Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1 degrees celcius.

At what temperature does Benzene boil when the external pressure is 440 torr?

Since...I didn't get the right answer. I'll just show my work and see if anyone can help me. Answer has to be in 2 sig figs.

Ln (440/760) = -30720/8.3145 * (1/x - 1/353.25)

I got x = 335.58. Since I was using Kelvin, the celcius is 63 degrees...something is off. What should I be getting?

Try removing the negative sign for delta Hvap and get back to us.

nevermind i had it right the whole time. thank you

To find the boiling point of benzene at a given external pressure, we can use the Clausius-Clapeyron equation:

ln(P₁/P₂) = ΔH_vap/R * (1/T₂ - 1/T₁)

where:
P₁ = vapor pressure at the normal boiling point (760 torr)
P₂ = external pressure (440 torr)
ΔH_vap = heat of vaporization (30.72 kJ/mol)
R = ideal gas constant (0.0083145 kJ/(mol·K))
T₁ = normal boiling point (80.1 °C + 273.15 K)
T₂ = boiling point at the given external pressure (unknown)

Let's substitute the given values into the equation and solve for T₂:

ln(440/760) = (30.72/0.0083145) * (1/T₂ - 1/353.25)

Simplifying the equation:

ln(0.5789) = 3698.8 * (1/T₂ - 0.002833)

Solving for (1/T₂ - 0.002833):

1/T₂ - 0.002833 = ln(0.5789) / 3698.8

1/T₂ = (ln(0.5789) / 3698.8) + 0.002833

Now, solve for T₂:

T₂ = 1 / [(ln(0.5789) / 3698.8) + 0.002833]

T₂ ≈ 57.7 °C

Therefore, at an external pressure of 440 torr, benzene boils at approximately 57.7 °C.

To determine the correct boiling point of benzene at an external pressure of 440 torr, you need to use the Clausius-Clapeyron equation, which relates the boiling point of a substance to its heat of vaporization and external pressure.

The Clausius-Clapeyron equation is expressed as:

ln(P₁/P₂) = ΔHvap/R * (1/T₂ - 1/T₁)

Where:
P₁ and P₂ are the initial and final pressures respectively.
ΔHvap is the heat of vaporization of the substance.
R is the ideal gas constant.
T₁ and T₂ are the initial and final temperatures respectively, expressed in Kelvin.

Given values:
P₁ (initial pressure) = 760 torr (standard atmospheric pressure)
P₂ (final pressure) = 440 torr
ΔHvap (heat of vaporization) = 30.72 kJ/mol
R = 8.3145 J/(mol·K) (note the unit conversion from kJ to J)
T₁ (initial temperature) = 373.25 K (converted from 80.1 degrees Celsius)

Now, let's solve the equation for T₂ (final temperature) using the given values:

ln(440/760) = (30.72 * 10^3 J/mol) / (8.3145 J/(mol·K)) * (1/T₂ - 1/373.25)

ln(440/760) = 3694.27(1/T₂ - 1/373.25)

ln(0.579) = 3694.27(1/T₂ - 0.002677)

0.3852 = 1/T₂ - 0.009904

1/T₂ = 0.3951

T₂ = 1 / (0.3951) ≈ 2.5307 K (Kelvin)

Converting back to Celsius:

T₂ ≈ 2.5307 K - 273.15 ≈ -270.62 °C

Rounding to 2 significant figures, the boiling point of benzene at an external pressure of 440 torr is approximately -271 °C