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Given the equation

$$\ce{A(g) + B(g) → C(g) + D(g)}$$

with reaction rate given by

$$v = k[\ce{A}]^2$$

How can I prove that reaction rate goes up with pressure? I tried substituting $k$ for $Ae^{-E_A/(RT)}$ which leaves me with

$$v = Ae^{-\frac{E_A}{RT}}[A]^2$$

If I use

$$PV = nRT \implies RT = \frac{1}{n}PV \implies RT = \frac{P}{C}$$

I end up with this:

$$v = Ae^{-E_A\frac{C}{P}}[A]^2$$

This shows there is a relation, but I don't know what to do with the concentration $C$, or whether it was correct to plug in the ideal gas law in the first place. Any help?

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    $\begingroup$ Pressure is directly proportional to concentration via the ideal gas law, so the rate $\sim kp^2$ . $\endgroup$ – porphyrin Apr 20 at 8:04
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The exponential term in the Arrhenius equation does not imply reaction with other molecules. It applies to ready to react unstable adduct, about to overcome the energy barrier $E_a$ to form the product. The concentration of this adduct is driven by concentration of reagents and therefore by the pressure.

Therefore the reaction rate equation implicitly involve 2 multiplicative terms:

  1. Concentration of reagent adduct

    $$\mathrm{const}\cdot [A]^2 = \mathrm{const}\cdot \left(\frac{p_{\rm A}}{RT}\right)^2$$

  2. Probability the adduct reaches the reaction energy barrier, from the Boltzmann distribution:

    $$\exp{\left(-\frac{E_A}{RT}\right)}$$

By other words, the constant $A$ in your equation

$$v = A \exp\left(-\frac{E_A}{RT}\right)$$

implicitly involves concentration, respectively pressure of the reagents.

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