentration of reactants |
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Answer» Correct Answer - B `t_(1//2)prop(1)/(a^(n-1)),` a= initial conc, n= order `t_(1//2)` (second order ) `prop(1)/(a)` |
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| 1649. |
Conisder the following elementary reaction, `2A + B + C rarr` Products. All reactants are present in the gaseous state and reactant `C` is taken in excess. What is the unit of rate constant of the reaction?A. `mol^(-1) time^(-1)`B. `time^(-1)`C. `mol^(-1) L^(2) time^(-1)`D. `mol^(-2) L^(2) time^(-1)` |
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Answer» Correct Answer - D Order of reaction `=3` Units of third order reaction `=[Conc]^(1+n)=[Conc]^(1-3)= mol^(-2)L^(2) time^(-1)` |
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| 1650. |
Conisder the following elementary reaction, `2A + B + C rarr` Products. All reactants are present in the gaseous state and reactant `C` is taken in excess. How will the rate change if the concentration of A is doubled and that of B is tripled ?A. The rate increases `12` times of original value.B. The rate increases `8` times of original value.C. The rate reduces `12` times of original value.D. The rate reduces `8` times of original value. |
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Answer» Correct Answer - A `r_(1)=k[A]^(2)[B]` `r_(2)=k[2A]^(2)[3B]` `(r_(2))/(r_(1))= 12impliesr_(2)= 12r_(1)` |
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