91Ó°ÊÓ

In the world of chemistry, the Equilibrium Constant, denoted as \(K_c\), is a crucial concept when studying reactions at equilibrium. It helps us understand the ratio between the concentrations of products and reactants in a balanced chemical equation. The value of \(K_c\) reveals whether the products or reactants are favored in a reversible reaction. A large \(K_c\) suggests that products are favored, while a small \(K_c\) indicates that reactants prevail at equilibrium.

For the reaction: \[2 \mathrm{BrF}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{Br}_{2}(\mathrm{~g})+5 \mathrm{~F}_{2}(\mathrm{~g})\]\(K_c\) can be expressed based on the concentrations of the substances involved: \[K_c = \frac{[\text{Br}_2][\text{F}_2]^5}{[\text{BrF}_5]^2}\]This expression indicates that the equilibrium constant is influenced by the stoichiometric coefficients of the reactants and products. For accurate calculations, it is vital to use the concentrations measured at the equilibrium state, ensuring consistency with the chemical equation.

The Reaction Quotient, represented as \(Q\), is an essential concept for checking how a reaction is progressing toward equilibrium. While \(K_c\) gives us the ratio of product and reactant concentrations at equilibrium, \(Q\) helps us determine the current state of the reaction at any given point. By comparing \(Q\) and \(K_c\), we can predict the direction the reaction will shift to reach equilibrium.

- If \(Q = K_c\), the system is at equilibrium.- If \(Q < K_c\), the reaction will proceed in the forward direction, making more products.- If \(Q > K_c\), the reaction will shift towards the reactants to achieve equilibrium.
In practice, you calculate \(Q\) using the same formula as \(K_c\), but with the initial concentrations instead of equilibrium ones. This knowledge allows chemists to control reactions and adjust conditions to favor the formation of desired products.

Calculating concentrations accurately is vital for determining both the Equilibrium Constant and the Reaction Quotient. When given concentrations at equilibrium, you can substitute them directly into the equilibrium expression to find \(K_c\). Let's revisit the given equilibrium concentrations for the reaction:

- \([\mathrm{BrF}_{5}] = 0.0064 \text{ mol/L}\)- \([\mathrm{Br}_{2}] = 0.0018 \text{ mol/L}\)- \([\mathrm{F}_{2}] = 0.0090 \text{ mol/L}\)
To calculate \(K_c\), insert these values into the equilibrium expression: \[K_c = \frac{(0.0018)(0.0090)^5}{(0.0064)^2}\]

First, compute the powers and products:- Calculate \((0.0090)^5 = 5.9049 \times 10^{-9}\)- Multiply by \(0.0018\) to find the numerator: \(1.063 \times 10^{-11}\)
The denominator is found by squaring \(0.0064\): \((0.0064)^2 = 4.096 \times 10^{-5}\).

Dividing the numerator by the denominator yields: \[K_c = \frac{1.063 \times 10^{-11}}{4.096 \times 10^{-5}} = 2.593 \times 10^{-7}\]
Concentration calculations must be precise as they feed directly into the computation of \(K_c\), affecting the accuracy of our equilibrium analysis.