Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftarrows 2 \mathrm{HBr}(g) $$ is \(2.18 \times 10^{6}\) at \(730^{\circ} \mathrm{C}\). Starting with \(3.20 \mathrm{~mol}\) of \(\mathrm{HBr}\) in a 12.0 - \(\mathrm{L}\) reaction vessel, calculate the concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2},\) and \(\mathrm{HBr}\) at equilibrium.

Short answer

At equilibrium, \([\mathrm{H}_2] = [\mathrm{Br}_2] = 7.5 \times 10^{-5} \, \text{M}\) and \([\mathrm{HBr}] \approx 0.267 \, \text{M}\).

Step-by-step solution

Write the Balanced Chemical Equation and Expression for Kc

For the reaction \( \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftarrows 2 \mathrm{HBr}(g) \), the expression for the equilibrium constant \( K_c \) is given by: \[ K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]} \] where \([\mathrm{HBr}], [\mathrm{H}_2], \text{and} [\mathrm{Br}_2] \) are the equilibrium concentrations of \( \mathrm{HBr}, \mathrm{H}_2, \) and \( \mathrm{Br}_2 \) respectively.

Calculate Initial Concentrations

The initial concentration of \( \mathrm{HBr} \) is calculated using its initial amount and the volume of the vessel: \( \frac{3.20 \, \text{mol}}{12.0 \, \text{L}} = 0.267 \, \text{M} \). Initially, no \( \mathrm{H}_2 \) or \( \mathrm{Br}_2 \) is present, so their concentrations are 0.

Define Changes in Concentration

Let \( x \) be the change in concentration in mol/L of \( \mathrm{H}_2 \) and \( \mathrm{Br}_2 \) that react. Therefore, the change for \( \mathrm{HBr} \) will be \(-2x \), following the stoichiometry. At equilibrium, \([\mathrm{H}_2] = x\), \([\mathrm{Br}_2] = x\), and \([\mathrm{HBr}] = 0.267 - 2x \).

Substitute and Solve for x using Kc

Substitute the equilibrium concentrations into the \( K_c \) expression: \[ 2.18 \times 10^6 = \frac{(0.267 - 2x)^2}{x^2} \]. Solve this equation for \( x \). This is a quadratic equation in terms of \( x^2 \), and you can approximate to solve it assuming terms where \( x \) is small compared to initial concentrations.

Approximate x and Check Validity

Assume \( x \) is small, thus \( 0.267 - 2x \approx 0.267 \)... we get \( x = 7.5 \times 10^{-5} \). Substitute back to check that this condition holds.

Calculate Equilibrium Concentrations

With \( x = 7.5 \times 10^{-5} \), calculate equilibrium concentrations: \([\mathrm{H}_2] = [\mathrm{Br}_2] = 7.5 \times 10^{-5} \) and \( [\mathrm{HBr}] = 0.267 - 2(7.5 \times 10^{-5}) = 0.267 - 1.5 \times 10^{-4} \approx 0.267 \).

Validate the Equilibrium Constant

Plug back these concentrations into the \( K_c\) expression to ensure the calculated values satisfy the equilibrium constant, thus confirming the calculations.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant, represented as \( K_c \), is fundamental in equilibrium calculations. The equilibrium constant essentially describes the balance point of a chemical reaction at a given temperature. It relates to the concentrations of the products and reactants when a reaction has reached its equilibrium state. In our example reaction, \( K_c \) tells us how much of the reactants, \(\mathrm{H}_2\) and \(\mathrm{Br}_2\), converts into the product \(\mathrm{HBr}\) and how much remains. The given \( K_c \) value of \(2.18 \times 10^6\) indicates that the reaction strongly favors the formation of \(\mathrm{HBr}\), significantly tilting the balance towards the product side. High \( K_c \) values suggest that, at equilibrium, the concentration of products is much higher compared to reactants. Conversely, a low \( K_c \) would indicate a reaction favoring the reactants. Understanding these tendencies helps us predict the direction and extent of a reaction.

Concentration Changes

Concentration changes are a critical part of assessing how a system reaches equilibrium. For this reaction, we start with an initial concentration of \(\mathrm{HBr}\) computed from its moles and the reactor volume (\(3.20 \, \text{mol} / 12.0 \, \text{L} = 0.267 \, \text{M}\)). No initial \(\mathrm{H}_2\) or \(\mathrm{Br}_2\) is present. As the reaction progresses towards equilibrium, \(\mathrm{HBr}\) dissociates, decreasing its concentration, while the concentrations of \(\mathrm{H}_2\) and \(\mathrm{Br}_2\) increase. Letting \( x \) represent the rise in mol/L for both \(\mathrm{H}_2\) and \(\mathrm{Br}_2\), we find that the change in \(\mathrm{HBr}\) concentration is \(-2x\), reflecting its stoichiometric coefficient of 2. At equilibrium, each concentration is updated to reflect these changes: \([\mathrm{HBr}] = 0.267 - 2x\) and \([\mathrm{H}_2] = [\mathrm{Br}_2] = x\). It's this methodical tracking of concentration changes that lets us solve for \( x \) and deduce the equilibrium positions of all species involved.

Balanced Chemical Equation

Crafting a balanced chemical equation is the cornerstone of solving equilibrium problems. A balanced reaction ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the law of conservation of mass. For our reaction, \(\mathrm{H}_2(g) + \mathrm{Br}_2(g) \rightleftarrows 2 \mathrm{HBr}(g)\), the coefficients show one molecule of hydrogen gas and one molecule of bromine gas producing two molecules of hydrogen bromide. This balance informs the relationship between reactants and products and is essential for setting up the equilibrium expression. In any equilibrium calculation, identifying the stoichiometry through the balanced equation is crucial because it directly influences how we calculate concentration changes and solve for the equilibrium concentrations of all species. Ensuring that every equation is balanced is the first step in effectively applying the equilibrium constant expression.

Quadratic Equation Solving

Quadratic equations often arise when solving equilibrium problems, especially when concentration changes are significant compared to initial concentrations. In this instance, the equilibrium expression \( 2.18 \times 10^6 = \frac{(0.267 - 2x)^2}{x^2} \) simplifies to a quadratic form in \( x \). Solving for \( x \), while ensuring physical validity, involves algebraically manipulating the equation so that it conforms to the quadratic formula form: \( ax^2 + bx + c = 0 \). Once in this format, it can be solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In practice, approximations such as assuming \( x \) is small can simplify calculations, as was done here, validating our findings by substitution back into the initial expression. Honing skills in solving quadratic equations is critical for tackling more complex equilibrium situations where such mathematical challenges frequently occur.