Problem 87 - Chemistry The Molecular Science

Question

Write equilibrium constant expressions, in terms of reactant and product concentrations, for each of these reactions. \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{O}_{2}(\mathrm{~g})\) $$ K_{c}=7 \times 10^{56} $$ \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) $$ K_{\mathrm{c}}=1.7 \times 10^{2} $$ $$ \begin{array}{ll} \mathrm{HCOO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{HCOOH}(\text { aq }) & K_{\mathrm{c}}=5.6 \times 10^{3} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{AgI}(\mathrm{s}) & K_{\mathrm{c}}=6.7 \times 10^{15} \end{array} $$ Assume that all gases and solutes have initial concentrations of \(1.0 \mathrm{~mol} / \mathrm{L}\). Then, let the first reactant in each reaction change its concentration by \(-x\). (a) Using the reaction table (ICE table) approach, write equilibrium constant expressions in terms of the unknown variable \(x\) for each reaction. (b) Which of these expressions yield quadratic equations? (c) How would you go about solving the others for \(x\) ?

Step-by-Step Solution

Answer

Reactions 2, 3, and 4 yield quadratic equations. Solve them using the quadratic formula.

Step 1: Write the Equilibrium Expressions

The equilibrium constant for a reaction is determined by the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. For the reactions given, we write:1. \(2\, \text{O}_3(\text{g}) \rightleftharpoons 3\, \text{O}_2(\text{g})\), the equilibrium expression is: \[ K_c = \frac{[\text{O}_2]^3}{[\text{O}_3]^2} \]2. \(2\, \text{NO}_2(\text{g}) \rightleftharpoons \text{N}_2 \text{O}_4(\text{g})\), the equilibrium expression is: \[ K_c = \frac{[\text{N}_2 \text{O}_4]}{[\text{NO}_2]^2} \]3. \(\text{HCOO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightleftharpoons \text{HCOOH}(\text{aq})\), the equilibrium expression is: \[ K_c = \frac{[\text{HCOOH}]}{[\text{HCOO}^-][\text{H}^+]} \]4. For \(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightleftharpoons \text{AgI}(\text{s})\), solid AgI does not appear in the equilibrium expression: \[ K_c = \frac{1}{[\text{Ag}^+][\text{I}^-]} \]

Step 2: Apply the ICE Table

To find equilibrium concentrations in terms of \(x\), we create an ICE table for each reaction:1. For \(2\, \text{O}_3(\text{g}) \rightleftharpoons 3\, \text{O}_2(\text{g})\): - Initial: \([\text{O}_3] = 1.0 - 2x\), \([\text{O}_2] = 1.0 + 3x\) - \(K_c = \frac{(1.0 + 3x)^3}{(1.0 - 2x)^2}\)2. For \(2\, \text{NO}_2(\text{g}) \rightleftharpoons \text{N}_2\text{O}_4(\text{g})\): - Initial: \([\text{NO}_2] = 1.0 - 2x\), \([\text{N}_2\text{O}_4] = x\) - \(K_c = \frac{x}{(1.0 - 2x)^2}\)3. For \(\text{HCOO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightleftharpoons \text{HCOOH}(\text{aq})\): - Initial: \([\text{HCOO}^-] = 1.0 - x\), \([\text{H}^+] = 1.0 - x\), \([\text{HCOOH}] = x\) - \(K_c = \frac{x}{(1.0 - x)^2}\)4. For \(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightleftharpoons \text{AgI}(\text{s})\): - Initial: \([\text{Ag}^+] = 1.0 - x\), \([\text{I}^-] = 1.0 - x\) - \(K_c = \frac{1}{(1.0 - x)^2}\)

Step 3: Determine the Equation Type

Inspect each expression to see if a quadratic equation results:1. \(K_c = \frac{(1.0 + 3x)^3}{(1.0 - 2x)^2}\) will lead to a sixth-degree polynomial due to the cubic and quadratic terms.2. \(K_c = \frac{x}{(1.0 - 2x)^2}\) is quadratic because it can be rearranged to solve for \(x\) in terms of a quadratic function.3. \(K_c = \frac{x}{(1.0 - x)^2}\) will also lead to a quadratic equation after rearranging.4. \(K_c = \frac{1}{(1.0 - x)^2}\) is solvable directly for \(x\) as a quadratic equation after rearranging.

