Tag: rate of chemical reaction

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Questions Related to rate of chemical reaction

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

The rate constant $k$, for the reaction
${N} _{2}{O} _{5}(g) \longrightarrow 2{NO} _{2}(g)+\cfrac{1}{2}{O} _{2}(g)$
is $1.3\times {10}^{-2}{s}^{-1}$. Which equation given below describes the change of $[{N} _{2}{O} _{5}]$ with time?
${[{N} _{2}{O} _{5}]} _{0}$ and ${[{N} _{2}{O} _{5}]} _{t}$ correspond to concentration of ${N} _{2}{O} _{5}$ initially and at time $t$.

  1. ${[{N} _{2}{O} _{5}]} _{t}={[{N} _{2}{O} _{5}]} _{0}+kt$
  2. ${[{N} _{2}{O} _{5}]} _{0}={[{N} _{2}{O} _{5}]} _{t}{e}^{kt}$
  3. $\log{{[{N} _{2}{O} _{5}]} _{t}}=\log{{[{N} _{2}{O} _{5}]} _{0}}+kt$
  4. $\ln{\cfrac{{[{N} _{2}{O} _{5}]} _{0}}{{[{N} _{2}{O} _{5}]} _{t}}}=kt$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the unit of rate constant is ${sec}^{-1}$, the reaction is first order reaction. 

${N} _{2}{O} _{5}(g) \longrightarrow 2{NO} _{2}(g)+\cfrac{1}{2}{O} _{2}(g)$
$k{t}=\ln{\cfrac{a}{(a-x)}}$ 
$kt=\ln{\cfrac { { [{ N } _{ 2 }{ O } _{ 5 }] } _{ 0 } }{ { [{ N } _{ 2 }{ O } _{ 5 }] } _{ t } } }$

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

Inversion of cane sugar in dilute acid is:

  1. bimolecular reaction

  2. pseudo-unimolecular reaction

  3. unimolecular reaction

  4. trimolecular reaction
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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. $4\cfrac{d}{dt}[{NO} _{2}]$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\displaystyle -\cfrac{1}{2}\cfrac{d[{N} _{2}{O} _{5}]}{dt}=\cfrac{1}{4}\cfrac{d[{NO} _{2}]}{dt}$

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

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 $0.1mol$ ${L}^{-1}$ :

  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. $9.6\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 ${NO} _{2}$
$=4\times 2.4\times {10}^{-4}=9.6\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$

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

$H _2O _2$ 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:

  1. $0.01M.min^{-1}$
  2. $0.002M.min^{-1}$
  3. $0.05M.min^{-1}$
  4. $0.06M.min^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

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

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 $2\times { 10 }^{ -4 }M{ s }^{ -1 }$?

  1. $6.67\times { 10 }^{ -3 }\ M$
  2. $2\ M$
  3. $4\ M$
  4. $0.08\ M$
Reveal answer Fill a bubble to check yourself
C Correct answer
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] $
$\left[ { H } _{ 2 }{ O } _{ 2 } \right] =4 M$

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

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:

  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 } } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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)  } $

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

The inversion of cane sugar into glucose and fructose is:

  1. $I$ order
  2. $II$ order
  3. $III$ order

  4. zero order
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $C _{12}H _{22}O _{11}$.

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

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:

  1. $6931$ sec
  2. $4000$ sec
  3. $2000$ sec
  4. $1000$ sec
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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:

  1. 0.446 ${\min ^{ - 1}}$
  2. 0.0446 ${\min ^{ - 1}}$
  3. 0.223 ${\min ^{ - 1}}$
  4. 0.112 ${\min ^{ - 1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.