Question

Methanethiol has a vapor pressure of 429 torr at −25 ∘c and a normal boiling point of 6.0 ∘c. find δhvap for methanethiol.

Answer 1

The enthalpy of vaporization for methanethiol can be calculated using the Clausius-Clapeyron equation. Plugging in the given values, we find that the ΔHvap of methanethiol is approximately -14.6 kJ/mol.

Step-by-step explanation:

The enthalpy of vaporization (ΔHvap) for a substance can be calculated using the Clausius-Clapeyron equation. The equation is:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

Where P1 and P2 are the vapor pressures of the substance at two different temperatures (T1 and T2), ΔHvap is the enthalpy of vaporization, and R is the gas constant. In this case, we know the normal boiling point of methanethiol is 6.0 ∘C (279.15 K) and its vapor pressure at -25 ∘C is 429 torr. To find ΔHvap, we can plug these values into the Clausius-Clapeyron equation and solve for ΔHvap.

ln(429 torr/P1) = (ΔHvap/8.314 J/mol∙K) * (1/279.15 K - 1/248.15 K)

Solving for ΔHvap gives us a value of approximately -14.6 kJ/mol.

Answer 2

Naturally, this type of thermodynamic problem calls for a formula. In this case, we need the Clausius-Clapeyron equation which is shown in the attached picture. Take note of the given units shown in the legend.

Tb = 6 + 273 = 279 K
T0 = -25 + 273 = 248 K
P0 = 429 torr * 101.325 kPa/706 torr = 61.57 kPa

Applying the equation,

279 = {[8.314(ln61.57 - ln101.325)/ΔHvap] + 1/248}⁻¹
Use a scientific calculator to find ΔHvap.
ΔHvap = 9,244.25 J/mol