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

##### 1 Answer
Aug 27, 2016

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:

$\textcolor{w h i t e}{m m m m m m m m m m m m m m} \text{Energy"color(white)(mmml)"HF"color(white)(mmmm)"HI}$
$\textcolor{w h i t e}{m m m m m m m m m m m m m m l l} \text{term" color(white)(mmm)"kJ/mol"color(white)(mm)"kJ/mol}$
"HX(g)" → "H·(g)" + "X·(aq)" stackrel(———————————————)(color(white)(mmll) "BE" color(white)(mmmml) 563.2 color(white)(mml) 298.7)
$\text{H·(g)" → "H"^+("g") + "e"^"-} \textcolor{w h i t e}{m m m m m} {I}_{1} \textcolor{w h i t e}{m m m m l} 1318.0 \textcolor{w h i t e}{m m} 1318.0$
$\text{H"^+("g") → "H"^+("aq") color(white)(mmmmmm) Δ_"soln"H_text(H) color(white)(ml) "-1091" color(white)(mml) "-1091}$

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

$\text{HX(aq)" → "HX(g)" color(white)(mmmmmm) "-"Δ_"soln} {H}_{\textrm{H X}} \textcolor{w h i t e}{m l l} 48.1 \textcolor{w h i t e}{m m m l l} 23.0$
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

$\textcolor{w h i t e}{m m m m m m m m m m m m m m} \text{Entropy"color(white)(mmml)"HF"color(white)(mmmml)"HI}$
$\textcolor{w h i t e}{m m m m m m m m m m m m m m l l} \text{term" color(white)(mmm)"J·K"^"-1""mol"^"-1"color(white)(m)"J·K"^"-1""mol"^"-1}$
"HX(g)" → "H·(g)" + "X·(g)" stackrel(———————————————)(color(white)(mmm) ΔS_text(BE) color(white)(mmmm) 99.6 color(white)(mmmm) 88.7)
$\text{H·(g)" → "H"^+("aq") + "e"^"-" color(white)(mmmml) ΔS_"H" color(white)(mmml) "-114.6" color(white)(mmm) "-114.6}$
$\text{X·(g)" + "e"^"-" → "X"^"-""(aq") color(white)(mmmmml) ΔS_"X" color(white)(mmml) "-168.2" color(white)(mmmll) "-71.1}$
$\text{HX(aq)" → "HX(g)" color(white)(mmmmmm) "-"Δ_"soln} {S}_{\textrm{H X}} \textcolor{w h i t e}{m m} 96.2 \textcolor{w h i t e}{m m m m} 83.7$
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 $\text{HF}$

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 $\text{HI}$

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 $\Delta G$

(a) For $\text{HF}$

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"

$K = {e}^{\text{-5.7" = 10^"-3}}$ (Experimental = 7.2 × 10^"-4")

(b) For $\text{HI}$

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 $\text{HF}$ is greater than that of $\text{HI}$.

Its effect is not completely cancelled by the greater enthalpy of hydration of $\text{F"^"-}$.

The major entropy difference is that the tighter solvation of $\text{F"^"-}$ 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$\text{HF}$.