Questions: Adsorption Thermodynamics and Surface Entropy

Question 1

A catalytic surface is operating at 200°C and is then heated to 600°C while exposed to the same gas pressure. What happens to surface coverage, and why?

• A. Coverage increases because higher temperature gives molecules more kinetic energy to reach the surface
• B. Coverage decreases because the −TΔS penalty grows large enough to make ΔG positive, favoring desorption
• C. Coverage stays the same because the adsorption enthalpy does not change with temperature
• D. Coverage first increases then decreases at the Sabatier optimum temperature

Answer: (B)

Adsorption has ΔH < 0 (favorable) and ΔS < 0 (unfavorable, adsorbate loses translational freedom). At 200°C the enthalpy term dominates and ΔG is negative; at 600°C the −TΔS term is much larger, making ΔG positive, so desorption is thermodynamically favored. This is why catalytic surfaces 'clean themselves' at high temperatures.

Question 2

Gas A adsorbs with ΔH_ads = −20 kJ/mol and Gas B with ΔH_ads = −80 kJ/mol. At a fixed temperature and pressure, which gas has higher equilibrium surface coverage?

• A. Gas A, because weaker binding allows faster equilibrium exchange with the gas phase
• B. Gas B, because the more negative ΔH shifts the adsorption equilibrium toward the surface-bound state
• C. They are equal because entropy terms dominate at all temperatures
• D. Cannot be determined without knowing the entropy of adsorption for each gas

Answer: (B)

More negative ΔH_ads makes ΔG more negative at a given temperature, shifting the equilibrium toward higher coverage. Gas B's stronger surface interaction produces a more stable adsorbed state. While entropy matters quantitatively, the question specifically contrasts enthalpy values — the thermodynamic driving force is clearly larger for Gas B.

Question 3

When a molecule adsorbs onto a solid surface, its entropy increases because it achieves a more ordered, lower-energy configuration.

• T. True
• F. False

Answer: False

The adsorbate's entropy DECREASES upon adsorption. A gas-phase molecule freely translates and rotates in three dimensions; once pinned to a surface it loses these degrees of freedom. ΔS_ads is negative. The favorable driving force for adsorption is the enthalpy release (bond formation), not entropy.

Question 4

A surface that binds reaction intermediates with a very large, negative ΔH_ads will typically be a more effective catalyst than one with moderate binding.

• T. True
• F. False

Answer: False

This violates Sabatier's principle. If the surface binds intermediates or products too strongly, they cannot desorb and the surface becomes poisoned — catalytic activity collapses. An effective catalyst must bind reactants strongly enough to lower the activation barrier but weakly enough for products to desorb. The optimal binding strength lies between too weak and too strong.

Question 5

Why does increasing temperature generally cause desorption from a surface, even though adsorption itself releases energy (ΔH < 0)?

Model answer: Because adsorption also involves a loss of entropy (ΔS < 0 for the adsorbate). The Gibbs free energy is ΔG = ΔH − TΔS. Even though ΔH is negative, ΔS is also negative, so the −TΔS term is positive and grows with temperature. At high enough temperature, −TΔS overwhelms ΔH, making ΔG positive and desorption thermodynamically favorable.

This is the central thermodynamic insight: adsorption is an enthalpy-entropy competition. At low T, enthalpy wins (adsorption spontaneous); at high T, entropy wins (desorption spontaneous). The crossover temperature depends on the ratio ΔH/ΔS and is directly related to the desorption temperature observed experimentally.