91Ó°ÊÓ

The equilibrium constant (\( K_p \)) is a vital concept in understanding chemical equilibrium. It defines the ratio of product pressures to reactant pressures at equilibrium. For a generic reaction \( aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) \), the equilibrium constant expression is given by:\[K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}\]where \(P_C, P_D, P_A,\) and \(P_B\) are the partial pressures of the gases involved.
For our specific reaction involving tetrachloromethane (\( \text{CCl}_4 \)), carbon (\( \text{C} \)), and chlorine (\( \text{Cl}_2 \)), the equilibrium constant equation simplifies to only include the gases:\[K_p = \frac{[P(\text{Cl}_2)]^2}{P(\text{CCl}_4)}\]This is because the partial pressure of the solid carbon is not included in the expression as its activity is constant. The value of \( K_p \) indicates how far the reaction tends to go towards the products side under equilibrium conditions. In this case, \( K_p = 0.76 \) suggests a moderate conversion of reactants to products.

The ICE table is a useful organizational tool for understanding changes in concentrations or pressures during a reaction that achieves equilibrium. "ICE" stands for Initial, Change, and Equilibrium.

• Initial: Represents the starting pressures or concentrations of the reactants and products.
• Change: Indicates how much the concentrations or pressures change as the system approaches equilibrium.
• Equilibrium: Shows the final pressures or concentrations when the reaction is at equilibrium.

For our reaction, initially, there are only reactants, so \( P(\text{CCl}_4) = 2.00 \) atm and both \( P(\text{C}) \) and \( P(\text{Cl}_2) \) are 0 atm. Once the reaction proceeds:

• \( \Delta \text{CCl}_4 = -x \)
• \( \Delta \text{C} = +x \)
• \( \Delta \text{Cl}_2 = +2x \)

This ultimately leads to the equilibrium pressures:

\( P(\text{CCl}_4) = 2.00 - x \)

\( P(\text{C}) = x \)

\( P(\text{Cl}_2) = 2x \)

Partial pressures are a crucial concept when dealing with gases in chemical equilibrium, representing the pressure exerted by a single type of gas in a mixture. In any system at equilibrium, the total pressure is the sum of the partial pressures of the gases present.
When calculating partial pressures during chemical reactions like the one in our exercise, we track the pressure changes using the ICE table's equilibrium row. For example:

• Before reaction: \( P(\text{CCl}_4) = 2.00 \) atm, \( P(\text{Cl}_2) = 0 \) atm
• At equilibrium: \( P(\text{CCl}_4) = 2.00 - x \) atm, and \( P(\text{Cl}_2) = 2x \) atm

These partial pressures help us to write the equilibrium expression:\( K_p = \frac{(2x)^2}{2.00-x} \)which allows solving for \( x \)and finding the equilibrium pressures.

In many instances, chemical equilibrium problems involving \( K_p \)and partial pressures require solving quadratic equations. This emerges because the equilibrium expression is a rational equation that can turn into a quadratic when rearranged, as shown below.
From our example, starting from:\[0.76 = \frac{(2x)^2}{2.00 - x} \]Reorganizing gives:\[0.76(2.00-x) = 4x^2\]Expand and rearrange the equation into standard quadratic form:\[4x^2 + 0.76x - 1.52 = 0\]To solve it, apply the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a=4 \), \( b=0.76 \), and \( c=-1.52 \). Calculate the square root and solve for the positive \( x \), since negative pressure isn’t realistic:\[x \approx 0.266 \,\text{atm}\]Solving quadratic equations often aids in finding unknowns crucial for determining equilibrium states.