Calculation of Standard Gibbs Energy Change

The chemical reaction is given as: $\text{A}_2\text{(g)} \rightleftharpoons 2\text{A(g)}$.

The standard Gibbs energy change for the reaction ($\Delta G^\circ_{rxn}$) is calculated using the standard Gibbs energies of formation ($\Delta G_f^\circ$) of the products and reactants:

$\Delta G^\circ_{rxn} = \sum \nu_p \Delta G_f^\circ(\text{products}) - \sum \nu_r \Delta G_f^\circ(\text{reactants})$

Given values:

• $\Delta G_f^\circ(\text{A}_2) = -100.00 \text{ kJ mol}^{-1}$
• $\Delta G_f^\circ(\text{A}) = -50.832 \text{ kJ mol}^{-1}$

Convert these values to Joules per mole for consistency with the gas constant R:

• $\Delta G_f^\circ(\text{A}_2) = -100.00 \times 1000 = -100000 \text{ J mol}^{-1}$
• $\Delta G_f^\circ(\text{A}) = -50.832 \times 1000 = -50832 \text{ J mol}^{-1}$

Calculate $\Delta G^\circ_{rxn}$:

$\Delta G^\circ_{rxn} = [2 \times \Delta G_f^\circ(\text{A})] - [1 \times \Delta G_f^\circ(\text{A}_2)]$

$\Delta G^\circ_{rxn} = [2 \times (-50832 \text{ J mol}^{-1})] - [1 \times (-100000 \text{ J mol}^{-1})]$

$\Delta G^\circ_{rxn} = -101664 \text{ J mol}^{-1} + 100000 \text{ J mol}^{-1}$

$\Delta G^\circ_{rxn} = -1664 \text{ J mol}^{-1}$

Calculation of Equilibrium Constant

The relationship between the standard Gibbs energy change and the equilibrium constant ($K_p$) is:

$\Delta G^\circ = -RT \ln K_p$

Given values:

• $R = 8 \text{ J mol}^{-1} \text{ K}^{-1}$
• $T = 300 \text{ K}$

Calculate the $RT$ term:

$RT = (8 \text{ J mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K}) = 2400 \text{ J mol}^{-1}$

Now, find $\ln K_p$:

$\ln K_p = -\frac{\Delta G^\circ_{rxn}}{RT} = -\frac{-1664 \text{ J mol}^{-1}}{2400 \text{ J mol}^{-1}}$

$\ln K_p = \frac{1664}{2400} = 0.69333...$

Calculate $K_p$:

$K_p = e^{0.69333...}$

The value of $e^{0.69333...}$ is approximately $2.0006$. Let's use $K_p \approx 2.0006$.

Calculation of Degree of Dissociation

Let $\alpha$ be the degree of dissociation of $\text{A}_2$. At equilibrium, for 1 mole of $\text{A}_2$ initially:

• Moles of $\text{A}_2 = 1 - \alpha$
• Moles of A = $2\alpha$
• Total moles = $(1 - \alpha) + 2\alpha = 1 + \alpha$

The partial pressures at equilibrium are:

• $p_{\text{A}_2} = \frac{1-\alpha}{1+\alpha} P_{total}$
• $p_{\text{A}} = \frac{2\alpha}{1+\alpha} P_{total}$

The expression for $K_p$ is:

$K_p = \frac{(p_{\text{A}})^2}{p_{\text{A}_2}} = \frac{\left(\frac{2\alpha}{1+\alpha} P_{total}\right)^2}{\frac{1-\alpha}{1+\alpha} P_{total}}$

$K_p = \frac{4\alpha^2}{(1+\alpha)(1-\alpha)} P_{total} = \frac{4\alpha^2}{1-\alpha^2} P_{total}$

Given $P_{total} = 1 \text{ bar}$, substitute $K_p \approx 2.0006$:

$2.0006 = \frac{4\alpha^2}{1-\alpha^2} \times 1$

$2.0006(1-\alpha^2) = 4\alpha^2$

$2.0006 - 2.0006\alpha^2 = 4\alpha^2$

$2.0006 = (4 + 2.0006)\alpha^2$

$2.0006 = 6.0006\alpha^2$

Solve for $\alpha^2$:

$\alpha^2 = \frac{2.0006}{6.0006} \approx 0.333396$

Determining the Value of x

The degree of dissociation is given in the format $(x \times 10^{-2})^{1/2}$. Therefore:

$\alpha = (x \times 10^{-2})^{1/2}$

Squaring both sides yields:

$\alpha^2 = x \times 10^{-2}$

Substitute the calculated value of $\alpha^2$:

$0.333396 = x \times 10^{-2}$

Solve for $x$:

$x = \frac{0.333396}{10^{-2}} = 0.333396 \times 100$

$x = 33.3396$

The question asks for the value of $x$ as the nearest integer.

The nearest integer to $33.3396$ is 33.