Question

Elemental mercury is a silver liquid at room temperature. Its normal freezing point is –38.9 °C, and its molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol. What is the entropy change of the system (in J/K) when 5.590 g of Hg(l) freezes at the normal freezing point?

Answer 1

ΔS = -0.272 J/K

Step-by-step explanation:

Step 1: Data given

Its normal freezing point is –38.9 °C

molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol

Mass of Hg = 5.590 grams

Step 2:

ΔG = ΔH - TΔS

At the normal freezing point, or any phase change in general,

ΔG =0

0 = Δ

Hfus

−

Tfus Δ

S

fus

Δ

S

fus = Δ

Hfus

/Tfus

Δ

S

fus = 2290 J/mol / 234.25 K

Δ

S

fus = 9.776 J/mol*K

Since fusion is from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K

Step 3: Calculate moles Hg

Moles Hg = 5.590 grams / 200.59 g/mol

Moles Hg = 0.02787 moles

Step 4: Calculate the entropy change of the system:

Δ

S = -9.776 J/mol*K * 0.02787 moles

ΔS = -0.272 J/K

Answer 2

`\Delta _fS=0.2724(J)/(K)`

Step-by-step explanation:

Hello,

In this case, we define the entropy change for such freezing process as:

`\Delta _fS=(n_(Hg)\Delta _fH)/(T_f)`

Thus, we compute the moles that are in 5.590 g of liquid mercury:

`n_(Hg)=5.590 gHg*(1molHg)/(200.59gHg) =0.02787molHg`

Hence, we compute the required entropy change, considering the temperature to be in kelvins:

`\Delta _fS=(0.02787mol*2.29(kJ)/(mol) )/((-38.9+273.15)K)\\\\\Delta _fS=2.724x10^(-4)(kJ)/(K) *(1000J)/(kJ) \\\\\Delta _fS=0.2724(J)/(K)`

Best regards.