Problem 32 - Chemistry and Chemical Reactivity

Question

The equilibrium constant, \(K_{c}\), for the following reaction is 1.05 at \(350 \mathrm{K}\) $$ 2 \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CCl}_{4}(\mathrm{g}) $$ If an equilibrium mixture of the three gases at \(350 \mathrm{K}\) contains \(0.0206 \mathrm{M} \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) and \(0.0163 \mathrm{M} \mathrm{CH}_{4},\) what is the equilibrium concentration of \(\mathrm{CCl}_{4} ?^{-}\)

Step-by-Step Solution

Answer

The equilibrium concentration of \( \text{CCl}_4 \) is 0.0273 M.

Step 1: Write the expression for the equilibrium constant

For the reaction \( 2\text{CH}_2\text{Cl}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{CCl}_4(g) \), the equilibrium constant \( K_c \) is given by the expression: \[ K_c = \frac{[\text{CH}_4][\text{CCl}_4]}{[\text{CH}_2\text{Cl}_2]^2} \] where the square brackets denote the concentrations of the species at equilibrium.

Step 2: Substitute known values into the Kc expression

Substitute the given equilibrium concentrations into the equilibrium expression: \( K_c = 1.05 \), \([\text{CH}_2\text{Cl}_2] = 0.0206 \text{ M} \), and \([\text{CH}_4] = 0.0163 \text{ M} \). This gives: \[ 1.05 = \frac{0.0163 \cdot [\text{CCl}_4]}{(0.0206)^2} \].

Step 3: Solve for the concentration of CCl4

Rearrange the expression to solve for \([\text{CCl}_4]\): \[ [\text{CCl}_4] = \frac{1.05 \cdot (0.0206)^2}{0.0163} \]. Calculate the value to find \([\text{CCl}_4]\).

Step 4: Calculate the value

Carry out the calculation: \( (0.0206)^2 = 0.00042436 \) and \( 0.00042436 \times 1.05 = 0.000445578 \). Finally, divide by 0.0163 to find \([\text{CCl}_4] = 0.02733 \text{ M} \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This concept is fundamental to understanding how chemical reactions can proceed to a point where they seem to "stop changing." At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This balance does not mean that the reactions cease altogether, but rather that they occur at the same rate, maintaining constant concentrations of the involved species.

To determine if a reaction has reached equilibrium, we use the equilibrium constant, represented as \( K_c \). This constant is specific for a given reaction at a certain temperature. It is calculated using the concentrations of the products and reactants at equilibrium. The equilibrium constant provides insight into the extent of the reaction, showing how much of the reactants have been converted to products. A larger \( K_c \) typically indicates that at equilibrium, the concentration of products is greater than that of the reactants, assuming the reaction proceeds to the right or forward direction.

Understanding chemical equilibrium is crucial for predicting how a change in conditions, like concentration or temperature, can influence the position of the equilibrium. This concept also explains why certain reactions might not proceed completely to form products.

Concentration Calculations

Concentration calculations involve determining the amounts of reactants and products in a chemical reaction. These calculations are essential for understanding how reactions progress and reach equilibrium. In a chemical context, concentration is usually expressed in moles per liter (Molarity, M), which helps in quantifying how much of a substance is present in a solution.

For equilibrium problems, like the one in the given exercise, concentration calculations are key to solving for unknowns using the equilibrium constant expression. By substituting known concentrations into this expression, we can find unknown concentrations, as was done for \( \[ \text{CCl}_4 \] \) in the exercise. This process often involves algebraic manipulation, where known variables are substituted into the equilibrium expression, and the equation is solved for the unknown variable.

For example, with the equilibrium constant \( K_c = \frac{[\text{CH}_4][\text{CCl}_4]}{[\text{CH}_2\text{Cl}_2]^2} \), knowing concentrations of some species allows calculation of others. If the initial concentrations and changes in concentration are also known, one might use an "ICE" table (Initial, Change, Equilibrium) to assist with the calculations. This technique is invaluable for systematically organizing given and derived information related to a chemical reaction.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to understand the quantitative relationships between reactants and products. This part of chemistry allows us to predict quantities needed or produced in a reaction. Stoichiometry is based on the balanced chemical equation which tells us the proportionate amounts of each substance involved.

In the exercise's reaction \( 2\text{CH}_2\text{Cl}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{CCl}_4(g) \), the stoichiometric coefficients (2:1:1) show that two moles of \( \text{CH}_2\text{Cl}_2 \) decompose to produce one mole of \( \text{CH}_4 \) and one mole of \( \text{CCl}_4 \). These coefficients are crucial when calculating the changes in concentrations as the reaction reaches equilibrium.

The stoichiometric ratios help in setting up the equilibrium expression and in doing concentration calculations, ensuring that the law of conservation of mass is followed. By applying stoichiometry, we can relate changes in concentration of one species to others based on their stoichiometric ratios. This ensures accurate calculation of quantities and provides a framework to understand how different amounts of substances in a chemical reaction relate to each other at equilibrium.