Questions: A chlorine (Cl) atom is adsorbed on a small patch of surface. This patch is known to contain 81 possible adsorption sites. The Cl atom has enough energy to move from site to site, so it could be on any one of them. Suppose a Br atom also becomes adsorbed onto the surface. Calculate the change in entropy. Round your answer to 2 significant digits, and be sure it has the correct unit symbol.

Solution Steps

Initially, only the Cl atom is adsorbed on the surface with 81 possible sites. The number of configurations for the Cl atom is 81.

When both Cl and Br atoms are adsorbed on the surface, each atom can occupy any of the 81 sites. Therefore, the number of configurations for both atoms is \(81 \times 81 = 81^2\).

The change in entropy (\(\Delta S\)) can be calculated using the formula for the change in entropy due to a change in the number of configurations:

\[ \Delta S = k \ln \left(\frac{\text{Final configurations}}{\text{Initial configurations}}\right) \]

where \(k\) is the Boltzmann constant (\(1.3807 \times 10^{-23} \, \text{J/K}\)).

Substituting the values:

\[ \Delta S = 1.3807 \times 10^{-23} \ln \left(\frac{81^2}{81}\right) = 1.3807 \times 10^{-23} \ln(81) \]

Calculate the natural logarithm of 81:

\[ \ln(81) = \ln(3^4) = 4 \ln(3) \approx 4 \times 1.0986 = 4.3944 \]

Substitute \(\ln(81)\) into the entropy change equation:

\[ \Delta S = 1.3807 \times 10^{-23} \times 4.3944 \approx 6.065 \times 10^{-23} \, \text{J/K} \]

Round the answer to 2 significant digits:

\[ \Delta S \approx 6.1 \times 10^{-23} \, \text{J/K} \]

Final Answer