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Chemical equilibrium occurs when the forward and reverse reactions in a chemical process happen at the same rate. At this point, the concentrations of reactants and products remain constant. In the reaction \[ \mathrm{SbCl}_5(g) \rightleftharpoons \mathrm{SbCl}_3(g) + \mathrm{Cl}_2(g), \]the decomposition of \( \mathrm{SbCl}_5 \) into \( \mathrm{SbCl}_3 \) and \( \mathrm{Cl}_2 \) reaches a balance where the rate of decomposition equals the rate of combination of \( \mathrm{SbCl}_3 \) and \( \mathrm{Cl}_2 \) back into \( \mathrm{SbCl}_5 \). This state is known as dynamic equilibrium because reactions do not stop; they continue to occur at the same rate in both directions.

• Equilibrium is dynamic, not static.
• The ratio of products to reactants is represented by the equilibrium constant.

By measuring the concentrations or pressures of reactants and products at equilibrium, we can calculate the equilibrium constant \( K_p \), which helps predict the direction and extent of reactions.

The ideal gas law is a fundamental equation that relates the four properties of gases: pressure, volume, temperature, and moles. It is expressed as \[ PV = nRT, \]where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin. This law is essential in chemistry for

• determining unknown properties of gases,
• understanding the behavior of gases under different conditions, and
• converting moles to gas pressures, which is crucial for equilibrium calculations.

In our example, we use the ideal gas law to find the partial pressures of \( \mathrm{SbCl}_5 \), \( \mathrm{SbCl}_3 \), and \( \mathrm{Cl}_2 \) in a 5.00 L vessel at 468 K, a step necessary to calculate \( K_p \).
This equation assumes gases behave ideally, meaning there are no interactions between molecules. Although real gases deviate slightly, the ideal gas law provides an excellent approximation in most cases.

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It combines the atomic masses (from the periodic table) of all atoms in a molecule. For \( \mathrm{SbCl}_5 \), the molar mass is calculated by adding up the atomic masses of antimony (Sb) and chlorine (Cl), noting that there are five Cl atoms involved. Knowing the molar mass allows us to convert between mass and moles, a critical step in chemical calculations.
To find the number of moles from mass, use the formula:\[ n = \frac{\text{mass}}{\text{molar mass}}. \]In our problem, with 65.4 g of \( \mathrm{SbCl}_5 \), and a molar mass of \( 299.0 \text{ g/mol} \), we calculate the initial moles of \( \mathrm{SbCl}_5 \).

• Molar mass provides a bridge between grams and moles.
• Critical for stoichiometry calculations in reactions.

Understanding molar mass is essential for converting quantities in chemical reactions and is foundational in stoichiometry.

In a gaseous mixture, each component gas exerts its pressure, called partial pressure. Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each constituent gas. Partial pressure is used in equilibrium calculations like the one for \( K_p \), since \( K_p \) involves the pressures of each gas at equilibrium.
To find the partial pressure of a gas, use the formula derived from the ideal gas law:\[ P = \frac{nRT}{V}. \]This calculation is straightforward using the number of moles, temperature, volume, and the universal gas constant \( R \). In the reaction \( \mathrm{SbCl}_5(g) \rightarrow \mathrm{SbCl}_3(g) + \mathrm{Cl}_2(g) \), each gas's partial pressure at equilibrium is necessary to find \( K_p \).

• Partial pressure depends on moles and temperature.
• Essential for calculating \( K_p \), especially in gaseous equilibria.

Understanding partial pressures is crucial for associating the equilibrium constant with gas reactions and predicting the behavior of gaseous systems.