# Relationship between rate of reaction and pressure

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?

• Pressure is directly proportional to concentration via the ideal gas law, so the rate $\sim kp^2$ . – porphyrin Apr 20 at 8:04

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:

$$\mathrm{const}\cdot [A]^2 = \mathrm{const}\cdot \left(\frac{p_{\rm A}}{RT}\right)^2$$
$$\exp{\left(-\frac{E_A}{RT}\right)}$$
By other words, the constant $$A$$ in your equation
$$v = A \exp\left(-\frac{E_A}{RT}\right)$$