Question

In a certain gaseous reaction between $X$ and $Y$ , $X + 3Y \to X{Y_3}$ . The initial rates are reported as follows:-

$[X]$ (M)	$[Y]$ (M)	Rates (M/S)
$0.1$	$0.1$	$0.002$
$0.2$	$0.1$	$0.002$
$0.3$	$0.2$	$0.008$
$0.4$	$0.3$	$0.018$

(A) $r = K[X]{[Y]^3}$
(B) $r = K{[X]^0}{[Y]^2}$
(C) $r = K[X][Y]$
(D) $r = K{[X]^0}{[Y]^3}$

Answer 1

Hint : Before heading ahead, let us recall first chemical kinetics. Chemical Kinetics is the study of the degree of rate of the reaction and factors affecting the rates of reactions. Whereas, rate of any reaction is defined as the change in the concentrations of the reactants and products given that there is no net change in the total concentration of the system.

Complete Step By Step Answer:
Rate of any given reaction is defined as the changes occurring in the concentrations of the reactants and products given that there is no net change in the total concentration of the system. That means in hypothetical manner that if $R \to P$ then the rate is calculated by $r = - \dfrac{{\vartriangle [R]}}{{\vartriangle t}} = \dfrac{{\vartriangle P}}{{\vartriangle t}}$ $\forall$ the sign have their usual meanings.
One can determine the rate law, the rate constant and the order of the reaction by two ways: -(i) The graphical method which uses the differential rate law equations. (ii) Initial rate method, in this method the stoichiometry is used in the calculations of the rate constant. (iii) Integrated rate law method, (iv) Half-life method, (v) Ostwald isolation method.
In the Initial rate method different experiments are carried out with variation in the concentration of the reactants and the initial rate of formation of the product is given. Then by the manipulation of the data we are supposed to calculate the rate law equation (in accordance to the question)
The given equation is $X + 3Y \to X{Y_3}$ and the concentration of reactants and the rate is as follows:-

$[X]$ (M)	$[Y]$ (M)	Rates (M/S)
$0.1$	$0.1$	$0.002$
$0.2$	$0.1$	$0.002$
$0.3$	$0.2$	$0.008$
$0.4$	$0.3$	$0.018$

Thus, the equation is $r = k{[X]^\alpha }{[Y]^\beta }$ .
Taking first and second rates the equation becomes
$0.002 = k{[0.1]^\alpha }{[0.1]^\beta }\, - (1)$
$0.002 = k{[0.2]^\alpha }{[0.1]^\beta }\, - (2)$
Now we need to calculate the values of the $\alpha$ and $\beta$ to get the rate law equation. Now dividing the above equations: -
$\dfrac{{0.002}}{{0.002}} = \dfrac{{k{{[0.1]}^\alpha }{{[0.1]}^\beta }}}{{k{{[0.2]}^\alpha }{{[0.1]}^\beta }}}$
$\Rightarrow 1 = {\left[ {\dfrac{1}{2}} \right]^\alpha }$ .
Anything raised to power $0$ is always equal to $1$ . Thus if $\alpha$ is equal to $0$ then $\dfrac{1}{2}$ will be equal to $1$ .
${[1]^0} = {\left[ {\dfrac{1}{2}} \right]^\alpha }$
$\Rightarrow 0 = \alpha$
Now taking the third and the fourth equation into account, the equations becomes as follows: -
$0.008 = k{[0.3]^\alpha }{[0.2]^\beta } - (3)$
$0.018 = k{[0.3]^\alpha }{[0.3]^\beta } - (4)$
Now we need to calculate the values of the $\alpha$ and $\beta$ to get the rate law equation. Now dividing the above equations, but $\alpha = 0$
$\dfrac{8}{{18}} = {\left[ {\dfrac{2}{3}} \right]^\beta }$
$\Rightarrow \beta = 2$
Thus putting the values of $\alpha$ and $\beta$ the equation becomes as $r = k{[X]^0}{[Y]^2}$ .

Note :
It is to remember that the constants $\alpha$ and $\beta$ are unit less. In case of rate of a reaction a negative sign is used to denote the decrease in the concentration of the reactants and positive sign is used to denote the formation of the product. However, the kinetics follows the le-chatelier’s principle that the rate of reaction increases with increase in concentration of the reactants and decreases when the concentration of reactants decreases.