Problem 67 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The correct option is (c) \((\mathrm{RT})^{1/2}\).

Step 1: Understanding the Reaction

The given reaction is \( \mathrm{CO} (\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2} (\mathrm{g}) \). This is a gaseous equilibrium reaction involving one mole of carbon dioxide being formed from one mole of carbon monoxide and half a mole of oxygen.

Step 2: Applying the Relation between \( K_p \) and \( K_c \)

The relation between the equilibrium constants \( K_p \) (pressure) and \( K_c \) (concentration) is given by the formula \( K_p = K_c (RT)^{\Delta n} \), where \( \Delta n \) is the change in moles of gas products minus moles of gas reactants.

Step 3: Calculating \( \Delta n \)

The change in moles, \( \Delta n \), is calculated as the difference between the moles of gaseous products and the moles of gaseous reactants. For this reaction, \( \Delta n = 1 - (1 + \frac{1}{2}) = 1 - 1.5 = -0.5 \).

Step 4: Finding the Ratio \( \frac{K_p}{K_c} \)

Substitute \( \Delta n = -0.5 \) into the relation \( K_p = K_c (RT)^{\Delta n} \). Thus, \( \frac{K_p}{K_c} = (RT)^{-0.5} = (RT)^{1/2} \).

Step 5: Selection of the Right Answer

Based on the calculation, \( \frac{K_p}{K_c} = (RT)^{1/2} \). Therefore, the correct option is (c) \((\mathrm{RT})^{1/2}\).

Key Concepts

Gaseous Equilibrium

In chemical reactions involving gases, equilibrium is reached when the rates of the forward and reverse reactions become equal. This state is referred to as gaseous equilibrium. At this point, the concentrations of all species remain constant over time, because the balanced rates mean there is no net change. For reactions involving gases, pressure can often influence equilibrium, as changes in pressure also affect concentration. It is crucial to understand that while proportions of reactants and products don’t change at equilibrium, individual molecules continue to react, shifting dynamically between reactants and products.

Equilibrium Constants

Equilibrium constants are vital in quantifying the position of equilibrium for a reaction. They take into account the concentrations of the reactants and products once equilibrium has been achieved.

• Kc is the equilibrium constant in terms of concentration, usually measured in moles per liter.
• Kp is the equilibrium constant in terms of partial pressure, specifically for reactions involving gases.

The value of these constants provides insights into the favorability of a reaction—in other words, whether products or reactants are prevalent at equilibrium. A large value of Kc or Kp suggests the formation of more products, while a small value implies mostly reactants remain.

Relation between Kp and Kc

The relationship between the equilibrium constants Kp and Kc is described by the expression: \[ K_p = K_c (RT)^{\Delta n} \]Here, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. The term \( \Delta n \) symbolizes the change in moles of gas, calculated as moles of gaseous products minus moles of gaseous reactants. This formula is particularly significant in determining how changes in conditions, like temperature, affect the equilibrium among gaseous molecules. Understanding this relationship helps predict how a reaction will respond when subjected to different pressures or temperatures.

Thermodynamics of Reactions

Thermodynamics is the study of energy and its transformations. In chemical reactions, it helps understand the feasibility and extent of reactions under given conditions. Key concepts include:

• Enthalpy (\( \Delta H \)): Measures the heat change at constant pressure. Exothermic reactions release heat, while endothermic reactions absorb it.
• Entropy (\( \Delta S \)): Describes the disorder or randomness in a system. An increase in entropy favors spontaneity.
• Gibbs Free Energy (\( \Delta G \)): Combines enthalpy and entropy to predict spontaneity: \(\Delta G = \Delta H - T\Delta S \). A negative \( \Delta G \) indicates a spontaneous reaction.

Thermodynamics allows predictions about a reaction's behavior, which is crucial for understanding both the direction and extent to which equilibrium will shift.