Given handbook std enthalpies of formation at 298K for C5H10 (liq)=-105.9 kJ/mol and C5H10 (g)=-77.2 kJ/mol, estimate the normal boiling point of cyclopentane.

I calculated delta H (vap or rxn) to be +28.7 kJ/mol. I am not sure how to proceed from here. Thank you very much for your help.

I figured it out....you use Trouton's rule of delta Svap is approx 88 J mol-1K-1 for normal bps of liquids.....you get 326K.

To estimate the normal boiling point of cyclopentane, you can use the Clausius-Clapeyron equation, which relates temperature, enthalpy change, and vapor pressure.

The Clausius-Clapeyron equation can be written as:

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)).

Since we are trying to estimate the boiling point, T2 will be the boiling point in Kelvin, and we know ΔHvap = +28.7 kJ/mol.

To proceed, we need to know the vapor pressure at a known temperature. Do you have the vapor pressure of cyclopentane at a specific temperature?

To estimate the normal boiling point of cyclopentane, you need to consider the enthalpy of vaporization (ΔHvap) and the boiling point information for the liquid and gaseous forms of the compound.

First, we're given the standard enthalpies of formation for the liquid (C5H10(liq)) and gaseous (C5H10(g)) forms of cyclopentane. These values are -105.9 kJ/mol and -77.2 kJ/mol, respectively.

The enthalpy change (ΔH) for vaporization can be calculated using the equation:

ΔHvap = ΔH(g) - ΔH(liq)

Substituting the given values:

ΔHvap = -77.2 kJ/mol - (-105.9 kJ/mol)
ΔHvap = -77.2 kJ/mol + 105.9 kJ/mol
ΔHvap = 28.7 kJ/mol

Now that you have the enthalpy of vaporization, you can estimate the normal boiling point using the Clausius-Clapeyron equation:

ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)

Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively,
R is the gas constant (8.314 J/(mol·K)),
T1 is the known boiling point of the liquid form (298 K),
T2 is the unknown boiling point of the gaseous form.

In this case, we'll assume the vapor pressures of the liquid and gaseous forms at their boiling points are equal. So we can simplify the equation to:

ln(1) = ΔHvap/R * (1/T1 - 1/T2)

Simplifying further:

0 = 28.7 kJ/mol / (8.314 J/(mol·K)) * (1/298 K - 1/T2)

Now, you need to solve this equation for T2, which represents the boiling point of the gaseous form (cyclopentane). Rearranging the equation:

(T2 - 298 K) / T2 = 28.7 kJ/mol / (8.314 J/(mol·K)) * (1/298 K)

After rearranging, the equation becomes:

T2^2 - 596 K * T2 + 2.8727 × 10^4 K^2 = 0

You can solve this quadratic equation to get the boiling point (T2) of cyclopentane.