91Ó°ÊÓ

In chemical reactions, the Equilibrium Constant, represented by \( K_c \), is crucial in determining the position of equilibrium. Specifically, it indicates the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. For the dissociation of dinitrogen tetroxide (\( \mathrm{N}_{2} \mathrm{O}_{4} \)) into nitrogen dioxide (\( \mathrm{NO}_{2} \)), the equilibrium expression is written as:

• \( K_c = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2} \mathrm{O}_{4}]} \)

Here, \( [\mathrm{NO}_{2}] \) and \( [\mathrm{N}_{2} \mathrm{O}_{4}] \) are the concentrations of \( \mathrm{NO}_{2} \) and \( \mathrm{N}_{2} \mathrm{O}_{4} \) at equilibrium. The given \( K_c \) value at 25°C is 0.125, suggesting that the concentrations of products and reactants are not equal. Understanding \( K_c \) helps predict how far the reaction will proceed towards products or remain as reactants under a given set of conditions, providing insight into the system's behavior at equilibrium.

The dissociation of dinitrogen tetroxide involves breaking it down into nitrogen dioxide, depicted in the equation:

• \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \)

Dissociation reactions like this one are reversible, meaning they can proceed in both forward and reverse directions. Initially, you have 0.0300 moles of \( \mathrm{N}_{2} \mathrm{O}_{4} \) in a 1.00 L flask. Over time, \( \mathrm{N}_{2} \mathrm{O}_{4} \) dissociates, producing \( \mathrm{NO}_{2} \). The system reaches equilibrium when the rate of forward dissociation equals the rate of reformation of \( \mathrm{N}_{2} \mathrm{O}_{4} \). At equilibrium, the concentration of \( \mathrm{N}_{2} \mathrm{O}_{4} \) decreases by \( x \), and the concentration of \( \mathrm{NO}_{2} \) increases by \( 2x \). Understanding dissociation reactions is essential in analyzing the transformation of substances in equilibrium.

In our dissociation problem, we encounter a quadratic equation when solving for equilibrium concentrations of \( \mathrm{N}_{2} \mathrm{O}_{4} \) and \( \mathrm{NO}_{2} \). This arises when we substitute the equilibrium concentrations:

• \( [\mathrm{N}_{2} \mathrm{O}_{4}] = 0.0300 - x \)
• \( [\mathrm{NO}_{2}] = 2x \)

into the equilibrium expression of \( K_c = \frac{(2x)^2}{0.0300 - x} = 0.125 \). After simplification, we get the quadratic equation:

• \( 4x^2 + 0.125x - 0.00375 = 0 \)

Solving this equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 0.125 \), and \( c = -0.00375 \), helps determine the change in concentration \( x \). Solving quadratic equations allows us to find the precise amounts of substances involved in the dissociation.

Percentage dissociation measures how much of a given substance has dissociated into its components, relative to its initial amount. In our example of \( \mathrm{N}_{2} \mathrm{O}_{4} \) dissociation:

• Initial concentration of \( \mathrm{N}_{2} \mathrm{O}_{4} = 0.0300 \text{ M} \)
• Amount dissociated is represented by \( x \)

The percentage dissociation is calculated using:

• \( \text{Percentage Dissociation} = \left( \frac{x}{0.0300} \right) \times 100\% \)

In the exercise, solving the quadratic equation resulted in \( x \approx 0.0307 \). When calculated, it mistakenly suggests over 100% dissociation which technically cannot occur. This outcome indicates all \( \mathrm{N}_{2} \mathrm{O}_{4} \) has dissociated and suggests a possible calculation or setup error, affirming the need for careful validation of assumptions and arithmetic throughout equilibrium computations.