Question

Consider the reaction and the associated equilibrium constant: $$a \mathrm{A}(g)+b \mathrm{B}(g) \rightleftharpoons c \mathrm{C}(g) \quad K_{\mathrm{c}}=5.0$$ Find the equilibrium concentrations of \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) for each value of \(a\) \(b,\) and \(c\) . Assume that the initial concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are each 1.0 \(\mathrm{M}\) and that no product is present at the beginning of the reaction. \begin{equation} \begin{array}{l}{\text { a. } a=1 ; b=1 ; c=2} \\ {\text { b. } a=1 ; b=1 ; c=1} \\ {\text { c. } a=2 ; b=1 ; c=1 \text { (set up equation for } x ; \text { don't solve) }}\end{array} \end{equation}

Short answer

For each case, the equilibrium concentrations of A, B, and C can be calculated by solving for x in their respective equilibrium expressions following the creation of ICE tables. Values of x are then used to find the equilibrium concentrations. For case c, the equilibrium equation is set up but not solved.

Step-by-step solution

Write the ICE Table for Case a

Set up an ICE (Initial, Change, Equilibrium) table to track the molar concentrations of reactants A and B and the product C. Initial concentrations of A and B are 1.0 M and C is 0 M. As the reaction proceeds, A and B decrease by x and C increases by 2x to reach equilibrium.

Write the Equilibrium Expression for Case a

The equilibrium expression based on the reaction stoichiometry for case a is:\[ K_c = \frac{[C]^{c}}{[A]^{a}[B]^{b}} = \frac{[C]^{2}}{[A][B]} \] Plugging in the equilibrium concentrations gives us:\[ K_c = 5.0 = \frac{(2x)^{2}}{(1-x)(1-x)} \]

Solve for x in Case a

Solve the quadratic equation:\[ 5.0 = \frac{4x^{2}}{(1-x)^{2}} \] This can be simplified to:\[ (1-x)^{2} = \frac{4x^{2}}{5} \] Leading to a solvable quadratic equation.

Calculate Equilibrium Concentrations for Case a

Once x is determined, calculate the equilibrium concentrations:\[ [A] = [B] = 1 - x \] and \[ [C] = 2x \]

Repeat Steps 1-4 for Case b

Set up an ICE table and equilibrium expression in a similar manner for case b, with reaction coefficients as given. Solve for x using the given equilibrium constant \( K_c = 5.0 \).

Set Up the Equilibrium Equation for Case c

For case c, the task is to set up the equation without solving. Write the ICE table, and the equilibrium expression is:\[ K_c = \frac{[C]^{c}}{[A]^{a}[B]^{b}} = \frac{[C]}{[A]^2[B]} \] Achieve the initial equation for x without solving it.

Key concepts

ICE Table

Understanding the ICE table is crucial for mastering chemical equilibrium problems. ICE stands for Initial, Change, and Equilibrium, and it's a simple yet powerful way to organize the data needed to solve equilibrium problems.

How to Construct an ICE Table

Initially, you list the starting concentrations of reactants and products. In the context of our exercise, reactants A and B both begin with 1.0 M concentration, while C starts at 0 M. Upon proceeding with the reaction, we record the changes in concentration as the reaction shifts towards equilibrium. The 'Change' row includes the amount decreased from reactants and increased for products, usually represented by a variable like 'x' tied to the stoichiometry of the balanced equation. Finally, we use these changes to calculate the 'Equilibrium' concentrations of each species. The beauty of an ICE table lies in its ability to visualize how concentrations shift, assisting students in transforming abstract equations into tangible values.

As for the exercise improvement advice, breaking down each part of the ICE table with examples or providing a template helps students to easily apply this method to various problems without confusion, hence making the concept more accessible.

Equilibrium Constant

The equilibrium constant, denoted as K_c, is a number that gives us insight into the ratio of products to reactants at equilibrium for a given chemical reaction. It's a reflection of the reaction’s tendency to occur and is dependent on temperature.

Understanding K_c

The value of K_c is calculated using the equilibrium concentrations of the reactants and products. For a general reaction with the form of the one in our exercise, we would express K_c mathematically as: \[ K_c = \frac{[C]^{c}}{[A]^{a}[B]^{b}} \] Each concentration is raised to the power of its respective coefficient in the balanced equation. A large K_c implies a reaction that heavily favors formation of products. Conversely, a small K_c suggests that reactants are predominantly present at equilibrium.

For students to solidify their understanding, incorporating examples with varying values of K_c can illustrate different reaction scenarios. This reinforces the conceptual link between the numeric value of K_c and the distribution of species at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry describes the quantitative relationship between reactants and products in a chemical reaction. It's the core tool for predicting the amounts of products formed from given amounts of reactants.

Applying Stoichiometry to Equilibrium

In an equilibrium context, stoichiometry dictates how the quantities of reactants and products shift when a system changes. For instance, in our exercise, if 'a' moles of A and 'b' moles of B react to form 'c' moles of C, the stoichiometry plays a direct role in determining how these moles convert and affect the equilibrium concentrations. Using stoichiometry, we set up ratios in our ICE table and directly link them to the 'x' variable that represents the shift in molar concentration towards equilibrium.

Emphasizing the mole-to-mole conversion aspect in examples and providing tips on keeping track of these ratios in calculations can help prevent common mistakes students make. Such an approach fosters a more intuitive grasp of how chemical reactions move forward and how we can predict their behavior at equilibrium.