Question

Question: Use the given data from Appendix D and F to estimate the temperature at which I₂(g) and I₂(s)are in equilibrium at a pressure of 1 atm. Can this equilibrium actually be achieved?

Short answer

$T=159.2C^{°}$

The dissociation energy of$I_2$ is $151{\text{kJmol}}^{-1}$. Equilibrium cannot be achieved because the bonds in$I_2$ are weak and show the tendency to dissociate into the atom |.

Step-by-step solution

Given data

$\Delta H_{\text{f}}^{°}$represents the standard enthalpy of molecules.

$\Delta H_{\text{f}}^{°}$ $\left[{\text{I}}_2\left(s\right)\right]$is 0.

$\Delta H_{\text{f}}^{°}$$\left[{\text{I}}_2\left(g\right)\right]$ is $62.4{\text{kJmol}}^{-1}$.

$\Delta S^{°}$represents standard entropy of the molecules.

$\Delta S^{°}$$\left[{\text{I}}_2\left(s\right)\right]$is $116.14{\text{JK}}^{-1}$.

$\Delta S^{°}$$\left[{\text{I}}_2\left(g\right)\right]$is $260.58{\text{JK}}^{-1}$.

Pressure = 1 atm.

Concept used

Enthalpy change is the amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in the chemical reaction.

It is represented by ΔH.

Entropy Change is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic system. It is related to the conversion of heat or enthalpy in work.

Calculate the change in enthalpy

Calculate the enthalpy change by using the given data as follows:

$\Delta H_{\text{f}}^{°}$represents the standard enthalpy of molecules.

$\begin{array}{c}\Delta H^{°}=\Delta H_{\text{f}}^{°}\left[{\text{I}}_2\left(s\right)\right]-\Delta H_{\text{f}}^{°}\left[{\text{I}}_2\left(g\right)\right]\\ \Delta H^{°}=0-1\text{mol}\times 62.4{\text{kJmal}}^{-1}\\ =-62.44\text{kJ}\end{array}$

Calculate the change in entropy

Calculate the entropy change by using the given data as follows:

$\begin{array}{c}\Delta S^{°}=S^{°}\left[{\text{I}}_2\left(s\right)\right]-S^{°}\left[{\text{I}}_2\left(g\right)\right]\\ \Delta S^{°}=116.14{\text{JK}}^{-1}-260.58{\text{JK}}^{-1}\\ =-144.44{\text{JK}}^{-1}\end{array}$

Conclusion

For a reaction at equilibrium, free energy change of the reaction is zero. So$\Delta G^{°}$is 0.

$\Delta G^{°}=\Delta H^{°}-T\Delta S^{°}$

$T=\frac{\Delta H^{°}}{\Delta S^{°}}$

$\begin{array}{c}T=\frac{-62.44\text{k}.\text{J}}{-\mathrm{144.44.}{\text{JK}}^{-1}}\times \frac{{10}^3\text{J}}{1\text{k}.\text{J}}\\ =432.3\text{K}\end{array}$

Change into Celsius:

$432.3-273.15=159.2C^{°}$

The dissociation energy of I₂is$151{\text{kJmol}}^{-1}$. Equilibrium cannot be achieved because the bonds in I₂. are weak and show tendency to dissociate into atom |.