Question

A sealed 2.0 -L container initially contains 0.12 mol each of \(\mathrm{H}_{2} \mathrm{O}(g), \mathrm{Cl}_{2} \mathrm{O}(g),\) and \(\mathrm{HClO}(g) ;\) the reaction mixture is allowed to come to equilibrium according to the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HClO}(g) \quad K_{\mathrm{c}}=0.090 $$ Calculate the equilibrium concentrations of all three compounds.

Short answer

[\mathrm{H}_{2} \mathrm{O}] = 0.049 \text{ M}, [\mathrm{Cl}_{2} \mathrm{O}] = 0.049 \text{ M}, [\mathrm{HClO}] = 0.082 \text{ M}

Step-by-step solution

Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the reaction \ \( \mathrm{H}_{2} \mathrm{O}(g) + \mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HClO}(g) \) can be written as: \ \[ K_{c} = \frac{[\mathrm{HClO}]^2}{[\mathrm{H}_{2} \mathrm{O}][\mathrm{Cl}_{2} \mathrm{O}]} \]

Set Up the Initial Concentrations

Given: The initial moles of each substance are 0.12 mol in a 2.0 L container. \ Initial concentrations are: \ \[ [\mathrm{H}_{2} \mathrm{O}] = \frac{0.12 \text{ mol}}{2.0 \text{ L}} = 0.060 \text{ M} \ [\mathrm{Cl}_{2} \mathrm{O}] = \frac{0.12 \text{ mol}}{2.0 \text{ L}} = 0.060 \text{ M} \ [\mathrm{HClO}] = \frac{0.12 \text{ mol}}{2.0 \text{ L}} = 0.060 \text{ M} \]

Define the Change in Concentrations at Equilibrium

Let the change in concentration of \(\mathrm{H}_{2} \mathrm{O}(g)\) and \(\mathrm{Cl}_{2} \mathrm{O}(g)\) be \( -x \). Then the change in concentration of \(\mathrm{HClO}(g)\) will be \( +2x \). \ The equilibrium concentrations will be: \ \[ [\mathrm{H}_{2} \mathrm{O}] = 0.060 - x \] \ \[ [\mathrm{Cl}_{2} \mathrm{O}] = 0.060 - x \] \ \[ [\mathrm{HClO}] = 0.060 + 2x \]

Substitute the Equilibrium Concentrations into the Equilibrium Constant Expression

Substitute the equilibrium concentrations into the expression for \(K_c\): \ \[ K_{c} = \frac{(0.060 + 2x)^2}{(0.060 - x)(0.060 - x)} = 0.090 \]

Solve the Quadratic Equation

Expand and simplify the equation to solve for \(x\): \ \[ \frac{(0.060 + 2x)^2}{(0.060 - x)^2} = 0.090 \] \ \[ (0.060 + 2x)^2 = 0.090(0.060 - x)^2 \] \ \[ 0.0036 + 0.24x + 4x^2 = 0.090(0.0036 - 0.12x + x^2) \] \ \[ 0.0036 + 0.24x + 4x^2 = 0.000324 - 0.0108x + 0.090x^2 \] \ \[ 4x^2 - 0.090x^2 + 0.24x + 0.0108x = 0.000324 - 0.0036 \] \ \[ 3.91x^2 + 0.2508x - 0.003276 = 0 \] \ Use the quadratic formula: \ \[ x = \frac{-0.2508 \pm \sqrt{(0.2508)^2 - 4 \cdot 3.91 \cdot (-0.003276)}}{2 \cdot 3.91} \] \ Solve for \( x = 0.011 \text{ M} \)

Calculate Equilibrium Concentrations

Using \( x = 0.011 \text{ M} \), calculate equilibrium concentrations: \ \[ [\mathrm{H}_{2} \mathrm{O}] = 0.060 - x = 0.060 - 0.011 = 0.049 \text{ M} \] \ \[ [\mathrm{Cl}_{2} \mathrm{O}] = 0.060 - x = 0.060 - 0.011 = 0.049 \text{ M} \] \ \[ [\mathrm{HClO}] = 0.060 + 2x = 0.060 + 2(0.011) = 0.082 \text{ M} \]

Key concepts

equilibrium constant

Chemical reactions often reach a state where the concentrations of reactants and products remain constant. This state is known as chemical equilibrium. The equilibrium constant, denoted by \(K_c\), quantifies this balance. It is a numerical value derived from the ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced chemical equation. For the reaction \(\mathrm{H}_{2} \mathrm{O}(g) + \mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HClO}(g)\), the equilibrium constant expression is written as:
\[ K_{c} = \frac{[\mathrm{HClO}]^2}{[\mathrm{H}_{2} \mathrm{O}][\mathrm{Cl}_{2} \mathrm{O}]} \]. Understanding the equilibrium constant helps in determining the extent of a reaction and predicting the concentrations of reactants and products at equilibrium.

