Problem 29 - Chemistry and Chemical Reactivity

Question

Suppose 0.086 mol of \(\mathrm{Br}_{2}\) is placed in a \(1.26-\mathrm{L}\). flask and heated to \(1756 \mathrm{K}\), a temperature at which the halogen dissociates to atoms. $$ \mathrm{Br}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{Br}(\mathrm{g}) $$ If \(\mathrm{Br}_{2}\) is \(3.7 \%\) dissociated at this temperature, calculate \(K_{\mathrm{c}}\).

Step-by-Step Solution

Answer

\(K_{c} = 0.000389\)

Step 1: Calculate Initial Concentration of Br2

The initial concentration of \( \mathrm{Br}_2 \) can be calculated using the formula \( c = \frac{n}{V} \), where \( n = 0.086 \text{ mol} \) and \( V = 1.26 \text{ L} \). Thus, \( c_{\mathrm{Br}_2, \text{initial}} = \frac{0.086}{1.26} = 0.0683 \text{ mol/L} \).

Step 2: Calculate Change in Concentration Due to Dissociation

Since \( \mathrm{Br}_2 \) is \(3.7\%\) dissociated, it changes by \( 0.0683 \times 0.037 = 0.00253 \text{ mol/L} \).

Step 3: Calculate Equilibrium Concentrations

At equilibrium, the concentration of \( \mathrm{Br}_2 \) is \( 0.0683 - 0.00253 = 0.06577 \text{ mol/L} \). The concentration of \( \mathrm{Br} \) is \( 2 \times 0.00253 = 0.00506 \text{ mol/L} \) since it produces 2 moles of \( \mathrm{Br} \) per mole of \( \mathrm{Br}_2 \).

Step 4: Calculate the Equilibrium Constant \(K_c\)

The equilibrium constant \( K_c \) is given by \( K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]} \). Substituting the concentrations found, \( K_c = \frac{(0.00506)^2}{0.06577} = 0.000389 \).

Key Concepts

Dissociation

Dissociation is a fundamental concept in chemistry, especially when discussing chemical reactions involving compounds breaking down into simpler forms. In our example, we see the dissociation of bromine gas (\( \mathrm{Br}_2 \)) into individual bromine atoms (\( 2 \mathrm{Br} \)). This process is crucial because it indicates how substances react under certain conditions like temperature and pressure.

In a chemical context, dissociation refers to a reversible change where molecules break apart into two or more atoms or ions. When calculating dissociation percentages, such as a 3.7% dissociation, it signifies that a certain fraction of the compound is splitting under specified conditions. This can be reflected in altered concentration levels within a reaction equilibrium.

• This knowledge helps us understand how much of a reactant remains as is and how much shifts into a new form.
• Dissociation contributes to determining the extent and direction of a chemical reaction over time.

Equilibrium Concentration

Equilibrium concentration is a fascinating aspect when dealing with chemical reactions. By definition, it is the concentration of reactants and products in a reaction mixture, in a state where the rate of the forward reaction equals the rate of the reverse reaction. This balanced state is what we call equilibrium.

In our exercise, we calculated equilibrium concentrations after considering the dissociation of \( \mathrm{Br}_2 \). We initially calculate the change in concentration due to dissociation and then find out how much remains at equilibrium.

• For example, if \( \mathrm{Br}_2 \) started at 0.0683 mol/L and changed by 0.00253 mol/L due to dissociation, then at equilibrium, it stands at 0.06577 mol/L.
• Conversely, the concentration of \( \mathrm{Br} \), which results from the dissociation, is measured as 0.00506 mol/L, doubled due to each \( \mathrm{Br}_2 \) yielding two \( \mathrm{Br} \) atoms.

Understanding equilibrium concentrations is pivotal for anticipating the amounts of each reactant and product at equilibrium, thus allowing for precise calculations in the context of equilibrium constants.

Chemical Equilibrium

Chemical equilibrium is one of the key principles in chemistry. It's the state in a reaction where the concentrations of reactants and products remain constant over time. At this point, the forward and reverse reaction rates are identical, but that doesn't mean reactions stop occurring. They continue, but cancel each other out in effect.

In the context of our example, chemical equilibrium explains why both gaseous \( \mathrm{Br}_2 \) and the resultant \( \mathrm{Br} \) atoms maintain certain concentration levels over time. This balance is crucial as it allows us to calculate important metrics like the equilibrium constant \( K_c \).

• Achieving equilibrium means reactants and products are produced at equal rates, hence their concentration tends not to change once balance is reached.
• Recognizing this allows chemists to predict how changing conditions like pressure or temperature might shift a reaction's balance.

In the larger picture, chemical equilibrium makes it possible to understand and control reactions across various processes and industries.

Br2 Dissociation

Bromine \(( \mathrm{Br}_2 )\) dissociation is an interesting case of how temperature influences chemical reactions. When the given flask is heated to 1756 Kelvin, it drives the dissociation of \( \mathrm{Br}_2 \) into bromine atoms.

This example illustrates the high temperatures at which molecules such as bromine gas are prompted to break bonds, producing individual atoms.

• It's important to note that in a lab setting, controlling temperature is critical as it significantly affects reaction dynamics.
• The temperature-induced shift in equilibrium concentration provides insight into the thermal behavior of gaseous compounds and their tendency to form simpler species.

Understanding \( \mathrm{Br}_2 \) dissociation, particularly at elevated temperatures, offers valuable insights into reaction mechanisms and kinetic behaviors under extreme thermal conditions, key for applications in industrial chemistry.