Question

The equilibrium constant \(K_{P}\) for the reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ the same temperature? (b) The very small value of \(K_{P}\) (and \(K_{\mathrm{c}}\) ) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

Short answer

High activation energy prevents reaction at room temperature despite a small \( K_P \).

Step-by-step solution

Write the Expression for Equilibrium Constant

For the given reaction \( 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \), the equilibrium constant \( K_P \) is expressed in terms of the partial pressures of the reactants and products as follows:\[K_P = \frac{{(P_{\mathrm{H}_2})^2 \cdot P_{\mathrm{O}_2}}}{{(P_{\mathrm{H}_2O})^2}}.\]

Relationship Between Equilibrium Constant and Reaction Direction

A very small value of \( K_P \) (and \( K_c \)) implies that, at equilibrium, the partial pressures of the reactants (water vapor) are much higher than those of the products (hydrogen and oxygen). This indicates that the reaction strongly favors the formation of \( \mathrm{H}_2O \) (water) at equilibrium.

Explain Stability at Room Temperature

Despite the small value of \( K_P \), a mixture of \( \mathrm{H}_2 \) and \( \mathrm{O}_2 \) gases can be stable at room temperature because the activation energy for the reaction is high. Without sufficient energy to overcome the activation barrier, the gases will not react, thus they can coexist without changing at room temperature.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_P\) when dealing with gases, is essential in understanding how a chemical reaction behaves at equilibrium. For any reaction, the equilibrium constant is determined by the partial pressures of the gases involved.

For the example reaction \( 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \), the equilibrium constant \( K_P \) is calculated using the formula:

• \[K_P = \frac{{(P_{\mathrm{H}_2})^2 \cdot P_{\mathrm{O}_2}}}{{(P_{\mathrm{H}_2O})^2}}.\]

This equation shows how the equilibrium is influenced by the concentration of each gas, helping to determine which side of the reaction is favored under certain conditions.

A very small \(K_P\) value tells us that, at equilibrium, the reaction lies heavily towards the reactants (in this case, water vapor), meaning the products \(\mathrm{H}_2\) and \(\mathrm{O}_2\) are present only in minor amounts.

Activation Energy

Activation energy is a critical concept in chemistry, especially when addressing why certain reactions do not happen spontaneously even if they are thermodynamically favored. Activation energy is the minimum energy that reacting molecules must possess for a reaction to take place.

Despite the very small equilibrium constant \(K_P\) for the given reaction, indicating that the formation of water is favored, the reaction between \(\mathrm{H}_2\) and \(\mathrm{O}_2\) gases doesn't occur easily at room temperature.

This is because the activation energy for this reaction is high, meaning that a significant amount of energy is required to initiate the reaction. At room temperature, the molecules do not have enough energy to overcome this barrier, so the mixture remains stable without reacting. This stability can be disrupted by supplying energy in the form of heat or a spark, often seen when igniting hydrogen gas.

Partial Pressure

Partial pressure is a measure of the concentration of a gas in a mixture of gases. It's calculated by assuming each gas in a mixture behaves independently, exerting its portion of pressure as if it occupies the whole volume by itself.

Understanding partial pressure is vital in the equilibrium expression for gas-phase reactions like \(K_P\). When the equilibrium constant \(K_P\) is very small, it indicates that the partial pressures of the reactants are significantly higher than those of the products at equilibrium, favoring the reactant side.

• In the example: \(P_{\mathrm{H}_2}\) and \(P_{\mathrm{O}_2}\), the partial pressures of the products are much lower compared to \(P_{\mathrm{H}_2O}\), the reactant.

This balance of partial pressures highlights how conditions are skewed toward the formation of water vapor over its decomposition into hydrogen and oxygen gases. Such details are important in predicting how gases behave under various pressures and conditions, leading to better control and manipulation of chemical reactions in practice.