The normal boiling point for acetone is 56.5°C. At an elevation of 6500 ft, the atmospheric pressure is 606 torr. What would be the boiling point of acetone (ΔHvap = 32.0 kJ/mol) at this elevation?
What would be the vapor pressure of acetone at 25.0°C at this elevation?
To calculate the boiling point of acetone at an elevation of 6500 ft, we need to consider the change in atmospheric pressure compared to the normal boiling point.
1. Determine the atmospheric pressure at the given elevation:
Atmospheric pressure decreases with increasing elevation. We are given that at 6500 ft, the atmospheric pressure is 606 torr.
2. Convert the atmospheric pressure to atm:
1 atm = 760 torr, so 606 torr ÷ 760 torr/atm = 0.797 atm.
3. Use the Clausius-Clapeyron equation to calculate the boiling point:
ln(P₁/P₂) = ΔHvap/R * (1/T₂ - 1/T₁)
Where:
P₁ = normal boiling point pressure = 1 atm
P₂ = new pressure at the given elevation = 0.797 atm
ΔHvap = enthalpy of vaporization = 32.0 kJ/mol
R = ideal gas constant = 0.008314 J/(K*mol)
T₁ = normal boiling point temperature in Kelvin = 56.5°C + 273.15 = 329.65 K
T₂ = unknown boiling point temperature in Kelvin
Rearranging the equation and solving for T₂:
ln(1/0.797) = (32,000 J/mol) / (0.008314 J/(K*mol)) * (1/T₂ - 1/329.65 K)
ln(1/0.797) = 3,848.69 K/mol * (1/T₂ - 1/329.65 K)
Now we can solve for T₂, the boiling point at the given elevation.
4. Calculate the boiling point of acetone at 6500 ft:
ln(1/0.797) = 3,848.69 K/mol * (1/T₂ - 1/329.65 K)
ln(1/0.797) = 3,848.69 K/mol * (329.65 K - T₂) / (329.65 K * T₂)
Taking the inverse logarithm (e^x) of both sides:
1/0.797 = e^(3,848.69 K/mol * (329.65 K - T₂) / (329.65 K * T₂))
0.797 = e^(3,848.69 K/mol * (329.65 K - T₂) / (329.65 K * T₂))
Now, solve for T₂.
To calculate the vapor pressure of acetone at 25.0°C (298.15 K) at this elevation, we can use the Clausius-Clapeyron equation in a slightly different form:
ln(P₁/P₂) = -ΔHvap/R * (1/T₂ - 1/T₁)
Where:
P₁ = unknown vapor pressure at 25.0°C
P₂ = atmospheric pressure at the given elevation = 0.797 atm
ΔHvap = enthalpy of vaporization = 32.0 kJ/mol
R = ideal gas constant = 0.008314 J/(K*mol)
T₁ = normal boiling point temperature in Kelvin = 56.5°C + 273.15 = 329.65 K
T₂ = temperature in Kelvin = 25.0°C + 273.15 = 298.15 K
Now, rearrange the equation and solve for P₁.
ln(P₁/0.797) = -32,000 J/mol / (0.008314 J/(K*mol)) * (1/298.15 K - 1/329.65 K)
ln(P₁/0.797) = -3,848.69 K/mol * (1/298.15 K - 1/329.65 K)
Taking the inverse logarithm (e^x) of both sides:
P₁/0.797 = e^[-3,848.69 K/mol * (1/298.15 K - 1/329.65 K)]
P₁ = 0.797 * e^[-3,848.69 K/mol * (1/298.15 K - 1/329.65 K)]
Now, calculate the vapor pressure of acetone at 25.0°C at an elevation of 6500 ft.
To find the boiling point of acetone at an elevation of 6500 ft, we need to take into account the change in atmospheric pressure. The boiling point of a liquid depends on the balance between the vapor pressure exerted by the liquid and the external atmospheric pressure.
Let's start by converting the atmospheric pressure at the given elevation from torr to atm. There are 760 torr in 1 atm, so:
Atmospheric pressure = 606 torr / 760 torr/atm = 0.798 atm
Next, we need to consider the change in boiling point due to the difference in atmospheric pressure. The boiling point of a liquid increases as the external pressure increases. This relationship can be described using Raoult's law:
∆T = (Kb * m * i) / (∆Hvap)
Where:
∆T = Change in boiling point
Kb = Molal boiling point elevation constant (depends on the solvent - for acetone it is 1.71 °C/m)
m = Molality of the solution (moles of solute per kg of solvent)
i = Van't Hoff factor (1 for acetone)
∆Hvap = Heat of vaporization
Since we are dealing with pure acetone, the molality of the solution is 0. We can neglect this term in our calculations.
Now, let's calculate the change in boiling point (∆T):
∆T = (1.71 °C/m * 0 * 1) / (32.0 kJ/mol)
∆T = 0°C
Since ∆T is 0°C and the normal boiling point of acetone is 56.5°C, the boiling point of acetone at an elevation of 6500 ft will still be 56.5°C.
Moving on to the second part of the question - finding the vapor pressure of acetone at 25.0°C at this elevation.
To determine the vapor pressure using the Antoine equation:
log(P) = A - (B / (T + C))
Where:
P = Vapor pressure of acetone
T = Temperature in Celsius
A, B, and C are constants specific to acetone.
For acetone, the Antoine equation constants are as follows:
A = 14.204
B = 2643.5
C = -13.572
Now, let's plug in the values to calculate the vapor pressure at 25.0°C:
log(P) = 14.204 - (2643.5 / (25 + (-13.572)))
log(P) = 14.204 - (2643.5 / 11.428)
log(P) = 14.204 - 231.349
log(P) = -217.145
To solve for P, we can take the antilog of both sides:
P = 10^(-217.145)
P = 2.884 x 10^(-218) atm
Therefore, the vapor pressure of acetone at 25.0°C at an elevation of 6500 ft is approximately 2.884 x 10^(-218) atm.