Q94CP - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the equilibrium constants for decomposition of the hydrogen halides at  :

 $\mathbf{\text{2HX(g)}}\mathbf{\rightleftharpoons }{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + X}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

What do these values indicate about the extent of decomposition of $\mathit{H}\mathit{X}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{298}\mathit{K}\mathbf{?}$ Suggest a reason for this trend.

Step-by-Step Solution

Answer

The equilibrium constants for the halides are,

Fluorine: $K=3.89\times {10}^{-97}$ , Chlorine:$K=3.89\times {10}^{-34}$  , Bromine:$K=4.96\times {10}^{-20}$  , lodine: $K=1.14\times {10}^{-3}$ . Because it is difficult to separate the hydrogen from more electronegative halogens, their $K$will be lower.

Step 1: Concept Introduction

Hydrogen halides are diatomic inorganic substances that act as Arrhenius acids in chemistry. The formula is  $\mathit{H}\mathit{X}$ where  $\mathit{X}$ is fluorine, chlorine, bromine, iodine, or astatine is, and $\mathit{Y}$  is one of the halogens.

Step 2: Calculating the equilibrium constant for Fluorine

The reaction is given as,

 $\text{2HX(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{(g) + X}}_{\text{2}}\text{(g)}$

Fluorine:   $\text{2HF(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{(g) + F}}_{\text{2}}\text{(g)}$

First, we have to solve for $∆G^{°}$  using the Gibb's free energy constants found in Appendix B.

 $$\begin{gathered} ∆G^{°}=n_p∆G_f^{°}\left(product\right)-n_r∆G_f^{°}\left(reac\tan t\right) \\ =\left[n∆G_f^{°}\left({\text{H}}_{\text{2}}\text{(g)}\right)+n_r∆G_f^{°}\left({\text{F}}_{\text{2}}\text{(g)}\right)\right]-\left[n∆G_f^{°}\text{(HF(g))}\right] \\ =\left(\left[1\left(0\right)+1\left(0\right)\right]-\left[2\left(-275\right)\right]\right)kJ/mol \\ ∆G^{°}=550kJ/mol \end{gathered}$$

Then solve for $K$ .

$$\begin{gathered} ∆G^{°}=-RTInK \\ K=e^{\frac{∆f°}{{\displaystyle \text{ - TT}}}} \\ \frac{∆G^{°}}{-RT}=\frac{550kJ/mol\times {\displaystyle \frac{1000J}{1kJ}}}{-8.314J/mol\times K\times 298K} \\ \frac{∆G^{°}}{-RT}=-222.0 \\ K=e^{-222.0} \\ K=3.89\times {10}^{-97} \end{gathered}$$

 

The equilibrium constant for fluorine is $3.89\times {10}^{-97}$ .

Step 3: Calculating the equilibrium constant for Chlorine

The reaction is, 

 $\text{2HCl(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{(g) + Cl}}_{\text{2}}\text{(g)}$

First, we have to solve for $∆G^{°}$  using the Gibb's free energy constants found in Appendix B.

 $$\begin{gathered} ∆G^{°}=n_p∆G_f^{°}\left(product\right)-n_r∆G_f^{°}\left(reac\tan t\right) \\ =\left[n∆G_f^{°}\left({\text{H}}_{\text{2}}\text{(g)}\right)+n∆G_f^{°}\left({\text{Cl}}_{\text{2}}\text{(g)}\right)\right]-\left[n∆G_f^{°}(\text{HCl(g))}\right] \\ =\left(\left[1\left(0\right)+1\left(0\right)\right]-\left[2\left(-95.30\right)\right]\right)kJ/mol∆G^{°} \\ =190.6kJ/mo\ln \end{gathered}$$

Then solve for $K$ .

$$\begin{gathered} ∆G^{°}=-RTInK \\ K=e^{\frac{∆G^{°}}{-RT}} \\ \frac{∆G^{°}}{-RT}=\frac{190.6kJ/mol\times {\displaystyle \frac{1000J}{1kJ}}}{-8.314J/mol\times K\times 298K} \\ \frac{∆G^{°}}{-RT}=-76.93 \\ K=e^{-76.93} \\ K=3.89\times {10}^{-34} \end{gathered}$$

 

The equilibrium constant for chlorine is $3.89\times {10}^{-34}$ .

