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Electrochemistry and chemical kinetics - class-XII

Description: electrochemistry and chemical kinetics
Number of Questions: 85
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Tags: chemistry electrochemistry and chemical kinetics chemical kinetics
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Rate constant in case of first order reaction is :

  1. Inversely proportional to the concentration units

  2. Independent of concentration units

  3. Directly proportional to concentration units

  4. Inversely proportional to the square of concentration units


Correct Option: B
Explanation:

For First order of reaction,
$Rate= k [A]$,


$k = \cfrac {mol/L}{sec\times {mol/L}}=sec^{-1}$

Option B is correct.

Fill in the blanks by choosing the correct option;
Order of the reaction is the $X$ of the powers to which concentration terms are raised in experimentally determined rate equation. The unit of first order rate constant is $Y$. The unit of first order rate constant when concentration is measured in terms of pressure and time in minutes is $Z$.

  1. $X\rightarrow product, Y\rightarrow mol\ L^{-1} time^{-1}, Z\rightarrow atm\ min^{-1}$

  2. $X\rightarrow sum, Y\rightarrow L\ mol^{-1}time^{-1}, Z\rightarrow atm\ min^{-1}$

  3. $X\rightarrow product, Y\rightarrow L\ mol^{-1}, Z\rightarrow atm\ min^{-1}$

  4. $X\rightarrow sum, Y\rightarrow time^{-1}, Z \rightarrow min^{-1}$


Correct Option: D
Explanation:

The order of a chemical reaction is defined as the sum of the powers of the concentration of the reactants in the rate equation of that particular chemical reaction.


General formula for the unit of rate constant = mole$^{(1-n)}$ L$^{(n-1)}$ min$^{-1}$.

For 1$^{st}$ order reaction, 
n = 1
Therefore, unit of rate constant $=$ min$^{-1}$

The unit of rate constant in terms of pressure and time:
  mol $^{( 1-n)}$  L $^{ (n-1)}$ min$^{ -1 }$ or  atm$^{( 1-n) }$ min $^{ -1 }$


For 1$^{st}$ order reaction, 
n = 1
Therefore, the unit is min$^{-1}.$

Hence, the correct answer is option $\text{D}$.

Match the rate law given in column I with the dimensions of rate constant 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) $Rate = k[C _{2}H _{5}Cl]$ (iv) $L^{1/2} mol^{-1/2} s^{-1}$
  1. $(A)\rightarrow (iv), (B) \rightarrow (iii), (C)\rightarrow (ii), (D) \rightarrow (i)$

  2. $(A)\rightarrow (i), (B) \rightarrow (ii), (C)\rightarrow (iii), (D) \rightarrow (iv)$

  3. $(A)\rightarrow (ii), (B) \rightarrow (i), (C)\rightarrow (iv), (D) \rightarrow (iii)$

  4. $(A)\rightarrow (i), (B) \rightarrow (ii), (C)\rightarrow (iv), (D) \rightarrow (iii)$


Correct Option: D
Explanation:
(A) $Rate= k[NH _3]^0$
It is zero order reaction.
$\therefore$ Units of rate constant are same as rate.
i.e $ mol\ l^{-1}s^{-1}$

(B) $Rate=k[H _2O _2]^1[I^-]^1$
It is second order reaction as order $=1+1=2$
$\therefore$ Units of Rate constant are $L \ mol^{-1}s^{-1}$

(C) $Rate=k[CH _3CHO]^{3/2}$
It is fractional order reaction with order $= \cfrac 32$
$\therefore$ Units of rate constant are $L^{\cfrac 12}mol^{-\cfrac 12}s^{-1}$

(D) $Rate=k[C _2H _5Cl]^{-1}$
It is first order reaction. So units of rate constant are $s^{-1}$

For the second order reaction, concentration $(x)$ of the product at time $t$ starting with initial concentration $[A] _0$ is:

  1. $\dfrac{kt[A _0]^2}{1 + kt[A _0]}$

  2. $\dfrac{k + [A _0]^2}{1 + kt}$

  3. $\dfrac{1 + kt[A _0]^2}{k + [A _0]^2}$

  4. none of these


Correct Option: A
Explanation:

A reaction said to be a second order when the overall order is $2$. The rate of second order reaction may be proportional to one concentration squared.

$R=K[A]^2$
For rate proportional to single concentration squared, the time dependance of concentration is given by
$\cfrac{1}{[A]}=\cfrac{1}{A _0}+Kt
Therefore, concentration of product after time $t=\cfrac{kt[A_0]^2}{1+kt[A_0]}$.

Units of rate constant of a first order reaction is :

  1. $mole.lit^{-1}$

  2. $lit. mole$

  3. $mole. sec^{-1}$

  4. $sec^{-1}$


Correct Option: D
Explanation:

A $\rightarrow$ product

For first order reaction, rate is dependent on single reactant A for example, rate = k[A]
$k=\frac{rate}{[A]}$

$=\frac{mole}{liter}sec\times \frac{liter}{mole}$
$=sec^{-1}$

A gaseous reaction, $A _{2}\left ( g \right )\rightarrow B\left ( g \right )+\frac{1}{2}\left ( g \right )$ 
Show increase in pressure from 40 mm to 120 mm in 5 minutes. the rate of disappearance of$A _{2}$ is ?

  1. 4 mm $min ^{-1}$

  2. 8mm $min^{-1}$

  3. 16 mm $min^{-1}$

  4. 2 mm $min^{-1}$


Correct Option: B

Consider the reaction $2A+B$ $\rightarrow$products,when the concentration of a alone was doubled, the half-life of the  reaction did not change.When the concentration of B alone was double,the rate was not altered.The unit of rate constant for this reaction is

  1. $S^{-1}$

  2. $L\ mol^{-1}\ s^{-1}$

  3. $mol\ L^{-1}\ s^{-1}$

  4. $mol^{-2}\ L^{5}\ S^{-1}$


Correct Option: B

For the reaction $A\rightarrow C+D$, the initial concentration of $A$ is $1000 M$. After $10^{2} sec$ concentration of $A$ is $100\ M$. The rate constant of the reaction has the numerical value of $9.0$. What is the unit of the reaction rate constant? 

  1. $M^{-1}s^{-1}$

  2. $Ms^{-1}$

  3. $s^{-1}$

  4. $M^{-1.5}s^{-1}$


Correct Option: B

The second order rate constant is usually expressed as :

  1. $Mol\ L\ s^{-1}$

  2. $Mol^{-1}\, L^{-1}\, s^{-1}$

  3. $Mol\, L^{-1}\, s^{-1}$

  4. $Mol^{-1}\, L\, s^{-1}$


Correct Option: D
Explanation:

As we know,
for second order reaction:
$r = k[A]^2$
so unit of rate constant is $Mol^{-1}\, L\, s^{-1}$.

The unit of rate constant obeying the rate expression, $r=k{ \left[ A \right]  }{ \left[ B \right]  }^{ { 2 }/{ 3 } }$ is:

  1. ${ mol }^{ { -2 }/{ 3 } }\ { litre }^{ { 2 }/{ 3 } }\ { time }^{ -1 }$

  2. ${ mol }^{ { 2 }/{ 3 } }\ { litre }^{ { -2 }/{ 3 } }\ { time }^{ -1 }$

  3. ${ mol }^{ { -5 }/{ 3 } }\ { litre }^{ { 5 }/{ 3 } }\ { time }^{ -1 }$

  4. none of the above


Correct Option: D
Explanation:

The unit of rate constant is ${ mol }^{ { -2 }/{ 3 } }\ { litre }^{ { 2 }/{ 3 } }\ { time }^{ 2/3 }$ and it does not match with any of the A, B, C options. So option D is correct

For the second order reaction, if the concentration of reactant changes from $0.08M$ to $0.04M$ in 10 minutes. Calculate the time at which concentration of reactant becomes $0.01M$.

