Expt No	Initial conc. (mol L-1)	Initial rate (mol L-1 s-1)
	[A]	[B]
1	0.03	0.03	0.3 X 10-4
2	0.06	0.06	1.2 X 10-4
3	0.06	0.09	2.7 X 10-4

For a reaction in which A and B form C, the data obtained from three experiments is given above. What is the rate equation of the reaction and what is the value of the rate constant?

Hint: rate of reaction depends on the concentration of one or more of the reactants. The rate equation is of the form Rate=k${\left[ A \right]^x}{\left[ B \right]^y}$ where k is the rate constant or velocity constant. unit of rate constant
K=$mol{e^{1 - n}}litr{e^{n - 1}}tim{e^{ - 1}}$ (where n is the order of the reaction )

Complete step by step answer:
Order of a reaction may be defined as the sum of powers to which concentration terms are raised in the rate law expression. It may be zero, fractional or integer. First calculate the rate of equation and order of reaction. We are given with initial concentration and value of rate constant according to every observation. The rate law for the three experiments can be written as
For the reaction
The rate equation for the reaction r = K ${\left[ A \right]^x}{\left[ B \right]^y}$
Where x is the order wrt A and y is the order wrt B
For the first experiment rate equation is $r_1$ = $0.3 \times {10^{ - 4}} = k{\left[ {0.03} \right]^x}{\left[ {0.03} \right]^y} = k{[0.03]^{x + y}}$
For the second experiment rate equation is $r_2$ = $1.2 \times {10^{ - 4}} = k{[0.06]^x}{[0.06]^y}$
For the third experiment rate equation is $r_3$ = $2.7 \times {10^{ - 4}} = {[0.06]^x}{[0.09]^y}$
On dividing the above equations we get
$\dfrac{{{r_2}}}{{{r_1}}} = \dfrac{{1.2 \times {{10}^{ - 4}}}}{{0.3 \times {{10}^{ - 4}}}} = {\left[ {\dfrac{{[0.06]}}{{[0.03]}}} \right]^{x + y}} = {2^{x + y}}
\dfrac{{{r_3}}}{{{r_1}}} = \dfrac{{2.7 \times {{10}^{ - 4}}}}{{0.3 \times {{10}^{ - 4}}}} = \dfrac{{{{[0.06]}^x}{{[0.09]}^y}}}{{{{[0.03]}^{x + y}}}} $
$\dfrac{{{r_2}}}{{{r_1}}}$=4= ${2^{x + y}}$, $\dfrac{{{r_3}}}{{{r_1}}}$=9=${2^x}{3^y}$
Now dividing the above two equations will give $r_2$/$r_3$ which is
 $\dfrac{{{r_3}}}{{{r_2}}} = \dfrac{9}{4} = \dfrac{{{2^x}}}{{{2^x}}} \times {\left[ {\dfrac{3}{2}} \right]^y}$
hence we observe the value of y=2. Put the value of y in eqn (r2/r1) we get
 $4 = {2^x}{2^y} = {2^x}{2^2}$
$1 = {2^x} $
Therefore values of x=0 and y =2 therefore the reaction is of second order. Now we can calculate the rate of the equation using rate law
$r = k{\left[ A \right]^0}{\left[ B \right]^2}$

$r = k{[B]^2} = 0.3 \times {10^{ - 4}} = k{[0.03]^2} $
$k = \dfrac{{0.3 \times {{10}^{ - 4}}}}{{{{[0.03]}^2}}} = {10^{ - 3}}litr{e^1}mo{l^{ - 1}}tim{e^{ - 1}} $

Hence the order of the reaction is 2 and one can put the value of n in the unit rate constant and the rate of the reaction is ${10^{ - 3}}litr{e^1}mo{l^{ - 1}}tim{e^{ - 1}}$

Note:
We can therefore conclude that the process of drawing a picture on the sheet could be reversible or irreversible depending upon the way the picture is drawn.