91Ó°ÊÓ

In chemical reactions, the equilibrium constant, often denoted as \(K_p\) or \(K_c\), provides insight into the balance between the forward and reverse reactions at a given temperature. It is a ratio of product concentrations to reactant concentrations, each raised to the power of their respective stoichiometric coefficients.For gas-phase reactions, such as \[\mathrm{H_2}(\mathrm{~g})+\mathrm{Br_2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{HBr}(\mathrm{g}),\]the equilibrium constant in terms of pressure, \(K_p\), is calculated using the partial pressures of the gases involved.An equilibrium constant tells us:

• If \(K > 1\), products are favored at equilibrium.
• If \(K < 1\), reactants are favored at equilibrium.
• If \(K = 1\), neither reactants nor products are favored.

In this exercise, with \(K_p = 1.6 \times 10^5\), it indicates a strong tendency for the formation of \(\mathrm{HBr}\) at equilibrium.

An ICE table is a valuable tool used in chemistry to keep track of the Initial concentrations, Changes in concentrations, and Equilibrium concentrations of the species involved in a chemical reaction. Constructing an ICE table can help simplify equilibrium problems, making them easier to solve.When setting up an ICE table:

• "I" stands for Initial concentrations or pressures. Initially, only \(\mathrm{HBr}\) had a measurable pressure of 10.0 bar.
• "C" stands for Change. For this reaction, as equilibrium is established, \(x\) bars of \(\mathrm{H_2}\) and \(\mathrm{Br_2}\) are generated while \(2x\) bars of \(\mathrm{HBr}\) are consumed.
• "E" stands for Equilibrium values of the pressures or concentrations.

By using the ICE table, we deduced that at equilibrium, the pressures of \(\mathrm{H_2}\) and \(\mathrm{Br_2}\) would both be \(x\) and that of \(\mathrm{HBr}\) would be \(10.0 - 2x\). This setup is crucial before proceeding to calculate exact equilibrium values.

Equilibrium pressure refers to the pressures of reactants and products once a chemical reaction has reached a state where the rates of the forward and reverse reactions are equal. At this point, the concentrations or pressures of reactants and products remain constant over time.In our problem, initially, we started with an \(\mathrm{HBr}\) pressure of 10.0 bar and no other gases. As the reaction progressed to equilibrium:

• \(\mathrm{H_2}\) and \(\mathrm{Br_2}\)'s pressures increased to \(x\) due to production.
• \(\mathrm{HBr}\)'s pressure decreased to \(10.0 - 2x\) as it was consumed.

To find these equilibrium pressures, we solved a quadratic equation derived from substituting into the equilibrium constant expression, ultimately yielding precise values for each gas at equilibrium.

The reaction quotient, \(Q\), is an important concept that helps determine which direction a reaction will proceed to reach equilibrium. It is calculated the same way as the equilibrium constant, using the current pressures or concentrations of the species involved instead of those at equilibrium.For the equation:\[\mathrm{H_2}(\mathrm{~g})+\mathrm{Br_2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{HBr}(\mathrm{g}),\]\(Q\) is determined by:\[Q = \frac{{P_{\mathrm{HBr}}^2}}{{P_{\mathrm{H_2}} \cdot P_{\mathrm{Br_2}}}}.\]Comparing \(Q\) and \(K_p\) provides insight into the system:

• If \(Q < K_p\), the reaction will proceed forward to produce more products.
• If \(Q > K_p\), it will reverse to form more reactants.
• If \(Q = K_p\), the system is at equilibrium.

Initially in our case, immediately after adding 10.0 bar of \(\mathrm{HBr}\) (and hence no \(\mathrm{H_2}\) or \(\mathrm{Br_2}\)), \(Q\) would be undefined due to zero pressures for the reactants. As the reaction proceeds, \(Q\) changes until it equals \(K_p\), indicating equilibrium has been reached.