Carbon disulfide,CS2, is a volatile, flammable liquid. It has a vapor pressure of 426.6 mmHg at 29.8C and 760.0 mmHg at 46.5C. What is the heat of vaporization of this substance?
Use the Clausius-Clapeyron equation. Post your work if you get stuck.
To find the heat of vaporization of carbon disulfide (CS2), we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and heat of vaporization.
The Clausius-Clapeyron equation is given as:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final vapor pressures,
T1 and T2 are the initial and final temperatures,
ΔHvap is the heat of vaporization, and
R is the ideal gas constant (8.314 J/mol·K).
Given data:
P1 = 426.6 mmHg
T1 = 29.8 °C
P2 = 760.0 mmHg
T2 = 46.5 °C
First, we need to convert the pressures from mmHg to atm:
P1 = 426.6 mmHg / 760 mmHg/atm = 0.561337 atm
P2 = 760.0 mmHg / 760 mmHg/atm = 1.0000 atm
Next, convert temperatures from °C to Kelvin (K):
T1 = 29.8 °C + 273.15 = 302.95 K
T2 = 46.5 °C + 273.15 = 319.65 K
Now we can substitute the values into the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
ln(1.0000/0.561337) = -(ΔHvap/8.314) * (1/319.65 - 1/302.95)
Solving for ΔHvap:
ΔHvap = -8.314 * ln(1.0000/0.561337) / (1/319.65 - 1/302.95)
Calculating the value, we find:
ΔHvap ≈ 30.57 kJ/mol
Therefore, the heat of vaporization of carbon disulfide (CS2) is approximately 30.57 kJ/mol.