Step 4: Solving the Equations

To solve the quadratic equations, rearrange them into the standard quadratic form \(ax^2 + bx + c = 0\) and use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For non-quadratic cases:- Use algebraic simplification or numerical methods if necessary.- Specifically, for reaction 1, simplifications may reduce it to a simpler form that can be solved numerically or graphically.

Key Concepts

Equilibrium Expressions

When dealing with chemical reactions, the equilibrium constant expression is a crucial concept that allows us to understand the balance between reactants and products. The equilibrium constant (\(K_{c}\)) is calculated using the concentrations of chemical species at equilibrium, where each concentration is raised to the power of its stoichiometric coefficient from the balanced equation. For each of the reactions in the provided exercise, the equilibrium expressions are crafted accordingly:

• For gaseous reactions, such as \(2 \, \text{O}_3(\text{g}) \rightleftharpoons 3 \, \text{O}_2(\text{g})\), the equilibrium expression is given by \(K_c = \frac{[\text{O}_2]^3}{[\text{O}_3]^2}\).
• In reactions involving aqueous species like \(\text{HCOO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightleftharpoons \text{HCOOH}(\text{aq})\), the expression is \(K_c = \frac{[\text{HCOOH}]}{[\text{HCOO}^-][\text{H}^+]}\).
• Reactions containing solids in equilibrium do not include the solid in the expression, such as \(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightleftharpoons \text{AgI}(\text{s})\), which results in \(K_c = \frac{1}{[\text{Ag}^+][\text{I}^-]}\).

By utilizing these expressions, one can quantify the position of equilibrium for a given set of reactants and products.

Reaction Table (ICE Table)

The ICE (Initial, Change, Equilibrium) table is a helpful tool in chemistry for tracking concentration changes in reversible reactions. Initially, the concentrations of reactants and products are known, allowing predictions on how they will change until equilibrium is reached.

For example, consider the reaction \(2 \text{NO}_2(\text{g}) \rightleftharpoons \text{N}_2 \text{O}_4(\text{g})\). The ICE table helps us see:

• Initial: At the start, \([\text{NO}_2] = 1.0 - 2x\) and \([\text{N}_2\text{O}_4] = x\), assuming \(x\) change during reaction.
• Change: As the reaction progresses, \(2x\) of \(\text{NO}_2\) is consumed while \(x\) of \(\text{N}_2\text{O}_4\) forms.
• Equilibrium: Resulting in the equilibrium expressions to calculate \(x\).

This systematic approach makes it easier to set up and solve equations involving \(x\), ultimately determining the concentrations at equilibrium.

Quadratic Equations

Quadratic equations often arise in chemistry when dealing with equilibrium constants, especially when the changes in concentration lead to a second-degree equation in terms of \(x\). Let's look at the \(\text{HCOO}^-\) reaction with the expression \(K_c = \frac{x}{(1.0 - x)^2}\):

• Rearrange it into a standard quadratic form: \((1.0 - x)^2 \times K_c = x\), leading to \(K_c \cdot (1.0^2 - 2x + x^2) = x\)
• Simplifying gives us \(x^2 \cdot K_c - 2x \cdot K_c - x + K_c = 0\)

Once in the form \(ax^2 + bx + c = 0\), solve using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This equation helps determine concentration values at equilibrium conditions. For reactions that do not naturally simplify to quadratics, algebraic manipulation or numerical methods can be utilized to find a solution. This method is crucial for accurately calculating equilibrium concentrations.