Problem 26 - Chemistry: An Atoms-Focused Approach

Question

At \(1045 \mathrm{K}\) the partial pressures of an cquilibrium mixture of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2},\) and \(\mathrm{O}_{2}\) are \(0.040,0.0045,\) and \(0.0030 \mathrm{atm}\) respectively. Calculate the value of \(K_{\mathrm{p}}\) at \(1045 \mathrm{K}\) $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$

Step-by-Step Solution

Answer

Answer: The equilibrium constant Kp at 1045 K is approximately 0.00009469.

Step 1: Write the Kp expression

Start by writing the Kp expression based on the given chemical equation: $$K_p = \frac{[\mathrm{H}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]^2}$$

Step 2: Substitute the given partial pressures

Substitute the given equilibrium partial pressures of H2O, H2, and O2 into the Kp expression: $$K_p = \frac{(0.0045)^2(0.0030)}{(0.040)^2}$$

Step 3: Calculate Kp

Calculate the equilibrium constant Kp at 1045 K by solving the equation: $$K_p = \frac{(0.0045)^2(0.0030)}{(0.040)^2} = 0.00009469$$ So, the value of Kp at 1045 K is approximately 0.00009469.

Key Concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_p\), is an essential concept in chemical equilibrium, particularly for gaseous reactions. It quantifies the ratio of the partial pressures of the products to the reactants at equilibrium. For a reaction involving gases, like \(2 \text{H}_2\text{O}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{O}_2(g)\), \(K_p\) is expressed in terms of partial pressures. Each substance's partial pressure is raised to the power of its stoichiometric coefficient in the balanced chemical equation.
\[K_p = \frac{[\text{H}_2]^2[\text{O}_2]}{[\text{H}_{2}\text{O}]^2}\]
In this expression, the partial pressures are denoted by brackets. Understanding \(K_p\) helps predict the direction of the reaction under different conditions. A high \(K_p\) value indicates that the reaction tends to favor the products at equilibrium, while a low \(K_p\) suggests the reactants are favored. In our given scenario, the \(K_p\) value is relatively low, meaning the equilibrium lies heavily towards the reactants side under these conditions.

Partial Pressure

Partial pressure is a crucial concept when dealing with gases in chemical reactions. It refers to the pressure exerted by a single type of gas in a mixture of gases. In the context of our chemical equation \(2 \text{H}_2\text{O}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{O}_2(g)\), each gas contributes to the total pressure of the system its own partial pressure.

• The partial pressure of \(\text{H}_2\) is 0.0045 atm.
• The partial pressure of \(\text{O}_2\) is 0.0030 atm.
• The partial pressure of \(\text{H}_2\text{O}\) is 0.040 atm.

These partial pressures are used in the formula to calculate \(K_p\).
Partial pressure is impacted by the volume, temperature, and amount of gas present, following Dalton’s Law of Partial Pressures, which states that total pressure of a mixture of gases is the sum of the partial pressures of each individual gas.

Gaseous Reactions

Gaseous reactions are chemical reactions in which all participating compounds are gases. They are often challenging to handle due to factors like pressure and volume changes. The reaction \(2 \text{H}_2\text{O}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{O}_2(g)\) illustrates a typical gaseous system where products and reactants are all in the gaseous state.
In gaseous reactions, the relationship between pressure, volume, and temperature often follows the ideal gas law, \(PV = nRT\). Here, each component's behavior in a mixture can be predictable if we understand the partial pressures and total pressure involved.

• Temperature and pressure can significantly influence the direction of the equilibrium.
• Pressure changes can shift the equilibrium position according to Le Chatelier's principle.

Understanding these factors is crucial for controlling and predicting the outcomes of gaseous reactions, making them significant in industrial and laboratory settings.