Tag: chemical kinetics

Questions Related to chemical kinetics

The reaction $2N _2O _5(g)\, \rightarrow\, 4NO _2(g)\, +\, O _2(g)$ is first order w.r.t. $N _2O _5$. Which of the following graphs would yield a straight line ?

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

  1. $log\, p _{N _2O _5}$ vs time with -ve slope

  2. $(p _{N _2O _5})^{-1}$ vs time

  3. $p _{N _2O _5}$ vs time

  4. $log\, p _{N _2O _5}$ vs time with +ve slope

Show answer
Correct Option: A
Explanation:

For a first order reaction, the graph of logarithm of the partial pressure of reactant to the time is a straight line with negative slope. 


Hence, $\displaystyle log\, p _{N _2O _5}$ vs time t will give a straight line.

When ethyl acetate was hydrolyzed in the presence of $0.1 M$ $HCl$, the constant was found to be $5.40\, \times\, 10^{-5}\, s^{-1}$. But when $0.1$ $M\, H _2SO _4$ was used for hydrolysis, the rate constant was found to be $6.20\, \times\,10^{-5}\, s^{-1}$. From these we can say that:

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

  1. $H _2SO _4$ is stronger than $HCl.$

  2. $H _2SO _4$ and $HCl$ are both of the same strength.

  3. $H _2SO _4$ is weaker than $HCl.$

  4. The data is insufficient to compare the strength of $HCl$ ad $H _2SO _4$.

Show answer
Correct Option: A
Explanation:

Since $k _{H _2SO _4}\, >\, k _{HCl},$ hence $H _2SO _4$ is stronger acid than HCl.

The rate law for the reaction : $:Ester+H^+\rightarrow Acid+Alcohol\,$ is
$V\,=\,k\;\left[ester \right]\;\left[H _3O^+ \right]^0$
What would be the new rate if
(a)$\;$conc. of ester is doubled
(b)$\;$conc. of $:H^{+}$ is doubled

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

  1.  (a)$\;v\;$ (b)$\;2v$

  2.  (a)$\;2v\;$ (b)$\;v$

  3.  (a)$\;2v\;$ (b)$\;2v$

  4. None of the above 

Show answer
Correct Option: B
Explanation:

$\upsilon=k[ester][H _3O^+]$


(a) Conc. of ester is doubled rate also double that is $2\upsilon$ because rate of the reaction depends upon ester concentration.

(b) Conc. of $H^+$ is doubled rate does not change that is $\upsilon$ because rate of the reaction does not depends on $H _3O^+$ concentration.

So answer is B.

In the presence of acid, the initial concentration of cane-sugar was reduced from 0.2 M to 0.1 in 5 hr and to 0.05 M in 10 hr. The reaction must be of :

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

  1. Zero order

  2. First order

  3. Second order

  4. Fractional order

Show answer
Correct Option: B
Explanation:

$\displaystyle 0.2\, M \underset{t _{1/2}\, =\, 5hr}{\rightarrow} 0.1\, M \underset{t _{1/2}\, =\, 5hr}{\rightarrow} 0.05\, M$

$From\, 0.2\, M \underset{t\, =\, 10hr}{\rightarrow}\, 0.05\, M$
So $t _{1/2}$ is constant which is characteristic of first order reaction.
Hence, $t _{1/2}\, \propto\, (a)^0$.

The decomposition of $H _2O _2$ can be followed by titration with $KMnO _4$ and is found to be a first order reaction. The rate constant is $4.5\, \times\, 10^{-2}$. In an experiment, the initial titrate value was 25 mL. The titrate value will be 5 mL after a lapse of :

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

  1. $4.5\, \times\, 10^{-2}\, \times\, 5\, min$

  2. $\displaystyle \frac{log _{e}5}{4.5\, \times\, 10^{-2}}\, min$

  3. $\displaystyle \frac{log _{e}5/4}{4.5\, \times\, 10^{-2}}\, min$

  4. None of the above

Show answer
Correct Option: B
Explanation:

As we know,
$\displaystyle t\, =\, \frac{2.303}{k}\, log\, \frac{V _0}{V _1}$

$\displaystyle =\, \frac{1}{k}\, ln\, \frac{V _0)}{V _1}$

$\displaystyle =\, \frac{1}{4.5\, \times\, 10^{-2}\, min^{-1}}\, In\, \frac{25mL}{5mL}$

$\displaystyle =\, \frac{log _{e}5}{4.5\, \times\, 10^{-2}}min$ 

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 minutes, the volume of $O _2$ produced is 9 $mL$ and at the end of the reaction is 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. $\;\displaystyle\frac{1}{15}log _e\displaystyle\frac{35}{26}$

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

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

  4. None of the above 

Show answer
Correct Option: A
Explanation:

For a first order reaction,

$KT= ln (a/a-x)$

So, for the following reaction:

$N _2O _5\rightarrow N _2O _4+1/2O _2$

$K\times15\,=\,ln\begin{pmatrix}\displaystyle\frac{35-0}{35-9}\end{pmatrix}$


$N _2O _5\rightarrow N _2O _4+1/2O _2$

$K= \dfrac{1}{15} \,ln\begin{pmatrix}\displaystyle\frac{35}{26}\end{pmatrix}$

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.