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

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

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

Units of $\mathrm k$ are (atm. h ) $^{-1}$

For gases :

$$ \mathrm{C}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}} / \mathrm{RT} \Rightarrow \mathrm P_{\mathrm A}=\mathrm {C_{A} R T} $$

$$ \begin{aligned} \frac{d \mathrm{ P_{A}}}{\mathrm {d t}}&=\mathrm {R T} \frac{\mathrm d \mathrm C_{A}}{\mathrm {d t}}\\ -\frac{d \mathrm {P_{A}}}{\mathrm {d t}}&=\mathrm k^{\prime} \mathrm {P_{A}^{2}}\\ -\mathrm{RT} \frac{\mathrm d \mathrm{C}_{\mathrm{A}}}{\mathrm{dt}} &=k^{\prime}\left(\mathrm{C}_{\mathrm{A}} \mathrm{RT}\right)^{2}\\ -\frac{\mathrm{dC}_{\mathrm{A}}}{\mathrm{dt}} &=(\mathrm{RT}) \mathrm{k}^{\prime} \mathrm{C}_{\mathrm{A}}^{2}, \mathrm{~mol}/(\mathrm{~l} \cdot \mathrm{h})\\ -\frac{\mathrm{dC}_{\mathrm{A}}}{\mathrm{dt}}&=\mathrm{kC}_{\mathrm{A}}^{2}\\ k&=(R T) k'\\ &=0.08206\left(\frac{l . a t m}{m o l . K}\right) \times(400 \mathrm{~K}) \times 3.66\left(\frac{1}{\mathrm{atm.h}}\right)\\ &=0.08206 \times 400 \times 3.66\left(\frac{l}{\mathrm{~mol} . \mathrm{h}}\right) \\ &=120.13~ \mathrm{l}/(\mathrm{mol.} \mathrm{h}) \end{aligned} $$

Value of rate constant is $120.13$ with units $l/(\mathrm{mol} \cdot \mathrm{h})$.

If rate is in $\mathrm{mol} /\left(\mathrm{m}^{3}.s\right)$ then $\mathrm{k}$ is

$$ \begin{aligned} k &=120.13 ~l/(\mathrm{~mol.} \mathrm{~h}) \\ &=\frac{120.13 \times 10^{-3} \mathrm{~m}^{3}}{(\mathrm{~mol.})(3600 \mathrm{~s})} \\ &=3.34 \times 10^{-3} \mathrm{~m}^{3} /(\mathrm{mol.s}) \end{aligned} $$