Question

The normal melting point of H2O is 273.15 K, and Hfusion= 6010 J mol–1. Calculate the decrease in the normal freezing point at 100 and 500 bar assuming that the density of the liquid and solid phases remains constant at 997 and 917 kg m–3, respectively.

Answer #1

using the clapeyron equation for the solid-liquid boundary,

p=p* + (d_{fus}H/d_{fus}V) &ln(T/T*)

solving it for T,

ln(T/T*)=(p-p*)X dfusV/dfusH

or T=T* e^((p-p*)*dV/dH)

dV= Vm(l) - Vm(s)

since Vm = M/density,

Vm(l) = 18.02*10^-3/997 = 1.8074*10^-5 m^3/mol

Vm(s)=18.02/917 = 1.9651*10^-5 m^3/mol

dfusV = -1.577*10^-6 m^3/mol

so,

for p=100, T=273.15

so,

putting the values in the equation, we get

T=269.6 K

so decrease = 273.15-269.6

= 3.55 K

so the melting point decreases by 3.55 K

At the normal melting point of H2O, ?_fus H_m = 6010 J mol-1. Calculate the melting point for
ice at 2300 bar if both ?_fus H_m and densities ?(s)= 917 kg m-3 and ?(l)= 997 kg m-3
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The normal melting point of substance A is 320C. Melting
heat is a function of temperature. Determine the melting
temperature of substance A at 10 bar.
Data :
Density of liquid A = 10200
kg/m3
Density of Solid A = 12000 kg/m3
For liquid A CpL =
30,0-3,1 x 10-3T (J/mol.K)
For solid A CpS = 20,0 +
9x10-3T (J/mol.K)
Tref =
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ΔHreferime = 4900
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MA = 250 kg/kmol

1. At the normal melting point of ice
∆Hfus= 6.007 kJ mol−1 and
∆Sfus= 22.00 J K−1
mol−1.
a) What is ∆Gfusat the normal melting
point? Look at your answer. Is it correct? Why or why not?
b) Determine ∆G for freezing water at 1 atm and
-10oC assuming that
∆Hfusand
∆Sfusdo not change much over the 0 →
−10oC temperature range.
c) Determine ∆G for freezing water at 1 atm and
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he normal freezing point of water is 0.0 degrees celsius. at
this temperature the density of liquid water is 1.000 g/mL and
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R= 8.3145 J/mol.K, T(K)=T(°C)+273.15.
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Constants
for mercury at 1 atm
heat capacity of Hg(l)
28.0 J/(mol·K)
melting point
234.32 K
enthalpy of fusion
2.29 kJ/mol
Calculate the heat energy released when 12.0 g of liquid mercury
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