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

$\ce{\frac {dθ} {dt} =kCN(1−θ)}$$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = kCN(1−\theta)$$

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

The reverse reaction is given by

$\ce{\frac{dθ}{dt} =−kNθ}$.$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = −kN\theta.$$

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

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

$\ce{\frac {dθ} {dt} =kCN(1−θ)}$

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

$\ce{\frac{dθ}{dt} =−kNθ}$.

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

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.

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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

$\ce{\frac {dθ} {dt} =kCN(1−θ)}$

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

$\ce{\frac{dθ}{dt} =−kNθ}$.

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