91Ó°ÊÓ

The equilibrium constant, denoted as \(K_c\), measures the ratio of the concentration of products to reactants at equilibrium for a given chemical reaction. For the reaction provided: $$ \text{H}_{2}(g) + \text{F}_{2}(g) \rightleftharpoons 2 \text{HF}(g) $$ The expression for \(K_c\) is written as: $$ K_c = \frac{[\text{HF}]^2}{[\text{H}_{2}][\text{F}_{2}]} $$ The concentration terms inside the brackets denote the molarity (moles per liter) of each substance at equilibrium. This expression tells you how much of each substance is present when the system has reached equilibrium. A higher \(K_c\) value, in this case, 115, indicates a greater concentration of products compared to the reactants when the reaction reaches equilibrium. Understanding the equilibrium constant helps in predicting the extent of the reaction and in calculating the concentrations of reactants and products.

When a chemical reaction reaches equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant over time. In the provided reaction: $$ \text{H}_{2}(g) + \text{F}_{2}(g) \rightleftharpoons 2 \text{HF}(g) $$ initially, you have 0.050 M of both \(\text{H}_{2}\) and \(\text{F}_{2}\) and 0.10 M of \(\text{HF}\). As the reaction proceeds towards equilibrium, \(\text{H}_{2}\) and \(\text{F}_{2}\) are consumed to produce more \(\text{HF}\). The key to solving this equilibrium problem is to set up an ICE table (Initial, Change, Equilibrium) and write the changes in concentrations in terms of a variable, \(x\). The changes are then used to express the equilibrium concentrations of all species involved. This process allows you to substitute these expressions into the equilibrium constant equation and solve for \(x\), thereby determining the equilibrium concentrations.

To determine the concentration changes at equilibrium, you start by identifying the initial concentrations and noting the changes that occur as the system reaches equilibrium. For our reaction: $$\text{H}_{2}(g) + \text{F}_{2}(g) \rightleftharpoons 2 \text{HF}(g)$$ Given initial concentrations:

• \([\text{H}_{2}]_{initial} = 0.050 \text{M}\)
• \([\text{F}_{2}]_{initial} = 0.050 \text{M}\)
• \([\text{HF}]_{initial} = 0.10 \text{M}\)

We denote the change in concentration of \(\text{HF}\) as \(+2x\) because 2 moles of \(\text{HF}\) are produced for every 1 mole of \(\text{H}_{2}\) and \(\text{F}_{2}\) consumed, which are reduced by \(x\). So, at equilibrium:

• \([\text{H}_{2}]_{eq} = 0.050 - x\)
• \([\text{F}_{2}]_{eq} = 0.050 - x\)
• \([\text{HF}]_{eq} = 0.10 + 2x\)

Finally, substituting these expressions into the \(K_c\) equation and solving for \(x\) allows calculation of the exact equilibrium concentrations. This approach ensures you account for all changes and correct equilibrium concentrations of the involved substances.