Tag: chemical kinetics

Questions Related to chemical kinetics

Identify the reaction order from each of the following rate constants.
(i) $k=2.3 \times 10^{-5} L \quad mol^{-1} \quad s^{-1}$
(ii) $k=3 \times 10^{-4} \quad s^{-1}$

  1. (i) First order (ii) Second order

  2. (i)Second order (ii) First order

  3. (i) Zero order (ii) First order

  4. None


Correct Option: B
Explanation:

(i) Rate of the reaction is given by,

$Rate=k\times [A]^x$, where x is the order of reaction
$(\dfrac{mol}{L})=2.3\times 10^{-5}L(mol^{-1})(s)^{-1}[A]^x$
$({\dfrac{mol}{L}})^2=constant(s^{-1})[\dfrac{mol}{L}]^x$
Therefore, $x=2$ and the reaction is second order
(ii)$Rate=k\times [A]^x$, where x is the order of reaction
$(\dfrac{mol}{L})=3\times 10^{-4}(s)^{-1}[A]^x$
$({\dfrac{mol}{L}})=constant(s^{-1})[\dfrac{mol}{L}]^x$
Therefore, $x=1$ and reaction is first order

Units of rate constant for the first and zero order reactions in terms of molarity M units are respectively:

  1. $sec^{-1},\;Msec^{-1}$

  2. $sec^{-1},\;M$

  3. $Msec^{-1},\;sec^{-1}$

  4. None of the above


Correct Option: A
Explanation:

The unit of the rate constant K is $\left (\text {Molarity} \right )^{1-n}time^{-1}.$


Here, n is the order of the reaction.

For first order reaction, $n=1.$

The unit of the rate constant K is $\left ( \text{Molarity} \right )^{1-1}sec^{-1}=sec^{-1}.$

For zero order reaction, $n=0.$

The unit of the rate constant K is $\left ( \text{Molarity} \right )^{1-0}sec^{-1}=\left ( \text{Molarity} \right )sec^{-1}=\left (M \right )sec^{-1}.$

Consider the reaction, $2A + B \rightarrow$ Products, When concentration of B alone was doubled, the rate did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is:

  1. $s^{-1}$

  2. $L\ mol^{-1}s^{-1}$

  3. Unitless

  4.  $mol\ L^{-1}s^{-1}$


Correct Option: A
Explanation:
When the concentration of B alone was doubled, the rate did not change. Hence the reaction is zero order in B.

When the concentration of A alone was doubled, the rate increased by two times. Hence, the reaction is of first-order in A.

$r=k\left [ A \right ]^{1}\left [ B \right ]^{0}$

The overall order of the reaction is 1.

For the first-order reaction, the unit of k is $sec^{-1}$.

Hence, the correct option is $\text{A}$

The following mechanism has been proposed for the reaction of $NO$ with $\displaystyle Br _{2}$ to form $NOBr$


$NO(g)+Br _{2}(g)\rightleftharpoons NOBr _{2}(g)$
$NOBr _{2}(g)+NO(g)\rightarrow 2NOBr(g)$


If the second step is the rate determining step,the order of the reaction with respect to $NO(g)$ is:
  1. 2

  2. 1

  3. 0

  4. 3


Correct Option: A

Taking the reaction $x+2y\rightarrow$ prodcuts, to be of second order, which of the following are the rate law expressions for the reaction :


(I) $\cfrac{dx}{dt}=K[x][y]$  (II) $\cfrac{dx}{dt}=K[x]{[y}]^{2}$

(III) $\cfrac{dx}{dt}=k{[x]}^{2}$  (IV) $\cfrac{dx}{dt}=K[x]+K{[y]}^{2}$

  1. I only

  2. I and III only

  3. I and II only

  4. I and IV only


Correct Option: C

The rate of formation of ${SO} _{3}$ in the reaction $2{SO} _{2}+{O} _{2}\rightarrow 2{SO} _{3}$ is $100g{min}^{-1}$. Hence, rate of disappearance of ${O} _{2}$ is

  1. $50g{min}^{-1}$

  2. $100g{min}^{-1}$

  3. $20g{min}^{-1}$

  4. $40g{min}^{-1}$


Correct Option: A

Reaction $A+B\longrightarrow C+D$ follows rate law, $r=k{ \left[ A \right]  }^{ 1/2 }{ \left[ B \right]  }^{ 1/2 }$ starting with $1M$ of $A$ and $B$ each. What is the time taken for concentration of $A$ become $0.1M$?
[Given $2.303\times { 10 }^{ -2 }sec^{ -1 }$].

  1. $10sec$

  2. $100sec$

  3. $1000sec$

  4. $434sec$


Correct Option: B
Explanation:

The rate constant is $2.303\times { 10 }^{ -2 }sec^{ -1 }$. The overall reaction order is 1.


$t = \dfrac {2.303}{k}log \dfrac {a}{a-x}$

$t = \dfrac {2.303}{2.303 \times 10^{-2}}log \dfrac {1}{0.1}=100$

Hence, the time taken to consume 90% of concentration is 100 sec.

The reaction, $CH _3COOC _2H _5+NaOH\rightarrow CH _3COONa+C _2H _5OH$ is:

  1. bimolecular reaction

  2. second order reaction

  3. third order reaction

  4. none of these


Correct Option: A,B
Explanation:

The reaction, $CH _3COOC _2H _5+NaOH\rightarrow CH _3COONa+C _2H _5OH$ is bimolecular and second order reaction,as it contains two diiferent types of molecules.
Rate = $ k$ [$CH _3COOC _2H _5$] [$NaOH$]

For the reaction: $2NO+Cl _2\rightarrow 2NOCl$, the following mechanism was proposed on the basis of experimental observation.
$NO+Cl _2\overset {Fast}{\rightleftharpoons}NOCl _2$
$NOCl _2+NO\xrightarrow {Slow}2NOBr$
The order of reaction is:

  1. 2nd order

  2. pseudo 2nd order

  3. 3rd order

  4. first order


Correct Option: C
Explanation:

From the slow step, the rate law for the reaction is rate $\displaystyle = k' [NOCl _2][NO]$.....(1)


 $\displaystyle NOCl _2$ is an intermediate and its concentration is obtained from fast step (equilibrium step)

 $\displaystyle [NOCl _2] = K[NO][Cl _2]$......(2)

Substitute (2) in (1)
 $\displaystyle rate = k'K [NO][Cl _2][NO]$

Hence, rate $\displaystyle = k [NO]^2[Cl]$

The reaction is of second order in NO and first order in $\displaystyle Cl _2$. The overall reaction order is  $\displaystyle 2 + 1 = 3$

The unit of rate constant for a given reaction is $M^{1-n}L^{n-1}t^{-1}$ where n is order of reaction.

  1. True

  2. False


Correct Option: A
Explanation:

The statement is true. The unit of rate constant for a given reaction is  $M^{1-n}L^{n-1}t^{-1}$ where $n$ is order of reaction.It is generally form for expressing rate constant. Rates are usually given in concentrations units over time units. The units of the rate constants depend upon the order of the reaction. The concentration is molarity and time is in seconds.