A chlorine $(Cl)$ and bromine $(Br)$ atom are adsorbed on a small patch of surface (see sketch at right). This patch is known to contain 49 possible adsorption sites. The Cl and bromine $(Br)$ atoms have enough energy to move from site to site, so they could be on any two of them.
Suppose the Br atom desorbs from the surface and drifts away. Calculate the change in entropy.
Round your answer to 2 significant digits, and be sure it has the correct unit symbol.
$-1.1 \times 10^{-25} \frac{J}{K}$
Answer (1)
Calculate the initial number of configurations: W i = 49 × 48 .
Calculate the final number of configurations: W f = 49 .
Calculate the change in entropy: Δ S = k B ln ( W f / W i ) .
Substitute the values and calculate: Δ S = − 5.3 × 1 0 − 23 J / K .
Explanation
Problem Analysis We are given a scenario where a chlorine (Cl) and bromine (Br) atom are adsorbed on a surface with 49 possible adsorption sites. The Br atom desorbs, and we need to calculate the change in entropy.
Initial Configurations Initially, both Cl and Br atoms are on the surface. The number of ways to arrange these two atoms on the 49 sites is given by the number of ways to choose two sites out of 49. Since the atoms are distinct (Cl and Br), the number of initial configurations, W i , is calculated as 49 × 48 .
Final Configurations After the Br atom desorbs, only the Cl atom remains on the surface. The number of ways to arrange the Cl atom on the 49 sites is simply 49. Thus, the number of final configurations, W f , is 49.
Entropy Change Formula The change in entropy, Δ S , is given by the formula Δ S = S f − S i = k B ln W f − k B ln W i = k B ln W i W f , where k B is the Boltzmann constant, which is approximately 1.38 × 1 0 − 23 J / K .
Calculating Entropy Change Substituting the values of W f and W i into the formula, we get: Δ S = k B ln 49 × 48 49 = k B ln 48 1 Using the value of k B = 1.38 × 1 0 − 23 J / K , we have: Δ S = ( 1.38 × 1 0 − 23 J / K ) × ln 48 1 Δ S = ( 1.38 × 1 0 − 23 J / K ) × ln ( 1/48 ) ≈ ( 1.38 × 1 0 − 23 J / K ) × ( − 3.8712 ) ≈ − 5.342 × 1 0 − 23 J / K Rounding to two significant figures, we get Δ S = − 5.3 × 1 0 − 23 J / K .
Examples
This concept is crucial in understanding how gases adsorb onto surfaces, which is vital in catalysis and surface chemistry. For instance, when designing catalysts, understanding the entropy changes associated with adsorption helps optimize the process for better efficiency and selectivity. The change in entropy affects the equilibrium and kinetics of surface reactions, influencing the overall performance of catalytic converters in vehicles or industrial chemical reactors.
