Question

Write the \(K_{\mathrm{P}}\) expression for the following gas-phase reaction: \(\mathrm{ClO}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \rightleftarrows \mathrm{ClO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)

Short answer

\( K_{\mathrm{P}} = \frac{P_{\mathrm{ClO}_2} \cdot P_{\mathrm{O}_2}}{P_{\mathrm{ClO}} \cdot P_{\mathrm{O}_3}} \)

Step-by-step solution

Identifying the Reaction Components

The reaction we are given is \( \mathrm{ClO} (\mathrm{g}) + \mathrm{O}_3 (\mathrm{g}) \rightleftarrows \mathrm{ClO}_2 (\mathrm{g}) + \mathrm{O}_2 (\mathrm{g}) \). We need to find the equilibrium constant, \( K_{\mathrm{P}} \), for this reaction in terms of the partial pressures of each gaseous component.

Writing the General Expression for Kp

The equilibrium constant \(K_{\mathrm{P}}\) for a gaseous reaction is defined in terms of the partial pressures of the products raised to their stoichiometric coefficients divided by the partial pressures of the reactants raised to their stoichiometric coefficients.

Substituting the Reaction Components into Kp Formula

Substitute the given reaction into the \(K_{\mathrm{P}}\) expression: \[ K_{\mathrm{P}} = \frac{P_{\mathrm{ClO}_2} \cdot P_{\mathrm{O}_2}}{P_{\mathrm{ClO}} \cdot P_{\mathrm{O}_3}} \] Here, \(P_{\mathrm{ClO}_2}\), \(P_{\mathrm{O}_2}\), \(P_{\mathrm{ClO}}\), and \(P_{\mathrm{O}_3}\) are the partial pressures of \(\mathrm{ClO}_2\), \(\mathrm{O}_2\), \(\mathrm{ClO}\), and \(\mathrm{O}_3\) respectively.

Key concepts

Partial Pressure

Partial pressure is a key concept in understanding gas-phase reactions and chemical equilibrium. The pressure exerted by a single type of gas in a mixture of gases is called its partial pressure. When gases mix in a container, each gas contributes to the total pressure. The contribution is proportional to the fraction of the total gas that it represents.

For example, if you have a mixture of nitrogen and oxygen in a container, the partial pressure of nitrogen is the pressure nitrogen itself would exert if it occupied the entire volume alone at the same temperature.

• The total pressure of a gas mixture is simply the sum of the partial pressures of all the gases present.
• Mathematically, for gases A, B, C etc., the total pressure, \( P_{total} \) can be represented as \( P_{total} = P_A + P_B + P_C + \dots \)

Understanding partial pressures is crucial because they are used to write expressions for equilibrium constants in gas-phase reactions.

Gas-phase Reaction

Gas-phase reactions involve reactants and products that are in the gaseous state. In these reactions, reactant gases interact to form product gases.

When analyzing these reactions, it is important to consider the behavior of gases under various conditions. These reactions follow the laws of gas behavior, such as the ideal gas law, which relates pressure, volume, temperature, and the number of moles of gas present.

• The ideal gas law is given by the equation: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
• Understanding gas behavior helps predict how gases will react and how changes in conditions might shift the equilibrium.

Gas-phase reactions can be evaluated using partial pressures of the involved gases, often leveraging tools like equilibrium constants for practice and comprehension.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. At this point, the concentrations of reactants and products remain constant - though not necessarily equal - over time. When dealing with gases, we often use the equilibrium constant based on partial pressures, denoted as \( K_P \).

• The value of \( K_P \) at a given temperature provides insight into the position of equilibrium.
• If \( K_P \) is large, the equilibrium is product-favored; if small, it is reactant-favored.

To find \( K_P \), you substitute the partial pressures of the products and reactants into the equilibrium constant expression, as shown in the original exercise solution. This expression is fundamental to predicting the extent of reaction and understanding how a system at equilibrium responds to external changes, such as pressure or temperature changes (as described by Le Chatelier's principle).

Equilibrium concepts are vital for chemistry students as they lay the groundwork for many areas of chemical research and industrial applications.