Reactions of Various Orders

Pseudo first order reaction

Reaction whose actual order is different from that expected using rate law expression are called pseudo – order reactions, eg.

$RCl+H_{2}O\to ROH+HCl$

Expected rate law :

$Rate=K[RCl][H_{2}O]$, Expected order = 1 + 1 = 2

Actual rate law :

$Rate=K^{\prime }[RCl]$, Actual order = 1

Water is taken in excess; therefore, its concentration may be taken constant. This reaction is therefore, pseudo first order. Similarly, the acid catalysed hydrolysis of ester, viz,

$RCOO{R}^{\prime }+H_{2}O\rightleftharpoons RCOOH+R^{\prime }OH$

follows first order kinetics. $Rate=K[RCOO{R}^{\prime }]$

It is also a pseudo – first order reaction.

Illustration 1. If the decomposition of nitrogen (V) oxide

$2N_2{O}_5\to 4N{O}_2+O_2$

Following a first order kinetics

(i) Calculate the rate constant for 0.04 M solution, if the instantaneous rate is

$1.4\times {10}^{-6}mol{\ell }^{-1}{s}^{-1}$.

(ii) Also calculate the rate of reaction when the concentration of N₂O₅ is 1.20 M.

(iii) What concentration of N₂O₅ would give a rate of$2.45\times {10}^{-5}mol{\ell }^{-1}{s}^{-1}$?

Solution: (i) As the given rate is of the first order, therefore, rate = $K\left[N_2{O}_5\right]$

$K=\frac{rate}{\left[N_2{O}_5\right]}=\frac{1.4\times {10}^{-6}}{0.04}=3.5\times {10}^{-5}{s}^{-1}$

(ii) Now if the concentration of N₂O₅ is 1.2 M, then

rate = $K\left[N_2{O}_5\right]$

\begin{align}  =3.5\times {{10}^{-5}}\times 1.20 \\  =4.2\times {{10}^{-5}}\,\,mol\,\,{{\ell }^{-1}}{{s}^{-1}} \\ \end{align}

(iii) To obtain concentration of N₂O₅ when the rate is $2.45\times {10}^{-5}mol{\ell }^{-1}{s}^{-1},$

$\left[N_2{O}_5\right]=\frac{Rate}{K}=\frac{2.45\times {10}^{-5}}{3.5\times {10}^{-5}}=0.7mol{\ell }^{-1}$

or 0.7 M

Difference between order and molecularity

(i) Order is an experimental property while molecularity is the theoretical property.

(ii) Order concerns with kinetics (rate law) while molecularity concerns with mechanism.

(iii) Order may be any number, fraction, integral or even zero whereas molecularity is always an integer expecting zero.

REACTIONS OF VARIOUS ORDERS

(i) Zero order reactions

A reaction is said to be zero order if its rate is independent of the conc. of the reactants, consider the general reactions: $A\to \mathrm{Pr}oducts$

If it is reac. of zero order

$Rate=\frac{-d[A]}{dt}=K{[A]}^o=K$ or d[A] = -K dt

Integrating both sides, we get

[A] = -Kt + I ………………….. (i)

where I is a constant of integration

At t = 0, $[A]={[A]}_o$

$\therefore {[A]}_o=I$

Substituting this value of I in equation (i), we get

$[A]=-Kt+{[A]}_o......................(ii)$

or $Kt={[A]}_o-[A]$

$K=\frac{1}{t}\left\{{\left[A\right]}_o-\left[A_t\right]\right\}................(iii)$

Some important characteristics of reaction of zero order.

(ii) Any reaction of zero order must obey equation (ii). As it is a equation of straight line (y = mx + c), the plot of [A] versus t will be a straight line with slope $=-K$and intercept on the conc. axis $={[A]}_o$ as shown in figure

(iii) Half reaction period:

Half life period $(t_{1/2})$ is the time in which half of the substance has reacted.

When $[A]=\frac{[A_o]}{2},t=t_{1/2}$ substituting those values in equation (iii), we get

$t_{1/2}=\frac{1}{K}\left[{[A]}_o-\frac{{[A]}_o}{2}\right]=\frac{{[A]}_o}{2K}$

$t_{1/2}=\frac{{[A]}_o}{2K}$

Unit of K:

$K=\frac{molarconc.}{Time}=\frac{mol{L}^{-1}}{time}=mole{L}^{-1}tim{e}^{-1}$

Examples of zero order reaction

(i) Photochemical reaction between hydrogen and chlorine

$H_{2(g)}+C{l}_{2(g)}\stackrel{h\nu }{\to }2HC{l}_{(g)}$

(ii) Decomposition of N₂O on hot platinum surface :

$N_{2}O\to N_2+\frac{1}{2}{O}_2$

(iii) Decomposition of NH₃ in presence of molybdenum or tungsten

$2N{H}_3\stackrel{[Mo]}{\to }{N}_2+3H_2$

Illustration 2. Write the integrated rate equation for a zero order reaction.