  1. $20min$

  2. $30min$

  3. $50min$

  4. $70min$


Correct Option: D
Explanation:

For a second order reaction we have 

$ \cfrac { 1 }{ { [A] } _{ t } } -\cfrac { 1 }{ [A] _{ o } } =kt$
 In $10$ minutes,  concentration changes from $0.08M $ to $0.04M$
$ \left( \cfrac { 1 }{ 0.04 } -\cfrac { 1 }{ 0.08 }  \right) \cfrac { 1 }{ 10 } =k{ \quad min }^{ -1 }\ 1.25{ min }^{ -1 }=k$ 
Substituting value of $k$, we get it become $ \left( \cfrac { 1 }{ 0.01 } -\cfrac { 1 }{ 0.08 }  \right) \cfrac { 1 }{ 1.25 } =t\ 70min=t$
 Hence answer is $[D]$

For a second order reaction rate at a particular time is $x$. Ifthe initial concentration is trapled, the rate will becomes?

  1. $3x$

  2. $9x^{2}$

  3. $9x$

  4. $27x$


Correct Option: C
Explanation:

Its a second Order reaction.


The rate at a particular time is x. 

$x = k{[A]}^{2}$

If the initial concentration is tripled, the rate becomes

$rate = k[{3[A]}^{2}]$ = $9x$

The given reaction 
$2FeCl _{3}+SnCl _{2}\rightarrow 2FeCl _{2}+SnCl _{4}$
Is an example of:

  1. Third order reaction

  2. First order reaction

  3. Second order reaction

  4. None of these


Correct Option: A
Explanation:

For the third order reaction, the general reaction can be written as,

$A+B+C\rightarrow Products\ r={ K } _{ 3 }\left[ A \right] ^{ 2 }\left[ B \right] ^{ 1 }$
$r=\left[ Fe{ Cl } _{ 2 } \right] ^{ 2 }\left[ Sn{ Cl } _{ 4 } \right] ^{ 1 }$
the order of products are third order.

The rate constant for forward and backward reaction of hydrolysis of ester are $1.1\times 10^{-2}$ and $1.5\times 10^{-3}$ per minute respectively.  


Equilibrium constant for the reaction is :


$CH _3COOC _2H _5 + H _2O\    \rightleftharpoons\ CH _3COOH+C _2H _5OH$ 

  1. 4.33

  2. 5.33

  3. 6.33

  4. 7.33


Correct Option: D
Explanation:

Rate constant of forward reaction $(K _f)=1.1\times 10^{-2}\ min^{-1}$

Rate constant of backward reaction $(K _b)=1.5\times 10^{-3}\ min^{-1}$ 
Equilibrium constant $(K _c)$$=\dfrac {K _f}{K _b}$

$=\dfrac {1.1\times 10^{-2}}{1.5\times 10^{-3}}$

$=7.33$
Hence, option $(D)$ is correct.

The dimensions of rate constant of a second order reaction involves:

  1. neither time nor concentration

  2. only time

  3. time and concentration

  4. time square and concentration


Correct Option: C
Explanation:

second order
$2A \rightarrow product$
$rate=K[A]^{2}$
$k=\frac{rate}{[A]^{2}}$
$=\frac{concentration}{time (concentration )^{2}}$
$=\frac{1}{time . concentration}$
So it involves both time & concentration.

Except for first ofder reactions, the unit of rate constant depends on the dimensions
($mol^{1-n}.lit^{-1}.sec^{-1}  $) concentration and time.
Where, n= order of the reaction.


Which is not true for a second order reaction ?

  1. It can have rate constant $1 \times 10^{-2} $ $L mol^{-1} s^{-1} $

  2. Its half - life is inversely propotional to its initial concentration

  3. Time to compelete $75 % $ reaction is twice of half - life

  4. $ T _{50} $ = $ \frac { 1 }{ K\quad \times \quad Initial\quad conc. } $


Correct Option: C
Explanation:

Unit of Rate Constant for a second order reaction is ${mol}^{-1}L{s}^{-1}$.


${t} _{1/2}\space \alpha\space {a}^{1-n}$. Here n =2 

${t} _{1/2}\space \alpha\space {a}^{-1}$. So Its Inversely Proportional to Half Life.

${t} _{1/2}$ = $\dfrac{1}{ak}$ where a is initial concentration

TIme to complete 75% reaction is twice to half life is false. Its is true for 1st Order Reaction

For a second order reaction rate at a particular time is $X$. If the initial concentration is tripled, the rate will become:

  1. $3X$

  2. $9X^2$

  3. $9X$

  4. $27X$


Correct Option: B
Explanation:

Its a second Order Reaction. Rate is given as X

$\therefore rate=[x]^2$
When the concentration is tripled then,
$rate=[3x]^2=9x^2$

Which of the following statements is incorrect?

  1. A second order reaction must be a bimolecular elementary reaction

  2. A bimolecular elementary reaction must be a second order reaction

  3. Zero order reaction must be a complex reaction

  4. First order reaction may be complex or elementary reaction.


Correct Option: B
Explanation:

A bimolecular elementary reaction may or may not be a Second Order Reaction whereas a Second Order reaction must be a bimolecular elementary reaction.

Ex: Decomposition of ${N} _{2}O _{5}$ is a 1st Order Reaction though its a bimolecular reaction.

The rate of solvolysis of tert-butyl bromide will be maximum in which of the following solvents?

  1. ${ Et } _{ 2 }O$

  2. ${CCl} _{4}$

  3. $EtOH$

  4. ${H} _{2}O$


Correct Option: D
Explanation:

The rate of solvolysis of Tert - butyl bromide will be maximum in Water.

tert-butyl bromide follows $SN1$ mechanism thus it prefers weak nucleophiles like water.

The inversion of cane sugar to produce glucose and fructose is represented by the reaction
     ${ C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 }+{ H } _{ 2 }O\quad \xrightarrow { { H }^{ + } } { C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }+{ C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }$
The reaction is:

  1. second order

  2. unimolecular

  3. pseudo uni-molecular

  4. bi molecular


Correct Option: C
Explanation:

Inversion of Cane Sugar to produce glucose and fructose is a Pseudo uni-molecular Reaction.

Here the rate of reaction is made first order by changing the concentration of sucrose ($C _{12}H _{22}O _{11}$).

Units of the rate constant of first and zero order reactions in terms of molarity M unit are respectively:

  1. $sec^{-1}, M sec^{-1}$

  2. $sec^{-1}, M$

  3. $M sec^{-1}, sec^{-1}$

  4. $M, sec^{-1}$


Correct Option: A
Explanation:

For first order, $M/sec=k[M].$
$\therefore k=sec^{-1}$
For zero order, $M/sec = k[M]^0.$
$\therefore k=M sec^{-1}.$

The unit of rate constant obeying the rate expression $r=K[A]^{1}[B]^{2/3}$ is:

  1. $mole^{-2/3}lit^{2/3}time^{-1}$

  2. $mole^{2/3}lit^{-2/3}time^{-1}$

  3. $mole^{-5/3}lit^{2/3}time^{-1}$

  4. none of the above


Correct Option: A
Explanation:

$r=K[A][B]^{2/3}$


$K=\dfrac{r}{[A][B]^{2/3}}$

$K=\dfrac{mol}{liter.sec}$$\left ( \dfrac{liter}{mole} \right )\left ( \dfrac{liter}{mole} \right )^{2/3}$

$K=mole^{-2/3} liter^{2/3} sec^{-1}$

The reaction, $2A+ B \rightarrow$ Products, follows the mechanism:


$2A \leftrightharpoons A _2$ (at equilibrium)
$A _2 + B \rightarrow$ Products (slow) 

The order of the reaction is:

  1. $2$

  2. $1$

  3. $3$

  4. $1\dfrac{1}{2}$


Correct Option: C
Explanation:

From the slow step, $rate = k [A _2][B]$ ...... (1)
From the equilibrium reaction, equilibrium  constant  $K = \dfrac {[A _2]} {[A]^2}$
$[A _2]=K [A]^2$......(2)


Substitute equation (2) in equation (1).
$rate = k K [A]^2[B]$

Thus the overall order of the reaction is $2+1=3$.

