# The normal boiling point of Br2(l) is 58.8 ∘C, and its molar enthalpy of vaporization is ΔHvap = 29.6 kJ/mol.?

Calculate the value of ΔS when 3.00 mol of Br2(l) is vaporized at 58.8 ∘C.

Express your answer to three significant figures.

### 1 Answer

- 2 months ago
Classical thermodynamics tells us that dS = dQ/T, where S is entropy, Q is heat, and T is temperature.

When we integrate over any given path, this gives us that ΔS = ∫1/T dQ.

(All this is saying is "the big change is the sum of all the little changes")

When we boil a liquid at its boiling point, the temperature doesn't change, so T is a constant. Due to the rules of integration, this means we can carry it out of the integral:

ΔS = ∫1/T dQ = 1/T * ∫dQ = Q/T

So now we just need to find the total heat required to vaporize the bromine and plug everything in:

Q = n * ΔHvap = 3.00 mol * 29.6 kJ/mol = 88.8 kJ

ΔS = Q/T = 88.8 kJ / 58.8 ∘C = 1.51 kJ.

Hope this helps!