Answer:
[tex]\Delta _fS=0.2724\frac{J}{K}[/tex]
Explanation:
Hello,
In this case, we define the entropy change for such freezing process as:
[tex]\Delta _fS=\frac{n_{Hg}\Delta _fH}{T_f}[/tex]
Thus, we compute the moles that are in 5.590 g of liquid mercury:
[tex]n_{Hg}=5.590 gHg*\frac{1molHg}{200.59gHg} =0.02787molHg[/tex]
Hence, we compute the required entropy change, considering the temperature to be in kelvins:
[tex]\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}[/tex]
Best regards.
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