Hence, option C is correct.

For the elementary reaction 2A $\rightarrow $ C ,the concentration of A after 30 minutes was found to be 0.01 mole/lit. If the rate constant of the reaction is $2.5 \times 10^{-2}$ lit mole$^{-1}$ sec$^{-1}$. The rate of the reaction at 30 minutes is:

  1. $2.5 \times 10^{-4}$ lit mole$^{-1}$ sec$^{-1}$

  2. $2.5 \times 10^{-6}$ lit mole$^{-1}$ sec$^{-1}$

  3. $2.5 \times 10^{-2}$ lit mole$^{-1}$ sec$^{-1}$

  4. $2.5 \times 10^{-8}$ lit mole$^{-1}$ sec$^{-1}$


Correct Option: B
Explanation:
Rate of Reaction = Rate Constant  x ${ [A] }^{ 2 }$

 $= 2.5 \times { 10 }^{ -2 } \times   { [0.01] }^{ 2 }$

$ =  2.5  \times   { 10 }^{ -2 }  \times   { 10 }^{ -4 }$

$=  2.5  \times   { 10 }^{ -6 }$  lit   mol $^{ -1 }$ sec$^{ -1 }$

Hence, the correct option is $\text{B}$

The specific rate of a reaction is $1.51 \times10^{-4}$ lit mole$^{-1}$ sec$^{-1}$. If the reaction is commenced with 0.2 mole lit$^{-1}$ of the reactant, the initial rate of the reaction in mole lit$^{-1}$ sec$^{-1}$ is:

  1. $1.5 \times 10^{-4}$

  2. $3 \times 10^{-5}$

  3. $6 \times 10^{-6}$

  4. $6 \times 10^{-5}$


Correct Option: C
Explanation:

For a general reaction of order $n$, the units of rate constant is given as $(mol/litre)^{1-n}sec^{-1}$.
So for our question, the unit of rate constant is $(mol/litre)^{-1}sec^{-1}$ so $ 1-n = -1$.


So, the value of $n =2$, so the reaction is of second order.

The rate is given as $r = k[A]^{2}$. Specific rate of equation is the rate for concentration of 1 $mol/litre$. 

From this, the value of k is $1.51 \times1 0^{-4}$.
$r = 1.51 \times 10^{-4}[A]^{2}$.
[A] =  $0.2 mol/litre$. 

Initial rate is $6 \times1 0^{-6}$.

Hence, option C is correct.

Read the following table and chose the appropriate option


Rate equation Units of K
I) rate $=$ k[A] a) mol lit$^{-1}$ sec $^{-1}$
II) rate $=$ k[A][B] b) mol$^{-2}$ lit$^{2}$ sec $^{-1}$
III) rate $=$ k[A][B]$^2$ c) sec $^{-1}$
IV) rate $=$ k d) lit mol$^{-1}$ sec $^{-1}$

  1. I - d, II - c, III - a, IV - b

  2. I - c, II - d, III - b, IV - a

  3. I - a, II - b, III - c, IV - d

  4. I - b, II - a, III - d, IV - c


Correct Option: B
Explanation:

Units of k $= mole^{1-n} lit^{n-1} sec^{-1}$ , where n is the order of reaction.


For option A, its first order reaction. Rate constant $=sec^{-1}$

For option B, its second order reaction. Rate constant $=mole^{-1} lit sec^{-1}$


For option C, its third order reaction. Rate constant $=mole^{-2} lit^{2} sec^{-1}$

For option D, its zero order reaction. rate constant $=mole/lit sec^{-1}$

Hence, the correct option is $\text{B}$

Identify the reaction order from each of the following rate constants.
(i) $k=2.3 \times 10^{-5} L \quad mol^{-1} \quad s^{-1}$
(ii) $k=3 \times 10^{-4} \quad s^{-1}$

  1. (i) First order (ii) Second order

  2. (i)Second order (ii) First order

  3. (i) Zero order (ii) First order

  4. None


Correct Option: B
Explanation:

(i) Rate of the reaction is given by,

$Rate=k\times [A]^x$, where x is the order of reaction
$(\dfrac{mol}{L})=2.3\times 10^{-5}L(mol^{-1})(s)^{-1}[A]^x$
$({\dfrac{mol}{L}})^2=constant(s^{-1})[\dfrac{mol}{L}]^x$
Therefore, $x=2$ and the reaction is second order
(ii)$Rate=k\times [A]^x$, where x is the order of reaction
$(\dfrac{mol}{L})=3\times 10^{-4}(s)^{-1}[A]^x$
$({\dfrac{mol}{L}})=constant(s^{-1})[\dfrac{mol}{L}]^x$
Therefore, $x=1$ and reaction is first order

Units of rate constant for the first and zero order reactions in terms of molarity M units are respectively:

  1. $sec^{-1},\;Msec^{-1}$

  2. $sec^{-1},\;M$

  3. $Msec^{-1},\;sec^{-1}$

  4. None of the above


Correct Option: A
Explanation:

The unit of the rate constant K is $\left (\text {Molarity} \right )^{1-n}time^{-1}.$


Here, n is the order of the reaction.

For first order reaction, $n=1.$

The unit of the rate constant K is $\left ( \text{Molarity} \right )^{1-1}sec^{-1}=sec^{-1}.$

For zero order reaction, $n=0.$

The unit of the rate constant K is $\left ( \text{Molarity} \right )^{1-0}sec^{-1}=\left ( \text{Molarity} \right )sec^{-1}=\left (M \right )sec^{-1}.$

Consider the reaction, $2A + B \rightarrow$ Products, When concentration of B alone was doubled, the rate did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is:

  1. $s^{-1}$

  2. $L\ mol^{-1}s^{-1}$

  3. Unitless

  4.  $mol\ L^{-1}s^{-1}$


Correct Option: A
Explanation:
When the concentration of B alone was doubled, the rate did not change. Hence the reaction is zero order in B.

When the concentration of A alone was doubled, the rate increased by two times. Hence, the reaction is of first-order in A.

$r=k\left [ A \right ]^{1}\left [ B \right ]^{0}$

The overall order of the reaction is 1.

For the first-order reaction, the unit of k is $sec^{-1}$.

Hence, the correct option is $\text{A}$

The following mechanism has been proposed for the reaction of $NO$ with $\displaystyle Br _{2}$ to form $NOBr$


$NO(g)+Br _{2}(g)\rightleftharpoons NOBr _{2}(g)$
$NOBr _{2}(g)+NO(g)\rightarrow 2NOBr(g)$


If the second step is the rate determining step,the order of the reaction with respect to $NO(g)$ is:
  1. 2

  2. 1

  3. 0

  4. 3


Correct Option: A

Taking the reaction $x+2y\rightarrow$ prodcuts, to be of second order, which of the following are the rate law expressions for the reaction :


(I) $\cfrac{dx}{dt}=K[x][y]$  (II) $\cfrac{dx}{dt}=K[x]{[y}]^{2}$

(III) $\cfrac{dx}{dt}=k{[x]}^{2}$  (IV) $\cfrac{dx}{dt}=K[x]+K{[y]}^{2}$

  1. I only

  2. I and III only

  3. I and II only

  4. I and IV only


Correct Option: C

The rate of formation of ${SO} _{3}$ in the reaction $2{SO} _{2}+{O} _{2}\rightarrow 2{SO} _{3}$ is $100g{min}^{-1}$. Hence, rate of disappearance of ${O} _{2}$ is

  1. $50g{min}^{-1}$

  2. $100g{min}^{-1}$

  3. $20g{min}^{-1}$

  4. $40g{min}^{-1}$


Correct Option: A

Reaction $A+B\longrightarrow C+D$ follows rate law, $r=k{ \left[ A \right]  }^{ 1/2 }{ \left[ B \right]  }^{ 1/2 }$ starting with $1M$ of $A$ and $B$ each. What is the time taken for concentration of $A$ become $0.1M$?
[Given $2.303\times { 10 }^{ -2 }sec^{ -1 }$].

