Chemical Kinetics - Class XII

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

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.

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.

A $\rightarrow$ product

The reaction is zero order because the rate of change of concentration is constant (1000M to 100M in 100s, rate = 9 M/s). The rate constant k for a zero-order reaction has units of concentration per time, which is M s^-1.

As we know,
for second order reaction:
$r = k[A]^2$
so unit of rate constant is $Mol^{-1}, L, s^{-1}$.

The unit of rate constant is ${ mol }^{ { -2 }/{ 3 } }\ { litre }^{ { 2 }/{ 3 } }\ { time }^{ 2/3 }$ and it does not match with any of the A, B, C options. So option D is correct

For a second order reaction we have 

Its a second Order reaction.

For the third order reaction, the general reaction can be written as,

Rate constant of forward reaction $(K _f)=1.1\times 10^{-2}\ min^{-1}$

second order
$2A \rightarrow product$
$rate=K[A]^{2}$
$k=\frac{rate}{[A]^{2}}$
$=\frac{concentration}{time (concentration )^{2}}$
$=\frac{1}{time . concentration}$
So it involves both time & concentration.

Unit of Rate Constant for a second order reaction is ${mol}^{-1}L{s}^{-1}$.

Its a second Order Reaction. Rate is given as X

A bimolecular elementary reaction may or may not be a Second Order Reaction whereas a Second Order reaction must be a bimolecular elementary reaction.

The rate of solvolysis of Tert - butyl bromide will be maximum in Water.

Inversion of Cane Sugar to produce glucose and fructose is a Pseudo uni-molecular Reaction.

For first order, $M/sec=k[M].$
$\therefore k=sec^{-1}$
For zero order, $M/sec = k[M]^0.$
$\therefore k=M sec^{-1}.$

$r=K[A][B]^{2/3}$

From the slow step, $rate = k [A _2][B]$ ...... (1)
From the equilibrium reaction, equilibrium  constant  $K = \dfrac {[A _2]} {[A]^2}$
$[A _2]=K [A]^2$......(2)

For a general reaction of order $n$, the units of rate constant is given as $(mol/litre)^{1-n}sec^{-1}$.
So for our question, the unit of rate constant is $(mol/litre)^{-1}sec^{-1}$ so $ 1-n = -1$.

Units of k $= mole^{1-n} lit^{n-1} sec^{-1}$ , where n is the order of reaction.

(i) Rate of the reaction is given by,

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

In the mechanism, the first step is a fast equilibrium: K = [NOBr2] / ([NO][Br2]). The second step is the rate-determining step: Rate = k2[NOBr2][NO]. Substituting the equilibrium expression: Rate = k2 * K * [NO][Br2] * [NO] = k' [NO]^2 [Br2]. The order with respect to NO is 2.

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

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$ [$CH _3COOC _2H _5$] [$NaOH$]

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

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.

$\displaystyle [O _3] = \frac {P}{RT} = \frac {0.76}{760 \times 0.0821 \times 298} = 4.08 \times 10^{-5} $
$\displaystyle [O] = \frac {rate}{k \times [O _3]} = \frac {0.15}{1.5 \times 10^{7} \times 4.08 \times 10^{-5} = 2.45 \times 10^{-4}  }$
The rate of formation of oxygen under this condition is equal to twice the rate of the reaction.
It is equal to $\displaystyle 2 \times 0.15 : mol litre^{-1} sec^{-1} = 0.30 : mol litre^{-1} sec^{-1} $

As we know,
For zero order reaction, $\frac {dx}{dt}=K$.
so, unit and value of rate constant and that of rate of reaction are same for zero order.

As we know,
For nth order; unit of rate constant may be derived by
$K=\frac {rate}{[reactant]^n}$
$=Mol^{1-n} litre^{n-1} sec^{-1}$

The overall order of the reaction is second order.
Reactions are proceeded in many steps, it does not means that no. of step will be equal to order of reaction. But rate is determined only by slow step of mechanism.

As we know,

For nth order; unit of rate constant may be derived by
$rate  = k[a]^n$
$K=\frac {rate}{[reactant]^n}$.

As we know,
$k, =, [conc]^{1-n}, min^{-1}$
For third order reaction = $[mol, L^{-1}]^{1-3}, min^{-1}$
$L^2, mol^{-2}, min^{-2}$

For zero order,
R = k[A]
Unit of rate = mol $L^{-1}, t^{-1}$
Unit of zero order = mol $L^{-1}, t^{-1}$

rate = k$[A]^{3/2}[B]^{-1}$
OR = $\displaystyle \frac{3}{2}, -, 1, =, \frac{1}{2}$

If a reaction involves gaseous reactants and products, the unit of its rate is $atm\ s^{-1}$. It represents the change in pressure (in atm units) in unit time ($1$ second).

$A + B + C$ $\rightarrow$ Products
$\displaystyle r, =, - \frac{dA}{dt}, = k[A]^{1/2}[B]^{1/3}[C]^{1/4}$
so order is
$\displaystyle \frac{1}{2}, +, \frac{1}{3}, +, \frac{1}{4}, =,\frac{13}{12}$

Since C is taken in excess, so its concentration does not change. So it is not taken in rate expression of the reaction.

As we know,
for second order reaction:
$r = k[A]^2$
so unit of rate constant is $Mol^{-1}, L, s^{-1}$.

