Tag: rate of chemical reaction

Questions Related to rate of chemical reaction

The rate constant, $\mathrm{k}$ for the reaction $\displaystyle \mathrm{N} _{2}\mathrm{O} _{5}(\mathrm{g})\rightarrow 2\mathrm{N}\mathrm{O} _{2}(\mathrm{g})+\frac{1}{2}\mathrm{O} _{2}(\mathrm{g})$ ls $2.3\times 10^{-2}\mathrm{s}^{-1}$. Which equation given below describes the change of $[\mathrm{N} _{2}\mathrm{O} _{5}]$ with time?

$[\mathrm{N} _{2}\mathrm{O} _{5}] _{0}$ and $[\mathrm{N} _{2}\mathrm{O} _{5}] _{\mathrm{t}}$ correspond to concentration of $\mathrm{N} _{2}\mathrm{O} _{5}$ initially and at time $\mathrm{t}$.

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. $log _{10}[N _{2}O _{5}] _{t}=log _{10}[N _{2}O _{5}] _{0}-kt$

  4. $ln\dfrac{[N _{2}O _{5}] _{0}}{[N _{2}O _{5}] _{t}}=kt$

Show answer
Correct Option: D
Explanation:

The decomposition of  $\mathrm{N} _{2}\mathrm{O} _{5}$ follows first order kinetics.


The integrated rate law expression is $ln\dfrac{[N _{2}O _{5}] _{0}}{[N _{2}O _{5}] _{t}}=kt$.

It can also be represented as $log _{10}[N _{2}O _{5}] _{t}=log _{10}[N _{2}O _{5}] _{0}-\dfrac {kt} {2.303}.$


It can also be represented as $[N _{2}O _{5}] _{t}=[N _{2}O _{5}] _{0}e^{-kt}.$

For the reaction ; $2H _2O _2(aq)\rightarrow 2H _2O(l)+O _2(g)$, rate of decomposition for $H _2O _2=k[H _2O _2]^2$
chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

For $2H _2O _2(aq)\rightarrow 2H _2O(l)+O _2(g)$ rate of decomposition for $H _2O _2=k[H _2O _2]$. It is a first order reaction.  It proceeds through following mechanism.

$\displaystyle H _2O _2 \xrightarrow {slow} H _2O + O $

$\displaystyle  O + O \xrightarrow {fast} O _2$

For the reaction; $2N _2O _5\rightarrow 4NO _2+O _2$, rate and rate constant are $1.02\times 10^{-4} M sec^{-1}$ and $3.4\times 10^{-5}  sec^{-1}$ respectively, then concentration of $N _2O _5$, at that time will be:

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

  1. $1.732\ M$

  2. $3\ M$

  3. $1.02\times 10^{-4} M$

  4. $3.5\times 10^{5} M$

Show answer
Correct Option: B
Explanation:
From the unit of rate constant we can identify the reaction as first order.

As we know,
$r=K[N _2O _5]$

$\therefore [N _2O _5]=\frac {r}{K}=\frac {1.02\times 10^{-4}}{3.4\times 10^{-5}}=3M$.

For the first order reaction:-
$2N _2O _5(g)\rightarrow 4NO _2(g)+O _2(g)$

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

  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

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

Option (A),(B),(D) are correct.
(C) : The half-life of the reaction is independent of the initial concentration of the reactant. Half-life for first order reaction is :$t _{1/2} = 0.693/k$
A first-order reaction has a rate proportional to the concentration of one reactant.
First-order rate constants have units of $sec^{-1}$. In other words, a first-order reaction has a rate law in which the sum of the exponents is equal to 1. 

Among the following unimolecular reaction is

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

  1. $C _{12}H _{22}O _{11}+H _2O \rightarrow C _6H _{12}O _6+C _6H _{12}O _6$

  2. $2NO+O _2 \rightarrow 2NO _2$

  3. $2H _2O _2 \rightarrow 2H _2O+O _2$

  4. $2NO _2+F _2 \rightarrow 2NO _2F$

Show answer
Correct Option: A

The reaction; $N _2O _5(g) \longrightarrow 2NO _2(g)+\frac {1}{2}O _2(g)$ is of first order for $N _2O _5$ with rate constant $6.2\times 10^{-4}s^{-1}$. What is the value of rate of reaction when $[N _2O _5]=1.25 \ mol L^{-1}$?

