Question

For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K_{\mathrm{p}}=23.2\) at \(600 \mathrm{K} .\) Explain which of the fol- lowing situations might equilibrium: (a) \(\quad P_{\mathrm{CO}}=P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{CO}_{2}}=P_{\mathrm{H}_{2}} ; \quad\) (b) \(\quad P_{\mathrm{H}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=\) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{CO}} ; \quad(\mathrm{c}) \quad\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2}}\right)=\left(P_{\mathrm{CO}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}^{2}\right)\) (d) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{H}_{2}} / P_{\mathrm{CO}}\)

Short answer

Only situation (c), where \((P_{CO2})(P_{H2}) = (P_{CO})(P_{H2O}^2)\) might be at equilibrium.

Step-by-step solution

Identify The Reaction and The Expression For Kp

The chemical reaction is \(CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)\). The expression for Kp would be \(K_p = \frac{[CO_2][H_2]}{[CO][H_2O]}\)

Evaluate situation (a)

In this situation, all gaseous substances have the same pressure, meaning \(P_{CO} = P_{H2O} = P_{CO2} = P_{H2}\). Plugging these values into the expression for Kp, it wouldn't be 23.2. Therefore, situation (a) is not at equilibrium.

Evaluate situation (b)

In situation (b), \( P_{H2} / P_{H2O} = P_{CO2} / P_{CO} \). This isn't equivalent to the Kp expression and hence the system wouldn't be in an equilibrium state.

Evaluate situation (c)

For situation (c), we have \((P_{CO2})(P_{H2}) = (P_{CO})(P_{H2O}^2)\). This situation is the only matching situation to the Kp expression for the given reaction. Therefore, situation (c) is possibly at equilibrium.

Evaluate situation (d)

In situation (d), \(P_{CO2} / P_{H2O} = P_{H2} / P_{CO}\). Like situation (b), it does not match the expression for Kp, therefore it isn't at equilibrium.

Key concepts

Reaction Quotient

In chemical equilibrium, the reaction quotient, represented as \( Q \), plays a critical role. It helps determine if a chemical reaction is at equilibrium, or if it will shift towards reactants or products to reach equilibrium status. The formula for \( Q \) is similar to that of the equilibrium constant, but it uses the initial concentrations or partial pressures instead of those at equilibrium.
To calculate \( Q \) for a gas-phase reaction, such as \( ext{CO(g) + H}_2 ext{O(g)} \rightleftharpoons ext{CO}_2 ext{(g) + H}_2 ext{(g)},\) you use:

• \( Q = \frac{[CO_2][H_2]}{[CO][H_2O]} \)
• If \( Q = K_p \), the system is at equilibrium.
• If \( Q < K_p \), the reaction will proceed towards products.
• If \( Q > K_p \), the reaction will shift towards reactants.

By comparing \( Q \) with the equilibrium constant \( K_p \), we can predict the behavior of the reaction and how it will shift to reach equilibrium.

Equilibrium Constant

The equilibrium constant \( K_p \) describes the balance of a reaction in the gas phase at a specific temperature, reflecting the ratio of the concentrations of products to reactants. It is calculated using the pressures of gases involved in the reaction. For instance, in our reaction:
\( ext{CO(g) + H}_2 ext{O(g)} \rightleftharpoons ext{CO}_2 ext{(g) + H}_2 ext{(g)},\)
We have the expression:

• \( K_p = \frac{P_{CO_2} imes P_{H_2}}{P_{CO} imes P_{H_2O}} = 23.2 \text{ at } 600 \text{ K}.\)

The equilibrium constant provides insights into the position of equilibrium and the extent to which a reaction proceeds:

• A large \( K_p \) (\( > 1 \)) indicates a greater concentration of products.
• A small \( K_p \) (\( < 1 \)) suggests that reactants are favored.

Understanding \( K_p \) helps students determine whether a given set of conditions satisfies the equilibrium requirements of a particular reaction.

Gas-Phase Reactions

Gas-phase reactions involve reactants and products in the gaseous state, often represented by their partial pressures. The behavior of these reactions can be studied using the principles of kinetics and thermodynamics. For the given reaction:
\( ext{CO(g) + H}_2 ext{O(g)} \rightleftharpoons ext{CO}_2 ext{(g) + H}_2 ext{(g)},\) each component's pressure is vital in calculations.
Some key points about gas-phase reactions include:

• Ideal gas behavior is often assumed for simplicity in calculations.
• The effect of temperature or pressure changes can be analyzed using Le Chatelier’s principle.
• Such reactions are often sensitive to changes in conditions, affecting equilibrium positioning.

In gas-phase equilibrium, adjustments to pressure, volume, or temperature can shift the reaction. For instance, increasing pressure by decreasing volume generally favors the side with fewer moles of gas. Understanding these principles is essential for predicting and controlling chemical reactions in the gaseous state.