91Ó°ÊÓ

In a chemical reaction, the equilibrium constant, denoted as \( K_c \), provides vital information about the concentration of products and reactants once equilibrium is reached. Understanding this concept helps in predicting the direction and extent of the reaction. In the reaction \( \mathrm{CO}_{2(g)} + \mathrm{C}_{(s)} \rightleftharpoons 2 \mathrm{CO}_{(g)} \), the equilibrium constant expression is derived from the balanced chemical equation. Only gaseous and aqueous species are included in this calculation. Hence, the equilibrium constant for this reaction is given by:

• \[ K_c = \frac{\left[ \mathrm{CO} \right]^2}{\left[ \mathrm{CO}_2 \right]} \]

The value of \( K_c \) signifies how far the reaction will proceed before reaching equilibrium. When \( K_c \) is much greater than 1, as in this case \( K_c = 14.0 \), it indicates a greater concentration of products at equilibrium. This understanding is crucial for predicting how much of each substance exists at equilibrium, aiding in determining how efficient a reaction process might be.

An ICE table is a simple tabular tool used to predict the changes in concentrations of reactants and products in a chemical reaction that reaches equilibrium. Its acronym stands for Initial, Change, Equilibrium because it allows you to track the concentrations through these three stages. For the reaction \( \mathrm{CO}_{2(g)} + \mathrm{C}_{(s)} \rightleftharpoons 2 \mathrm{CO}_{(g)} \), it helps visualize changes like this:

• Initial: The starting concentration of \( \mathrm{CO}_2 \) is 1 M, while \( \mathrm{CO} \) starts at 0 M.
• Change: Let \( x \) represent the change in concentration. \( \mathrm{CO}_2 \) decreases by \( x \), leading to its concentration as \( 1 - x \), and \( 2x \) represents the increase in \( \mathrm{CO} \), since two moles of \( \mathrm{CO} \) are generated per mole of \( \mathrm{CO}_2 \) decomposed.
• Equilibrium: Therefore, at equilibrium, \([\mathrm{CO}_2] = 1 - x\) and \([\mathrm{CO}] = 2x\).

By setting up these stages, the ICE table makes it easier to plug values into the equilibrium constant expression and solve for \( x \), eventually figuring out the equilibrium concentrations of all involved species.

Solving equilibrium problems often requires manipulating the \( K_c \) expression into an algebraic form, and sometimes, this becomes a quadratic equation. In the case of our reaction, we formed a quadratic equation from the equilibrium expression:

• \[ 4x^2 + 14x - 14 = 0 \]

The quadratic formula provides the means to find \( x \), which represents the change in concentration:

• Formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substitute \( a = 4 \), \( b = 14 \), and \( c = -14 \) from the equation, we calculate:

• \[ x = \frac{-14 \pm \sqrt{14^2 - 4 \times 4 \times (-14)}}{2 \times 4} \]

Upon calculating, only the positive solution of \( x \approx 0.811 \) is feasible, as concentrations cannot be negative. Using this value of \( x \), you determine the equilibrium concentrations back in your ICE table, completing the problem's solution. This technique is not only applicable here but is also widely used in various equilibrium calculations in chemistry.