Chemistry - Chemical Kinetics Question with Solution | TestHub

The reaction is given by:

$\mathrm{A}(\mathrm{g})⟶\mathrm{P}(\mathrm{s})$

For a reaction of order $n$, the half-life $t_{1/2}$ is proportional to the initial concentration $a_0$ raised to the power of $(1-n)$. We can write this as:

$$t_{1/2}\propto a_0^{1-n}\Rightarrow \frac{(t_{1/2}{)}_1}{(t_{1/2}{)}_2}={\left(\frac{a_{0_1}}{a_{0_2}}\right)}^{1-n}={\left(\frac{P_{0_1}}{P_{0_2}}\right)}^{1-n}$$

Given the following data:

If $P_{0_1}=20bar$, then $(t_{1/2}{)}_1=20min$ (from $20bar$ to $10bar$).

If $P_{0_2}=10bar$, then $(t_{1/2}{)}_2=60-20=40min$ (from $10bar$ to $5bar$).

Substituting these values into the equation:

$$\frac{20}{40}={\left(\frac{20}{10}\right)}^{1-n}$$

$$\frac{1}{2}=(2{)}^{1-n}$$

Since $\frac{1}{2}=2^{-1}$, we have:

$$2^{-1}=2^{1-n}$$

$$-1=1-n$$

$$n=2$$

Therefore, the reaction is a second-order reaction.