Question

The vapor pressure of liquid chloroform, CHCl3, is 400.0 torr at 24.1°C and 100.0 torr at -6.3°C. What is ΔHvap of chloroform?

Answer 1

ΔHvap of CHCl3: 29,767 J/mol

To calculate the enthalpy of vaporization (ΔHvap) of chloroform (CHCl3), we can use the Clausius-Clapeyron equation:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

Where:

P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.

ΔHvap is the enthalpy of vaporization.

R is the ideal gas constant (8.314 J/(mol·K)).

Let's plug in the given values and solve for ΔHvap:

P1 = 100.0 torr

P2 = 400.0 torr

T1 = -6.3°C + 273.15 = 266.85 K

T2 = 24.1°C + 273.15 = 297.25 K

R = 8.314 J/(mol·K)

ln(400.0/100.0) = -(ΔHvap/8.314) * (1/297.25 - 1/266.85)

First, let's simplify the equation:

ln(4) = -(ΔHvap/8.314) * (0.003361 - 0.003746)

ln(4) = -(ΔHvap/8.314) * (-0.000385)

Now, we can solve for ΔHvap by rearranging the equation:

ΔHvap/8.314 = ln(4) / 0.000385

ΔHvap = (ln(4) / 0.000385) * 8.314

Calculating the value:

ΔHvap = (1.3863 / 0.000385) * 8.314

ΔHvap ≈ 29,767 J/mol

Therefore, the enthalpy of vaporization (ΔHvap) of chloroform is approximately 29,767 J/mol.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

To determine the enthalpy of vaporization (ΔHvap) of chloroform (CHCl3), we can use the Clausius-Clapeyron equation, which relates the vapor pressures of a substance at two different temperatures to the enthalpy of vaporization. The equation is as follows:

ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)

Where:

P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.

ΔHvap is the enthalpy of vaporization.

R is the ideal gas constant (8.314 J/(mol·K)).

T1 and T2 are the temperatures in Kelvin.

First, let's convert the temperatures given from Celsius to Kelvin:

T1 = 24.1°C + 273.15 = 297.25 K

T2 = -6.3°C + 273.15 = 266.85 K

Now we can plug in the values into the equation:

ln(100.0 torr / 400.0 torr) = (-ΔHvap / (8.314 J/(mol·K))) * (1/266.85 K - 1/297.25 K)

Simplifying the equation:

ln(0.25) = (-ΔHvap / 8.314) * (0.00375 - 0.00336)

ln(0.25) = (-ΔHvap / 8.314) * (0.00039)

Now, solve for ΔHvap by isolating it:

ΔHvap = -ln(0.25) * (8.314 / 0.00039)

Using a calculator, we find:

ΔHvap ≈ 31,090 J/mol

Therefore, the enthalpy of vaporization (ΔHvap) of chloroform is approximately 31,090 J/mol.