Question

The ratio of \(\mathrm{K}_{\mathrm{p}} / \mathrm{K}_{\mathrm{c}}\) for the reaction: \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) is (a) 1 (b) RT (c) \((\mathrm{RT})^{1 / 2}\) (d) \((\mathrm{RT})^{-1 / 2}\)

Short answer

The ratio \( K_p/K_c \) is \((RT)^{-1/2}\). Option (d) is correct.

Step-by-step solution

Identify Reaction Details

The given reaction is \( \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g}) \). This reaction involves 1 mole of CO and 0.5 moles of O_2 converting into 1 mole of CO_2, suggesting a change in the number of moles of gas of \(-0.5\).

Recall Relationship between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by the equation \( K_p = K_c (RT)^\Delta n \), where \( \Delta n \) is the change in moles of gas during the reaction, and \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin.

Determine \( \Delta n \)

For the reaction \( \Delta n = \text{{total moles of gaseous products}} - \text{{total moles of gaseous reactants}} = 1 - (1 + 0.5) = -0.5 \).

Apply the \( K_p \) and \( K_c \) Relationship

Substitute \( \Delta n = -0.5 \) into the relationship equation: \[ K_p = K_c (RT)^{-0.5} \]. Therefore, \[ \frac{K_p}{K_c} = (RT)^{-0.5} \].

Select the Correct Option

The provided options are (a) 1, (b) RT, (c) \((RT)^{1/2}\), and (d) \((RT)^{-1/2}\). Since we found \( \frac{K_p}{K_c} = (RT)^{-0.5} \), the correct option is (d).

Key concepts

Relationship between Kp and Kc

In chemical equilibrium, two essential constants are often considered: the equilibrium constant calculated with partial pressures, denoted as \( K_p \), and the equilibrium constant calculated with concentrations, \( K_c \). These constants are critical in understanding how gases behave in various reactions where pressures and concentrations play a role.
Whenever a gaseous reaction is involved, there's a simple yet powerful relationship that connects \( K_p \) and \( K_c \): \[ K_p = K_c (RT)^{\Delta n} \] where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas from the reactants to the products.
Here’s how it works: * If \( \Delta n \) equals zero, there’s no change in moles of the gas during the reaction, making \( (RT)^{0} = 1 \), implying \( K_p = K_c \). * If \( \Delta n \) is positive, it means more gas moles are produced than consumed, resulting in \( K_p > K_c \). * Conversely, if \( \Delta n \) is negative, fewer moles of gas are produced, leading to \( K_p < K_c \).
Understanding these relationships allows chemists to predict how reactions will behave when pressures and concentrations shift. This equation elegantly ties together the parameters affecting chemical equilibrium in gases.

Change in moles of gas (Δn)

The concept of \( \Delta n \), or the change in moles of gas, is pivotal in predicting how gases behave at equilibrium under different conditions.
For calculations involving gaseous equilibria, \( \Delta n \) is determined by subtracting the total moles of gaseous reactants from those of products. The result tells us how much the quantity of gas changes during the reaction:

• If \( \Delta n = 0 \), the moles of gaseous reactants equal those of the products.
• If \( \Delta n > 0 \), more moles of gas are produced than what reacted, indicating an expansion in the gaseous phase.
• Conversely, \( \Delta n < 0 \) implies a contraction, with fewer gaseous moles produced than reacted.

In our specified reaction, \( \Delta n \) was found to be \(-0.5\). This indicates a reduction in the total moles of gas, as 1 mole of carbon monoxide and 0.5 moles of oxygen are converted into 1 mole of carbon dioxide. Because temperature and the ideal gas law factor into this calculation, \( \Delta n \) has direct consequences for the relationship between \( K_p \) and \( K_c \), as explained above.

Ideal gas law in equilibrium calculations

The ideal gas law is a fundamental principle useful for equilibrium calculations in chemistry, especially when dealing with gases. The ideal gas law is represented as \( PV = nRT \), where:

• \( P \) stands for pressure,
• \( V \) for volume,
• \( n \) for moles of gas,
• \( R \) for the ideal gas constant,
• and \( T \) for temperature in Kelvin.

In equilibrium calculations, this law is particularly valuable because it links pressure with the amount of gas and temperature — all key factors when considering \( K_p \) and \( \Delta n \).
In the context of \( \Delta n \) and chemical equilibria, the ideal gas law allows chemists to make accurate predictions and adjustments. By knowing how moles of gas change (\( \Delta n \)), and applying the ideal gas law, you can determine how these changes affect pressure and, hence, \( K_p \). Thus, the ideal gas law serves as a bridge connecting microscopic molecular changes to macroscopic observable properties like pressure and volume, ensuring that your equilibrium calculations are grounded in consistent physical principles.