Tag: chemistry

Questions Related to chemistry

The reaction $N _{2}O _{5}$ (in $CCl _{4}$) $\rightarrow 2NO _{2}+1/2O _{2}(g)$ is the first order in $N _{2}O _{5}$ with rate constant $6.2\times 10^{-4}S^{-1}$. 


What is the value of the rate of reaction when $N _2O _5=1.25:mole:L^{-1}$ ?

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

  1. $7.75\times 10^{-4}mol:L^{-1}S^{-1}$

  2. $6.35\times 10^{-3}mol:L^{-1}S^{-1}$

  3. $5.15\times 10^{-5}mol:L^{-1}S^{-1}$

  4. $3.85\times 10^{-4}mol:L^{-1}S^{-1}$

Show answer
Correct Option: A
Explanation:
For the first-order reaction, the rate of the reaction is given by the expression

Rate $\displaystyle  = k [N _2O _5]$ where k is the rate constant.

Substitute values in the above expression

Rate $\displaystyle  = 6.2\times 10^{-4}S^{-1} \times 1.25\:mole\:L^{-1} = 7.75\times 10^{-4}mol\:L^{-1}S^{-1}$

So, the correct option is $A$

The half life of decomposition of $N _2O _5$ is a first order reaction represented by
$N _2O _5\, \rightarrow\, N _2O _4\, =\, 1/2O _2$
After 15 min the volume of $O _2$ produced is $9mL$ and at the end of the reaction $35 mL$. The rate constant is equal to :

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

  1. $\displaystyle \frac{1}{15}\, log\frac{35}{26}$

  2. $\displaystyle \frac{1}{15}\log\frac{44}{26}$

  3. $\displaystyle \frac{1}{15}\, log\frac{35}{36}$

  4. None of the above 

Show answer
Correct Option: A
Explanation:

$\displaystyle k\, =\, \frac{2.303}{t}\, log\, \frac{V _{\infty}}{V _{\infty}\, -\, V _t}$

$\displaystyle =\, \frac{1}{t}\, log _e\, \frac{V _{\infty}}{V _{\infty}\, -\, V _t}$

$\displaystyle \frac{1}{15}\, log _e\, \frac{35mL}{(35\, -\, 9)\, mL}\, =\, \frac{1}{15}\, log _e\, \frac{35}{26}$

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 $t$.

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

  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. $\ln{\cfrac{{[{N} _{2}{O} _{5}]} _{0}}{{[{N} _{2}{O} _{5}]} _{t}}}=kt$

Show answer
Correct Option: D
Explanation:

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

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

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 $C _{12}H _{22}O _{11}$.