  1. $10sec$

  2. $100sec$

  3. $1000sec$

  4. $434sec$


Correct Option: B
Explanation:

The rate constant is $2.303\times { 10 }^{ -2 }sec^{ -1 }$. The overall reaction order is 1.


$t = \dfrac {2.303}{k}log \dfrac {a}{a-x}$

$t = \dfrac {2.303}{2.303 \times 10^{-2}}log \dfrac {1}{0.1}=100$

Hence, the time taken to consume 90% of concentration is 100 sec.

The reaction, $CH _3COOC _2H _5+NaOH\rightarrow CH _3COONa+C _2H _5OH$ is:

  1. bimolecular reaction

  2. second order reaction

  3. third order reaction

  4. none of these


Correct Option: A,B
Explanation:

The reaction, $CH _3COOC _2H _5+NaOH\rightarrow CH _3COONa+C _2H _5OH$ is bimolecular and second order reaction,as it contains two diiferent types of molecules.
Rate = $ k$ [$CH _3COOC _2H _5$] [$NaOH$]

For the reaction: $2NO+Cl _2\rightarrow 2NOCl$, the following mechanism was proposed on the basis of experimental observation.
$NO+Cl _2\overset {Fast}{\rightleftharpoons}NOCl _2$
$NOCl _2+NO\xrightarrow {Slow}2NOBr$
The order of reaction is:

  1. 2nd order

  2. pseudo 2nd order

  3. 3rd order

  4. first order


Correct Option: C
Explanation:

From the slow step, the rate law for the reaction is rate $\displaystyle = k' [NOCl _2][NO]$.....(1)


 $\displaystyle NOCl _2$ is an intermediate and its concentration is obtained from fast step (equilibrium step)

 $\displaystyle [NOCl _2] = K[NO][Cl _2]$......(2)

Substitute (2) in (1)
 $\displaystyle rate = k'K [NO][Cl _2][NO]$

Hence, rate $\displaystyle = k [NO]^2[Cl]$

The reaction is of second order in NO and first order in $\displaystyle Cl _2$. The overall reaction order is  $\displaystyle 2 + 1 = 3$

The unit of rate constant for a given reaction is $M^{1-n}L^{n-1}t^{-1}$ where n is order of reaction.

  1. True

  2. False


Correct Option: A
Explanation:

The statement is true. The unit of rate constant for a given reaction is  $M^{1-n}L^{n-1}t^{-1}$ where $n$ is order of reaction.It is generally form for expressing rate constant. Rates are usually given in concentrations units over time units. The units of the rate constants depend upon the order of the reaction. The concentration is molarity and time is in seconds.

A 22.4 litre flask contains 0.76 mm of ozone at $25^oC$. Calculate:
(i) the concentration of oxygen atom needed so that the reaction, $O+O _3\rightarrow 2O _2$ having rate constant equal to $1.5\times 10^7$ litre $mol^{-1} sec^{-1}$ can proceed with a rate of 0.15 mol $litre^{-1} sec^{-1}$
(ii) the rate of formation of oxygen under this condition.

  1. (i) $2.45\times 10^{-4}$, (ii) $0.30 mol L^{-1} t^{-1}$

  2. (i) $4.9\times 10^{-4}$, (ii) $0.60 mol L^{-1} t^{-1}$

  3. (i) $2.45\times 10^{-5}$, (ii) $0.30 mol L^{-1} t^{-1}$

  4. None of these


Correct Option: A
Explanation:

$\displaystyle [O _3] = \frac {P}{RT} = \frac {0.76}{760 \times 0.0821 \times 298} = 4.08 \times 10^{-5} $
$\displaystyle [O] = \frac {rate}{k \times [O _3]} = \frac {0.15}{1.5 \times 10^{7} \times 4.08 \times 10^{-5} = 2.45 \times 10^{-4}  }$
The rate of formation of oxygen under this condition is equal to twice the rate of the reaction.
It is equal to $\displaystyle 2 \times 0.15 : mol litre^{-1} sec^{-1} = 0.30 : mol litre^{-1} sec^{-1} $

The unit and value of rate constant and that of rate of reaction are same for:

  1. first order

  2. zero order

  3. second order

  4. all are wrong


Correct Option: B
Explanation:

As we know,
For zero order reaction, $\frac {dx}{dt}=K$.
so, unit and value of rate constant and that of rate of reaction are same for zero order.

The rate constant of $n^{th}$ order has units:

  1. $Litre^{-1} mol^{1-n} sec^{-1}$

  2. $Mol^{1-n} litre^{1-n} sec^{-1}$

  3. $Mol^{1-n} litre^{n-1} sec^{-1}$

  4. $Mol^{1-n^{2}} litre^{n^{2}} sec^{-1}$


Correct Option: C
Explanation:

As we know,
For nth order; unit of rate constant may be derived by
$K=\frac {rate}{[reactant]^n}$
$=Mol^{1-n} litre^{n-1} sec^{-1}$

A reaction proceeds in three stages. The first stage is slow and involves two molecules of reactants. The second and third stage are fast. The overall order of the reaction is:

  1. first order

  2. second order

  3. third order

  4. zero order


Correct Option: B
Explanation:

The overall order of the reaction is second order.
Reactions are proceeded in many steps, it does not means that no. of step will be equal to order of reaction. But r
ate is determined only by slow step of mechanism.

The rate of the reaction, $A+B+C\rightarrow P$; is given by; $r=-\frac {d[A]}{dt}=K[A]^{1/2}[B]^{1/2}[C]^{1/4}$. The order of the reaction is:

  1. 1

  2. 2

  3. 1/2

  4. 5/4


Correct Option: D
Explanation:

As we know,

For a reaction, $A+B\rightarrow Product$, if rate $=K[A]^m[B]^n$, the order of reaction $=m+n$.

So here for the given reaction, order of reaction $= 1/2+1/2+1/4 = 5/4$

The rate constant of nth order has units:

  1. $Litre^{-1} mol^{1-n} sec^{-1}$

  2. $Mol^{1-n} litre^{1-n} sec^{-1}$

  3. $Mol^{1-n} litre^{n-1} sec^{-1}$

  4. none of the above


Correct Option: C
Explanation:

For nth order; unit of rate constant may be derived by
$rate  = k[a]^n$
$K=\frac {rate}{[reactant]^n}$.

If the concentration is measured in mol $L^{-1}$ and time in minutes, the unit for the rate constant of a third order reaction is:

  1. mol $L^{-1}\, min^{-1}$

  2. $L^{2}\, mol^{-2}\, min^{-1}$

  3. $L\, mol^{-1}\, min^{-1}$

  4. $min^{-1}$


Correct Option: B
Explanation:

As we know,
$k\, =\, [conc]^{1-n}\, min^{-1}$
For third order reaction = $[mol\, L^{-1}]^{1-3}\, min^{-1}$
$L^2\, mol^{-2}\, min^{-2}$

For which of the following reactions, the units of rate constant and rate of reaction are same ?

