Tag: chemical kinetics

Quizzes (10) Questions (179)

Questions Related to chemical kinetics

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a 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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\dfrac{1}{a}=\dfrac{1}{a _{0}}+kt$
$\dfrac{1}{0.04}=\dfrac{1}{0.2}+0.002t$
$t=10000sec$
Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The rate law is given as Rate = k[P_NO]^2 [P_H2]. Since the initial concentration of NO (1000 mm Hg) is much larger than the concentration of H2 (10 mm Hg), the concentration of NO remains effectively constant throughout the reaction. Thus, the reaction behaves as pseudo-first order with respect to H2.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Saponification of ethyl acetate with equal concentrations of ester and alkali is a classic second-order reaction. The data provided would show a linear relationship for 1/[concentration] vs time, confirming second-order kinetics.

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

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. $mol Ls,\quad \displaystyle mol^{\frac{1}{2}}L^{\frac{1}{2}}s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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. $mole^{n-1} l^{n} s^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

The unit of rate of a first order reaction is:

  1. $mol\ lit^{-1}$
  2. $l\ mol^{-1} \ s^{-1}$
  3. $s^{-1}$
  4. $l^2 \ mol^{-2} \ s^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $9 \times 10^{-3} moleslitre^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 }\\ $
Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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$={ t } _{ 1/2 }=\cfrac { 1 }{ K[A] } =\cfrac { 1 }{ 0.622\times [4.1\times { 10 }^{ -2 }] } \\ =0.3921\times { 10 }^{ 2 }\\ =39.21\quad minutes.$

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

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. $bar min^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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