# Is it possible for the forward and reverse reactions to have different reaction orders?

For example, in the case of non-dissociative adsorption, the rate of the forward process is given by

$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = kCN(1−\theta)$$

where $C$ is the concentration, $N$ is the number of surface sites, and $\theta$ is the fraction of filled sites.

The reverse reaction is given by

$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = −kN\theta.$$

Hence, it seems like the first reaction is second-order, while the second is first order. This seems strange to me.

• But $\mathrm N$ is constant? – DHMO Oct 16 '16 at 13:55
• N is constant, but $\theta$ is not – Marcel Oct 16 '16 at 14:10
• The usage of \ce should be limited to typesetting chemicals and chemical equations - please don't use it to enclose mathematical expressions - these ought to be in italics. – orthocresol Oct 16 '16 at 19:29
• @orthocresol Thanks! What does the \mathrm you used mean? – Marcel Oct 16 '16 at 21:04
• Math, roman type; i.e. no italics. In this case, it is only suitable for the $\mathrm{d}$ in differentials, which should be upright. – orthocresol Oct 16 '16 at 21:04

$$\ce{2A <=>[k_1][k_{-1}] A_2}$$
Forward rate is second order: $k_1[\ce{A}]^{2}$.
Reverse rate is first order $k_{-1}[\ce{A2}]$.