Methyl salicylate, C8H8O3, the odorous constituent of oil of wintergreen, has a vapour pressure of 1.00 torr at 54.3oC and 10.0 torr at 95.3oC.
a) What is its vapour pressure at 25oC?
b) What is the minimum number of liters of air that must pass over a sample of the compound at 25oC to vaporize 1.0 mg of it?
1. Use the Arrhenius equation. Use p1 and p2 and T1 and T2 and solve for Ea (actgivation energy) in Joules, then plug that back into the Arrhenius equation and either p1 or p2 to calculate p at 25C.
To solve this problem, we can use the Clausius-Clapeyron equation:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final vapor pressures
ΔHvap is the enthalpy of vaporization
R is the gas constant (0.0821 L*atm/(mol*K))
T1 and T2 are the initial and final temperatures in Kelvin
a) To calculate the vapor pressure at 25°C (298 K), we will use the given data for the vapor pressures at 54.3°C (327.45 K) and 95.3°C (368.45 K).
Using the equation, we have:
ln(P1/1.00) = (ΔHvap/0.0821) * (1/327.45 - 1/298)
We can rearrange and solve for P1:
P1 = 1.00 * e^((ΔHvap/0.0821) * (1/327.45 - 1/298))
b) To determine the minimum number of liters of air required to vaporize 1.0 mg of methyl salicylate, we need to convert the mass of the compound to moles.
First, let's calculate the molar mass of methyl salicylate:
C: 8 * 12.01 g/mol = 96.08 g/mol
H: 8 * 1.01 g/mol = 8.08 g/mol
O: 3 * 16.00 g/mol = 48.00 g/mol
Molar mass of methyl salicylate = 96.08 + 8.08 + 48.00 = 152.16 g/mol
Next, convert 1.0 mg to grams:
1.0 mg = 0.001 g
Now, we can calculate the number of moles of methyl salicylate:
Number of moles = Mass / Molar mass = 0.001 g / 152.16 g/mol
Finally, we can use the ideal gas law to find the volume of air required:
PV = nRT
Assuming the air pressure is 1 atm (since it is not given), and the temperature is 298 K (25°C):
V = (nRT) / P = [(0.001 g / 152.16 g/mol) * 0.0821 L*atm/(mol*K) * 298 K] / 1 atm
Simplifying the equation will give you the minimum number of liters of air required.
To determine the vapor pressure of methyl salicylate at 25oC, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (∆Hvap/R)((1/T1) - (1/T2))
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
R is the ideal gas constant.
∆Hvap is the enthalpy of vaporization.
T1 and T2 are the initial and final temperatures, respectively.
We are given the vapor pressures of methyl salicylate at two temperatures, 54.3oC and 95.3oC. Let's use this data to find the value of ∆Hvap.
Using the given data, we have:
P1 = 1.00 torr at 54.3oC
P2 = 10.0 torr at 95.3oC
T1 = 54.3 + 273.15 = 327.45 K
T2 = 95.3 + 273.15 = 368.45 K
R = 0.0821 L·atm/(mol·K) (Ideal Gas Constant)
Now we can rearrange the Clausius-Clapeyron equation to solve for ∆Hvap:
ln(P2/P1) = (∆Hvap/R)((1/T1) - (1/T2))
∆Hvap/R = (ln(P2/P1))/((1/T1) - (1/T2))
∆Hvap = R × ((ln(P2/P1))/((1/T1) - (1/T2)))
Plugging in the values:
∆Hvap = 0.0821 L·atm/(mol·K) × ((ln(10.0/1.00))/((1/327.45) - (1/368.45)))
Now we can calculate the value of ∆Hvap:
∆Hvap = 12.01 kJ/mol
a) To find the vapor pressure of methyl salicylate at 25oC, we need to rearrange the Clausius-Clapeyron equation again:
ln(P2/P1) = (∆Hvap/R)((1/T1) - (1/T2))
P2 = P1 × exp((∆Hvap/R)((1/T1) - (1/T2)))
P1 = 1.00 torr at 54.3oC
T1 = 54.3 + 273.15 = 327.45 K
T2 = 25 + 273.15 = 298.15 K
R = 0.0821 L·atm/(mol·K) (Ideal Gas Constant)
Now we can plug in the values to calculate the vapor pressure at 25oC:
P2 = 1.00 torr × exp((12.01 kJ/mol)/(0.0821 L·atm/(mol·K)) × ((1/327.45) - (1/298.15)))
Calculating the expression:
P2 ≈ 0.032 torr
Therefore, the vapor pressure of methyl salicylate at 25oC is approximately 0.032 torr.
b) To determine the minimum number of liters of air that must pass over a sample of the compound at 25oC to vaporize 1.0 mg, we need to use the ideal gas law:
PV = nRT
Where:
P is the pressure in atm
V is the volume in liters
n is the number of moles
R is the ideal gas constant
T is the temperature in Kelvin
First, let's calculate the number of moles of methyl salicylate in 1.0 mg:
n = mass/molar mass
Given:
mass = 1.0 mg
molar mass of methyl salicylate (C8H8O3) = 152.15 g/mol
n = (1.0 × 10^(-3) g)/(152.15 g/mol)
n = 6.57 × 10^(-6) mol
Now we can rearrange the ideal gas law to solve for the volume:
V = (nRT)/P
Given:
P = 0.032 torr (or 0.032/760 atm, since 1 atm = 760 torr)
R = 0.0821 L·atm/(mol·K) (Ideal Gas Constant)
T = 25 + 273.15 K
Plugging in the values:
V = (6.57 × 10^(-6) mol × 0.0821 L·atm/(mol·K) × (25 + 273.15) K)/(0.032/760 atm)
Now calculate the expression:
V ≈ 0.0095 L
Therefore, the minimum number of liters of air that must pass over a sample of methyl salicylate at 25oC to vaporize 1.0 mg of it is approximately 0.0095 L.