water heat capacity = 75.3 J/mol K ice heat capacity = 36.2 1. (35%) Using only
ID: 1019958 • Letter: W
Question
water heat capacity = 75.3 J/mol K
ice heat capacity = 36.2
1. (35%) Using only the values in Table 2.4 of the text and a heat of fusion of ice of 6010 J/mol at 273.15K, -2.5°C. You must treat the heat of fusion as a function of temperature. Things you may assume: the volume change involved is constant; the heat capacities are constant, the heat of fusion is independent of pressure. calculate the pressure in bar that must be applied to ice to drop its melting point from O°C to 2. (25%) For the mechanism below, show the equation for the rate of formation of hydrogen iodide in terms of the intermediates. Do not solve it terms of the reactants. Also, show the two equations thatExplanation / Answer
We require to relate Freezing Pressure and Tempature freezing.
This is a special point, it can be modeling the euqilibrium line of solid-liquid for water.
Recall that in solid-liquid equilibirum, the P-T diagram will form a straight line in the equilibrium.
This stright line can be modeled as a straight line since, PvsT increases almost linearly
then; the model that best suits this is Clasius Clapeyron Equation.
The CCE:
ln(P2/P1) = H/R*(1/T1 - 1/T2)
note that we assume heat of fussion is independent of P and is constant for V change.
That is, H is constant
Hice = 6010 J/mol (at 273.15K and P = 1 atm)
R = 8.314 J/molK
question:
for T from 0C to -2.5C... use Abs T
T1 = 0+273.15 = 273.15 K
P1 = 1 atm = 1.01325 bar
T2 = -2.5+273.15 = 270.65
P2 = ? atm = ? bar
Substitute in Equation
ln(P2/P1) = H/R*(1/T1 - 1/T2)
ln(P2/1.01325) = 6010/8.314*(1/273.15 - 1/270.65)
Now,
solve for P2
P2/1.01325 = exp(6010/8.314*(1/273.15 - 1/270.65))
P2 = 1.01325*0.9758510
P2 = 0.988781025 bar
Then
this must be the Pressure in which T = -2.5 C will be the freezing point...
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