# 朗缪尔方程

$\theta =\frac{\alpha \cdot P}{1+\alpha \cdot P}$

$\theta$是表面覆盖范围分数，$P$是气体压力或浓度，$\alpha$为常数。

## 公式推导

$S^* + P \rightleftharpoons SP$

$K =\frac{[SP]}{[S^*][P]}$

$\alpha =\frac{\theta}{(1-\theta)p}$

$\theta = \alpha(1-\theta)p$
$\theta = p\alpha - p\theta\alpha$
$\theta + p\theta\alpha = p\alpha$
$\theta (1 + p\alpha) = p\alpha$

$\theta =\frac{\alpha \cdot p}{1+\alpha \cdot p}$

## 统计学推导

1. 假设有M个活性位点以供N个微粒结合。
2. 一个活性位点只能被一个微粒所占有。
3. 所有活性位点都是相互独立的。一个位点被占有的概率是与邻近位点的状态无关的。

N个微粒被吸附到M个位点之系统的配分函数（假设位点较微粒数量多）为：

$Q(N,M,T)=\frac{M!}{N!(M-N)!}(q\lambda)^N$

$q=q_v(T)^3$$\lambda=e^{\beta \mu}$.

$\Xi(\mu,M,T) = \sum_{N=0}^M Q(N,M,T)= \sum_{N=0}^M \binom{M}{N} (q\lambda)^N=(1+q\lambda)^M$

$\xi = 1+q\lambda$

$\langle N \rangle=\frac{\partial{\ln{\Xi(\mu,M,T)}}}{\partial{\beta \mu}}=M\frac{\partial{\ln{\xi(\mu,M,T)}}}{\partial{\beta \mu}}$

$\langle s \rangle=\frac{}{M}=\frac{\partial{\ln{\xi(T)}}}{\partial{\beta \mu}}=\lambda\frac{\partial\ln{\xi(T)}}{\partial\lambda}$

$\langle s \rangle=\frac{q\lambda}{1+q\lambda}$

## 方程拟合

${\Gamma} = \Gamma_{max} \frac{K c}{1 + K c}$

${\Gamma(c=K^{-1})} = \Gamma_{max} \frac{K K^{-1}}{1 + K K^{-1}} = \frac{\Gamma_{max}}{2}$

$\frac{1}{\Gamma} = \frac{1}{\Gamma_{max}} + \frac{1}{\Gamma_{max}Kc}$

$\Gamma = \Gamma_{max} - \frac{\Gamma}{Kc}$

$\frac{\Gamma}{c} = K\Gamma_{max} - K\Gamma$

$\frac{c}{\Gamma} = \frac{c}{\Gamma_{max}} + \frac{1}{K\Gamma_{max}}$

## 参考文献

• The constitution and fundamental properties of solids and liquids. part i. solids. Irving Langmuir; J. Am. Chem. Soc. 38, 2221-95 1916 Langmuir, I. Journal of the American Chemical Society. 1916, 38 (11): 2221–2295. doi:10.1021/ja02268a002. 编辑