Problem Statement
Derive the Langmuir adsorption isotherm for a gas adsorbing on a surface with $N_{sites}$ identical sites.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Let $\theta$ = fraction of sites occupied. Adsorption rate $\propto (1-\theta)\cdot p$ (pressure × available sites). Desorption rate $\propto \theta$. At equilibrium:
Step 2 — Apply the relevant physical law or equation: $$k_a p(1-\theta) = k_d\theta \implies \theta = \frac{K p}{1+Kp}$$
Step 3 — Solve algebraically for the unknown: where $K = k_a/k_d = K_0 e^{E_{ads}/k_BT}$ is the adsorption equilibrium constant.
Step 4 — Substitute numerical values with units: Limits:
Step 5 — Compute and check the result:
- Low pressure ($Kp\ll1$): $\theta \approx Kp$ (Henry’s law — linear adsorption)
- High pressure ($Kp\gg1$): $\theta \to 1$ (monolayer saturation)
Step 6: Surface excess: $\Gamma = N_{sites}\theta/A_{total}$. The Langmuir isotherm is widely used in heterogeneous catalysis and biosensing.
Worked Calculation
$$k_a p(1-\theta) = k_d\theta \implies \theta = \frac{K p}{1+Kp}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{k_a p(1-\theta) = k_d\theta \implies \theta = \frac{K p}{1+Kp}}$$
Let $\theta$ = fraction of sites occupied. Adsorption rate $\propto (1-\theta)\cdot p$ (pressure × available sites). Desorption rate $\propto \theta$. At equilibrium:
$$k_a p(1-\theta) = k_d\theta \implies \theta = \frac{K p}{1+Kp}$$
where $K = k_a/k_d = K_0 e^{E_{ads}/k_BT}$ is the adsorption equilibrium constant.
Limits:
- Low pressure ($Kp\ll1$): $\theta \approx Kp$ (Henry’s law — linear adsorption)
- High pressure ($Kp\gg1$): $\theta \to 1$ (monolayer saturation)
Surface excess: $\Gamma = N_{sites}\theta/A_{total}$. The Langmuir isotherm is widely used in heterogeneous catalysis and biosensing.
Answer
$$\boxed{k_a p(1-\theta) = k_d\theta \implies \theta = \frac{K p}{1+Kp}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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