Solution: $k=\frac{{\left[R\right]}_0-\left[R\right]}{t}$

where k = rate constant

[R]₀ = initial concentration of the reactant

[R] = concentration of the reactant at time, t.

First order reaction

A reaction is said to be of the first order if the rate of the reaction depends upon one conc. term only.

Consider the reaction

$A\to \mathrm{Pr}oducts$

Let 'a' be the conc. of A at the start and after time t, the conc. becomes (a – x), i.e., x has been changed into products. The rate of reaction after time 't' is given by the expression

$\frac{dx}{dt}=K(a-x)$

or $\frac{dx}{(a-x)}=Kdt$

Upon integration of above equation,

$\int \frac{dx}{(a-x)}=K\int dt$ or $-{\log }_e(a-x)=Kt+c$

where c is integration constant

when t = 0, x = 0

$\therefore c=-{\log }_{e}a$

Putting the value of 'c',

$-{\log }_e(a-x)=Kt-{\log }_{e}a$

$or{\log }_{e}a-{\log }_e(a-x)=Kt$

${\log }_e\frac{a}{(a-x)}=Kt$

$K=\frac{2.303}{t}{\log }_{10}\frac{a}{(a-x)}$

If the initial concentration is ${[A]}_o$ and the concentration after time t is [A], then putting $a={[A]}_o$ and (a – x) = [A], equation becomes

$K=\frac{2.303}{t}\log \frac{{[A]}_o}{[A]}$

This equation can be written in the exponential form as

$\frac{{[A]}_o}{{[A]}_t}=e^{kt}or\frac{[A]}{{[A]}_o}=e^{-Kt}$ or $[A]={[A]}_o{e}^{-Kt}$

Illustration 3. Give any one example of

(i) zero order reaction

(ii) first order reaction

Solution: (i) $2N{H}_3\left(g\right)\stackrel{Pt}{\to }{N}_2\left(g\right)+3H_2\left(g\right)$

r = k[NH₃]^°

(ii) $N{H}_{4}N{O}_2\left(s\right)\to N_2\left(g\right)+H_{2}O\left(\ell \right)$

r = k[NH₄NO₂]

Some important characteristics of first order reaction

(a) A change in conc. unit will not change the numerical value of K. Let the new unit be n times the first one.

So $K=\frac{2.303}{t}{\log }_{10}\frac{na}{n(a-x)}=\frac{2.303}{t}{\log }_{10}\frac{a}{(a-x)}$

(b) Graphical method:

$K=\frac{2.303}{t}{\log }_{10}\frac{{[A]}_o}{[A]}$

${\log }_{10}{[A]}_o-{\log }_{10}[A]=\frac{Kt}{2.303}$

${\log }_{10}[A]=\frac{-Kt}{2.303}+{\log }_{10}{[A]}_o$

Comparing it with y = mx + c

Slope $m=\frac{-K}{2.303},\mathrm{int}erceptc={\log }_{10}{[A]}_o$

(c) Half life period:

The time taken for any fraction of the reaction to be completed is independent of the initial concentration.

$t=\frac{2.303}{K}{\log }_{10}\frac{{[A]}_o}{[A]}$

when the half reaction is completed $[A]=\frac{{[A]}_o}{2}$

$t_{1/2}=\frac{2.303}{K}\log \frac{{[A]}_o}{\frac{{[A]}_o}{2}}=\frac{2.303}{K}\log 2$

$t_{1/2}=\frac{0.6932}{K}$

Half – Life of a n^th Order Reaction:

To find out the t_½ for a n^th order reaction where n ¹1.

$\therefore$ $\frac{-d\left[A\right]}{dt}=k{\left[A\right]}^n$ $\Rightarrow$ x$\frac{-\text{d}\left[\text{A}\right]}{{\left[\text{A}\right]}^{\text{n}}}=\text{kdt}$. $\Rightarrow$$-\underset{\left[{\text{A}}_{\text{o}}\right]}{\overset{{}^{\left[A_o\right]}/{}_2}{\int }}\frac{\text{d}\left[\text{A}\right]}{{\left[\text{A}\right]}^{\text{n}}}=k\underset{\text{0}}{\overset{{\text{t}}_{\text{1/2}}}{\int }}\text{dt}$.$\underset{{}^{\left[A_o\right]}/{}_2}{\overset{\left[{\text{A}}_{\text{o}}\right]}{\int }}{\left[\text{A}\right]}^{-\text{n}}\text{d}\left[\text{A}\right]=k{t}_{\text{1/2}}$$or{\left[\frac{{\left[\text{A}\right]}^{\text{1}-\text{n}}}{\text{1}-\text{n}}\right]}_{{}^{{\text{A}}_{\text{0/2}}}}^{{\text{A}}_{\text{0}}}=k{\text{t}}_{\text{1/2}}$

