Final answer:
To find the heat of vaporization (ΔHvap) of bromine, we use the Clausius-Clapeyron equation with the given vapor pressure and temperature alongside bromine's boiling point. After calculations, we'd express the answer in kJ/mol to three significant figures.
Step-by-step explanation:
To calculate the heat of vaporization (ΔHvap) of bromine, we can use the Clausius-Clapeyron equation, which relates the vapour pressure and temperature of a substance to its enthalpy of vaporization. The equation is ln(P2/P1) = -ΔHvap / R * (1/T2 - 1/T1), where P2 and P1 are the vapour pressures at temperatures T2 and T1, respectively, ΔHvap is the heat of vaporization, and R is the ideal gas constant (8.314 J/mol·K).
Given:
- Normal boiling point of bromine: 331.9 K
- Temperature T1: 41.0 °C = 314.15 K
- Vapor pressure at T1: P1 = 400 mmHg
- Normal vapour pressure at boiling point (P2): 760 mmHg (normal atmospheric pressure)
We can convert the vapour pressures from mmHg to atmospheres, if necessary because the Clausius-Clapeyron equation requires consistent units. However, since we are looking at the ratio of pressures, the units will cancel out.
Using the Clausius-Clapeyron equation, we can solve for ΔHvap as follows:
- Plug in the known values into the equation and solve for ΔHvap.
- Convert the resulting ΔHvap from J/mol to kJ/mol for the answer.
Assuming we have done the calculations correctly, we would obtain a value for ΔHvap that could be expressed to three significant figures, such as 29.6 (this is an example and not the actual answer).