Question

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 3.00-L container at \(395^{\circ} \mathrm{C}\), the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). If \(K_{c}=0.802\), what are the concentrations of each substance in the equilibrium mixture?

Short answer

The equilibrium concentrations are CO2: 0.283 M, H2: 0.283 M, CO: 0.217 M, and H2O: 0.217 M.

Step-by-step solution

Write the equilibrium expression for the reaction

For the reaction: \[CO_{2}(g) + H_{2}(g) \rightleftharpoons CO(g) + H_{2}O(g)\] The equilibrium constant expression can be written as: \[K_c = \frac{[CO][H_2O]}{[CO_2][H_2]}\]

Calculate the initial concentrations of each substance

We are given that 1.50 moles of CO2 and 1.50 moles of H2 are placed in a 3.00 L container. Therefore, the initial concentrations are: \[ [CO_2]_0 = \frac{1.50\, \text{mol}}{3.00\, \text{L}} = 0.50\, \text{M} \] \[ [H_2]_0 = \frac{1.50\, \text{mol}}{3.00\, \text{L}} = 0.50\, \text{M} \] Initially, there is no CO and H2O in the container since the reaction has not occurred yet, so: \[ [CO]_0 = [H_2O]_0 = 0 \, \text{M} \]

Set up an ICE table

Now, we can set up an ICE table to keep track of the change in concentrations as the reaction occurs and reaches equilibrium: ``` | CO2 H2 CO H2O ---------------------------------------------- Initial | 0.50 M 0.50 M 0 M 0 M Change | -x -x +x +x Equilibrium|0.50-x 0.50-x x x ```

Use the equilibrium expression to solve for the unknown equilibrium concentrations

We can now substitute the equilibrium concentrations from the ICE table into the equilibrium constant expression and solve for x: \[K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} = \frac{x \cdot x}{(0.50 - x)(0.50 - x)} = 0.802\] Solving this quadratic equation for x: \[ x^2 = 0.802 (0.50 - x)^2 \] We will find that x has only one physically meaningful solution: \[x = 0.217\, \text{M}\] Now we can calculate the equilibrium concentrations of all substances: \[ [CO_2]_{eq} = [H_2]_{eq} = 0.50 - x = 0.50 - 0.217 = 0.283\, \text{M}\] \[ [CO]_{eq} = [H_2O]_{eq} = x = 0.217\, \text{M} \] The equilibrium concentrations are CO2: 0.283 M, H2: 0.283 M, CO: 0.217 M, and H2O: 0.217 M.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\), is a vital concept in understanding chemical equilibrium. In simple terms, it provides a mathematical relationship that allows us to predict the concentrations of reactants and products at equilibrium. For the reaction \(\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)\), the equilibrium constant expression is given by the formula:
\[\begin{aligned}K_c = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]}\end{aligned}\]
This ratio compares the concentrations of the products to the reactants, each raised to the power of their coefficients in the balanced chemical equation.

• A large \(K_c\) indicates a reaction with more products than reactants at equilibrium.
• A small \(K_c\) suggests few products compared to reactants.

Understanding \(K_c\) helps chemists to determine how far a reaction will proceed and to calculate the concentrations of substances at equilibrium.

ICE Table

An ICE table is a straightforward yet powerful tool used to organize and calculate the changes in concentration of species as a chemical reaction moves to equilibrium. The acronym "ICE" stands for Initial, Change, and Equilibrium:

• Initial: The concentrations of all reactants and products before the reaction starts.
• Change: The shift in concentration, usually represented by \(x\), as the reaction progresses.
• Equilibrium: The concentrations after the reaction has reached equilibrium.

For example, in the reaction \(\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)\), the ICE table helps us keep track of concentrations:

- Initial: \([\text{CO}_2]_0 = 0.50 \, \text{M}\), \([\text{H}_2]_0 = 0.50 \, \text{M}\), while initially, \([\text{CO}]_0 = 0 \, \text{M}\), \([\text{H}_2\text{O}]_0 = 0 \, \text{M}\).
- Change: \(-x\) for \(\text{CO}_2\) and \(\text{H}_2\), \(+x\) for \(\text{CO}\) and \(\text{H}_2\text{O}\).
- Equilibrium: \(0.50-x\) for \(\text{CO}_2\) and \(\text{H}_2\), \(x\) for \(\text{CO}\) and \(\text{H}_2\text{O}\).The ICE table is crucial for setting up the equations needed to find \(x\), which in turn gives us the equilibrium concentrations.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is an equation resembling the equilibrium constant expression but uses the initial concentrations instead of the equilibrium concentrations. It is a snapshot of the reaction at any stage, not just at equilibrium. Considering the reaction \(\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)\), the reaction quotient \(Q\) is formulated as:
\[\begin{aligned}Q = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]}\end{aligned}\]
By comparing \(Q\) to \(K_c\):

• If \(Q = K_c\), the system is at equilibrium.
• If \(Q < K_c\), the forward reaction is favored, going towards more products.
• If \(Q > K_c\), the reverse reaction is favored, forming more reactants.

Thus, \(Q\) helps predict the direction in which the reaction will proceed to reach equilibrium. Understanding \(Q\) further aids in anticipating the behavior of a reaction under different initial conditions.