  1. First order reaction

  2. Second order reaction

  3. Third order reaction

  4. Zero order reaction


Correct Option: D
Explanation:

For zero order,
R = k[A]
Unit of rate = mol $L^{-1}\, t^{-1}$
Unit of zero order = mol $L^{-1}\, t^{-1}$

What is the order of reaction which has a rate expression as follows:
rate = k$[A]^{3/2}[B]^{-1}$

  1. $3/2$

  2. $1/2$

  3. Zero

  4. None of these


Correct Option: B
Explanation:

rate = k$[A]^{3/2}[B]^{-1}$
OR = $\displaystyle \frac{3}{2}\, -\, 1\, =\, \frac{1}{2}$

If a reaction involves gaseous reactants and products, the units of its rate are:

  1. $atm$

  2. $atm$ $s$

  3. $atm$ $s^{-1}$

  4. $atm^2s^2$


Correct Option: C
Explanation:

If a reaction involves gaseous reactants and products, the unit of its rate is $atm\ s^{-1}$. It represents the change in pressure (in atm units) in unit time ($1$ second).

The rate of certain hypothetical reaction A + B + C $\rightarrow$ Products, is given by $\displaystyle r\, =\, - \frac{dA}{dt}\, = k[A]^{1/2}[B]^{1/3}[C]^{1/4}$ The order of a reaction is given by:

  1. $1$

  2. $\displaystyle \frac{1}{2}$

  3. $2$

  4. $\displaystyle \frac{13}{12}$


Correct Option: D
Explanation:

$A + B + C$ $\rightarrow$ Products
$\displaystyle r\, =\, - \frac{dA}{dt}\, = k[A]^{1/2}[B]^{1/3}[C]^{1/4}$
so order is
$\displaystyle \frac{1}{2}\, +\, \frac{1}{3}\, +\, \frac{1}{4}\, =\,\frac{13}{12}$

Consider the following elementary reaction,

    $2A + B + C \rightarrow Products$.

All reactant are present in the gaseous state and reactant C is taken in excess.

What is the unit of rate constant of the reaction?

  1. $mol\, L^{-1}\, time^{-1}$

  2. $time^{-1}$

  3. $mol^{-1}\, L^{2}\, time^{-1}$

  4. $mol^{-2} L^{2} time^{-1}$


Correct Option: D
Explanation:

Since C is taken in excess, so its concentration does not change. So it is not taken in rate expression of the reaction.


and rate law is given by,

$Rate = k[A]^{2}[B]$

Order of the reaction $= 3$

 For the third-order reaction, the unit of the rate constant is given by -

      $= [conc.]^{1 - n} {time}^{-1}= [conc.]^{1 - 3} {time}^{-1}=\, mol^{-2} L^{2} time^{-1}$

The second order rate constant is usually expressed as:

  1. Mol L $s^{-1}$

  2. $Mol^{-1}\, L^{-1}\, s^{-1}$

  3. $Mol\, L^{-1}\, s^{-1}$

  4. $Mol^{-1}\, L\, s^{-1}$


Correct Option: D
Explanation:

As we know,
for second order reaction:
$r = k[A]^2$
so unit of rate constant is $Mol^{-1}\, L\, s^{-1}$.

For a reaction having order equal to $3/2$, the units for rate constant are ${ sec }^{ -1 }$.
  1. True

  2. False


Correct Option: B
Explanation:

Rate $r = k[A]^{3/2}$

Unit of rate constant $k = [A]^{1-3/2}[time]^{-1}$
$k = [A]^{-1/2}[time]^{-1}$

The rate of certain hypothetical reaction $A+B+C\rightarrow $ products is given by, $\displaystyle r=-\frac{\mathrm{d} [A]}{\mathrm{d} t}=K[A]^{1/2}:K[B]^{1/3}:K[C]^{1/4}$. The order of the reaction:

  1. 1

  2. $\displaystyle \frac{1}{2}$

  3. 2

  4. $\displaystyle \frac{13}{12}$


Correct Option: D
Explanation:

$r=-\dfrac { d[A] }{ dt } =K[A{ ] }^{ \cfrac { 1 }{ 2 }  }K[B{ ] }^{ \cfrac { 1 }{ 3 }  }K[C{ ] }^{ \cfrac { 1 }{ 4 }  }\ $


$rate=\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 4 } \ $

$rate=\dfrac { 13 }{ 12 }$

Hence, the answer is D.

The rate constant of third order reaction is:

  1. mol lit$^{-1}$ s$^{-1}$

  2. mol$^{-2}$ L$^2$ T$^{-1}$

  3. mol$^{-1}$ lit s$^{-1}$

  4. mol$^{-2}$ lit$^{-1}$ s$^{-1}$


Correct Option: B
Explanation:

Units of rate constant for $n$th order = $[(mol\  lit^{-1})]^{1-n} t^{-1}$
For third order reaction $n = 3$
$\therefore $ Units are $(mol  lit^{-1})^{1- 3} t^{-1} = mol\ l^{-2} lit^{-2} t^{-1}$

The rate constant of a reaction depends on:

  1. Temperature

  2. Initial concentration ofthe reaction

  3. Time ofreaction

  4. Extent of reaction


Correct Option: A
Explanation:

The rate constant of a reaction depends only on two factors
(i) Temperature and (ii) Catalyst.

The rate constant of a first-order reaction is $3 \times 10^{-6}$ per second and initial concentration is 0.10 M. Then the initial rate of reaction is:

  1. $3 \times 10^{-6} Ms^{-1}$

  2. $3 \times 10^{-8} Ms^{-1}$

  3. $3 \times 10^{-7} Ms^{-1}$

  4. $3 \times 10^{-9} Ms^{-1}$


Correct Option: C
Explanation:

Rate = k [concentration] for first order reaction.
Given $k = 3 \times 10^{-6}/sec,$ [concentration] = 0.1 M
$\therefore rate = 3 \times 10^{-6} \times 0.1  = 3 \times 10^{-7} ms^{-1}$

What is the unit for the rate constant of a second order reaction?

  1. $\displaystyle { s }^{ -1 }$

  2. mol $\displaystyle { L }^{ -1 }$

  3. mol $\displaystyle { L }^{ -1 }{ s }^{ -1 }$

  4. L $\displaystyle { mol }^{ -1 }{ s }^{ -1 }$

  5. $\displaystyle { mol }^{ 2 }{ L }^{ -2 }{ s }^{ -2 }$


Correct Option: D
Explanation:

The unit for the rate constant of a second order reaction is $\displaystyle \displaystyle L { mol }^{ -1 }{ s }^{ -1 } $. 
For a second order reaction, $\displaystyle rate = k [A]^2 $
$\displaystyle mol { L }^{ -1 }{ s }^{ -1 }  = k (mol { L }^{ -1 })^2$
$\displaystyle k = L { mol }^{ -1 }{ s }^{ -1 }  $

Statement I : In a second order reaction doubling [A] quadruples the rate
Because
Statement II : The rate equation is $\displaystyle r={ k\left[ A \right]  }^{ 2 }$ for such a reaction

  1. Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1 .

  2. Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

  3. Statement 1 is correct but Statement 2 is not correct.

  4. Statement 1 is not correct but Statement 2 is correct.

  5. Both the Statement 1 and Statement 2 are not correct.


Correct Option: A
Explanation:

Secon order rate expression is given as $r$$=$$k[A]^2$ so if we double the concentration of a then rate increases by 4 times, hence both statements are correct and statement 2 is correct explanation of statement 1.

For the reaction $A + B \rightarrow C$, determine the order of the reaction with respect to $B$ from the information given below.

$\displaystyle { \left[ A \right]  } _{ \circ  }$ $\displaystyle { \left[ B \right]  } _{ \circ  }$ Initial rate (M/s)
1.00 1.00 2.0
1.00 2.00 8.1
2.00 2.00 15.9
  1. Zero order

  2. First order

  3. Second order

  4. Third order

  5. Fourth order


Correct Option: C
Explanation:

Analyzing first and second raw,when concentration of $B$ is doubled then initial rate increased by four times,so it is increasing by square of concentration.Hence it is a second order reaction.


