Tag: kinetic study of some first order reactions

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

In the following reaction $2H _2O _2\rightarrow2H _2O+O _2$ rate of formation of $O _2$ is 3.6 M min$^{-1}$.


(a) What is rate of formation of $H _2O\ ?$        
(b) What is rate of disappearance of $H _2O _2$?

  1. (i) $7.2$ mol litre$^{-1}$ min$^{-1},$ (ii) $7.2$ mol litre$^{-1}$ min$^{-1}$
  2. (i) $3.6$ mol litre$^{-1}$ min$^{-1},$ (ii) $3.6$ mol litre$^{-1}$ min$^{-1}$
  3. (i) $14.4$ mol litre$^{-1}$ min$^{-1},$ (ii) $14.4$ mol litre$^{-1}$ min$^{-1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ (a)2H _{2}O _{2}\rightarrow 2H _{2}O+O _{2} $

We know, $ \dfrac{-1}{2}\dfrac{d[H _{2}O _{2}]}{dt} = \dfrac{-1}{2}\dfrac{d[H _{2}O]}{dt} = \dfrac{d[O _{2}]}{dt} $

$ \dfrac{d[H _{2}O]}{dt} = 2\dfrac{d[O _{2}]}{dt} = 2\times 3.6\,M\,min^{-1} $

$ \dfrac{d[H _{2}O]}{dt} = 7.2\,M\,min^{-1} $

$(b) \dfrac{-d[H _{2}O _{2}]}{dt} = \dfrac{2}{2}\dfrac{d[H _{2}O]}{dt} = \dfrac{-d[H _{2}O _{2}]}{dt} = 7.2\,M\,min^{-1} $ 

Option A is correct.
Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

Dinitropentaoxide decomposes as follows :
    $N _2O _5:(g)\rightarrow2:NO _2(g)+\frac{1}{2}O _2:(g)$
Given that         

$ _d:[N _2O _5]:/:dt=k _1[N _2O _5]$
$d:[NO _2]:/:dt=k _2[N _2O _5]$
$d:[O _2]:/:dt=k _3[N _2O _5]$
What is the relation between $k _1,:k _2:and:k _3$?

  1. $2k _1=k _2=4k _3$
  2. $2k _2=k _1=4k _3$
  3. $2k _3=k _2=4k _1$
  4. $2k _1=k _2=4k _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle N _{2}O _{5}:(g)\rightarrow2:NO _{2}:(g)+\frac{1}{2}O _{2}:(g)$
$\displaystyle -d:[N _{2}O _{5}]/\mathrm{d} t=k _{1}[N _{2}O _{5}]$
$\displaystyle d:[NO _{2}]/\mathrm{d} t=k _{2}[N _{2}O _{5}]$
$\displaystyle d[O _{2}]/\mathrm{d} t=k _3[N _{2}O _{5}]$
$-\displaystyle \frac{\mathrm{d} N _{2}O _{5}}{\mathrm{d} t}=\frac{1}{2}\frac{\mathrm{d} NO _{2}}{\mathrm{d} t}
=2\frac{\mathrm{d} O _{2}}{\mathrm{d} t}$
$\displaystyle k _{1}=\frac{k _{2}}{2}=2k _{3}$
$\displaystyle 2k _{1}=k _{2}=4k _{3}$

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

The following data were obtained in experiment on inversion of cane sugar.
Time (minutes)         0        60        120      180      360     $\infty $
Angle of rotation  +13.1   +11.6   +10.2   +9.0   +5.87  -3.8
   (degree)
Determine total time ?

  1. 966 min

  2. 483 min

  3. 1932 min

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The integrated rate law expression for the inversion of can sugar (assuming first order kinetics) is as shown.
$\displaystyle k = \frac {2.303}{t} log \frac {r _0 - r _{\infty}}{r _t - r _{\infty}} $
For 60 minutes
$\displaystyle k = \frac {2.303}{60} log \frac {13.1 - (-3.8)}{11.6 - (3.8)} = 0.001549 $
For 120 minutes
$\displaystyle k = \frac {2.303}{120} log \frac {13.1 - (-3.8)}{10.2 - (3.8)} = 0.001569 $
For 180 minutes
$\displaystyle k = \frac {2.303}{180} log \frac {13.1 - (-3.8)}{9.0 - (3.8)} = 0.001544 $
For 360 minutes
$\displaystyle k = \frac {2.303}{360} log \frac {13.1 - (-3.8)}{5.87 - (3.8)} = 0.001551 $
Since, the value of k is constant, the reaction follows first order reaction.
The average value of k is $\displaystyle  \frac {0.001549+0.001569+0.001544+0.001551}{4} = \frac {0.0062135}{4} = 0.001553 : min^{-1}$
To determine the total time, substitute $\displaystyle r _t = 0 $ in the above expression.
$\displaystyle t = \frac {2.303}{k} log \frac {r _0 - r _{\infty}}{r _t - r _{\infty}} $
$\displaystyle t = \frac {2.303}{0.001553} log \frac {13.1 - (-3.8)}{0 - (-3.8)} = 966 : min $

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

Derive an expression for the Rate (k) of reaction :
$2N _{2}O _{5}(g)\rightarrow 4NO _{2}(g)+O _{2}(g)$


With the help of following mechanism:

$N _{2}O _{5}\overset{K _a}{\rightarrow}NO _{2}+NO _{3}$
$NO _{3}+NO _{2}\overset{K _{-a}}{\rightarrow}N _{2}O _{5}$
$NO _{2}+NO _{3}\overset{K _b}{\rightarrow}NO _{2}+O _{2}+NO$
$NO+NO _{3}\overset{K _c}{\rightarrow}2NO _{2}$

  1. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+2k _{b}}[N _{2}O _{5}]$
  2. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}-2k _{b}}[N _{2}O _{5}]$
  3. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+k _{b}}[N _{2}O _{5}]$
  4. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{2k _{-a}-2k _{b}}[N _{2}O _{5}]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rate $\displaystyle = k _b[NO _2][NO _3] $ .....(1)

But $\displaystyle \dfrac {[NO _2][NO _3]}{[N _2O _5]}=  \dfrac {K _a}{K _{-a} + 2k _b}$

Hence $\displaystyle [NO _2][NO _3]  =\dfrac {K _a}{K _{-a}+2k _b} [N _2O _5]$......(2)

Substitute equation (2) in equation (1):

$\displaystyle \displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+2k _{b}}[N _{2}O _{5}]$

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, ${ N } _{ 2 }{ O } _{ 5 }\left( g \right) \longrightarrow 2N{ O } _{ 2 }\left( g \right) +\dfrac { 1 }{ 2 } { O } _{ 2 }\left( g \right) $, is $2.3\times { 10 }^{ -2 }\ { sec }^{ -1 }$. Which equation given below describes the change of $\left[ { N } _{ 2 }{ O } _{ 5 } \right] $ with time, ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }$ and ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }$ corresponds to concentration of ${ N } _{ 2 }{ O } _{ 5 }$ initially and time $t$ respectively?

  1. ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t }{ e }^{ kt }$
  2. $\log _{ e }{ \dfrac { { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 } }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t } } } =kt$
  3. $\log _{ 10 }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t } } =\log _{ 10 }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 } } -kt$
  4. ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 }+kt$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }{ e }^{ -kt }\ { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }{ e }^{ kt }\ \ln { \left( \cfrac { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 } }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t } }  \right)  } ={ e }^{ kt }\ \ln { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t } } =\ln { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 } } -kt$