Tag: chemistry

Questions Related to chemistry

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.