Diethyl ether was one of the first chemicals used as an anesthetic. At 34.6°C, diethyl ether has a vapor pressure of 760. torr, and at 18.5°C, it has a vapor pressure of 413 torr. What is the ΔH of vaporization for diethyl ether?
Use the Clasius Clapeyron equation.
To find the enthalpy of vaporization (ΔH) of diethyl ether, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔH/R * (1/T2 - 1/T1)
where:
P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively,
ΔH is the enthalpy of vaporization,
R is the ideal gas constant (8.314 J/(mol·K)),
T1 and T2 are the temperatures in Kelvin.
First, let's convert the given temperatures from Celsius to Kelvin:
T1 = 34.6°C + 273.15 = 307.75 K
T2 = 18.5°C + 273.15 = 291.65 K
Now, let's substitute the values into the Clausius-Clapeyron equation:
ln(413/760) = -ΔH/(8.314) * (1/291.65 - 1/307.75)
Now we can solve this equation to find ΔH. Rearranging the equation to solve for ΔH:
ΔH = -R * (ln(413/760)) / (1/291.65 - 1/307.75)
Now, we can plug in the values and calculate ΔH.