The ΔHvap of a certain compound is 30.22 kJ·mol–1 and its ΔSvap is 60.63 J·mol–1·K–1. What it the boiling point of this compound?
dG = dH - TdS
Set dG = 0, substitute and solve for T.
To find the boiling point of a compound using its enthalpy of vaporization (ΔHvap) and entropy of vaporization (ΔSvap), we can use the following equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy (equal to zero at the boiling point),
ΔH is the enthalpy of vaporization,
T is the temperature in Kelvin,
ΔS is the entropy of vaporization.
At the boiling point, ΔG = 0, so we can rearrange the equation to solve for T:
T = ΔH / ΔS
Now, let's substitute the given values into the equation:
T = (30.22 kJ·mol–1) / (60.63 J·mol–1·K–1)
Note that we need to convert units for ΔH from kJ to J since the units for ΔS are in J.
1 kJ = 1000 J
T = (30.22 kJ·mol–1) / (60.63 J·mol–1·K–1) = 30220 J·mol–1 / 60.63 J·mol–1·K–1
T = 498.3 K
Therefore, the boiling point of this compound is 498.3 Kelvin (K).