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_{\text{sys}}[/tex] for the vaporization of 5.0 g of ethanol at 79.0°C?

Answer :

Final answer:

The question inquires about the change in entropy for the vaporization of 5.0 g of ethanol. This involves thermodynamics concepts in chemistry and calculations related to the heat of vaporization, number of moles, and absolute temperature. The result is 11.85 J/mol K.

Explanation:

The subject of your question is focused on calculating the change in entropy (ΔS_sys) which is a concept from Thermodynamics in Chemistry.

We have the heat of the vaporization of ethanol as 38.6 kJ/mol. We first need to convert that to Joules by multiplying by 1000, which gives us 38600 J/mol.

Next, we need to find the number of moles in 5 g of ethanol. Using the molecular weight (MW) of ethanol (46.07 g/mol), we have 5 g / 46.07 g/mol = 0.108 mol.

Finally, we use the formula ΔS_sys = q_rev / T where q_rev is the heat transferred in the vaporization process and T is the absolute temperature. The heat transfer is calculated by multiplying the number of moles by the molar heat of vaporization, 0.108 mol * 38600 J/mol = 4168.8 J. Then, we convert the temperature to Kelvin by adding 273.15 to the celsius temperature, 79.0 °C + 273.15 = 352.15 K.

Calculating, ΔS_sys = 4168.8 J / 352.15 K = 11.85 J/mol K.

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