Consider the reaction
PCl3(ℓ) → PCl3(g)
at 298 K. If ∆H◦
is 32.5 kJ/mol, ∆S
◦
is
93.3 J/K mol, and ∆G
◦
is 4.7 kJ/mol, what
would be the boiling point of PCl3 at one
atmosphere?
To determine the boiling point of PCl3 at one atmosphere, we need to use the Clausius-Clapeyron equation, which relates the boiling point of a substance to its enthalpy of vaporization (∆Hvap), the gas constant (R), the temperature (T), and the pressure (P).
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)
Where:
- P1 and T1 are the initial pressure and boiling point, respectively.
- P2 and T2 are the final pressure and boiling point, respectively.
- R is the gas constant (8.314 J/(mol*K)).
In this case, we know that the reaction is taking place at 298 K (T1) and one atmosphere pressure (P1 = 1 atm).
The enthalpy change (∆H) for the reaction is 32.5 kJ/mol. To use it in the equation, we need to convert it to J/mol:
∆H = 32.5 kJ/mol = 32.5 * 1000 J/mol = 32500 J/mol
Now, we can rearrange the Clausius-Clapeyron equation to solve for T2 (boiling point at one atmosphere):
T2 = (1/(∆Hvap/R)) * (ln(P2/P1)) + 1/T1
To proceed, we also need to calculate the enthalpy of vaporization (∆Hvap) using the given values of ∆H◦, ∆S◦, and ∆G◦. The relationship between ∆H, ∆S, and ∆G is given by:
∆G = ∆H - T∆S
Substituting the provided values:
4.7 kJ/mol = ∆H - (298 K * 93.3 J/(K mol))
Let's calculate ∆Hvap and then substitute it back into the Clausius-Clapeyron equation to find the boiling point at one atmosphere.
dGo = dHo - TdSo
Substitute for dH and dS, set dGo = 0, and solve for T (in kelvin).
dGo = 0 because that is that point that the liquid phase/gas phase is in equilibrium and dGop = 0.
The number you obtain is based on dHo and dSo not changing very much from the value you get from the tables at 298 K.