Using data below, calculate the vapour pressure (in torr) of liquid bromine at 72.3°C. Assume that ΔHo and ΔSo do not vary with temperature. Express your answer in scientific notation.
Br2(l) -> Br2(g)
Br2(l) ΔHfo= 0 kJ/mol
To calculate the vapor pressure of liquid bromine at 72.3°C, we need to use the Clausius-Clapeyron equation. The equation is:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 = Initial pressure (in this case, it is atmospheric pressure)
P2 = Final pressure (vapor pressure)
ΔHvap = Enthalpy of vaporization (ΔHf for Br2)
R = Gas constant (8.314 J/(mol·K))
T1 = Initial temperature (in this case, it is the boiling point of bromine)
T2 = Final temperature (72.3°C)
First, let's convert the given temperature of 72.3°C to Kelvin:
T2 = 72.3°C + 273.15 = 345.45 K
The enthalpy of vaporization (ΔHvap) for bromine is given as ΔHfo, which is the standard enthalpy of formation at 298 K. Since ΔHfo is 0 kJ/mol, we can assume ΔHvap is also 0 kJ/mol.
Substituting the values into the Clausius-Clapeyron equation:
ln(P2/1 atm) = (0 kJ/mol / 8.314 J/(mol·K)) * (1/345.45 K - 1/T1)
Simplifying the equation with the given values:
ln(P2/1) = 0 * (1/345.45 - 1/T1)
ln(P2/1) = 0
P2 = 1 atm
Therefore, the vapor pressure of liquid bromine at 72.3°C is 1 atm or 760 torr.