Question

Write the \(K\) p expression for each reaction. a) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) b) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\)

Short answer

Kp for (a) is \( \frac{(P_{\text{H}_2\text{O}})^2}{(P_{\text{H}_2})^2 \cdot (P_{\text{O}_2})} \); for (b) it's \( \frac{(P_{\text{H}_2\text{O}})^2 \cdot P_{\text{O}_2}}{(P_{\text{H}_2\text{O}_2})^2} \)."

Step-by-step solution

Identify the Reaction Equilibrium

For each chemical reaction, we need to determine the equilibrium state. This involves recognizing the reactants and products in the gaseous (g) state because only gases are considered in the partial pressure equilibrium expression. In both reactions, we focus on gases: (a) For the first reaction, it is: - Reactants: 2H₂(g), O₂(g) - Products: 2H₂O(g) (b) For the second reaction, it is: - Reactants: 2H₂O₂(g) - Products: 2H₂O(g), O₂(g).

Constructing Kp using Partial Pressures

The equilibrium constant for gas-phase reactions is expressed in terms of partial pressures, denoted as Kp. The general formula is:\[ K_p = \frac{(P_{\text{products}})^{\text{stoichiometry}}}{(P_{\text{reactants}})^{\text{stoichiometry}}} \]We apply this to each reaction.

Apply Kp Expression for Reaction (a)

For the reaction 2H₂(g) + O₂(g) \(\rightleftarrows\) 2H₂O(g), the Kp expression is:\[ K_{pa} = \frac{(P_{\text{H}_2\text{O}})^2}{(P_{\text{H}_2})^2 \cdot (P_{\text{O}_2})} \]This expression shows the ratio of the partial pressure of the product (H₂O) squared, divided by the partial pressures of each reactant raised to their respective stoichiometries.

Apply Kp Expression for Reaction (b)

For the reaction 2H₂O₂(g) \(\rightleftarrows\) 2H₂O(g) + O₂(g), the Kp expression is:\[ K_{pb} = \frac{(P_{\text{H}_2\text{O}})^2 \cdot P_{\text{O}_2}}{(P_{\text{H}_2\text{O}_2})^2} \]Here, the partial pressures of each product (H₂O and O₂) are multiplied (with H₂O squared), and the result is divided by the squared partial pressure of the reactant (H₂O₂).

Final Expression Validation

We ensure these expressions accurately reflect the principles of equilibrium in terms of the pressures involved and are consistent with the stoichiometry of each reaction. By evaluating and confirming, we finalize:\[ K_{pa} = \frac{(P_{\text{H}_2\text{O}})^2}{(P_{\text{H}_2})^2 \cdot (P_{\text{O}_2})} \]\[ K_{pb} = \frac{(P_{\text{H}_2\text{O}})^2 \cdot P_{\text{O}_2}}{(P_{\text{H}_2\text{O}_2})^2} \]

Key concepts

Partial Pressure

Partial pressure is a crucial concept when discussing gas-phase chemical reactions. It refers to the pressure that each gas in a mixture would exert if it occupied the entire volume by itself. Hence, the total pressure of the mixture is the sum of the partial pressures of all the gases present.
In chemical equilibrium, partial pressures are used to express the concentrations of gases. Since pressure is directly proportional to the number of moles of gas (at constant temperature and volume), it serves as a convenient measure for reactions involving gases.
For example, in the reaction: - \( 2 \mathrm{H}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \), the partial pressures of \( \mathrm{H}_2 \), \( \mathrm{O}_2 \), and \( \mathrm{H}_2\mathrm{O} \) are considered.
Each gas's partial pressure contributes to the overall equilibrium expression, which is vital for calculating the equilibrium constant, \( K_p \). Thus, understanding and calculating partial pressures accurately is essential for predicting the behavior of gas mixtures at equilibrium.

Equilibrium Constant

In chemical reactions, the equilibrium constant \(K_p\) for gaseous reactions is a vital parameter. It provides insights into the ratio of the concentrations, or more accurately, partial pressures of products to reactants at equilibrium.
The expression for \(K_p\) involves the partial pressures of the gases involved in the reaction. It is crucial to note that the stoichiometric coefficients from the balanced chemical equation are used as exponents for these pressures.
For instance, in the reaction: - \( 2 \mathrm{H}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \), the equilibrium constant expression is:\[K_{pa} = \frac{(P_{\mathrm{H}_2\mathrm{O}})^2}{(P_{\mathrm{H}_2})^2 \cdot (P_{\mathrm{O}_2})}\]This formula indicates that the product’s concentration, \(\mathrm{H}_2\mathrm{O}\), is squared because of its coefficient in the chemical equation; similarly, reactants also have their partial pressures raised to the power of their respective coefficients. Equilibrium constants can predict the direction of the reaction under new conditions, highlighting their importance in chemistry.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the relative quantities of reactants and products in chemical reactions. It provides a way to quantitatively interpret the coefficients in a balanced written equation. These coefficients imply the molar relationship between different substances.
When expressing equilibrium in terms of gas-phase reactions, stoichiometry directly affects how we set up and solve for the \( K_p \) values. It determines the exponents used in calculating the equilibrium constant, which is based on the reaction's balanced equation.
For the given reactions:

• \( 2 \mathrm{H}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \)
• \( 2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \)

Note that each component's coefficient is directly applied as an exponent in the \(K_p\) calculation:

• In the first reaction, \(\mathrm{H}_2\) and \(\mathrm{O}_2\) affect the power of the partial pressures in the dominator.
• In the second reaction, coefficients influence the exponents of \( \mathrm{H}_2\mathrm{O} \) and \( \mathrm{O}_2 \) in the expression.

Understanding stoichiometry ensures that chemical equations are properly balanced and maintained throughout calculations, making it invaluable in the context of chemical equilibria.