Question

Determine the equilibrium concentrations for this chemical reaction with the given \(K_{\text {eq. }}\) $$ \begin{array}{l} \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \quad K_{\mathrm{eq}}=16.0 \\ 0.750 \mathrm{M} \quad 0.750 \mathrm{M} \end{array} $$

Short answer

At equilibrium: \([CO] = 0.150\, M\), \([H_2O] = 0.150\, M\), \([CO_2] = 0.6\, M\), \([H_2] = 0.6\, M\).

Step-by-step solution

Write the Equilibrium Expression

The first step is to write the equilibrium constant expression for the reaction: \[ K_{eq} = \frac{[CO_2][H_2]}{[CO][H_2O]} \] where \( [CO], [H_2O], [CO_2], \text{ and } [H_2] \) are the concentrations of each species at equilibrium.

Determine Changes in Concentrations

Assume initial concentrations for \( [CO_2] \) and \( [H_2] \) are 0 M since they are products starting from 0, and the concentrations of \( CO \) and \( H_2O \) are initially both 0.750 M. Let \( x \) be the change in concentration of \( CO \) and \( H_2O \): \[ \begin{align*} [CO] &= 0.750 - x \ [H_2O] &= 0.750 - x \ [CO_2] &= x \ [H_2] &= x \end{align*} \]

Substitute into Equilibrium Expression

Substitute the concentrations from Step 2 into the equilibrium expression: \[ K_{eq} = \frac{x \cdot x}{(0.750-x)(0.750-x)} = 16.0 \] This can be simplified to: \[ \frac{x^2}{(0.750-x)^2} = 16.0 \]

Solve for x

Take the square root of both sides to simplify: \[ \frac{x}{0.750-x} = 4 \] Solve for \( x \) by cross-multiplying: \[ x = 4(0.750-x) \] This expands to: \[ x = 3.0 - 4x \]\[ 5x = 3.0 \] \[ x = 0.6 \]

Calculate Equilibrium Concentrations

Use the value of \( x = 0.6 \) to find the equilibrium concentrations: \[ \begin{align*} [CO] &= 0.750 - 0.6 = 0.150 \, M \ [H_2O] &= 0.750 - 0.6 = 0.150 \, M \ [CO_2] &= 0.6 \, M \ [H_2] &= 0.6 \, M \end{align*} \]

Key concepts

Equilibrium Constant (Keq)

The equilibrium constant, also known as \(K_{eq}\), is a crucial concept in understanding chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium. This ratio helps determine the extent to which a reaction occurs before reaching equilibrium.

For the given reaction, the equilibrium constant expression is as follows:

• \( K_{eq} = \frac{[CO_2][H_2]}{[CO][H_2O]} \)

Where:

• [CO] and [H₂O] are the equilibrium concentrations of reactants.
• [CO₂] and [H₂] are the equilibrium concentrations of products.

A higher \(K_{eq}\) value, such as 16 in this exercise, means the reaction tends to favor the formation of products at equilibrium. Understanding the magnitude of \(K_{eq}\) allows us to predict the direction in which a chemical reaction is more inclined to proceed.

Equilibrium Concentration Calculation

To calculate the equilibrium concentration, you must first determine the change in concentration of reactants and products as the system reaches equilibrium. For this exercise, we assumed that the initial concentrations of the products, \([CO_2]\) and \([H_2]\), start at 0 M because the reaction hadn't started yet.

Using algebraic methods, let \(x\) represent the amount by which the concentration changes as the system approaches equilibrium:

• \( [CO] = 0.750 - x \)
• \( [H_2O] = 0.750 - x \)
• \( [CO_2] = x \)
• \( [H_2] = x \)

These expressions are then substituted back into the \(K_{eq}\) equation to solve for \(x\). Once \(x\) is determined, it provides a simple computational path to find all equilibrium concentrations, allowing you to map out the equilibrium state of your reaction.

Chemical Reaction

A chemical reaction involves the transformation of reactants to products, and understanding it in the context of equilibrium is essential. In this exercise, the chemical reaction involves carbon monoxide and water vapor reacting to form carbon dioxide and hydrogen gas. These reactions can reach a balance point, or equilibrium, where the concentrations of reactants and products remain constant over time.

Understanding the dynamic nature of chemical equilibrium involves appreciating that even at equilibrium, reactions continue happening in both directions, forward and reverse, at the same rate. This dynamic balance is crucial for predicting reaction behavior and for manipulating conditions to favor either reactant or product side, if required.

• This particular reaction can be influenced by factors such as temperature, pressure, and concentration changes.
• In practical applications, shifting the equilibrium position can help in maximizing desired product yield.

Grasping these concepts can significantly improve your ability to predict and control chemical reactions in both academic and real-world settings.