High School

It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (MW: 46.07 g/mol). What will be [tex]\Delta S_{sys}[/tex] for the vaporization of 9.50 g of ethanol at 78.4 °C?

Answer :

Final answer:

To calculate the entropy change (ΔS) for the vaporization of ethanol, use the formula ΔSsys = q/T, where q is the heat exchanged, and T is the temperature.

Explanation:

The change in entropy of the system (ΔSsys) for the vaporization of ethanol can be calculated using the equation ΔSsys = q/T, where q is the heat absorbed or released during the process and T is the temperature.

First, calculate the heat (q) using the formula q = m×Hv, where m represents the mass of ethanol being vaporized and Hv is the molar heat of vaporization.

Convert the given mass of 9.50 g into moles by dividing it by the molar mass of ethanol. Then, use the calculated moles to determine the heat. Finally, divide the heat by the temperature in Kelvin to find the change in entropy of the system.

In this case, the molar heat of vaporization (Hv) is given as 38.6 kJ/mol.

The molar mass (MW) of ethanol is 46.07 g/mol. Therefore, the moles of ethanol in 9.50 g can be calculated as (9.50 g / 46.07 g/mol).

Next, calculate the heat using q = (moles of ethanol) × (Hv).

Finally, divide the calculated heat by the given temperature of 78.4 °C converted to Kelvin (78.4 + 273.15 K) to find the change in entropy of the system.

Learn more about entropy change here:

https://brainly.com/question/32484278

#SPJ11