Final answer:
Without the specific enthalpy of vaporization or condensation for HBr(g), we cannot accurately calculate the entropy change for the condensation of 3.30 mol of HBr(g). However, we can infer that this entropy change would be negative, reflecting a decrease in disorder.
Step-by-step explanation:
To determine entropy change when 3.30 mol of HBr(g) condenses at atmospheric pressure, we would need the enthalpy of vaporization or condensation of HBr(g), which is not provided. We can, however, look at the enthalpy of vaporization for another substance (Br2) as an analogous process. For Br2, the enthalpy of vaporization (ΔHvap) is 15.4 kJ/mol. Since we are considering condensation, this value would be negative when applied to the entropy change of HBr. The actual numeric value provided for Br2's enthalpy change during condensation is ΔH = −6.43 kJ, but without further information, we cannot directly apply this to HBr.
Additionally, the entropy change associated with condensation is typically negative, as the gas transitions to a liquid, resulting in a decrease in disorder. The formulas and constants needed to calculate the specific entropy change for HBr(g) condensation are not provided, so an accurate calculation cannot be performed based on the information given.