$Rate$$=$$k[A]^x [B]^2$  Using this rate expression and keeping $[A]$ constant,if we double $[B]$ then rate increases four times. 

Statement 1: In a second-order reaction with respect to $A$, when you double [$A$], the rate is quadrupled.
Statement 2: The rate equation is $r = k[A]^2$ for such a reaction.

  1. Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.

  2. Both the Statement 1 and Statement 2 are correct, but Statement 2 is not the correct explanation of Statement 1.

  3. Statement 1 is correct, but Statement 2 is not correct.

  4. Statement 1 is not correct, but Statement 2 is correct.


Correct Option: A
Explanation:

$\bullet \quad $If rate = ${ K\left[ A \right]  }^{ 2 }$ for a reaction.

       then it is a second order reaction with respect to A.
$\bullet \quad $Let ${ \gamma  } _{ 1 }={ K\left[ A \right]  } _{ t }^{ 2 }$
       if ${ \left[ A \right]  } _{ t }$ is doubled.
             ${ \gamma  } _{ 2 }={ K\left( { 2\left[ A \right]  } _{ t } \right)  }^{ 2 }$
   $\Rightarrow \quad { \gamma  } _{ 2 }={ 4\gamma  } _{ 1 }$.
Hence, statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1.

The unit for the rate constant is calculated from the rate law.
For the given rate law, determine the units of the rate constant for rate $= k[A]^{2} [B]$.

  1. $s^{-1}M^{-3}$

  2. $s^{-1}M^{-2}$

  3. $s^{-1}M^{-1}$

  4. $s^{-1}$


Correct Option: B
Explanation:

Since rate constant is given by $k[A]^2[B]$.
Where 1+2=3, so it is probably third order reaction.
So rate constant should have units of $L^2 mol^{-2} s^{-1} \implies M^{-2}s^{-1}$

A graph of concentration versus time data for a second-order reaction gives a straight line in which of the following plots of the data?

  1. $[A] _{t} = -kt + [A] _{0}$

  2. $ln [A] _{t} = -kt + ln [A] _{0}$

  3. $\dfrac {1}{[A _{t}]} = kt + \dfrac {1}{[A _{0}]}$

  4. All of the above

  5. None of the above


Correct Option: A
Explanation:

Let the concentration at $t=0$ be $[{ A } _{ o }]$ & at $t=t$ be $ [{ A } _{ t }]$

 By graph,
$ [{ A } _{ t }]=Kt+[{ A } _{ o }]$
 For negative slop,
$ [{ A } _{ t }]=-Kt+[{ A } _{ o }]$

The rate of the reaction, $C{ Cl } _{ 3 }CHO+NO\longrightarrow CH{ Cl } _{ 3 }+NO+CO$, is given by the equation, rate $=k\left[ C{ Cl } _{ 3 }CHO \right] \left[ NO \right] $. If concentration is expressed in ${mol}/{litre}$, the unit of $k$ is:

  1. ${ mol }^{ -2 }{ L }^{ 2 }{ s }^{ -1 }$

  2. $mol$ ${ L }^{ -1 }{ s }^{ -1 }$

  3. $L$ ${ mol }^{ -1 }{ s }^{ -1 }$

  4. ${ s }^{ -1 }$


Correct Option: C
Explanation:

Since it is second order reaction as clear by rate equation so unit of $k$ is given by $M^{-1} s^{-1}$ or $L\space mol^{−1}s^{−1}$

Units of rate constants for first and zero order reactions in terms of molarity $M$ unit are respectively:

  1. ${ sec }^{ -1 },M\ { sec }^{ -1 }$

  2. ${ sec }^{ -1 },M$

  3. $M\ { sec }^{ -1 },{ sec }^{ -1 }$

  4. $M,{ sec }^{ -1 }$


Correct Option: A
Explanation:

$K=\frac{dx}{dt[A]} $ for Ist order = $sec^{−1}$
$ K = \frac{dx}{dt}$ for zero order = $mol \space litre^{−1}sec^{−1}$

Hence A is the correct answer.

Consider the reaction, $2A+B\longrightarrow $ Products. When the concentration of $B$ alone was doubled, the half-life did not change. When the concentration of $A$ alone was doubled, the rate increased by two times. The unit of the rate constant for this reaction is:

  1. ${ s }^{ -1 }$

  2. $L$ ${ mol }^{ -1 }{ s }^{ -1 }$

  3. Unitless

  4. $mol$ ${ L }^{ -1 }{ s }^{ -1 }$


Correct Option: B
Explanation:

Concentration change in '$B$' does not change half life, it means the reaction is first order with respect to $B$. When concentration of only '$A$' is doubled, the rate of reaction becomes double, thus order with respect to $A$ will also be one.
Overall order of reaction $=2$
Unit of rate constant $=L$ ${ mol }^{ -1 }{ s }^{ -1 }$

Consider following two reactions:
$A\longrightarrow $ Product,         $-\dfrac { d\left[ A \right]  }{ dt } ={ k } _{ 1 }{ \left[ A \right]  }^{ 0 }$
$B\longrightarrow $ Product,         $-\dfrac { d\left[ B \right]  }{ dt } ={ k } _{ 2 }{ \left[ B \right]  }^{ 1 }$
${ k } _{ 1 }$ and ${ k } _{ 2 }$ are expressed in terms of molarity $\left( mol\ { L }^{ -1 } \right) $ and time $\left( { s }\right) $ as:

  1. ${ s }^{ -1 },M{ s }^{ -1 }{ L }^{ -1 }$

  2. $M{ s }^{ -1 },M{ s }^{ -1 }$

  3. ${ s }^{ -1 },{ M }^{ -1 }{ s }^{ -1 }$

  4. $M{ s }^{ -1 },{ s }^{ -1 }$


Correct Option: D
Explanation:

The units of rate of reaction is $Ms^{-1}$.

For zero order reaction, unit of K, will be $Ms^{-1}$
For 1st order reaction, unit of $K _2$ will be $s^{-1}$

Rate law expression of a reaction is:
            Rate $=k{ \left[ A \right]  }^{ { 2 }/{ 3 } }\left[ B \right] $
Which of the following are correct about the corresponding reaction?

  1. Order of reaction $=\dfrac { 2 }{ 3 } +1=\dfrac { 5 }{ 3 } $

  2. Unit of rate constant $={ L }^{ { 2 }/{ 3 } }{ mol }^{ { -2 }/{ 3 } }{ sec }^{ -1 }$

  3. Unit of rate constant $={ atm }^{ { -2 }/{ 3 } }{ sec }^{ -1 }$

  4. Unit of rate constant $=mol$ ${ L }^{ -1 }{ sec }^{ -1 }$


Correct Option: A,B,C
Explanation:

Overall order=$\frac{2}{3}+1=\frac{5}{3}$

Unit of rate constant=$mol^{1-n}L^{n-1}sec^{-1}$
where n=order of reaction
For given reaction, unit of rate constant=$mol^{1-5/3}L^{5/3-1}sec^{-1}$
=$L^{2/3}mol^{-2/3}sec^{-1}$
For gaseous reaction, concentration term= pressure in atm
unit of rate constant=$atm^{-2/3}sec^{-1}$

Mechanism of a hypothetical reaction $X _2+Y _2\rightarrow 2XY$ is given below;
(i) $ X _2\rightarrow X+X$ (fast)
(ii) $X+Y _2\rightleftharpoons XY+Y$ (slow)
(iii) $ X+Y \rightarrow XY$ (fast)
The overall order of the reaction will be: 

  1. $2$

  2. $0$

  3. $3$

  4. $1$


Correct Option: A
Explanation:

Order of Reaction is given by the slowest step of Reaction. In slowest step we 

have 2 reactants so, $2^{nd}$ order reaction. 

