What is the change in entropy in the system in J/K when 112.2 grams of ethanol at 1 atm condenses to a liquid at the normal boiling point? The normal boiling point of ethanol (C₂H₅OH)is 78.3 °C and its molar enthalpy of vaporization is 38.56 kJ/mol.

Question

Quizplus · Accepted Answer

We can use the formula:

ΔS = q/T

Where q is the heat transferred, T is the temperature at which the transfer occurs, and ΔS is the change in entropy.

First, we need to calculate the heat transferred in the process of ethanol condensing at its normal boiling point.

ΔHvap = 38.56 kJ/mol

The molar mass of ethanol (C2H5OH) is:

2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.07 g/mol

So, 1 mol of ethanol has a mass of 46.07 g, and we have 112.2 g, which is:

n = m/M = 112.2 g / 46.07 g/mol = 2.435 mol

Therefore, the amount of heat transferred, q, is:

q = ΔHvap n = 38.56 kJ/mol x 2.435 mol = 93.8 kJ

Now we can calculate the change in entropy:

ΔS = q/T = 93.8 kJ / (78.3 + 273.15) K = -0.266 kJ/K = -266 J/K

Therefore, the answer is (C) -267.

What Is the Change in Entropy in the System in J/K