Step 3: Calculating the equilibrium constant for Bromine

 The reaction is,$\text{2HBr(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{(g) + Br}}_{\text{2}}\text{(g)}$

 

First, we have to solve for  $∆G^{°}$ using the Gibb's free energy constants found in Appendix B.

 $$\begin{gathered} \text{2HBr(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{(g) + Br}}_{\text{2}}\text{(g)} \\ ∆G^{°}=n_p∆G_f^{°}\left(product\right)-n_r∆G_f^{°}\left(reac\tan t\right) \\ =\left[n∆G_f^{°}\left({\text{H}}_{\text{2}}\text{(g)}\right)+n∆G_f^{°}\left({\text{Br}}_{\text{2}}\text{(g)}\right)\right]-\left[{\text{n∆G}}_f^{°}\text{(HBr(g))}\right] \\ =\left(\left[1\left(0\right)+1\left(3.13\right)\right]-\left[2\left(-53.5\right)\right]\right)kJ/mol \\ ∆G^{°}=110.13kJ/mol \end{gathered}$$
Then solve for $K$ .

$$\begin{gathered} ∆G^{°}=-RTInK \\ K=e^{\frac{∆G^{°}}{-RT}} \\ \frac{∆G^{°}}{-RT}=\frac{110.13kJ/mol{\displaystyle \frac{1000J}{1kJ}}}{-8.314J/mol\times K\times 298K} \\ \frac{∆G^{°}}{-RT}=-44.45 \\ K=e^{-44.45} \\ K=4.96\times {10}^{-20} \end{gathered}$$ 

The equilibrium constant for bromine is$4.96\times {10}^{-20}$  .

Step 4: Calculating the equilibrium constant for Iodine

The reaction is,$\text{2HI(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{(g) + I}}_{\text{2}}\text{(g)}$

 

First, we have to solve for $∆G^{°}$  using the Gibb's free energy constants found in Appendix B.

 $$\begin{gathered} ∆G^{°}=n_p∆G_f^{°}\left(product\right)-n_r∆G_f^{°}\left(reac\tan t\right) \\ =\left[n∆G_f^{°}\left({\text{H}}_{\text{2}}\text{(g)}\right)+n∆G_f^{°}\left({\text{I}}_{\text{2}}\text{(g)}\right)\right]-\left[n∆G_f^{°}\text{(HI(g))}\right] \\ =\left(\left[1\left(0\right)+1\left(19.38\right)\right]-\left[2\left(1.3\right)\right]\right)kJ/mol \\ ∆G^{°}=16.78kJ/mol \end{gathered}$$

Then solve for $K$ .

 $$\begin{gathered} ∆G^{°}=-RTInK \\ K=e^{\frac{∆G^{°}}{{\displaystyle \text{TT}}}} \\ \frac{∆G^{°}}{-RT}=\frac{16.78kJ/mol{\displaystyle \frac{1000J}{1kJ}}}{-8.314J/mol\times K\times 298K} \\ \frac{∆G^{°}}{-RT}=-6.77 \\ K=e^{-6.77} \\ K-1.14\times {10}^{-3} \end{gathered}$$

The equilibrium constant for iodine is $1.14\times {10}^{-3}$ .

As   moves down the periodic table, the $K$  of each  $H X$ grows.$HF < HCl < HBr < HI$ . This is due to the fact that the electronegativity diminishes as the period progresses. If the electronegativity is low, the halogen, like iodine, will easily release the hydrogen. Because it is the most electronegative element, $HF$ has an extremely low$K$ .