A reaction, which is second-order, has a rate constant of $0.002  L\, mol^{-1}\, s^{-1}$. If the initial conc. of the reactant is 0.2 M, how long will it take for the concentration to become 0.0400 M?

  1. 1000 sec

  2. 400 sec

  3. 200 sec

  4. 10,000 sec


Correct Option: D
Explanation:
$\dfrac{1}{a}=\dfrac{1}{a _{0}}+kt$
$\dfrac{1}{0.04}=\dfrac{1}{0.2}+0.002t$
$t=10000sec$

For the reaction $CO(g)+2{ H } { 2 }(g)\rightleftharpoons { CH } _{ 3 }OH(g)$. If active mass of $CO$ is kept constant and active mass of ${H} _{2}$ is tripled, the rate of forward reaction will become _____ of its initial value.

  1. three times

  2. six times

  3. eight times

  4. nine times


Correct Option: D
Explanation:

For the following reaction, Rate is defined as


$rate = k[CO][{{H} {2}}]^{2}$
Now, the active mass of CO is kept constant and active mass of {H}{2} is tripled. Now the rate is,

${rate}^{'} =k[CO][{3 \times {H} _{2}}]^{2}$
${rate}^{'} = 9 \times rate$
 
So, the rate of forward reaction will become nine times of its initial value.

The reaction $2{NO} _{(g)}+{H} _{2(g)}\longrightarrow {N} _{2}{O} _{(g)}+{H} _{2}{O} _{(g)}$ follows the rate law $\cfrac { d{ P } _{ \left( { N } _{ 2 }O \right)  } }{ dt } =k{ \left( { P } _{ NO } \right)  }^{ 2 }{ p } _{ { H } _{ 2 } }$. If the reaction is initiated with ${P} _{NO}=1000mm$ $Hg$ and ${ p } _{ { H } _{ 2 } }=10mm$ $Hg$, then the reaction will follow:

  1. third order kinetics

  2. second order kinetics

  3. first order kinetics

  4. zero order kinetics


Correct Option: C

The following data were obtained for the saponification of ethyl acetate using equal concentrations of ester and alkali. The reaction order is:

Time(min) 0 4 10 20
Vol. of acid(mL)  8.04 5.30 3.50 2.22
  1. 0

  2. 1

  3. 2

  4. 3


Correct Option: C

For a reaction $r=K{[CH _3COCH _3]}^{3/2}$. The unit of rate of reaction and rate constant respectively is:

  1. $mol \displaystyle L^{-1}s^{-1},\quad mol^{-\frac{1}{2}}L^{\frac{1}{2}}s^{-1}$

  2. $\displaystyle mol^{-1}L^{-1}s^{-1},\quad mol^{-\frac{1}{2}}L^{-\frac{1}{2}}s^{-1}$

  3. $\displaystyle mol L^{-1}s^{-1},\quad mol^{\frac{1}{2}}L^{\frac{1}{2}}s^{-1}$

  4. $mol Ls,\quad \displaystyle mol^{\frac{1}{2}}L^{\frac{1}{2}}s$


Correct Option: A
Explanation:

For $1.5$ order rate law the units are $molL^{-1}s^{-1}$ for the rate while the [rate constant]$=\cfrac{molL^{-1}s^{-1}}{mol^{3/2}L^{-3/2}}$

$=mol^{-1/2}L^{1/2}s^{-1}$

Which of the following corresponds to the units of rate constant for n$^{th}$ order reaction ?

  1. $mole^{n-1} l^{1-n} s^{-1}$

  2. $mole^{n-1} l^{n-1} s^{-1}$

  3. $mole^{1-n} l^{n-1} s^{-1}$

  4. $mole^{n-1} l^{n} s^{-1}$


Correct Option: C
Explanation:

$ r= K\left [ A \right ]^{n}$

$K = \dfrac{r}{\left [ A \right ]^{n}}= \dfrac{mole \ l^{-1} \ sec^{-1}}{mole^{n} \ l ^{-n}}$ $= mole^{1-n} 1^{n-1} sec^{-1}$

The unit of rate of a first order reaction is:

  1. $mol\  lit^{-1}$

  2. $l\  mol^{-1} \ s^{-1}$

  3. $s^{-1}$

  4. $l^2 \ mol^{-2} \ s^{-1}$


Correct Option: C
Explanation:

For a first order reaction; rate law can be wriiten as; $r = k[A]^{1}$
Therefore k = $\dfrac{r}{[A]} = \dfrac{mol \times l^{-1} \times  s^{-1}}{mol \times l^{-1}}$ = $s^{-1}$ where concentration of $A =$ moles per litre and rate of reaction; r = change in concentration of $A$ with time.

For a particular $A+B \rightarrow C$ was studied at $25^{\circ}C$. The following results are obtained.


              [A]              [B]           [C]
    (mole/lit)       (moles/lit)  (mole  lit $^{-1} sec^{-2}$)  
$9 \times 10^{-5}$ $1.5 \times 10^{-2}$           $0.06$
$9 \times 10^{-5}$ $3 \times 10^{-3}$            $0.012$
$3 \times 10^{-5}$ $3 \times 10^{-3}$            $0.004$
$6 \times 10^{-5}$            x           $0.024$


Then the value of x is :

  1. $6 \times 10^{-3} moles litre^{-1}$

  2. $3 \times 10^{-3} moleslitre^{-1}$

  3. $4.5 \times 10^{-3} moleslitre^{-1}$

  4. $9 \times 10^{-3} moleslitre^{-1}$


Correct Option: D
Explanation:
$A+B\rightarrow C$

$ rate=k\left[ A \right] \left[ B \right] $

$Experiment \  3\& 2 \  chosen \  for \  value \  of \  k \  as\left[ B \right] is \  same \  in \  both$ 

$\dfrac { { r } _{ 3 } }{ { r } _{ 2 } } =\dfrac { 0.004 }{ 0.012 } =k\dfrac { \left[ { 3\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  }{ \left[ { 9\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  } $

$k=1 \ using \  this \  rate \  constant \  value \  in \  finding \  x\\$
$ \dfrac { { r } _{ 4 } }{ { r } _{ 3 } } =\dfrac { 0.024 }{ 0.004 } =k\dfrac { \left[ { 6\times 10 }^{ -5 } \right] \left[ x \right]  }{ \left[ { 3\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  } $

$\\ \left[ x \right] ={ 9\times 10 }^{ -3 }\\ $

Compound $A$ and $B$ react to form $C$ and $D$ in a reaction that was found to be second-order over all and second-order in $A$. The rate constant -at ${ 30 }^{ 0 }C$ is $0.622$ L ${ mol }^{ -1 }{ min }^{ -1 }$. What is the half-life of A when $4.10\times { 10 }^{ -2 }$ M of A is mixed with excess $B$?

  1. $40$ min

  2. $39.21$ min

  3. $28.59$ min

  4. None of these


Correct Option: B
Explanation:

$A+B\longrightarrow C+D$


 rate$=k{ [A] }^{ 2 }$ (given)

$ =0.622{ [4.10\times { 10 }^{ -2 }] }^{ 2 }$

$ =0.001$  is the rate of reaction initially

 Half-life$={ t } _{ 1/2 }=\cfrac { 1 }{ K[A] } =\cfrac { 1 }{ 0.622\times [4.1\times { 10 }^{ -2 }] } \\ =0.3921\times { 10 }^{ 2 }\\ =39.21\quad minutes.$

The decomposition of dimethyl ether leads to the formation of $CH _4, H _2$ and CO and the reaction rate is given by $Rate=k[CH _3OCH _3]^{\frac {3}{2}}$
The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., $Rate=k(P _{CH _3OCH _3})^{\frac {3}{2}}$
If the pressure is measured in bar and time in minutes, then the unit of rate constants is:

  1. $bar^{\frac {1}{2}} min$

  2. $bar^{\frac {3}{2}} min^{-1}$

  3. $bar^{-\frac {1}{2}} min^{-1}$

  4. $bar min^{-1}$


Correct Option: C
Explanation:

As $Rate=k(P _{CH _3OCH _3})^{\frac {3}{2}}$
$bar/min=k(bar)^{\frac {3}{2}}$
$\therefore$ unit of k$=bar^{-\frac {1}{2}}min^{-1}$

Taking the reaction, $A + 2B\rightarrow Products$, to be of the second order, which of the following may be the correct rate law expressions?