$\Rightarrow$ $\frac{1}{1-n}$ $\left({\left[\text{A}\right]}_{\text{0}}^{\text{1}-\text{n}}-{\left[\frac{{\left[\text{A}\right]}_{\text{0}}}{\text{2}}\right]}^{\text{1}-\text{n}}\right)=k{\text{t}}_{\text{1/2}}$ $\Rightarrow$ $\frac{{\left[\text{A}\right]}_{\text{0}}^{\text{1}-\text{n}}}{\text{1}-\text{n}}$ $\left[\text{1}-{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{1}-n}\right]=\text{k}{\text{t}}_{\text{1/2}}$

$\Rightarrow$ $\frac{\text{1}}{\left(\text{1}-\text{n}\right){\left[{\text{A}}_{\text{o}}\right]}^{\text{n}-\text{1}}}\left[\text{1}-{\text{2}}^{\text{n-1}}\right]=\text{k}{\text{t}}_{\text{1/2}}$ $\Rightarrow$ $\frac{\left(2^{n-1}-1\right)}{\left(n-1\right){\left[A_o\right]}^{n-1}}=k{t}_{1/2}$ ( order n$\ne$1)

Therefore for a nth order reaction, the half life is inversely related to the initial concentration raised to the power of (n–1).

Note:

It can be noted that for a zero order reaction t_1/2 =$\frac{{\left[A\right]}_{\text{o}}}{\text{2k}}$.

Illustration 4. Define half-life of a reaction.

Solution: The half-life of a reaction is defined as the time required for the reaction to be 50% complete.

It is also defined as the time during which the concentration of reactant is reduced to one half of its initial concentration.

Examples of first order reaction

(i) Decomposition of H₂O₂ in aqueous solution

$H_2{O}_2\to H_{2}O+\frac{1}{2}{O}_2$

(ii) $2HN{O}_3\to 2NO+H_{2}O+\frac{3}{2}{O}_2$

(iii) $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$

Unit of rate constant

$Unitofratecons\tan t={\left[\frac{1}{unitofconcentration}\right]}^{n-1}\times tim{e}^{-1}$

$={\left[\frac{1}{mole/lit}\right]}^{n-1}\times tim{e}^{-1}$

$={\left[\frac{lit}{mole}\right]}^{n-1}\times tim{e}^{-1}$

where n = order of reaction

Illustration 5. The t_1/2 of a first order reaction is 60 minutes. What percentage will be left after 240 minutes?

Solution: No. of half life = $\frac{240}{60}=4$

$\therefore$ Amount left [A] = $\frac{{\left[A\right]}_0}{16}$

= 0.0625 of [A]₀ = 6.25 %

SOME COMPLEX FIRST ORDER REACTIONS

Parallel Reactions:

In such reactions (mostly organic) a single reactant gives two products B and C with different rate constants. If we assume that both of them are first order, we get.

$\frac{-d\left[A\right]}{dt}=k_1\left[A\right]+k_2\left[A\right]=\left(k_1+k_2\right)\left[A\right]$ … (1)

$\frac{d\left[B\right]}{dt}=k_1\left[A\right]$ … (2)

and$\frac{d\left[C\right]}{dt}=k_{\text{2}}\left[\text{A}\right]$ …(3)

Let us assume that in a time interval, dt, x moles / lit of B was produced and y moles / lit of C was produced.

$\therefore$$\frac{d\left[B\right]}{dt}=\frac{x}{dt}$ and $\frac{d\left[C\right]}{dt}=\frac{y}{dt}$$\therefore$ $\frac{\frac{d\left[B\right]}{dt}}{\frac{d\left[C\right]}{dt}}$ =;$\frac{x}{y}$. We can also see that from (2) and (3), $\frac{\frac{d\left[B\right]}{dt}}{\frac{d\left[C\right]}{dt}}$$=\frac{k_1}{k_2}$$\therefore$$\frac{x}{y}=\frac{k_1}{k_2}$. This means that irrespective of how much time is elapsed, the ratio of concentration of B to that of C from the start (assuming no B and C in the beginning) is a constant equal to k₁/k₂.

Sequential Reactions:

A$\stackrel{k_1}{\to }$B$\stackrel{k_2}{\to }$C. In this A decomposes to B which in turn decomposes to C.

$\therefore$ $\frac{-\text{d}\left[\text{A}\right]}{\text{dt}}$ = k₁[A] …(1)

$\frac{\text{d}\left[\text{B}\right]}{\text{dt}}$= k₁ [A] –k₂[B] …(2)

$\frac{\text{d}\left[\text{C}\right]}{\text{dt}}$= k₂ [B] …(3)

$\therefore$t_max =$\frac{1}{K_1-K_2}In\frac{K_1}{K_2}$ … (4)

Substituting equation (4) in equation (3)

B_max = ${\left[A\right]}_o{\left[\frac{k_2}{k_1}\right]}^{k_2/k_1-k_2}$