Question

Methane (CH4) has a boiling point of −164°C at 1.00 atm and a vapor pressure of 42.8 atm at −100°C. What is the heat of vaporization of CH4?

I know that the answer is 9210 J/mol. I just do not understand the steps I take to solve the problem.

Answer 1

The heat of vapourization is `"9.22 kJ·mol"^"-1"`

Explanation:

Chemists often use the Clausius-Clapeyron equation to determine the enthalpy of vapourization of pure liquids:

`color(blue)(bar(ul(|color(white)(a/a) ln(p_2/p_1) = (Δ_"vap"H)/R(1/T_1- 1/T_2)color(white)(a/a)|)))" "`

where

`p_1` and `p_2` are the vapour pressures at temperatures
`T_1` and `T_2`

`Δ_"vap"H` = the enthalpy of vaporization of the liquid

`Rcolor(white)(mmll)` = the Universal Gas Constant

In your problem,

`p_1 = "1 atm"; color(white)(ml)T_1 = "-164 °C" = "109.15 K"`

`p_2 = "42.8 atm"; T_2 = "-100 °C" = "173.15 K"`

`R = "8.314 J·K"^"-1""mol"^"-1"`

`ln((42.8color(red)(cancel(color(black)("atm"))))/(1 color(red)(cancel(color(black)("atm"))))) = (Δ_"vap"H)/("8.314 J"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1")(1/(109.15 color(red)(cancel(color(black)("K")))) - 1/(173.15 color(red)(cancel(color(black)("K")))))`

`3.757 = (Δ_"vap"H)/("8.314 J·mol"^"-1") × 3.386 × 10^"-3"`

`"31.23 J·mol"^"-1" = 3.386 ×10^"-3"Δ_"vap"H`

`Δ_"vap"H = ("31.23 J·mol"^"-1")/(3.386 ×10^"-3") = "9220 J·mol"^"-1" = "9.22 kJ·mol"^"-1"`