For a gas reaction at of 400 K the rate is reported as

$$ \frac{-d\mathrm P_A}{dt}=3.66 ~\mathrm P^2_A (atm~h)$$

(a) What are the units of rate constant?

(b) What is the value of the rate constant for this reaction if the rate equation is written as

i) $ -\mathrm{r}_{\mathrm{A}}=\frac{-1}{\mathrm{~V}} \frac{\mathrm{d \textrm {N } _ { \mathrm { A } } }}{\mathrm{dt}}=k \mathrm{C}_{\mathrm{A}}^{2}, (\mathrm{~mol} / \mathrm{L} \cdot \mathrm{h} )$

ii) $ -\mathrm{r}_{\mathrm{A}}=k \mathrm{C}_{\mathrm{A}}^{2} (\mathrm{mol} / \mathrm{m}^{3} \text { s) } \\ $

Solution :

A gas phase reaction at $400 \mathrm{~K}$.

The rate is given as

$$ \begin{aligned} -\frac{\mathrm d \mathrm {P_{A}}}{\mathrm {d t}} &=3.66 \mathrm {P_{A}}^{2} \\ -\frac{\mathrm d \mathrm P_{A}}{\mathrm {d t}} &=k \mathrm {P_{A}}^{2} \end{aligned} $$

$$\begin{aligned}-\frac{\mathrm d \mathrm P_{A}}{\mathrm {d t}} &=3.66\left(\frac{1}{\text{atm .h}}\right) \mathrm P^2_{A} \\ \mathrm k …