Tag: rate of chemical reaction

Questions Related to rate of chemical reaction

Inversion of a sugar folllows first order rate equation which can be followed by nothing the change in rotation of the plane of polarization of light in the 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. $\;k\,=\,\displaystyle\frac{1}{t}log\displaystyle\frac{r _t-r _{\infty}}{r _0-r _{\infty}}$

  2. $\;k\,=\,\displaystyle\frac{1}{t}\,ln\displaystyle\frac{r _0-r _{\infty}}{r _t-r _0}$

  3. $\;k\,=\,\displaystyle\frac{1}{t}\,ln\displaystyle\frac{r _{\infty}-r _0}{r _{\infty}-r _t}$

  4. None of these

Show answer
Correct Option: C
Explanation:

$\underset{d-Sucrose}{C _{12}H _{22}O _{11}}+H _2O\xrightarrow{H+}\underset{d-Glucose}{C _6H _{12}O _6}+\underset{l-Fructose}{C _6H _{12}O _6}$


Initially               a                Excess                  0                0 
After time t        a-x            Constant                x                x
At infinity           0               Constant               a                 a
If $r _0,r _t$ and $r _{\infty}$ be the observed angle of rotations of the sample at zero times, times $t$ and infinity respectively, and $k _1,k _2$ and $k _3$ proportionate in terms of sucrose,glucose and fructose, respectively.
Then,
$r _0=k _1a$
$r _t=k _1(a-x)+k _2x+k _3x$
$r _{\infty}=k _2a+k _3a$
From these equations it can be shown that
$\dfrac{a}{a-x}=\dfrac{r _0-r _{\infty}}{r _t-r _{\infty}}$
So, the expression for the rate constant rate of this reaction in terms of the optical rotational data may be 
put as $k=\dfrac{2.303}{t}\log \dfrac{r _0-r _{\infty}}{r _t-r _{\infty}}$

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 $H _2O _2$?

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

  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

Show answer
Correct Option: A
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.

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 $k _1,:k _2:and:k _3$?

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

  1. $2k _1=k _2=4k _3$

  2. $2k _2=k _1=4k _3$

  3. $2k _3=k _2=4k _1$

  4. $2k _1=k _2=4k _2$

Show answer
Correct Option: A
Explanation:

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

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 ?

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

  1. 966 min

  2. 483 min

  3. 1932 min

  4. None of these

Show answer
Correct Option: A
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 $

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:

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

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

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

Show answer
Correct Option: A
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):

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

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?

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

  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. ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 }+kt$

Show answer
Correct Option: A,B,C
Explanation:

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