Tag: chemical kinetics

Questions Related to chemical kinetics

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