What is the entropy change for freezing 2.71 g of C2H5OH at 158.7 K? ∆H = - 4600 J/mol. Answer in units of J/K.

-1.71 J/K

Explanation:

To solve this problem we use the formula

ΔS = n*ΔH/T

Where n is mol, ΔH is enthalpy and T is temperature.

ΔH and T are already given by the problem, so now we calculate n:

Molar Mass C₂H₅OH = 46 g/mol

2.71 g C₂H₅OH : 46g/mol = 0.0589 mol

Now we calculate ΔS:

ΔS = 0.0589 mol * - 4600 J/mol / 158.7 K

ΔS = - 1.71 J/K