Problem 86 - Chemistry The Molecular Science

Question

Write equilibrium constant expressions, in terms of reactant and product concentrations, for each of these reactions. $$ \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \quad K_{\mathrm{c}}=1.0 \times 10^{-14} $$ \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) $$ \begin{array}{c} K_{\mathrm{c}}=1.8 \times 10^{-5} \\ \mathrm{~N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \end{array} $$ 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

Expression 2 yields a quadratic equation, solved using the quadratic formula. Other expressions require approximation due to small Kc values.

Step 1: Write the equilibrium expression for the first reaction

The reaction is \( \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \). The equilibrium constant \( K_c \) expression for this reaction is given by the products' concentrations divided by the reactant's concentration. However, since water is a pure liquid, its concentration is not included in the expression:\[ K_c = [\mathrm{H}^+] [\mathrm{OH}^-] \]Since \( K_c = 1.0 \times 10^{-14} \), this expression is already in terms of the equilibrium concentrations of the ions.

Step 2: Write the equilibrium expression for the second reaction

The reaction is \( \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \). The equilibrium constant \( K_c \) expression for this reaction is:\[ K_c = \frac{[\mathrm{CH}_3 \mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3 \mathrm{COOH}]} \]Let the initial concentration of \( \mathrm{CH}_{3} \mathrm{COOH} \) be \( 1.0 - x \), and the concentrations of products \( [\mathrm{CH}_3 \mathrm{COO}^-] = x \) and \( [\mathrm{H}^+] = x \). The equilibrium expression in terms of \( x \) is:\[ 1.8 \times 10^{-5} = \frac{x \cdot x}{1.0 - x} = \frac{x^2}{1.0 - x} \]

Step 3: Write the equilibrium expression for the third reaction

The reaction is \( \mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g}) \). The equilibrium constant \( K_c \) expression is:\[ K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \]Let the changes be \([\mathrm{N}_2] = 1.0 - x\), \([\mathrm{H}_2] = 1.0 - 3x\), and \([\mathrm{NH}_3] = 2x\). The expression in terms of \( x \) is:\[ K_c = \frac{(2x)^2}{(1.0-x)(1.0-3x)^3} \]

Step 4: Identify which expressions yield quadratic equations

The expression from Step 2, \( \frac{x^2}{1.0 - x} = 1.8 \times 10^{-5} \), is quadratic because when you set up the equation \( x^2 = 1.8 \times 10^{-5} (1.0 - x) \), you get a quadratic form \( x^2 + (1.8 \times 10^{-5})x - 1.8 \times 10^{-5} = 0 \).

Step 5: Solve non-quadratic expressions for \( x \)

In the first reaction \( K_c = 1.0 \times 10^{-14} = [\mathrm{H}^+][\mathrm{OH}^-] \), since \( x \approx 0 \) as \( K_c \) is extremely small, we can safely assume small changes, solve for \( x \) using direct substitution: \( x \approx \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \).In the third reaction, note that it may lead to a cubic equation without simplification or approximations: analyze further, take possible approximations depending on relative sizes of \( K_c \).

Step 6: Solve quadratic equation

To solve \( x^2 + (1.8 \times 10^{-5})x - 1.8 \times 10^{-5} = 0 \), use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this case, \( a = 1 \), \( b = 1.8 \times 10^{-5} \), and \( c = -1.8 \times 10^{-5} \). Plug the values into the quadratic formula to solve for \( x \).

Key Concepts

ICE Table Approach

The ICE Table is a cornerstone tool in chemical equilibrium calculations, often used to keep track of concentration changes. **ICE** stands for Initial, Change, and Equilibrium, the steps which help in systematically approaching equilibrium problems.

• **Initial**: Define initial concentrations of reactants and products before any reaction takes place. For our exercises, each reactant begins at 1.0 mol/L, and products initially at 0 mol/L.
• **Change**: With a variable (let's say **x**), describe how these concentrations change. For a reactant decreasing by **x** and products increasing, common changes might look like: Reactant = 1.0 - **x**, Product = **x**.
• **Equilibrium**: Combine initial concentrations and changes to get equilibrium concentrations. Utilize these to set up equilibrium expressions.
  

By structuring the problem into easily traceable parts, the ICE table not only organizes your thoughts but directly guides you toward setting up correct expressions for equilibrium constant calculations.

Quadratic Equations in Chemistry

When dealing with equilibrium constant expressions, finding the exact concentration changes often involves solving quadratic equations. This occurs when the equation describes changes in concentrations that are proportional, resulting in a quadratic form like: \[ 0 = ax^2 + bx + c \] Such is the case in our example of acetic acid dissociation: \[ x^2 + (1.8 \times 10^{-5})x - 1.8 \times 10^{-5} = 0 \] Here, the task is to solve for **x** using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In chemistry, solving these equations can tell us how much of a reactant has been consumed or how much of a product has formed at equilibrium. The values of **a**, **b**, and **c** directly arise from the structure of the equilibrium expression, and manipulating them into a solvable quadratic is often the key step in solving equilibrium problems.

Chemical Equilibrium Calculations

Equilibrium calculations provide insight into how chemical systems stabilize over time. Chemical equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction, causing concentrations of reactants and products to remain constant.**Key Concepts**:

• **Equilibrium Constant (K)**: It is a ratio that compares the concentrations of products over reactants raised to the power of their stoichiometric coefficients. These constants, denoted as \(K_c\), indicate the extent of a reaction at equilibrium.
• **Types of Equilibrium**: Soluble and gaseous reactions have equilibrium constants unlike liquids and solids which do not appear in the mathematical expression.
• **Solving for concentration changes (x)**: Employ derived expressions from **ICE tables** to substitute into equilibrium expressions. Utilizing the variable "x" as the change provides a straightforward approach to solving intricate chemical equilibria, often highlighted by small magnitude constants which require approximations for solvability.

Chemical equilibrium calculations are a vital tool that help predict the outcomes of chemical reactions, understanding shifts through Le Châtelier's Principle, and quantitatively describing system stability.