  1. $\frac {dx}{dt}=k[A][B]$

  2. $\frac {dx}{dt}=k[A][B]^2$

  3. $\frac {dx}{dt}=k[A]^2$

  4. $\frac {dx}{dt}=k _1[A]+k _2[B]^2$


Correct Option: A,C
Explanation:

option A and C are correct as the sum of their exponents equals to 2

The rate constant of a second order reaction is $10^{-2} lit.mole ^{-1}.sec^{-1}$. The rate constant when expressed as $cc. \ molecule^{-1} .\ min^{-1}$ is:

  1. $9.96\times 10^{-22}$

  2. $9.96\times 10^{-23}$

  3. $9.96\times 10^{-21}$

  4. $9.96\times 10^{-24}$


Correct Option: A
Explanation:

The rate constant of a second order reaction is $10^{-2} lit.mole ^{-1}.sec^{-1}$.
$1L=1000cc$


$1 mole = 6.023\times 10^{23}$molecules
$1min=60sec$
Hence, rate constant $=10^{-2} lit.mole ^{-1} sec^{-1}\times \dfrac {1000cc}{1L} \times \dfrac {1mole} {6.023\times 10^{23}molecules} \times \dfrac {60 sec} {1 min}$
$=9.96\times 10^{-22}cc\ molecule^{-1} \ min^{-1}$

In a certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours and so on. Dimension of the velocity constant are:

  1. hour$^{-1}$

  2. mole litre$^{-1}$ hour$^{-1}$

  3. litre mol$^{-1}$ hour$^{-1}$

  4. mole sec$^{-1}$


Correct Option: B
Explanation:

For the zero-order reaction, the time taken for the decomposition of the reactant is independent of initial concentration which is the case here. 


The dimension of the velocity constant for the zero-order reaction is = mole litre$^{-1}$ hour$^{-1}$

Option B is correct.

When ethyl acetate was hydrolysed in presence of 0.1 N HCl, the rate constant was found to be $5.40\times 10^{-5}sec^{-1}$. But when 0.1 N $H _2SO _4$ was used for hydrolysis, the rate constant was found to be $6.25\times 10^{-5} sec^{-1}$. Thus, it may be concluded that:

  1. $H _2SO _4$ is stronger than HCl

  2. $H _2SO _4$ is weaker than HCl

  3. $H _2SO _4$ and HCl both have the same strength

  4. the data are not sufficient to compare the strength of $H _2SO _4$ and HCl


Correct Option: A
Explanation:

Option (A) is correct.  $H _2SO _4$ is stronger than $HCl$.
Relative strength in favour of $H _2SO _4$
$=\frac {\text {rate constant of reaction catalysted by }H _2SO _4}{\text {rate constant of reaction catalysted by HCl}}$.

The rate constant (K) for the reaction, $2A+B\rightarrow$ Product was found to be $2.5\times 10^{-5}$ litre $mol^{-1} sec^{-1}$ after 15 sec, $2.60\times 10^{-5} litre\  mol^{-1} sec^{-1}$ after 30 sec and $2.55\times 10^{-5} litre \ mol^{-1}sec^{-1}$ after 50 sec. The order of reaction is:

  1. $2$

  2. $3$

  3. $0$

  4. $1$


Correct Option: A
Explanation:

 The order of reaction is 2.


K does not change with time; thus its value remains unchanged during the reaction.

By seeing the equation, it can be said that reaction is a third order but the unit of K suggest it to be 2nd order.

Assertion : In a second-order reaction with respect to A, when you double [A], the rate is quadrupled.
Reason : The rate equation is $\displaystyle r={ k\left[ A \right]  }^{ 2 }$ for such a reaction.

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

  3. Assertion is true but Reason is false

  4. Assertion is false but Reason is true

  5. Both Assertion and Reason are false


Correct Option: A
Explanation:

Rate equation is as follows:
$Older Rate = k [A]^{2}$                       equation -1
This shows that reaction is $2nd $ order.
So if we double the concentration of the A , then new reaction rate can be written as follows:
$New Rate = k [2A]^{2}$
$New Rate = k\times 4[A]^{2}$
$New Rate = 4 k [A]^{2}$                     equation -2
Now, comparing equation -1 and 2 we get :
New Rate = $4\times Older Rate$


For a gaseous reaction, $A\left( g \right) \longrightarrow $ Product, which one of the following is correct relation among $\dfrac { dP }{ dt } ,\dfrac { dn }{ dt }$ and $\dfrac { dc }{ dt } $?
($\dfrac { dP }{ dt } =$ Rate of reaction in $atm$ ${ sec }^{ -1 }$; $\dfrac { dc }{ dt } =$ Rate of reaction in molarity ${ sec }^{ -1 }$; $\dfrac { dn }{ dt } =$ Rate of reaction in $mol$ ${ sec }^{ -1 }$)

  1. $\dfrac { dc }{ dt } =\dfrac { dn }{ dt } =-\dfrac { dP }{ dt } $

  2. $-\dfrac { dc }{ dt } =-\dfrac { 1 }{ V } \dfrac { dn }{ dt } =-\dfrac {1}{RT}\dfrac{ dP }{ dt } $

  3. $\dfrac { dc }{ dt } =\dfrac { V }{ RT } \dfrac { dn }{ dt } =\dfrac { dP }{ dt } $

  4. None of the above


Correct Option: B
Explanation:
$PV=nRT$
$\frac{dP}{dt}.V=RT.\frac{dn}{dt}$=> $\frac{1}{V}\frac{dn}{dt}=\frac{1}{RT}\frac{dP}{dt}$-(1)
Now, PV=nRT or $P=\frac{n}{V}RT$
Concentration $c=\frac{n}{V}$
P=cRT
$\frac{dP}{dt}=\frac{dc}{dt}.RT$=> \frac{dc}{dt}=\frac{1}{RT}\frac{dP}{dt}$
$-\frac{dc}{dt}=-\frac{1}{V}\frac{dn}{dt}=-\frac{1}{RT}\frac{dP}{dt}$

The rate constant for the reaction is $2 10^{-4} s^{-1}.$ The reaction is :

  1. First order

  2. Second order

  3. Third order

  4. Zero order


Correct Option: A

For a second-order reaction of the type rate=$k[A]^2$ the plot of $\dfrac{1}{[A] _1}$ versus t is linear with a: 

  1. positive slope and zero intercept

  2. positive slope and non-zero intercept

  3. negative slope and zero intercept

  4. negative slope and non-zero intercept


Correct Option: B
Explanation:

Given that for a second order reaction

$Rate= K[A]^2$

We know that rate of the reaction is decrease in the concentration of the reactant with respect to time.

$Rate= \cfrac {-d[A]}{dt}$
$\Rightarrow$ $\cfrac {-d[A]}{dt}=K[A]^2$
$\Rightarrow \cfrac {-d[A]}{[A]^2}=+Kdt$

Integrating both sides we have:-
$\Rightarrow -\int \cfrac {d[A]}{[A]^2}=+\int Kdt$
$\Rightarrow - \left[\cfrac {-1}{[A]}\right]=+Kt+C$
$\Rightarrow Kt+C= \cfrac {1}{[A]}$      $- (i)$

Now, from the equation $(i)$ we can see that graph of $\cfrac {1}{[A]}$ $V/s$ $t$ has a positive slope $K$ and non-zero intercept $C$.

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