Problem 59 - Chemistry The Molecular Nature of Matter

Question

At high temperature, \(2.00 \mathrm{~mol}\) of \(\mathrm{HBr}\) was placed in a 4.00 L container where it decomposed in the reaction: $$ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{Br}_{2}\) was measured to be \(0.0955 \mathrm{M}\). What is \(K_{\mathrm{c}}\) for this reaction at this temperature?

Step-by-Step Solution

Answer

The equilibrium constant \(K_{\mathrm{c}}\) for this reaction at this temperature is approximately 0.0951.

Step 1: Write the expression for the equilibrium constant, Kc

For the reaction given, which is in the form of \(aA \rightleftharpoons bB + cC\), the equilibrium constant expression is written as, \( K_c = \frac{[B]^b[C]^c}{[A]^a} \). Therefore, for the reaction \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g)\), the expression for Kc is \(K_c = \frac{[\mathrm{H}_2][\mathrm{Br}_2]}{[\mathrm{HBr}]^2}\).

Step 2: Calculate moles of HBr initially

The initial moles of HBr placed in the container is 2.00 mol, as stated in the problem. This will be used to calculate the initial concentration of HBr.

Step 3: Determine the equilibrium concentrations

Let us denote the initial concentration of HBr as \([\mathrm{HBr}]_{initial}\), and the change in concentration upon reaching equilibrium as \(x\). Since the reaction shows a stoichiometry of 2:1:1, when \(2x\) moles of HBr decompose, \(x\) moles of \(\mathrm{H}_2\) and \(\mathrm{Br}_2\) are formed. Therefore, at equilibrium, \([\mathrm{HBr}] = [\mathrm{HBr}]_{initial} - 2x\), \([\mathrm{H}_2] = x\), and \([\mathrm{Br}_2] = x\). Given that the concentration of \(\mathrm{Br}_2\) at equilibrium is 0.0955 M, then \(x = 0.0955\).

Step 4: Find initial concentration of HBr

The initial concentration of HBr is found by dividing the number of moles of HBr by the volume of the container: \([\mathrm{HBr}]_{initial} = \frac{2.00 \, \mathrm{mol}}{4.00 \, \mathrm{L}} = 0.50 \, \mathrm{M}\).

Step 5: Calculate the equilibrium concentration for HBr

Using the value of \(x\) and the initial concentration, the equilibrium concentration of HBr is \([\mathrm{HBr}] = 0.50 - 2(0.0955) = 0.50 - 0.1910 = 0.309 \, \mathrm{M}\).

Step 6: Calculate the equilibrium constant Kc

Substitute the equilibrium concentrations into the equilibrium expression to solve for Kc: \(K_c = \frac{[\mathrm{H}_2][\mathrm{Br}_2]}{[\mathrm{HBr}]^2} = \frac{(0.0955)(0.0955)}{(0.309)^2}\).

Step 7: Compute Kc value

After calculating the expression given in the previous step, \(K_c = \frac{(0.0955)(0.0955)}{(0.309)^2} \approx 0.0951\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. It's important to know that reaching equilibrium does not mean that reactants and products are in equal concentrations, but rather that their concentrations have stabilized in a fixed ratio that does not change.

In the provided exercise, hydrogen bromide (\text{HBr}) is in equilibrium with hydrogen gas (\text{H}_2) and bromine gas (\text{Br}_2). To find the equilibrium constant, known as \text{\(K_c\)}, we need to establish the balanced chemical reaction and note that at equilibrium, no further changes in concentrations of reactants or products will occur. The equilibrium constant is a ratio of the concentration of the products raised to their stoichiometric coefficients over the concentration of the reactants raised to their stoichiometric coefficients.

Reaction Quotient

The reaction quotient, \text{\(Q\)}, serves as a predictor of which direction a reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant, \text{\(K_c\)}, but with the initial concentrations of the reactants and products instead of the equilibrium concentrations. \text{\(Q\)} is particularly useful for determining reaction direction: if \text{\(Q < K_c\)}, the reaction will proceed forward to generate more products. If \text{\(Q > K_c\)}, the reaction will shift towards the reactants. If \text{\(Q = K_c\)}, the system is already at equilibrium.

When applying this to our exercise, we calculate the reaction quotient using the initial and equilibrium concentrations to predict how the reaction shifts. Understanding \text{\(Q\)} helps us analyze the system before it has reached equilibrium, offering insights into how changes in conditions can affect the reaction's progression towards equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is a qualitative tool in chemistry that predicts how a change in conditions (such as concentration, temperature, or pressure) will affect the position of equilibrium in a chemical reaction. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change.

For example, increasing the concentration of a reactant or decreasing the concentration of a product will cause the equilibrium to shift towards the products (forward direction), thereby using up the increased reactant or replacing the decreased product. Conversely, increasing the product concentration or decreasing a reactant concentration will shift the equilibrium toward the reactants (reverse direction).

In the context of our exercise, if the temperature increased, the equilibrium would shift in the direction that absorbs heat, as per Le Chatelier's Principle. For endothermic reactions, this would mean favoring the forward reaction, while for exothermic reactions, it would mean favoring the reverse reaction. Understanding this principle allows us to manipulate a reaction to favor the formation of a desired product or reactant.