Question

In a certain gaseous reaction between A and B, ${\text{A + 3B}} \to {\text{A}}{{\text{B}}_{\text{3}}}$ . The initial rates are reported as follows: The rate law is ${\text{r = k}}{\left[ {\text{A}} \right]^{\text{x}}}{\left[ {\text{B}} \right]^{\text{y}}}$ , value of ${\text{x + y}}$ is:

$\left[ {\text{A}} \right]$	$\left[ {\text{B}} \right]$	Rate
${\text{0}}{\text{.1M}}$	${\text{0}}{\text{.1M}}$	${\text{0}}{\text{.002M}}{{\text{s}}^{{\text{ - 1}}}}$
${\text{0}}{\text{.2M}}$	${\text{0}}{\text{.1M}}$	${\text{0}}{\text{.002M}}{{\text{s}}^{{\text{ - 1}}}}$
${\text{0}}{\text{.3M}}$	${\text{0}}{\text{.2M}}$	${\text{0}}{\text{.008M}}{{\text{s}}^{{\text{ - 1}}}}$
${\text{0}}{\text{.4M}}$	${\text{0}}{\text{.3M}}$	${\text{0}}{\text{.018M}}{{\text{s}}^{{\text{ - 1}}}}$

Answer 1

Hint: The rate law for any reaction cannot be predicted by simply looking at the balanced chemical equation. It has to be determined experimentally.
The rate law is actually expressed in terms of molar concentrations of the reactants with each term raised to some power, which may or may not be the same as the stoichiometric coefficient of that reactant in the balanced chemical equation.

Complete step by step answer:
Given that the gaseous reaction between A and B is ${\text{A + 3B}} \to {\text{A}}{{\text{B}}_{\text{3}}}$ .
The rate law is given to be ${\text{r = k}}{\left[ {\text{A}} \right]^{\text{x}}}{\left[ {\text{B}} \right]^{\text{y}}}$ .
We need to find out the value of the sum of x and y.
From the first two rows of given data, it can be observed that when the concentration of the reactant B is kept constant and the concentration of the reactant A is doubled, the rate of the reaction does not change, it remains constant.
Thus, the rate law for the reaction is not dependent on the concentration of the reactant A. It is only dependent on the concentration of the reactant B.
Thus, the order of the reaction with respect to A is zero, that is the value of x is equal to zero.
So, now the rate law for the reaction can be expressed as ${\text{r = k}}{\left[ {\text{B}} \right]^{\text{y}}}$ since anything raised to the power 0 is 1.
From the second row of the table showing initial rates, the rate is:
$0.002 = {\text{k}}{\left[ {0.1} \right]^{\text{y}}}$
And from the third row of the table showing initial rates, the rate is:
$0.008 = {\text{k}}{\left[ {0.2} \right]^{\text{y}}}$
Divide the rate obtained from the third row by the rate from the second row.
$
  \dfrac{{0.008}}{{0.002}} = \dfrac{{{\text{k}}{{\left[ {0.2} \right]}^{\text{y}}}}}{{{\text{k}}{{\left[ {0.1} \right]}^{\text{y}}}}} \\
   \Rightarrow 4 = {2^{\text{y}}} \\
   \Rightarrow {2^2} = {2^{\text{y}}} \\
   \Rightarrow {\text{y = 2}} \\
 $
Therefore the rate law is ${\text{r = k}}{\left[ {\text{B}} \right]^2}$ and the value of ${\text{x + y}}$ is ${\text{0 + 2 = 2}}$ .

Note:
Higher value of rate constant indicates the faster rate of reaction.
Each reaction has a definite value of rate constant at a particular temperature.
The value of the rate constant is independent of the concentration of the reactants.