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".
