There are two steps to this problem:
- Use the Clausius-Clapeyron equation to find the enthalpy of vaporization of ethanol
- Use the new information in the Clausius-Clapeyron equation to calculate the normal boiling point of ethanol
The Clausius-Clapeyron equation is:
#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
Part 1. Calculate the enthalpy of vaporization
#p_1 = "0.016 bar"; T_1 = "0 °C" = "273.15 K"#
#p_2 = "0.470 bar"; T_2 = "60 °C" = "333.15 K"#
#R = "8.314 J·K"^"-1""mol"^"-1"#
#ln((0.470color(red)(cancel(color(black)("atm"))))/(0.016 color(red)(cancel(color(black)("atm"))))) = (Δ_"vap"H)/("8.314 J"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1")(1/(273.15 color(red)(cancel(color(black)("K")))) - 1/(333.15 color(red)(cancel(color(black)("K")))))#
#3.380 = (Δ_"vap"H)/("8.314 J·mol"^"-1") × 6.593 × 10^"-4"#
#"28.10 J·mol"^"-1" = 6.593 ×10^"-4"Δ_"vap"H#
#Δ_"vap"H = ("28.10 J·mol"^"-1")/(6.593 ×10^"-4") = "42 600 J·mol"^"-1" = "42.6 kJ·mol"^"-1"#
Part 2. Calculate the normal boiling point
#p_1 = "0.470 bar"; color(white)(mmmmmll)T_1 = "60 °C" = "333.15 K"#
#p_2 = "1 atm = 1.013 bar";color(white)(ml) T_2 = ?#
#R = "8.314 J·K"^"-1""mol"^"-1"; Δ_text(vap)H = "42.6 kJ·mol"^"-1"#
#ln((1.013color(red)(cancel(color(black)("bar"))))/(0.470 color(red)(cancel(color(black)("bar"))))) = ("42 600" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1"))))(1/("333.15 K") - 1/T_2)#
#ln2.155 = ("5124 K")(1/("333.15 K") - 1/T_2) = 5124/333.15 - "5124 K"/T_2)#
#0.7679 = 15.38 - "5124 K"/T_2#
#"5124 K"/T_2 = "15.38-0.7679 = 14.61"#
#T_2 = "5124 K"/14.61 = "350.7 K = 77.6 °C"#
