Tag: chemical kinetics

Questions Related to chemical kinetics

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:

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

Rate constant in case of first order reaction is :

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. Inversely proportional to the concentration units

  2. Independent of concentration units

  3. Directly proportional to concentration units

  4. Inversely proportional to the square of concentration units

Show answer
Correct Option: B
Explanation:

For First order of reaction,
$Rate= k [A]$,


$k = \cfrac {mol/L}{sec\times {mol/L}}=sec^{-1}$

Option B is correct.

Fill in the blanks by choosing the correct option;
Order of the reaction is the $X$ of the powers to which concentration terms are raised in experimentally determined rate equation. The unit of first order rate constant is $Y$. The unit of first order rate constant when concentration is measured in terms of pressure and time in minutes is $Z$.

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $X\rightarrow product, Y\rightarrow mol\ L^{-1} time^{-1}, Z\rightarrow atm\ min^{-1}$

  2. $X\rightarrow sum, Y\rightarrow L\ mol^{-1}time^{-1}, Z\rightarrow atm\ min^{-1}$

  3. $X\rightarrow product, Y\rightarrow L\ mol^{-1}, Z\rightarrow atm\ min^{-1}$

  4. $X\rightarrow sum, Y\rightarrow time^{-1}, Z \rightarrow min^{-1}$

Show answer
Correct Option: D
Explanation:

The order of a chemical reaction is defined as the sum of the powers of the concentration of the reactants in the rate equation of that particular chemical reaction.


General formula for the unit of rate constant = mole$^{(1-n)}$ L$^{(n-1)}$ min$^{-1}$.

For 1$^{st}$ order reaction, 
n = 1
Therefore, unit of rate constant $=$ min$^{-1}$

The unit of rate constant in terms of pressure and time:
  mol $^{( 1-n)}$  L $^{ (n-1)}$ min$^{ -1 }$ or  atm$^{( 1-n) }$ min $^{ -1 }$


For 1$^{st}$ order reaction, 
n = 1
Therefore, the unit is min$^{-1}.$

Hence, the correct answer is option $\text{D}$.

Match the rate law given in column I with the dimensions of rate constant given in column II and mark the appropriate choice.

Column I Column II
(A) $Rate = k[NH _{3}]^{0}$ (i) $mol\ L^{-1} s^{-1}$
(B) $Rate = k[H _{2}O _{2}][I^{-}]$ (ii) $L\ mol^{-1} s^{-1}$
(C) $Rate = k[CH _{3}CHO]^{3/2}$ (iii) $s^{-1}$
(D) $Rate = k[C _{2}H _{5}Cl]$ (iv) $L^{1/2} mol^{-1/2} s^{-1}$
chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $(A)\rightarrow (iv), (B) \rightarrow (iii), (C)\rightarrow (ii), (D) \rightarrow (i)$

  2. $(A)\rightarrow (i), (B) \rightarrow (ii), (C)\rightarrow (iii), (D) \rightarrow (iv)$

  3. $(A)\rightarrow (ii), (B) \rightarrow (i), (C)\rightarrow (iv), (D) \rightarrow (iii)$

  4. $(A)\rightarrow (i), (B) \rightarrow (ii), (C)\rightarrow (iv), (D) \rightarrow (iii)$

Show answer
Correct Option: D
Explanation:
(A) $Rate= k[NH _3]^0$
It is zero order reaction.
$\therefore$ Units of rate constant are same as rate.
i.e $ mol\ l^{-1}s^{-1}$

(B) $Rate=k[H _2O _2]^1[I^-]^1$
It is second order reaction as order $=1+1=2$
$\therefore$ Units of Rate constant are $L \ mol^{-1}s^{-1}$

(C) $Rate=k[CH _3CHO]^{3/2}$
It is fractional order reaction with order $= \cfrac 32$
$\therefore$ Units of rate constant are $L^{\cfrac 12}mol^{-\cfrac 12}s^{-1}$

(D) $Rate=k[C _2H _5Cl]^{-1}$
It is first order reaction. So units of rate constant are $s^{-1}$

For the second order reaction, concentration $(x)$ of the product at time $t$ starting with initial concentration $[A] _0$ is:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $\dfrac{kt[A _0]^2}{1 + kt[A _0]}$

  2. $\dfrac{k + [A _0]^2}{1 + kt}$

  3. $\dfrac{1 + kt[A _0]^2}{k + [A _0]^2}$

  4. none of these

Show answer
Correct Option: A
Explanation:

A reaction said to be a second order when the overall order is $2$. The rate of second order reaction may be proportional to one concentration squared.

$R=K[A]^2$
For rate proportional to single concentration squared, the time dependance of concentration is given by
$\cfrac{1}{[A]}=\cfrac{1}{A _0}+Kt
Therefore, concentration of product after time $t=\cfrac{kt[A_0]^2}{1+kt[A_0]}$.

Units of rate constant of a first order reaction is :

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $mole.lit^{-1}$

  2. $lit. mole$

  3. $mole. sec^{-1}$

  4. $sec^{-1}$

Show answer
Correct Option: D
Explanation:

A $\rightarrow$ product

For first order reaction, rate is dependent on single reactant A for example, rate = k[A]
$k=\frac{rate}{[A]}$

$=\frac{mole}{liter}sec\times \frac{liter}{mole}$
$=sec^{-1}$

A gaseous reaction, $A _{2}\left ( g \right )\rightarrow B\left ( g \right )+\frac{1}{2}\left ( g \right )$ 
Show increase in pressure from 40 mm to 120 mm in 5 minutes. the rate of disappearance of$A _{2}$ is ?

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. 4 mm $min ^{-1}$

  2. 8mm $min^{-1}$

  3. 16 mm $min^{-1}$

  4. 2 mm $min^{-1}$

Show answer
Correct Option: B