For Ethanol, change in H vapor = 38.7 kj/mol at 78 degrees Celsius, it's boiling point. Calculate change in S for the vaporization of 1.00 mol ethanol at 78 degrees celsius and 1.00 atm pressure

To calculate the change in entropy (ΔS) for the vaporization of 1.00 mol of ethanol at 78 degrees Celsius and 1.00 atm pressure, we can use the equation:

ΔS = (ΔH/T)

where ΔH is the change in enthalpy and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin:

T = 78 + 273.15 = 351.15 K

Next, we can substitute the values into the equation:

ΔS = (38.7 kJ/mol) / (351.15 K)

Now we just need to perform the calculation:

ΔS = 0.110 kJ/(mol·K)

Therefore, the change in entropy for the vaporization of 1.00 mol of ethanol at 78 degrees Celsius and 1.00 atm pressure is 0.110 kJ/(mol·K).

To calculate the change in entropy (ΔS) for the vaporization of 1.00 mol of ethanol at 78 degrees Celsius and 1.00 atm pressure, we can use the Clausius-Clapeyron equation. The equation is as follows:

ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)

Where:
P1 = Initial pressure (1.00 atm)
P2 = Final pressure (1.00 atm)
ΔHvap = Change in enthalpy of vaporization (38.7 kJ/mol)
R = Ideal gas constant (8.314 J/(mol·K))
T1 = Initial temperature (78 degrees Celsius, convert to Kelvin: 78 + 273.15 = 351.15 K)
T2 = Final temperature (also 78 degrees Celsius, convert to Kelvin: 78 + 273.15 = 351.15 K)

Now, we can substitute the values into the equation and solve for ΔS:

ln(1.00/1.00) = (38.7 kJ/mol) / (8.314 J/(mol·K)) * (1/351.15 K - 1/351.15 K)

Since ln(1.00/1.00) equals 0, we can simplify the equation:

0 = (38.7 kJ/mol) / (8.314 J/(mol·K)) * (0)

Therefore, the change in entropy (ΔS) for the vaporization of 1.00 mol of ethanol at 78 degrees Celsius and 1.00 atm pressure is 0 J/(mol·K).

Please note that the entropy change in this scenario is zero due to the isothermal nature of the process (the temperature remains constant).

delta H vap = Tvap x delta S vap. Substitute and solve for delta S vap.