Key Points:

• At a given temperature, \(K_c\) is constant for a particular reaction.
• A large \(K_c\) (>1) means products are favored.
• A small \(K_c\) (<1) means reactants are favored.

To find the equilibrium concentrations, we adjust the initial concentrations based on the stoichiometry of the reaction and solve for the equilibrium concentrations using the given \(K_c\).

chemical equilibrium

Chemical equilibrium occurs when a reaction's forward and reverse rates equalize. At this point, the concentrations of the reactants and products remain constant because the rate at which the reactants are converted to products is equal to the rate at which products revert to reactants. For the given reaction \(\mathrm{H}_{2} \mathrm{O}(g) + \mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HClO}(g)\), the system reaches equilibrium when the amounts of \(\mathrm{H}_{2} \mathrm{O}\), \(\mathrm{Cl}_{2} \mathrm{O}\), and \(\mathrm{HClO}\) no longer change over time.

To analyze equilibrium, you need to:

• Write the balanced chemical equation for the reaction.
• Develop the equilibrium constant expression.
• Determine the initial concentrations of all species involved.
• Let 'x' represent the change in concentrations due to the reaction proceeding to equilibrium.

From there, you can calculate the equilibrium concentrations by solving the equilibrium expression for 'x' using the quadratic formula if necessary. This approach requires you to account for how each reactant and product concentration changes as the reaction reaches equilibrium.

quadratic equation

When dealing with equilibrium expressions, it is common to end up with a quadratic equation. In our case, substituting the equilibrium concentrations into the \(K_c\) expression for \(\mathrm{H}_{2} \mathrm{O}(g) + \mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HClO}(g)\) results in the quadratic equation:

\[ \frac{(0.060 + 2x)^2}{(0.060 - x)(0.060 - x)} = 0.090 \]

This quadratic equation needs to be expanded and simplified as shown in the solution steps, leading to:

\[ 3.91x^2 + 0.2508x - 0.003276 = 0 \]

To solve this, you can use the quadratic formula given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 3.91\), \(b = 0.2508\), and \(c = -0.003276\). Plugging in these values, we find:

\[ x = 0.011 \text{ M} \].

Solving the quadratic equation correctly is critical to finding the accurate equilibrium concentrations of all species involved in the reaction.

initial concentrations

The initial concentrations refer to the molarity of reactants and products in the reaction mixture before equilibrium is established. For the given reaction, the initial amounts of \(\mathrm{H}_{2} \mathrm{O}(g)\), \(\mathrm{Cl}_{2} \mathrm{O}(g)\), and \(\mathrm{HClO}(g)\) are all 0.12 moles in a 2.0 L container, leading to initial concentrations of:

\[ [\mathrm{H}_{2} \mathrm{O}] = \frac{0.12 \text{ mol}}{2.0 \text{ L}} = 0.060 \text{ M} \] \[ [\mathrm{Cl}_{2} \mathrm{O}] = \frac{0.12 \text{ mol}}{2.0 \text{ L}} = 0.060 \text{ M} \] \[ [\mathrm{HClO}] = \frac{0.12 \text{ mol}}{2.0 \text{ L}} = 0.060 \text{ M} \]

These initial concentrations are essential for setting up the equilibrium calculations. They form the starting point from which we track the changes in concentration as the reaction shifts towards equilibrium. By defining how much each concentration changes (denoted as 'x'), we can express the equilibrium concentrations and solve for them using the equilibrium constant expression.

reaction mixture

A reaction mixture is composed of the reactants and products present in a closed system where a chemical reaction takes place. In our example, the reaction mixture initially includes \(\mathrm{H}_{2} \mathrm{O}(g)\), \(\mathrm{Cl}_{2} \mathrm{O}(g)\), and \(\mathrm{HClO}(g)\), all in a 2.0-L container. As the reaction proceeds, the concentrations of these species change according to the reaction stoichiometry until equilibrium is reached.

Key Points to Remember:

• The reaction mixture is crucial for studying chemical equilibrium.
• The system must be closed, meaning no reactants or products can enter or leave.
• Equilibrium concentrations depend on the initial amounts and the reaction's equilibrium constant (\(K_c\)).

The study of reaction mixtures helps predict the final concentrations of the species involved using the equilibrium constant expression and enables us to understand the dynamics of the reaction process more deeply.