Tag: dependence of reaction rate on concentration of reactants

Questions Related to dependence of reaction rate on concentration of reactants

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.

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


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 $B$?

  1. $40$ min

  2. $39.21$ min

  3. $28.59$ min

  4. None of these


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:

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

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

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

  4. $bar min^{-1}$


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?

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

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

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

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


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:

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

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

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

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


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:

  1. hour$^{-1}$

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

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

  4. mole sec$^{-1}$


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.

When ethyl acetate was hydrolysed in presence of 0.1 N HCl, the rate constant was found to be $5.40\times 10^{-5}sec^{-1}$. But when 0.1 N $H _2SO _4$ was used for hydrolysis, the rate constant was found to be $6.25\times 10^{-5} sec^{-1}$. Thus, it may be concluded that:

  1. $H _2SO _4$ is stronger than HCl

  2. $H _2SO _4$ is weaker than HCl

  3. $H _2SO _4$ and HCl both have the same strength

  4. the data are not sufficient to compare the strength of $H _2SO _4$ and HCl


Correct Option: A
Explanation:

Option (A) is correct.  $H _2SO _4$ is stronger than $HCl$.
Relative strength in favour of $H _2SO _4$
$=\frac {\text {rate constant of reaction catalysted by }H _2SO _4}{\text {rate constant of reaction catalysted by HCl}}$.