91Ó°ÊÓ

Chemical equilibrium is a state in which the rates of the forward and reverse reactions are equal, meaning the concentrations of the reactants and products remain constant over time. This does not imply that the concentrations are equal, just that they are stable. Consider the reaction \( 2 \text{NOBr(g)} \rightleftharpoons 2 \text{NO(g)} + \text{Br}_2\text{(g)} \): at equilibrium, the forward and reverse processes occur at the same rate.
The significance of achieving chemical equilibrium in a reaction is crucial for predicting the concentrations of substances at any given time. Understanding equilibrium helps chemists to control reactions safely and efficiently.
In the given exercise, chemical equilibrium is reached at a temperature of \(100^{\circ} \text{C}\). At this point, the partial pressures are specific and constant, allowing us to calculate the equilibrium constant \(K_p\), a crucial term in describing equilibrium.

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. In a gaseous equilibrium, each component contributes to the total pressure according to its mole fraction.
For the reaction \( 2 \text{NOBr(g)} \rightleftharpoons 2 \text{NO(g)} + \text{Br}_2\text{(g)} \), partial pressures for \( \text{NOBr} \), \( \text{NO} \), and \( \text{Br}_2 \) are given as 4 atm, 4 atm, and 2 atm respectively.
Understand that these values are measured at equilibrium. These partial pressures play an essential role in calculating the equilibrium constant \(K_p\), which characterizes the system's equilibrium state.
In essence, if the total pressure in the reaction system differs or changes, partial pressures will adjust to maintain the equilibrium constant, reflecting their dynamic balance.

Stoichiometry refers to the quantitative relationship between the reactants and products in a chemical reaction. It's governed by balancing the chemical equation and helps in understanding the proportions of substances consumed and produced. For the reaction \( 2 \text{NOBr(g)} \rightleftharpoons 2 \text{NO(g)} + \text{Br}_2\text{(g)} \), stoichiometric coefficients are crucial to determine changes in concentrations or pressures involved.

• Coefficients of \(2\) for both \(\text{NOBr}\) and \(\text{NO}\) imply equal amounts are involved in the forward and reverse reactions.
• The coefficient for \(\text{Br}_2\) is \(1\), indicating fewer molecules are produced compared to \(\text{NO}\).

This stoichiometric understanding aids in correctly using the partial pressures in the formula for \(K_p\). This way, by substituting the correct powers of partial pressures, the calculation aligns with the reaction's stochiometry.

The reaction quotient \(Q\) is similar to the equilibrium constant, but it applies to systems that are not at equilibrium. It is calculated in the same way as the equilibrium constant, using the current concentrations or partial pressures of the reactants and products.In our problem, calculating \(Q\) would involve using the same form of the equilibrium constant equation: \[ Q = \frac{(P_{NO})^2(P_{Br2})}{(P_{NOBr})^2} \]If \(Q\) is equal to \(K_p\), then the system is in equilibrium. If \(Q\) is less than \(K_p\), the forward reaction will proceed to form more products. Conversely, if \(Q\) is greater, the reverse reaction will proceed to form more reactants.In evaluating the problem at hand, the given partial pressures result in a \(Q\) that equals \(K_p\), confirming the state of equilibrium. This confirms the equilibrium scenario described, with partial pressures given at equilibrium.