Can you provide the thermodynamic data to show why #"HF"# is such a weak acid compared to #"HI"#?

1 Answer

Answer:

Warning! Long answer. Here are the numbers I found.

Explanation:

1. Calculate #"ΔH# values

Use Hess's Law (the Born-Haber cycle).

The target equation is: #"H-X(aq)" → "H"^+("aq") + "X"^"-"("aq")#

Here are the steps:

#color(white)(mmmmmmmmmmmmmm)"Energy"color(white)(mmml)"HF"color(white)(mmmm)"HI"#
#color(white)(mmmmmmmmmmmmmmll)"term" color(white)(mmm)"kJ/mol"color(white)(mm)"kJ/mol"#
#"HX(g)" → "H·(g)" + "X·(aq)" stackrel(———————————————)(color(white)(mmll) "BE" color(white)(mmmml) 563.2 color(white)(mml) 298.7)#
#"H·(g)" → "H"^+("g") + "e"^"-" color(white)(mmmmm) I_1 color(white)(mmmml) 1318.0 color(white)(mm) 1318.0#
#"H"^+("g") → "H"^+("aq") color(white)(mmmmmm) Δ_"soln"H_text(H) color(white)(ml) "-1091" color(white)(mml) "-1091"#

#"X·(g)" + "e"^"-" → "X"^"-""(g") color(white)(mmmmmm) "EA" color(white)(mmmm) "-343.1" color(white)(mml) "-316.7"#
#"X"^"-"("g") → "X"^"-"("aq") color(white)(mmmmmmm) Δ_"soln"H_text(X) color(white)(mll) "-506.9" color(white)(mml) "-291.0"#

#"HX(aq)" → "HX(g)" color(white)(mmmmmm) "-"Δ_"soln"H_text(HX) color(white)(mll) 48.1 color(white)(mmmll) 23.0#
#stackrel(————————————————————)("H-X(aq)" → "H"^+("aq") + "X"^"-"("aq") color(white)(l) Δ_"diss"H color(white)(mmm) "-12" color(white)(mmmm) "-59"#

2. Calculate #ΔS# values

The entropy changes are

#color(white)(mmmmmmmmmmmmmm)"Entropy"color(white)(mmml)"HF"color(white)(mmmml)"HI"#
#color(white)(mmmmmmmmmmmmmmll)"term" color(white)(mmm)"J·K"^"-1""mol"^"-1"color(white)(m)"J·K"^"-1""mol"^"-1"#
#"HX(g)" → "H·(g)" + "X·(g)" stackrel(———————————————)(color(white)(mmm) ΔS_text(BE) color(white)(mmmm) 99.6 color(white)(mmmm) 88.7)#
#"H·(g)" → "H"^+("aq") + "e"^"-" color(white)(mmmml) ΔS_"H" color(white)(mmml) "-114.6" color(white)(mmm) "-114.6"#
#"X·(g)" + "e"^"-" → "X"^"-""(aq") color(white)(mmmmml) ΔS_"X" color(white)(mmml) "-168.2" color(white)(mmmll) "-71.1"#
#"HX(aq)" → "HX(g)" color(white)(mmmmmm) "-"Δ_"soln"S_text(HX) color(white)(mm) 96.2 color(white)(mmmm) 83.7#
#stackrel(————————————————————)("H-X(aq)" → "H"^+("aq") + "X"^"-"("aq") color(white)(l) Δ_"diss"S color(white)(mmml) "-87" color(white)(mmmmll) "-13"#

3. Calculate #ΔG# values

(a) For #"HF"#

#Δ G = ΔH -TΔS = "-12 000 J·mol"^"-1" + 298.15 color(red)(cancel(color(black)("K"))) × "87 J·"color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1" = "14 000 J·mol"^"-1"#

(b) For #"HI"#

#ΔG = ΔH -TΔS = "-59 000 J·mol"^"-1" + 298.15 K × "13.3 J·K"^"-1""mol"^"-1" =" -55 000 J·mol"^"-1"#

4. Calculate #K# values from #DeltaG#

(a) For #"HF"#

#ΔG = -RTlnK#

#lnK = -(ΔG)/(RT) = ("-14 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K")))) = "-5.7"#

#K = e^"-5.7" = 10^"-3"# (Experimental = #7.2 × 10^"-4"#)

(b) For #"HI"#

#lnK = -(ΔG)/(RT) = ("55 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K")))) = 22#

#K = e^22 = 10^9# (Experimental = #2 × 10^9#)

5. Explanation of differences

The bond dissociation energy of #"HF"# is greater than that of #"HI""#.

Its effect is not completely cancelled by the greater enthalpy of hydration of #"F"^"-"#.

The major entropy difference is that the tighter solvation of #"F"^"-"# due to H-bonding makes the arrangement of the water molecules much less random,

This leads to a highly negative and unfavourable #ΔS_"X"# for#"HF"#.