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

  1. $5.15\times 10^{-5}mol L^{-1}s^{-1}$

  2. $6.35\times 10^{-3}mol L^{-1}s^{-1}$

  3. $7.75\times 10^{-4}mol L^{-1}s^{-1}$

  4. $3.85\times 10^{-4}mol L^{-1}s^{-1}$

Show answer
Correct Option: C
Explanation:

As we know,
$r=K[N _2O _5]=6.2\times 10^{-4}\times 1.25=7.75\times 10^{-4} mol L^{-1} s^{-1}$.

The concentration of acetate ions in $1 M$ acetic acid $(K _{a} = 2 \times 10^{-5})$ solution containing $0.1 M - HCl$ is

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

  1. $2 \times 10^{-1}$

  2. $2 \times 10^{-3}$

  3. $2 \times 10^{-4}$

  4. $4.4 \times 10^{-3}$

Show answer
Correct Option: C

The hydrolysis of an ester was carried out with 0.1 M $H _2SO _4$ and 0.1 M HCl separately. Which of the following expressions between the rate consists is expected? The rate expression being rate = $k[H^{\oplus}][ester]$ 

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

  1. $k _{HCl}\, =\, k _{H _2SO _4}$

  2. $k _{HCl}\, >\, k _{H _2SO _4}$

  3. $k _{HCl}\, <\, k _{H _2SO _4}$

  4. $k _{ H _2SO _4}\, =\, k _{HCl}$

Show answer
Correct Option: B
Explanation:

$[H _2SO _4]\, =\, 0.1\, M\, =\, 0.1\, \times\, 2\, =\, 0.2 N$

$[HCl]$ = $0.1 N$

In case of $[H _2SO _4]$ 

$r _1\, =\, k[H^{\oplus}][Ester]$ 

$\displaystyle k _{H _2SO _4}\, = \frac{r _1}{2\, N\, \times\, [Ester]}$ 

In case of HCl, $r _1\, =\, k[H^{\oplus}]\, [Ester]$ 

$\displaystyle k _{HCl}\, =\, \frac{r _2}{1\, N\, [Ester]}$ 

Hence $K _{HCl}\, >\, K _{H _2SO _4}$

$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$

If $\displaystyle -\, \frac{d[N _2O _5]}{dt}\, =\, k _1[N _2O _5]$

$\displaystyle \frac{d[NO _2]}{dt}\, =\, k _2[N _2O _5]$

$\displaystyle \frac{d[O _2]}{dt}\, =\, k _3[N _2O _5]$
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. $k _1\, =\, k _2\, =\, k _3$

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

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

  4. None

Show answer
Correct Option: B
Explanation:

As we know,
for a reaction:
$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$
$\displaystyle -\, \frac{1}{2}\, \frac{d[N _2O _5]}{dt}\, =\, \frac{1}{4}\, \frac{d[NO _2]}{dt}\, =\, \frac{d[O _2]}{dt}$
So
$2k _1\, =\, k _2\, =\, 4k _3$

The inversion of cane sugar proceeds with the half-life of 500 min at pH 5 for any concentration of sugar. However, if pH = 6, the half life changes to 50 min. The rate law expression for the sugar inversion can be written as:

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

  1. $r\, =\, k[sugar]^2[H]^6$

  2. $r\, =\, k[sugar]^1[H]^0$

  3. $r\, =\, k[sugar]^0[H^{\oplus}]^6$

  4. $r\, =\, k[sugar]^0[H^{\oplus}]^1$

Show answer
Correct Option: B
Explanation:

Given,
Since $t _{1/2}$ does not depends upon the sugar concentration means it is first order w.r.t [sugar]
$\therefore t _{1/2}\, \propto\, [sugar]^{1}$
$t _{1/2}\, \times\, a^{n\, -\, 1}\, =\, k$
$\displaystyle \frac{(t _1/2) _1}{(t _{1/2) _2}}\, =\, \frac{[H^{\oplus}] _1^{1-n}}{[H^{\oplus}] _2^{1-n}}$

$\displaystyle \frac{500}{50}\, =\, \left ( \frac{10^{-5}}{10^{-6}}\right )^{1-n}$

10 = $(10)^{1-n}\, \Rightarrow\, n\, =\, 0$