based on the following thermodynamic data, calculate the boiling pressure of ethanol in degrees celcius Substance Delta Hf ( Kj/mol) Delta Sf ( J/K*mol) C2H5OH (L) --- -277.7 ----- 160.6 C2H5OH (G) --- -235.1 ------- 282.6

To calculate the boiling pressure of ethanol, we can use the Clausius-Clapeyron equation derived from thermodynamic principles, which relates the temperature, pressure, and the change in enthalpy and entropy upon phase change:

\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

where \( P_1 \) and \( P_2 \) are the pressures at temperatures \( T_1 \) and \( T_2 \) respectively, \( \Delta H_{vap} \) is the heat of vaporization (enthalpy change for the phase change), \( R \) is the gas constant, which is 8.314 J/(K⋅mol).

We know \( \Delta H_{vap} \) is the difference between the enthalpy of the gas and the enthalpy of the liquid. From your thermodynamic data,

ΔHf(ethanol gas) is -235.1 kJ/mol, ΔHf(ethanol liquid) is -277.7 kJ/mol.

Thus, the enthalpy change of vaporization (ΔH_vap) is:

ΔH_vap = ΔHf(g) - ΔHf(l) = -235.1 kJ/mol - (-277.7 kJ/mol) = 42.6 kJ/mol = 42600 J/mol,

Similarly, the change in entropy for vaporization (ΔS_vap) is the difference between the entropy of the gas and the liquid:

ΔS_vap = S(g) - S(l) = 282.6 J/K⋅mol - 160.6 J/K⋅mol = 122 J/K⋅mol.

The boiling point occurs when the vapor pressure equals the external pressure, typically the atmospheric pressure at 1 atm or 101.3 kPa. We can take \( P_1 \) as 1 atm and \( T_1 \) as the temperature at which we want to find the boiling pressure. To find this temperature, we’ll rearrange the Clausius-Clapeyron equation:

\[ \frac{1}{T_2} = -\frac{R}{\Delta H_{vap}} \ln\left(\frac{P_2}{P_1}\right) + \frac{1}{T_1} \]

Since we know at normal boiling point, \( P_2 \) must be equal to 1 atm and ΔG = 0, thus:

\[ \Delta G = \Delta H_{vap} - T_1 \Delta S_{vap} = 0 \]

\[ T_1 = \frac{\Delta H_{vap}}{\Delta S_{vap}} = \frac{42600 J/mol}{122 J/(K⋅mol)} \]

So,

\[ T_1 = 349.18 K \]

Since we are asked for the boiling pressure in degrees Celsius, subtract 273.15 from the Kelvin temperature:

\[ T_1^\circ C = T_1 K - 273.15 \]

\[ T_1^\circ C = 349.18 K - 273.15 \]

\[ T_1^\circ C = 76.03^\circ C \]

Therefore, the boiling temperature of ethanol at 1 atm pressure is 76.03^\circ C.