Problem 82 - Chemistry The Molecular Nature of Matter

Question

The reaction \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) has a \(K_{\mathrm{c}}=\) \(2.0 \times 10^{9}\) at \(25^{\circ} \mathrm{C}\). If \(0.100 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) and \(0.200 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) are placed in a \(10.0 \mathrm{~L}\) container, what will all the equilibrium concentrations be at \(25^{\circ} \mathrm{C}\) ? (Hint: Where does the position of equilibrium lie when \(K\) is very large?)

Step-by-Step Solution

Answer

At equilibrium, the concentrations are approximately: \( [\mathrm{HBr}] = 0.0200 \, \mathrm{M} \), \( [\mathrm{H}_2] \approx 0 \, \mathrm{M} \), and \( [\mathrm{Br}_2] \approx 0 \, \mathrm{M} \).

Step 1: Write the equilibrium expression

The equilibrium constant expression for the reaction is given by the equation: \( K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]} \).

Step 2: Determine the initial concentrations

Initial concentrations are \( [\mathrm{H}_2]_0 = 0.100 \, \mathrm{mol}/10.0 \, \mathrm{L} = 0.0100 \, \mathrm{M} \) and \( [\mathrm{Br}_2]_0 = 0.200 \, \mathrm{mol}/10.0 \, \mathrm{L} = 0.0200 \, \mathrm{M} \). Initially, \( [\mathrm{HBr}] = 0 \) because it has not been formed yet.

Step 3: Use the large value of K to predict the direction of the reaction

Since \( K_c \) is very large (\( 2.0 \times 10^{9} \)), the reaction will proceed nearly to completion to the right, meaning virtually all of the \( \mathrm{H}_2 \) and \( \mathrm{Br}_2 \) will be converted to \( \mathrm{HBr} \).

Step 4: Calculate change in concentration

Let the change in concentration of \( \mathrm{H}_2 \) be \( -x \) mol/L. Then change for \( \mathrm{Br}_2 \) is also \( -x \) mol/L since they react in a 1:1 ratio, and the change for \( \mathrm{HBr} \) is \( +2x \) mol/L.

Step 5: Find the equilibrium concentrations

At equilibrium, \( [\mathrm{H}_2] = 0.0100 - x \approx 0 \) M and \( [\mathrm{Br}_2] = 0.0200 - x \approx 0 \) M, because the reaction goes nearly to completion. Thus, \( [\mathrm{HBr}] = 0 + 2x = 2(0.0100) = 0.0200 \) M since \( x \) is approximately equal to the initial \( \mathrm{H}_2 \) concentration.

Step 6: Validate equilibrium constant with calculated concentrations

Since the reaction goes to completion, \( x \) is close to the initial concentration of \( \mathrm{H}_2 \), we expect \( [\mathrm{H}_2] \) and \( [\mathrm{Br}_2] \) to be negligible compared to \( [\mathrm{HBr}] \), and no calculation of \( K_c \) is needed to confirm equilibrium.

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant, often represented as K_c, is crucial when dealing with chemical reactions. It's a ratio that tells us how far a reaction goes toward completion before reaching a state of balance, or equilibrium. This ratio is determined by the concentrations of the reactants and products at equilibrium.

It's calculated using the formula: \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] where A and B are the reactant species, C and D are the product species, and a, b, c, and d are their respective coefficients in the balanced equation.

In the given exercise, a very large value of K_c indicates the formation of product is heavily favored. Hence, we can understand why in the final state of equilibrium, the quantities of reactants are negligible, and the concentration of the product, HBr, is maximized.

Reaction Quotient

Closely related to the equilibrium constant is the reaction quotient (\bQ\bx). The reaction quotient is a measure of the relative quantities of products and reactants present during a reaction at a given moment, before the reaction has reached equilibrium.

It's calculated using the same formula as the equilibrium constant: \[ Q = \frac{[products]}{[reactants]} \] but with the concentrations at any point in time, not necessarily at equilibrium. When comparing Q to K_c, if Q < K_c, the reaction will proceed forward to produce more products. Conversely, if Q > K_c, the reaction will shift backward to produce more reactants. When Q = K_c, the reaction is at equilibrium, and no further net change occurs in the concentrations of reactants and products.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a system at equilibrium responds to external changes such as concentration, temperature, and pressure. It states that if a dynamic equilibrium is disturbed, the system adjusts to diminish the change and re-establish equilibrium.

For example, adding more reactants to a system will cause the reaction to shift towards the products to re-establish equilibrium. Similarly, increasing the temperature for an exothermic reaction will cause the system to shift towards the reactants, as the reaction will 'consume' heat. This principle is pivotal when predicting how a change in conditions can alter the composition of an equilibrium mixture.

Equilibrium Concentration Calculations

To determine equilibrium concentrations in a reaction, we start with the initial concentrations and calculate how they change as the system reaches equilibrium. We often use an ICE table, which stands for Initial, Change, and Equilibrium, to keep track of these changes.

In our exercise, we calculated the change required to reach equilibrium based on the stoichiometry of the balanced equation. By understanding that the reaction nearly goes to completion because K_c is very large, we simplified the calculation. We assumed the change in concentration of reactants to be roughly equal to their initial concentrations, while the change in product concentration was double that amount, thanks to the 1:1:2 stoichiometry in the reaction: \[H2 + Br2 \rightleftharpoons 2HBr\]. Thereby, we quickly derived the final concentrations without the need for complex calculations due to the overwhelmingly large equilibrium constant.