The heat of vaporization AH vap, of carbon tetrachloride, CCl4, at 250C: CCL4(1) CCL4 (g); Hvap=39.4 kJ/mol If 0.5 mol of liquid carbon tetrachloride at 25*C has an entropy of 108 J/K, what is the entropy of 0.25 mol of the vapor in equilibrium with the liquid at this temperature?

To solve this problem, we need to use the formula for the change in entropy (ΔS) during a phase transition, which is given by:

ΔS = ΔH/T

where ΔH is the heat of vaporization and T is the absolute temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin. The conversion is T(K) = T(C) + 273.15, so T = 25 + 273.15 = 298.15 K.

Next, we need to convert the heat of vaporization from kJ/mol to J/mol. The conversion is 1 kJ = 1000 J, so ΔH = 39.4 kJ/mol * 1000 J/kJ = 39400 J/mol.

Now we can calculate the change in entropy for one mole of CCl4:

ΔS = ΔH/T = 39400 J/mol / 298.15 K = 132.1 J/mol·K

This is the entropy change for one mole of CCl4. However, the problem asks for the entropy of 0.25 mol of the vapor. Since entropy is an extensive property, it scales with the amount of substance. Therefore, the entropy of 0.25 mol of CCl4 vapor is:

ΔS = 0.25 mol * 132.1 J/mol·K = 33.03 J/K

Finally, we need to add this entropy change to the initial entropy of the liquid to find the final entropy of the vapor. The initial entropy of 0.5 mol of liquid CCl4 is given as 108 J/K, so the entropy of 0.25 mol of liquid CCl4 is half of this, or 54 J/K. Therefore, the final entropy of the vapor is:

S(vapor) = S(liquid) + ΔS = 54 J/K + 33.03 J/K = 87.03 J/K

So, the entropy of 0.25 mol of CCl4 vapor in equilibrium with the liquid at 25°C is 87.03 J/K.