Dissociation of a gas $A_{2}$ takes place according to the following chemical reaction. At equilibrium, the…
Dissociation of a gas $A_{2}$ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K.
$A_{2}(g) \rightleftharpoons 2 A(g)$
The standard Gibbs energy of formation of the involved substances has been provided below:
$\begin{aligned}
\begin{array}{|c|c|}
\hline
\text{Substance} & \Delta G_{f}^{\circ} / \text{kJ mol}^{-1} \\
\hline
A_{2} & -100.00 \\
\hline
A & -50.832 \\
\hline
\end{array}
\end{aligned}$
The degree of dissociation of $A_{2}(g)$ is given by $\left(x \times 10^{-2}\right)^{\frac{1}{2}} where x= \_\_\_\_$. (Nearest integer).
[Given: $R=8 \text{J mol}^{-1} \text{K}^{-1}, \log 2=0.3010, \log 3=0.48$ ]
Assume degree of dissociation is not negligible.
Solution
For the dissociation reaction $A_2(g) \rightleftharpoons 2A(g)$, first calculate the standard Gibbs free energy: $\Delta G°_{rxn} = 2(-50.832) - (-100.00) = -1.664$ kJ/mol. Using $\Delta G° = -RT \ln K_p$: $-1664 = -8 \times 300 \times \ln K_p$, giving $K_p = e^{0.6933} = 2$. For dissociation with degree of dissociation $\alpha$, starting with 1 mole of $A_2$ at 1 bar total pressure: At equilibrium, moles are $(1-\alpha)$ for $A_2$ and $2\alpha$ for $A$, with total $(1+\alpha)$ moles. $K_p = \frac{4\alpha^2}{1-\alpha^2} = 2$. Solving: $2(1-\alpha^2) = 4\alpha^2$ gives $\alpha^2 = 1/3$ or $\alpha = 0.577$. Since degree of dissociation is $(x \times 10^{-2})^{1/2}$: $(x \times 10^{-2})^{1/2} = 0.577$ gives $x \times 10^{-2} = 0.333$, so $x = 33.3 \approx 33$.