Enthalpy of Vaporization of SO2 at -25°C

Introduction
The enthalpy of vaporization is the amount of energy required to change a substance from its liquid state to its gaseous state at a given temperature. In this case, we will calculate the enthalpy of vaporization of SO2 (sulfur dioxide) at -25°C.

Formula
The enthalpy of vaporization can be calculated using the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively,
ΔHvap is the enthalpy of vaporization,
R is the ideal gas constant, and
T1 and T2 are the initial and final temperatures respectively.

Given Data
Temperature, T1 = -25°C = 248.15 K (converted to Kelvin)

Step 1: Finding Vapor Pressure at T1
The first step is to find the vapor pressure of SO2 at -25°C. We can use the Antoine equation to calculate the vapor pressure:

log10(P) = A - (B / (T + C))

Using the Antoine coefficients for SO2:
A = 7.286
B = 1364.0
C = -42.66

Plugging in the temperature T1 = 248.15 K into the equation, we can solve for log10(P):

log10(P) = 7.286 - (1364.0 / (248.15 - 42.66))
log10(P) = 7.286 - (1364.0 / 205.49)
log10(P) = 7.286 - 6.639
log10(P) = 0.647
P = 10^0.647
P = 4.57 kPa (approximately)

Step 2: Calculating Vapor Pressure at T2
In this case, T2 is not given. We can assume T2 to be the boiling point of SO2, which is -10°C = 263.15 K.

Using the Antoine equation again with the same coefficients, we can calculate the vapor pressure at T2:

log10(P) = 7.286 - (1364.0 / (263.15 - 42.66))
log10(P) = 7.286 - (1364.0 / 220.49)
log10(P) = 7.286 - 6.181
log10(P) = 1.105
P = 10^1.105
P = 11.94 kPa (approximately)

Step 3: Calculating Enthalpy of Vaporization
Now that we have the vapor pressures at T1 and T2, we can substitute these values into the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

ln(11.94/4.57) = (ΔHvap/8.314) * (1/248.15 - 1/263.15)

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