Question

Diethyl ether has a H°vap of 29.1 kJ/mol and a vapor pressure of 0.703 atm at 25.0°C. What is its vapor pressure at 60.0°C?

Answer 1

To estimate the vapor pressure of diethyl ether at 60.0°C, we can use the Clausius-Clapeyron equation. Plugging in the given values, we find that the vapor pressure of diethyl ether at 60.0°C is estimated to be 1.316 atm.

Step-by-step explanation:

To estimate the vapor pressure of diethyl ether at 60.0°C, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature.

To calculate the vapor pressure at 60.0°C, we can rearrange the equation as follows:

ln(P2/P1) = (∆Hvap/R)((1/T1) - (1/T2))

Where P1 is the vapor pressure at 25.0°C, P2 is the vapor pressure at 60.0°C, ∆Hvap is the enthalpy of vaporization, R is the ideal gas constant, T1 is the initial temperature, and T2 is the final temperature.

Using the given values, P1 = 0.703 atm and T1 = 25.0°C (298 K), and solving for P2 at T2 = 60.0°C (333 K), we can find the value of ∆Hvap.

Plugging in the values, we get:

ln(P2/0.703) = (29.1 kJ/mol / (8.314 J/(mol·K)))(1/298 K - 1/333 K)

After solving the equation, we find that P2 = 1.316 atm, which is the estimated vapor pressure of diethyl ether at 60.0°C.

Answer 2

The vapor pressure at 60.0°C is 2.416 atm

Step-by-step explanation:

To solve this problem, we use Clausius-Clapeyron equation

`ln(P_2)/(P_1) = (-\delta H)/(R)[(1)/(T_2)-(1)/(T_1)]= (\delta H)/(R)[(1)/(T_1)-(1)/(T_2)]`

where;

Initial pressure P₁ = 0.703 atm

Initial Temperature T₁ = 25+273 = 298K

Final temperature T₂ = 60+273 = 333K

Change in enthalpy of vaporization ΔH = 29.1 KJ/mol = 29100J/mol

R is Boltzman constant = 8.314 J/K.mol

`ln(P_2)/(P_1) = (29100)/(8.314)[(1)/(298)-(1)/(333)] =1.23449`

`(P_2)/(P_1) = e^(1.23449)` ⇒
`(P_2)/(P_1) = 3.43663`

P₂ = P₁ (3.43663) = (0.703 atm)(3.43663) = 2.416 atm

P₂ = 2.416 atm

Therefore, the vapor pressure at 60.0°C is 2.416 atm.