High School

It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (molecular weight: 46.07 g/mol). What will be [tex]\Delta S_{sys}[/tex] for the vaporization of 12.5 g of ethanol at 79.4 °C?

Answer :

Final answer:

To find the δssys for the vaporization of 12.5 g of ethanol, convert the mass to moles. Use the given information to calculate δssys using the equation δssys = q / T. The final value of δssys is 0.0376 kJ/K.

Explanation:

To find the δssys for the vaporization of 12.5 g of ethanol, we need to convert the mass to moles. The molecular weight (Mw) of ethanol is 46.07 g/mol, so 12.5 g of ethanol is equal to 12.5 g / 46.07 g/mol = 0.271 mol. We can then use the given information to calculate δssys using the equation δssys = q / T, where q is the heat absorbed and T is the temperature.

We know that it takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol. Using this, we can find the heat absorbed by multiplying the molar heat of vaporization (38.6 kJ/mol) by the number of moles of ethanol (0.271 mol): q = 38.6 kJ/mol x 0.271 mol = 10.4796 kJ. The temperature (T) is given as 79.4 °C.

Putting it all together, we can calculate δssys = 10.4796 kJ / (79.4 + 273.15) K = 0.0376 kJ/K.

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