Activation energy depends on the form of the rate law

Question

I was going through the following example, on p. 73, in a textbook called Chemical Reaction Engineering (3rd ed.) by Octave Levenspiel:

Experimental studies of a specific decomposition of A in a batch reactor using pressure units show exactly the same rate at two different temperatures:

At $\ce {400K}$, $\ce {-r_A = 2.3 p_A^{2}}$, while at $\ce {500K}$, $\ce {-r_A = 2.3 p_A^{2}}$, where the reaction rate is measured in $\ce {mol m^{-3} s^{-1}}$ and the partial pressure of A is measured in $\ce {atm}$. Right away, we can conclude that the activation energy of this reaction, based on the fact that the rate constants determined at two different temperatures are the same, is zero. Activation energy is a measure of temperature dependence of the reaction rate, so if reaction rate does not change with temperature, then activation energy is zero. The same conclusion can be arrived at using the Arrhenius' equation for the activation energy.

Then, the example proceeds to change the rate laws to be based on concentration instead. The ideal gas law is made use of and after some unit cancellations, we obtain the following rate laws at $\ce {400 K}$ and $\ce {500 K}$:

At $\ce {400K}$, $\ce {-r_A = 0.0025 C_A^{2}}$, while at $\ce {500K}$, $\ce {-r_A = 0.0039 C_A^{2}}$, where the reaction rate is measured in $\ce {mol m^{-3} s^{-1}}$ and the concentration of A is measured in $\ce {mol m^{-3}}$. This would suggest that the activation energy is no longer zero and in fact, by the Arrhenius' equation, we obtain a value of $\ce {7394 J mol^{-1}}$.

The textbook does not explain why this change in the value of $\ce {E_a}$ occurs but my best guess is that in the original rate laws based on pressure, the temperature dependence of reaction rate was hidden within the partial pressure of A. On the other hand, after converting to concentrations, which are independent of temperature, the temperature dependence of the reaction would then be transferred to the rate constants. Is my explanation for why this happens correct?

Also, the textbook claims that the activation energies obtained using the rate laws based on pressure are incorrect and that it is only correct to derive activation energies based on the rate law involving concentrations.

Edit: From what I recall learning in high school, if a rate law is written in terms of pressure, the rate will be expressed in units of pressure per unit time (e.g. $\ce {atm s^{-1}, Pa s^{-1}}$), so in the example above, reporting rate as $\ce {mol m^{-3} s^{-1}}$ while writing the rate law based on pressures, seems a bit strange.

Answer 1

The textbook does not explain why this change in the value of $\ce {E_a}$ occurs but my best guess is that in the original rate laws based on pressure, the temperature dependence of reaction rate was hidden within the partial pressure of A. On the other hand, after converting to concentrations, which are independent of temperature, the temperature dependence of the reaction would then be transferred to the rate constants. Is my explanation for why this happens correct?

I will do an example to show one way to arrive at this situation.

Imagine we have the following data for a constant-volume batch reactor at two different temperatures:

\begin{array}{c|c|c} t \, (\pu{s}) & c_\mathrm{A} \, (\pu{M}) \quad (T = \pu{350 K}) & c_\mathrm{A} \, (\pu{M}) \quad (T = \pu{495 K}) \\ \hline 0 & 1000 & 500 \\ 100 & 500 & 250 \\ 200 & 333 & 166 \\ 300 & 250 & 125 \\ 400 & 200 & 100 \\ \end{array}

We will explore if the data responds to a second-order reaction. The amount balance for species $A$ leads to the well-known formula \begin{equation} \frac{\mathrm{d}c_A}{\mathrm{d}t} = -kc_A^2 \rightarrow \frac{1}{c_A} = \frac{1}{c_{A0}} + kt \tag{1} \end{equation} So, we check the linearity for $1/c_A$ vs $t$ and obtain the following ($m$ is the slope and $b$ the y-intercept):

\begin{array}{c|c|c|c} T \, (\pu{K}) & m \, (\pu{M^-1 s-1}) & b \, (\pu{M^-1}) & R^2 \\ \hline 300 & \pu{1.00 \times 10^{-5}} & \pu{1.00 \times 10^{-3}} & \approx 1 \\ 495 & \pu{2.00 \times 10^{-5}} & \pu{2.00 \times 10^{-5}} & \approx 1 \\ \end{array}

Thus, the rate constants are \begin{equation} k(\pu{350 K}) = \pu{1.00 \times 10^{-5} M^-1 s^-1} \hspace{2 cm} k(\pu{495 K}) = \pu{2.00 \times 10^{-5} M^-1 s^-1} \end{equation}

Now, we put the rate law in terms of the partial pressure of species $A$ \begin{equation} -r_\mathrm{A} = kc_\mathrm{A}^2 = k\left(\frac{p_A}{RT}\right)^2 = \frac{k}{(RT)^2}p_A^2 \rightarrow -r_\mathrm{A} =k'p_\mathrm{A}^2 \tag{2} \end{equation} where $k' \equiv k/(RT)^2$. If we calculate this quantity we get \begin{equation} k'(\pu{350 K}) = \pu{1.21 \times 10^{-8} M s^-1 atm^-2} \hspace{2 cm} k'(\pu{495 K}) = \pu{1.21 \times 10^{-8} M s^-1 atm^-2} \end{equation} and you will report the following rate laws \begin{align} -r_\mathrm{A} &= (\pu{1.21 \times 10^{-8} M s^-1 atm^-2}) p_A^2 \quad \pu{(350 K)} \\ -r_\mathrm{A} &= (\pu{1.21 \times 10^{-8} M s^-1 atm^-2}) p_A^2 \quad \pu{(495 K)} \\ \end{align}

Thus, if we calculated the activation energy with these two rate constants, it will lead to $E_a \approx 0$. According to the author, since the transition theory is derived using concentration units, the rate constants must be reported with the rate law put in terms of the concentration. This was rather a surprise to me, and it seems to be true.

Nonetheless, if you just do the whole derivation you will quickly see that you are searching for the 'real' $k$ and no confusion will arise.