Question

The conversion of the molecule $\mathrm{X}\mathrm{To}\mathrm{Y}$follows second-order kinetics. If the concentration $\mathrm{X}$ is increased to three times, How will it affect the rate of formation of $\mathrm{Y}$?

Answer 1

Order of a reaction:

• The order of a reaction is generally determined by summing up the powers of the concentration of reactants that forms the product.
• The order of a reaction is basically known to be the power dependence of the rate on the concentrations of each of the reactants.

Step 1: Rate law for second-order reaction:
As the order of the reaction is given to be a second-order reaction, the rate law can be written as :
$\mathrm{Rate}=\mathrm{k}{\left[\mathrm{X}\right]}^2={\mathrm{r}}_1$……(1)

Step 2: Rate law when the concentration is increased three times:
If the concentration will be increased to three times the rate law can be given as :
${\mathrm{r}}_2=\mathrm{k}{\left[3\mathrm{X}\right]}^2=9{\left[\mathrm{X}\right]}^2$……..(2)

Step 3: Affect on the rate of formation of $\mathrm{Y}$:

We can consider the change in the rate of formation of $\mathrm{Y}$, by taking the ratio of the above-mentioned equations, which can be given as:

$\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=\frac{9\mathrm{k}{\left[\mathrm{X}\right]}^2}{\mathrm{k}{\left[\mathrm{X}\right]}^2}$
$$\begin{gathered} \Rightarrow \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=9 \\ \Rightarrow {\mathrm{r}}^2=9{\mathrm{r}}^1 \end{gathered}$$
Hence, the rate of formation of $\mathrm{Y}$will increase by nine times.