Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction: $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.2 at \(1650^{\circ} \mathrm{C}\). Initially \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.80 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are injected into a 5.0-L flask. Calculate the concentration of each species at equilibrium.

Short answer

[H2] = [CO2] = 0.053 mol/L, [H2O] = [CO] = 0.107 mol/L after equilibrium.

Step-by-step solution

Write the balanced equation and expression for Kc

The balanced chemical equation is: \[\text{H}_2(g) + \text{CO}_2(g) \rightleftharpoons \text{H}_2\text{O}(g) + \text{CO}(g)\]The equilibrium constant expression is written as:\[K_c = \frac{[\text{H}_2\text{O}][\text{CO}]}{[\text{H}_2][\text{CO}_2]}\]

Determine initial concentrations

The initial moles of \( \text{H}_2 \) and \( \text{CO}_2 \) are both 0.80 mol. The volume of the flask is 5.0 L, so their initial concentrations are:\[[\text{H}_2]_0 = [\text{CO}_2]_0 = \frac{0.80 \, \text{mol}}{5.0 \, \text{L}} = 0.16 \, \text{mol/L}\]There are no \( \text{H}_2\text{O} \) or \( \text{CO} \) initially, so their concentrations are 0 mol/L.

Set up the change variable (ICE table)

Set up the Initial, Change, Equilibrium (ICE) table for the reaction.| Species | \([\text{H}_2]\) | \([\text{CO}_2]\) | \([\text{H}_2\text{O} ]\) | \([\text{CO}]\) ||-------------|------------------|-------------------|-------------------|----------------|| Initial | 0.16 | 0.16 | 0 | 0 || Change | -x | -x | +x | +x || Equilibrium | 0.16-x | 0.16-x | x | x |

Substitute concentrations into Kc expression

Substitute the equilibrium concentrations from the ICE table into the \(K_c\) expression:\[K_c = \frac{x \cdot x}{(0.16-x) \cdot (0.16-x)} = 4.2\]This simplifies to:\[\frac{x^2}{(0.16-x)^2} = 4.2\]

Solve for x

To solve for \(x\), take the square root of both sides:\[\frac{x}{0.16-x} = \sqrt{4.2}\]Calculate \( \sqrt{4.2} \approx 2.05 \). Thus, \[2.05 \cdot (0.16-x) = x\]This becomes:\[2.05 \times 0.16 - 2.05x = x\]\[0.328 = 3.05x\]\[x = \frac{0.328}{3.05} \approx 0.107 \text{ mol/L}\]

Calculate equilibrium concentrations

Using \(x = 0.107\), determine equilibrium concentrations:For \(\text{H}_2\) and \(\text{CO}_2\):\[[\text{H}_2] = [\text{CO}_2] = 0.16 - 0.107 = 0.053 \text{ mol/L}\]For \(\text{H}_2\text{O}\) and \(\text{CO}\):\[[\text{H}_2\text{O}] = [\text{CO}] = 0.107 \text{ mol/L}\]

Key concepts

ICE Table

The "ICE table" is a useful tool for tracking the concentrations of chemical species during a reaction reaching equilibrium. "ICE" stands for Initial, Change, and Equilibrium. It allows you to systematically organize information about the reactants and products to help solve equilibrium problems.

• "Initial" refers to the concentrations of the reactants and products before the reaction has reached equilibrium. In the given example, the reaction between \(\text{H}_2\) and \(\text{CO}_2\) begins with concentrations determined by the initial moles and volume of the container.
• "Change" indicates how the concentrations shift as the reaction moves toward equilibrium. Typically, changes are represented by the variable \(x\) for reactions involving reactants decreasing in concentration by \(-x\) and products increasing by \(+x\).
• "Equilibrium" shows the concentrations when the reaction has reached equilibrium. These are the values used in the equilibrium constant expression.

This organized approach is especially useful for complex reactions as it shows how the concentrations of all species relate during the equilibrium state.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a value calculated using the same expression as the equilibrium constant, \(K_c\), but with the current concentrations of reactants and products. It's a snapshot of the system at any point before, after, or during the achievement of equilibrium.

• If \(Q < K_c\), this indicates the reaction will proceed forward, favoring the formation of products to reach equilibrium.
• If \(Q > K_c\), this suggests the reverse reaction is favored, shifting towards the formation of more reactants.
• When \(Q = K_c\), the system is at equilibrium, and no net change is occurring in the concentration of reactants and products.

In our example, because we're asked to find equilibrium concentrations, we deal directly with the \(K_c\) value using the equilibrium concentrations we predict.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products in a chemical reaction at equilibrium. At this stage, the rate of the forward reaction equals the rate of the reverse reaction. This means the concentrations of the reactants and products remain constant.

In our example, after finding \(x\), we substitute back to find the equilibrium concentrations:

• For both \([\text{H}_2]\) and \([\text{CO}_2]\), we calculated \(0.053\ \text{mol/L}\).
• For \([\text{H}_2\text{O}]\) and \([\text{CO}]\), the value is \(0.107\ \text{mol/L}\).

These values illustrate the balance achieved when the reaction has stabilized, reflecting the system's dynamics under the given conditions.

Chemical Equilibrium

Chemical equilibrium is a state reached in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction. During equilibrium, the concentrations of both reactants and products remain constant over time, although they are not necessarily equal.

Key features of chemical equilibrium include:

• The system is dynamic, meaning reactions are still occurring, but with no net change in concentration.
• It's dependent on temperature, pressure (for gases), and concentrations.
• It does not mean reactants and products are present in equal amounts, but their ratios remain constant at equilibrium.

In the example provided, at \(1650^{\circ} \text{C}\), the system reaches a state where the concentrations of \(\text{H}_2\), \(\text{CO}_2\), \(\text{H}_2\text{O}\), and \(\text{CO}\) remain consistent according to the calculated values, achieving a proportion consistent with the \(K_c\) of 4.2.