Tag: kinetic study of some first order reactions

Questions Related to kinetic study of some first order reactions

Inversion of cane sugar in dilute acid is:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. bimolecular reaction

  2. pseudo-unimolecular reaction

  3. unimolecular reaction

  4. trimolecular reaction

Show answer
Correct Option: B
Explanation:

${ C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 }+{ H } _{ 2 }O\xrightarrow [  ]{ \quad { H }^{ + }\quad  } { C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }+{ C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }$
Rate $=k\left[ { C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 } \right] \left[ { H } _{ 2 }O \right] $
When water is in excess, its concentration will be constant.
$\therefore $ Rate $={ k }^{ ' }\left[ { C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 } \right] $
The reaction is, therefore, pseudo first order or pseudo unimolecular reaction.

For the reaction, $2{N} _{2}{O} _{5}\longrightarrow 4{NO} _{2}+{O} _{2}$ the rate of reaction is:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $\cfrac{1}{2}\cfrac{d}{dt}[{N} _{2}{O} _{5}]$

  2. $2\cfrac{d}{dt}[{N} _{2}{O} _{5}]$

  3. $\cfrac{1}{2}\cfrac{d}{dt}[{NO} _{2}]$

  4. $4\cfrac{d}{dt}[{NO} _{2}]$

Show answer
Correct Option: C
Explanation:

For the reaction,  $\displaystyle 2{N} _{2}{O} _{5}\longrightarrow 4{NO} _{2}+{O} _{2} $  the rate of reaction is  $\displaystyle \cfrac{1}{2}\cfrac{d}{dt}[{NO} _{2}]$


 Rate of reaction $\displaystyle -\cfrac{1}{2}\cfrac{d[{N} _{2}{O} _{5}]}{dt}=\cfrac{1}{4}\cfrac{d[{NO} _{2}]}{dt}$

The rate constant for the reaction,

$2{N} _{2}{O} _{5}\longrightarrow 4{NO} _{2}+{O} _{2}$ is $3.0\times {10}^{-4}{s}^{-1}$.

 If start made with $1.0$ $mol$ ${L}^{-1}$ of ${N} _{2}{O} _{5}$, calculate the rate of formation of ${NO} _{2}$ at the moment of the reaction when concentration of ${O} _{2}$ is $0.1mol$ ${L}^{-1}$ :

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $2.7\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$

  2. $2.4\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$

  3. $4.8\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$

  4. $9.6\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$

Show answer
Correct Option: D
Explanation:

$Mol$ ${L}^{-1}$ of ${N} _{2}{O} _{5}$ reacted $=2\times 0.1=0.2$

$[{N} _{2}{O} _{5}]$ left $=1.0-0.2=0.8mol$ ${L}^{-1}$
Rate of reaction $=k\times [{N} _{2}{O} _{5}]$
$=3.0\times {10}^{-4}\times 0.8$
$=2.4\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$
Rate of formation of ${NO} _{2}$
$=4\times 2.4\times {10}^{-4}=9.6\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$

$H _2O _2$ decomposes with first order kinetics in a 3 lit. container. If the pressure developed in 10 min. is 380 mm, the average rate at $27^oC$ is:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $0.01M.min^{-1}$

  2. $0.002M.min^{-1}$

  3. $0.05M.min^{-1}$

  4. $0.06M.min^{-1}$

Show answer
Correct Option: A
Explanation:

$t=A.{ e }^{ -kt }\ So,\quad (A-{ A } _{ o })=A.({ e }^{ -kt }-1)\ \therefore 380=A.({ e }^{ -kt }-1)\ { A } _{ o }=\cfrac { 380 }{ { e }^{ -10k }-1 } $
 Otherewise,
$ { P } _{ o }=[{ A } _{ o }]RT\ { P } _{ o }={ [{ A }] } _{ 10 }RT\ \cfrac { { P } _{ 10 }-{ P } _{ o } }{ 7 } =\cfrac { ({ A } _{ 10 }-{ A } _{ o })RT }{ 7 } \ \cfrac { \cfrac { 760 }{ 380 }  }{ 10 } =\vartheta .RT\ \cfrac { 2 }{ 10RT } =\vartheta $
$ \vartheta \sim 0.01{ M. }{ min }^{ -1 }\longrightarrow$ Option (A)

The rate constant of the reaction, $2{ H } _{ 2 }{ O } _{ 2 }\left( aq. \right) \rightarrow 2{ H } _{ 2 }O\left( l \right) +{ O } _{ 2 }\left( g \right) $, is $3\times { 10 }^{ -3 }{ min }^{ -1 }$.
At what concentration of ${ H } _{ 2 }{ O } _{ 2 }$, the rate of the reaction will be $2\times { 10 }^{ -4 }M{ s }^{ -1 }$?

