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

Answer:

The vapor pressure at 60.0°C is 2.416 atm

Explanation:

To solve this problem, we use Clausius-Clapeyron equation

$ln\frac{P_2}{P_1} = \frac{-\delta H}{R}[\frac{1}{T_2}-\frac{1}{T_1}]= \frac{\delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

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

$ln\frac{P_2}{P_1} = \frac{29100}{8.314}[\frac{1}{298}-\frac{1}{333}] =1.23449$

$\frac{P_2}{P_1} = e^{1.23449}$ ⇒ $\frac{P_2}{P_1} = 3.43663$

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.

Answer 2

Diethyl ether has a H°vap of 29.1 kJ/mol and vapor pressure of 0.703 atm at 25.0°C. its vapor pressure at 60.0°C - 2.14 atm.

In such case to find vapor pressure at a specific temperature we use Clausius-Clapeyron equation

$\ln\dfrac{P_2}{P_1}=\dfrac{-\delta H}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\\\ln\dfrac{P_2}{P_1}=\dfrac{\delta H}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$

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

$ln\dfrac{P_2}{P_1}=\dfrac{\delta H}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)\\ln\dfrac{P_2}{P_1}=\dfrac{\29100}{8.314}\left(\dfrac{1}{298}-\dfrac{1}{333}\right)$

$ln\frac{P2}{0.703} = 1.23449\\ln\frac{P2}{0.703} = e^{1.23449}\\ln\frac{P2}{0.703} = 3.43663$

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

= 2.416 atm

Thus, vapor pressure at 60.0°C P₂ = 2.416 atm.

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