Final answer:
The boiling temperature at 1 atm can be determined by using the equation T = ΔH/ΔS with the given ΔH = 30 kJ/mol and ΔS = 75 J K⁻¹mol⁻¹, resulting in a boiling temperature of 400 K.
Step-by-step explanation:
The question at hand involves calculating the boiling temperature of a substance given the enthalpy (ΔH) and the entropy (ΔS) changes of vaporization at 1 atm pressure. This is a classic thermodynamics problem that requires an understanding of the Clausius-Clapeyron equation, which relates the pressure and temperature during a phase change.
To find the boiling point, we use the thermodynamic equation ΔG = ΔH - TΔS, where ΔG (the change in Gibbs free energy) is zero at the boiling point because it is a phase equilibrium. Setting ΔG to zero and rearranging for the temperature (T) gives us T = ΔH/ΔS. Substituting the given values (ΔH = 30 kJ/mol, ΔS = 75 J K⁻¹mol⁻¹) and converting ΔH to J/mol for consistency yields T = (30 x 10³ J/mol) / (75 J K⁻¹mol⁻¹) = 400 K.
Therefore, the boiling temperature at 1 atm for the substance with the given thermodynamic values would be 400 K, or approximately 127°C when converted from Kelvin to Celsius (T°C = T(K) - 273.15).