Tag: chemistry

Questions Related to chemistry

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.