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $6.67\times { 10 }^{ -3 }\ M$

  2. $2\ M$

  3. $4\ M$

  4. $0.08\ M$

Show answer
Correct Option: C
Explanation:

Rate $=k{ \left[ { H } _{ 2 }{ O } _{ 2 } \right]  }^{ 1 }$
$2\times { 10 }^{ -4 }=\dfrac { 3\times { 10 }^{ -3 } }{ 60 } \times \left[ { H } _{ 2 }{ O } _{ 2 } \right] $
$\left[ { H } _{ 2 }{ O } _{ 2 } \right] =4 M$

Inversion of a sugar follows first order rate equation which can be followed by noting the change in rotation of the plane of polarisation of light in a polarimeter. If ${ r } _{ \infty  },{ r } _{ t }$ and ${ r } _{ 0 }$ are the rotations at $t=\infty , t=t$ and $t=0$, then first order reaction can be written as:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $k=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ t }-{ r } _{ \infty } }{ { r } _{ 0 }-{ r } _{ \infty } } } $

  2. $k=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ 0 }-{ r } _{ \infty } }{ { r } _{ t }-{ r } _{ 0 } } } $

  3. $k=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ \infty }-{ r } _{ 0 } }{ { r } _{ \infty }-{ r } _{ t } } } $

  4. $k=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ \infty }-{ r } _{ t } }{ { r } _{ \infty }-{ r } _{ 0 } } } $

Show answer
Correct Option: B
Explanation:

$({ r } _{ t }-{ r } _{ 0 })=({ r } _{ 0 }-{ r } _{ \infty  }){ e }^{ -kt }\ \ln { \left( \cfrac { { r } _{ t }-{ r } _{ 0 } }{ { r } _{ 0 }-{ r } _{ \infty  } }  \right)  } =-kt\ k=\cfrac { 1 }{ t } \ln { \left( \cfrac { { r } _{ t }-{ r } _{ 0 } }{ { r } _{ 0 }-{ r } _{ \infty  } }  \right)  } $

The inversion of cane sugar into glucose and fructose is:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $I$ order

  2. $II$ order

  3. $III$ order

  4. zero order

Show answer
Correct Option: A
Explanation:

Inversion of cane sugar follow Ist order reaction while its molecularity is 2 and reaction is given by
$ \implies C _{12}H _{22}O _{11} +H _2O \rightarrow C _6 H _{12}O _6 + C _6H _{12}O _6$

Here the rate of reaction is dependent on only $C _{12}H _{22}O _{11}$.

The rate constant for the hydrolysis reaction of an ester by dilute acid is $0.6931\times { 10 }^{ -3 }\ { s }^{ -1 }$. The time required to change the concentration of ester from $0.04$ $M$ to $0.01$ $M$ is:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $6931$ sec

  2. $4000$ sec

  3. $2000$ sec

  4. $1000$ sec

Show answer
Correct Option: C
Explanation:

$k=0.06931{ s }^{ -1 }$

 So,
$ t=\cfrac { \ln { \left( \cfrac { 0.04 }{ 0.01 }  \right)  }  }{ 0.06931 } \ =2000{ s }^{ -1 }$

Benzene diazonium chloride (A) decomposes into chloro-benzene (B) and ${{\text{N}} _{\text{2}}}\left( {\text{g}} \right)$ in first order reaction volume of ${{\text{N}} _2}$ collected after 5 min and at the complete decomposition of A are 10 ml and 50 ml respectively. The rate constant for the reaction is:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. 0.446 ${\min ^{ - 1}}$

  2. 0.0446 ${\min ^{ - 1}}$

  3. 0.223 ${\min ^{ - 1}}$

  4. 0.112 ${\min ^{ - 1}}$

Show answer
Correct Option: B
Explanation:

t = 0

                              $\mathop {\text{A}}\limits _{50}  \to \mathop {\text{B}}\limits _0  + \mathop {{{\text{N}} _2}}\limits _0 $
At t = 5 min.    50 - 10          10 ml
t = complete                          50 ml
          $\ln  = \left( {\dfrac{{50}}{{40}}} \right) = {\kappa _1} \times 5$
  $\dfrac{{0.223}}{5} = {\kappa _1} \Rightarrow 0.0446\,{\min ^{ - 1}}$
Hence, option (B) is correct.

For the reaction of first order, $2{ N } _{ 2 }{ O } _{ 5 }\left( g \right) \rightleftharpoons 4N{ O } _{ 2 }\left( g \right) +{ O } _{ 2 }\left( g \right) $, which of the following statements are correct?

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. The concentration of the reactant decreases exponentially with time.

  2. The half-life of the reaction decreases with increasing temperature.

  3. The half-life of the reaction depends on the initial concentration of the reactant.

  4. The reaction proceeds to $99.6$% completion in eight half-life duration.

Show answer
Correct Option: A,B,D
Explanation:
For 1st order reaction,
$a _t=a _0e^{-kt}$
Also, $t _{1/2}=\frac{In2}{k}$ or $t _{1/2}\alpha \frac{1}{k}$
As the temperature increases, value of k also increases due to which $t _{1/2}$ decreases.
For 99.6% completion, $a _t=(\frac{100-99.6}{100})a _0=\frac{4a _0}{1000}$
$t=\frac{1}{k}In\frac{a _0}{4a-0/1000}=\frac{1}{k}In\frac{1000}{4}$
$=(\frac{t _{1/4}}{In2}).In250$
$t=8t _{1/2}$