Tag: endothermic and exothermic reactions

Quizzes (1) Questions (104)

Questions Related to endothermic and exothermic reactions

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.

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

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?

  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.
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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, $a _t=(\frac{100-99.6}{100})a _0=\frac{4a _0}{1000}$
$t=\frac{1}{k}In\frac{a _0}{4a-0/1000}=\frac{1}{k}In\frac{1000}{4}$
$=(\frac{t _{1/4}}{In2}).In250$
$t=8t _{1/2}$
Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

For the reaction, ${\text{2}}{{\text{N}} _{\text{2}}}{{\text{O}} _{\text{5}}} \to {\text{4N}}{{\text{O}} _{\text{2}}} + {{\text{O}} _{\text{2}}}$, the value of rate and rate constant are $1.02\times 10^{-4} M/s$ and $3.4 \times {10^{ - 3}}{\sec ^{ - 1}}$ respectively. The concentration of ${{\text{N}} _{\text{2}}}{{\text{O}} _{\text{5}}}$ at that time will be: (in terms of molarity)

  1. $1.732$
  2. $3$
  3. ${\text{1}}{\text{.02}} \times {\text{1}}{{\text{0}}^{ - 4}}$
  4. $3.4 \times {10^4}$
Reveal answer Fill a bubble to check yourself
A Correct answer