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pashok25 [27]
3 years ago
15

Elemental mercury is a silver liquid at room temperature. Its normal freezing point is –38.9 °C, and its molar enthalpy of fusio

n 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?
Chemistry
2 answers:
Lilit [14]3 years ago
8 0

Answer:

ΔS = -0.272 J/K

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

Butoxors [25]3 years ago
4 0

Answer:

\Delta _fS=0.2724\frac{J}{K}

Explanation:

Hello,

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

\Delta _fS=\frac{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*\frac{1molHg}{200.59gHg} =0.02787molHg

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

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

Best regards.

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