Rate $r = k[A]^{3/2}$

$r=-\dfrac { d[A] }{ dt } =K[A{ ] }^{ \cfrac { 1 }{ 2 }  }K[B{ ] }^{ \cfrac { 1 }{ 3 }  }K[C{ ] }^{ \cfrac { 1 }{ 4 }  }\ $

Units of rate constant for $n$th order = $[(mol  lit^{-1})]^{1-n} t^{-1}$
For third order reaction $n = 3$
$\therefore $ Units are $(mol  lit^{-1})^{1- 3} t^{-1} = mol\ l^{-2} lit^{-2} t^{-1}$

The rate constant of a reaction depends only on two factors
(i) Temperature and (ii) Catalyst.

Rate = k [concentration] for first order reaction.
Given $k = 3 \times 10^{-6}/sec,$ [concentration] = 0.1 M
$\therefore rate = 3 \times 10^{-6} \times 0.1  = 3 \times 10^{-7} ms^{-1}$

The unit for the rate constant of a second order reaction is $\displaystyle \displaystyle L { mol }^{ -1 }{ s }^{ -1 } $. 
For a second order reaction, $\displaystyle rate = k [A]^2 $
$\displaystyle mol { L }^{ -1 }{ s }^{ -1 }  = k (mol { L }^{ -1 })^2$
$\displaystyle k = L { mol }^{ -1 }{ s }^{ -1 }  $

Secon order rate expression is given as $r$$=$$k[A]^2$ so if we double the concentration of a then rate increases by 4 times, hence both statements are correct and statement 2 is correct explanation of statement 1.

Analyzing first and second raw,when concentration of $B$ is doubled then initial rate increased by four times,so it is increasing by square of concentration.Hence it is a second order reaction.

$\bullet \quad $If rate = ${ K\left[ A \right]  }^{ 2 }$ for a reaction.

Since rate constant is given by $k[A]^2[B]$.
Where 1+2=3, so it is probably third order reaction.
So rate constant should have units of $L^2 mol^{-2} s^{-1} \implies M^{-2}s^{-1}$

Let the concentration at $t=0$ be $[{ A } _{ o }]$ & at $t=t$ be $ [{ A } _{ t }]$

Since it is second order reaction as clear by rate equation so unit of $k$ is given by $M^{-1} s^{-1}$ or $L\space mol^{−1}s^{−1}$

$K=\frac{dx}{dt[A]} $ for Ist order = $sec^{−1}$
$ K = \frac{dx}{dt}$ for zero order = $mol \space litre^{−1}sec^{−1}$

Hence A is the correct answer.

Concentration change in '$B$' does not change half life, it means the reaction is first order with respect to $B$. When concentration of only '$A$' is doubled, the rate of reaction becomes double, thus order with respect to $A$ will also be one.
Overall order of reaction $=2$
Unit of rate constant $=L$ ${ mol }^{ -1 }{ s }^{ -1 }$

The units of rate of reaction is $Ms^{-1}$.

Overall order=$\frac{2}{3}+1=\frac{5}{3}$

Order of Reaction is given by the slowest step of Reaction. In slowest step we 

For the following reaction, Rate is defined as

The rate law is given as Rate = k[P_NO]^2 [P_H2]. Since the initial concentration of NO (1000 mm Hg) is much larger than the concentration of H2 (10 mm Hg), the concentration of NO remains effectively constant throughout the reaction. Thus, the reaction behaves as pseudo-first order with respect to H2.

Saponification of ethyl acetate with equal concentrations of ester and alkali is a classic second-order reaction. The data provided would show a linear relationship for 1/[concentration] vs time, confirming second-order kinetics.

For $1.5$ order rate law the units are $molL^{-1}s^{-1}$ for the rate while the [rate constant]$=\cfrac{molL^{-1}s^{-1}}{mol^{3/2}L^{-3/2}}$

$ r= K\left [ A \right ]^{n}$

$K = \dfrac{r}{\left [ A \right ]^{n}}= \dfrac{mole \ l^{-1} \ sec^{-1}}{mole^{n} \ l ^{-n}}$ $= mole^{1-n} 1^{n-1} sec^{-1}$

For a first order reaction; rate law can be wriiten as; $r = k[A]^{1}$
Therefore k = $\dfrac{r}{[A]} = \dfrac{mol \times l^{-1} \times  s^{-1}}{mol \times l^{-1}}$ = $s^{-1}$ where concentration of $A =$ moles per litre and rate of reaction; r = change in concentration of $A$ with time.

$A+B\longrightarrow C+D$

As $Rate=k(P _{CH _3OCH _3})^{\frac {3}{2}}$
$bar/min=k(bar)^{\frac {3}{2}}$
$\therefore$ unit of k$=bar^{-\frac {1}{2}}min^{-1}$

option A and C are correct as the sum of their exponents equals to 2

The rate constant of a second order reaction is $10^{-2} lit.mole ^{-1}.sec^{-1}$.
$1L=1000cc$

For the zero-order reaction, the time taken for the decomposition of the reactant is independent of initial concentration which is the case here. 

Option (A) is correct.  $H _2SO _4$ is stronger than $HCl$.
Relative strength in favour of $H _2SO _4$
$=\frac {\text {rate constant of reaction catalysted by }H _2SO _4}{\text {rate constant of reaction catalysted by HCl}}$.

 The order of reaction is 2.

Rate equation is as follows:
$Older Rate = k [A]^{2}$                       equation -1
This shows that reaction is $2nd $ order.
So if we double the concentration of the A , then new reaction rate can be written as follows:
$New Rate = k [2A]^{2}$
$New Rate = k\times 4[A]^{2}$
$New Rate = 4 k [A]^{2}$                     equation -2
Now, comparing equation -1 and 2 we get :
New Rate = $4\times Older Rate$

The unit of the rate constant is s^-1. This unit is characteristic of a first-order reaction, where the rate constant k has dimensions of time^-1.

Given that for a second order reaction