What is the change in entropy?

Question

What is the change in entropy when 0.140 mol of potassium freezes at 67.8°C (ΔHfus = 2.39 kJ/mol)?

J/K

Answer 1

I believe this question asks for the change in entropy.

The entropy change is found by dividing the quantity of heat

given off (DQ) by the absolute temperature at which the

change occurs, that is: DS = DQ/T

The value of T is 67.8°C + 273.1 = 340.9°K

The quantity of heat is found by multiplying the heat of fusion

(2390 joules per mole) by the number of moles:

-2390 * 0.14 = -334.6 joules

(remember it is negative since the heat was evolved)

The entropy change DS equals -334.6/340.9

which is -0.9815 joule per degree.

Hope this helps

Answer 2

Assuming that the freezing potassium is in equilibrium with melting potassium, then DG (Gibbs free energy) is zero.  DG is zero for a system at equilibrium.

When freezing takes place the DH is negative since heat is being released, and since the substance is going to a solid the entropy must be decreasing.

DGo = DHo -TDSo

0 = DHo - TDSo

Then

DHo = TDSo

or

DSo = DHo / T

DSo = (-2.39 kJ / mol) / 340.8 K = -0.00701 kJ/mol K

or

-7.01 J/mol K

Then for 0.140 mol ....

DS = (-7.01 J/mol K) x 0.140 mol = -0.981J/K