A Langmuir adsorption isotherm shows that surface coverage θ approaches 1.0 asymptotically at high pressure. What physical assumption produces this saturation behavior?
AAdsorption sites become more energetically favorable as pressure increases
BThe surface has a finite number of equivalent, independent sites that each accommodate only one adsorbate molecule
CAt high pressure, adsorbate molecules stack into multiple layers above the first
DSurface area decreases as coverage increases, limiting further adsorption
The Langmuir saturation ceiling arises directly from the monolayer assumption: there are only so many binding sites, each can be occupied once, and all sites are equivalent. As pressure rises, more sites fill, but θ can never exceed 1 (100% coverage). This is mathematically analogous to enzyme kinetics (Michaelis-Menten) for the same reason — both describe a saturable process. Option C describes what the BET isotherm models, not Langmuir.
Question 2 Multiple Choice
Catalyst A binds oxygen with very high adsorption energy; Catalyst B binds oxygen very weakly. In an oxidation reaction that requires surface oxygen, which catalyst shows higher activity?
ACatalyst A, because strong binding ensures maximum surface coverage at all times
BCatalyst B, because weak binding lets the reaction proceed faster
CNeither extreme; a catalyst with intermediate binding energy sits at the volcano plot peak, balancing coverage against desorption rate
DCatalyst A, because higher surface coverage always means more product
This is the Sabatier principle. Catalyst A binds so strongly that products cannot desorb — the surface stays covered but the active sites are blocked. Catalyst B binds so weakly that reactants never accumulate to meaningful coverage. The optimal catalyst sits at the peak of the volcano plot: strong enough binding to achieve useful coverage, weak enough to release products promptly. Options A and D ignore the desorption half of the kinetic cycle.
Question 3 True / False
The Freundlich isotherm predicts a maximum surface coverage (saturation), just like the Langmuir isotherm — it simply reaches that maximum more gradually.
TTrue
FFalse
Answer: False
The Freundlich isotherm (θ ∝ P^(1/n)) is empirical and has no saturation limit — it predicts continuously increasing coverage with increasing pressure, which is physically unrealistic at high pressures. This is a known limitation. Langmuir achieves saturation via its explicit monolayer assumption and finite site count; Freundlich lacks this and fits heterogeneous surfaces well only over intermediate pressure ranges.
Question 4 True / False
At adsorption equilibrium, the rate of adsorption equals the rate of desorption, and this kinetic balance mathematically recovers the Langmuir isotherm.
TTrue
FFalse
Answer: True
Setting the adsorption rate (proportional to P × (1 − θ)) equal to the desorption rate (proportional to θ) and solving for θ yields θ = KP/(1 + KP) — exactly the Langmuir isotherm, where K = k_ads/k_des. This is a satisfying self-consistency check: the equilibrium isotherm and the kinetic rate equations are two faces of the same physical model.
Question 5 Short Answer
Explain why a catalyst that binds its reactants too strongly is ineffective, using the kinetics of adsorption and desorption.
Think about your answer, then reveal below.
Model answer: If adsorption energy is very high, the desorption activation energy E_des is also very high (Arrhenius: rate ∝ exp(−E_des/RT)). The desorption rate becomes vanishingly slow — products and reaction intermediates remain stuck on the surface, blocking active sites from accepting new reactants. Even though surface coverage is high, turnover is negligible because the surface never frees up. Catalysis requires a cycle: bind, react, release. A catalyst that binds too strongly gets stuck at the 'release' step.
This is the mechanistic basis of the Sabatier principle and the volcano plot. The optimal binding energy maximizes the product of coverage (favors strong binding) and turnover (favors weak binding). Too strong: surface poisoning by products or intermediates. Too weak: insufficient coverage to drive the reaction. The volcano plot visualizes this tradeoff across real catalysts.