Tag: chemical kinetics

Questions Related to chemical kinetics

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?

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. 1000 sec

  2. 400 sec

  3. 200 sec

  4. 10,000 sec

Show answer
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.

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. three times

  2. six times

  3. eight times

  4. nine times

Show answer
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:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. third order kinetics

  2. second order kinetics

  3. first order kinetics

  4. zero order kinetics

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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
chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. 0

  2. 1

  3. 2

  4. 3

Show answer
Correct Option: C

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

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

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

Show answer
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 ?

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a 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. $mole^{n-1} l^{n} s^{-1}$

Show answer
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:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $mol\  lit^{-1}$

  2. $l\  mol^{-1} \ s^{-1}$

  3. $s^{-1}$

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

Show answer
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 :

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

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

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

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

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

Show answer
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 $B$?

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $40$ min

  2. $39.21$ min

  3. $28.59$ min

  4. None of these

Show answer
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:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

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

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

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

  4. $bar min^{-1}$

Show answer
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$=bar^{-\frac {1}{2}}min^{-1}$