Tag: order of reactions

Questions Related to order of reactions

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

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

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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. $mole^{n-1} l^{n} s^{-1}$

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

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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

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

Taking the reaction, $A + 2B\rightarrow Products$, to be of the second order, which of the following may be the correct rate law expressions?

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

  1. $\frac {dx}{dt}=k[A][B]$

  2. $\frac {dx}{dt}=k[A][B]^2$

  3. $\frac {dx}{dt}=k[A]^2$

  4. $\frac {dx}{dt}=k _1[A]+k _2[B]^2$

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Correct Option: A,C
Explanation:

option A and C are correct as the sum of their exponents equals to 2

The rate constant of a second order reaction is $10^{-2} lit.mole ^{-1}.sec^{-1}$. The rate constant when expressed as $cc. \ molecule^{-1} .\ min^{-1}$ is:

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

  1. $9.96\times 10^{-22}$

  2. $9.96\times 10^{-23}$

  3. $9.96\times 10^{-21}$

  4. $9.96\times 10^{-24}$

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Correct Option: A
Explanation:

The rate constant of a second order reaction is $10^{-2} lit.mole ^{-1}.sec^{-1}$.
$1L=1000cc$


$1 mole = 6.023\times 10^{23}$molecules
$1min=60sec$
Hence, rate constant $=10^{-2} lit.mole ^{-1} sec^{-1}\times \dfrac {1000cc}{1L} \times \dfrac {1mole} {6.023\times 10^{23}molecules} \times \dfrac {60 sec} {1 min}$
$=9.96\times 10^{-22}cc\ molecule^{-1} \ min^{-1}$

In a certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours and so on. Dimension of the velocity constant are:

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

  1. hour$^{-1}$

  2. mole litre$^{-1}$ hour$^{-1}$

  3. litre mol$^{-1}$ hour$^{-1}$

  4. mole sec$^{-1}$

Show answer
Correct Option: B
Explanation:

For the zero-order reaction, the time taken for the decomposition of the reactant is independent of initial concentration which is the case here. 


The dimension of the velocity constant for the zero-order reaction is = mole litre$^{-1}$ hour$^{-1}$

Option B is correct.