# 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

## 1 Answer

Why is this strange?

Consider:

$$\ce{2A <=>[k_1][k_{-1}] A_2}$$

Forward rate is second order: $k_1[\ce{A}]^{2}$.

Reverse rate is first order $k_{-1}[\ce{A2}]$.

• IMHO it is very problematic to speak about 'the reaction order.' If you just say 'the order' of the reaction without specification against what the order is specified, this will lead to confusion very quickly. After all, there can be several reactants/products involved in the reaction. And at equilibrium the order of the reaction and reverse reaction should be the same against every component, respectively. – logical x 2 Oct 16 '16 at 20:33
• It's not that bad when there's only one reactant, but yes, in general, we should specify the reaction order w.r.t. some reactant. – Zhe Oct 17 '16 at 14:57