Problem 123 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The correct answer is (b) \((RT)^{-1/2}\).

Step 1: Understand the Reaction and Equilibrium Constants

In this step, we identify the synthesis reaction: \(\mathrm{CO} + \frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\). Here, \(K_p\) is the equilibrium constant related to partial pressures, while \(K_c\) is related to concentrations. We need to find the relationship between them.

Step 2: Use the Ideal Gas Law to Relate Pressure and Concentration

The ideal gas law provides \(PV = nRT\). Therefore, for concentration \(C\), we have \(C = \frac{P}{RT}\). This allows transformation between pressure-based \(K_p\) and concentration-based \(K_c\).

Step 3: Apply the Formula for \(K_p\) and \(K_c\) Relationship

The relationship between \(K_p\) and \(K_c\) is given by:\[K_p = K_c (RT)^{\Delta n}\]where \(\Delta n\) is the change in moles of gas between reactants and products.

Step 4: Calculate \(\Delta n\) for the Reaction

For the reaction \(\mathrm{CO} + \frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\), calculate \(\Delta n\):\[\Delta n = (1) - (1 + \frac{1}{2}) = -\frac{1}{2}\]

Step 5: Substitute \(\Delta n\) into the Formula

Substitute \(\Delta n = -\frac{1}{2}\) into the relationship formula:\[K_p = K_c (RT)^{-\frac{1}{2}}\]

Step 6: Determine the Correct Answer

The expression \(K_p = K_c (RT)^{-\frac{1}{2}}\) implicates that the ratio \(\frac{K_p}{K_c} = (RT)^{-\frac{1}{2}}\). Hence, the correct answer is option (b).

Key Concepts

Kp and Kc Relationship

In chemical equilibrium, both pressure and concentration come into play when defining equilibrium constants. The constants for gaseous reactions can be expressed either in terms of partial pressure

• Kp

, or in terms of molar concentration

• Kc

. While these constants are both crucial, they don't have the same value for a given reaction unless the change in moles of gas,

• \(\Delta n\)

, is zero.
The fundamental relationship connecting

• Kp \( = Kc (RT)^{\Delta n} \)

shows how Kp can convert to Kc if the conditions of temperature and gas constant are considered together with the change in moles of gas.
This understanding is vital as it helps predict behavior in large-scale industrial reactions, environmental systems, or even in everyday applications such as cooking or breathing.

Ideal Gas Law

The Ideal Gas Law is a cornerstone in understanding gases and their behaviors. This equation links several physical properties of an ideal gas:

• Pressure \(P\)
• Volume \(V\)
• Moles of gas \(n\)
• Temperature \(T\)
• Gas constant \(R\)

The formula itself is a simple one, expressed as

• \( PV = nRT \)
. This law implies that when a gas behaves ideally, these properties are proportionally and predictably related.
In practice, when solving for chemical equilibria, the Ideal Gas Law helps transform between pressures and concentrations, as they might be needed to convert equations in terms of Kp or Kc.
Applying the Ideal Gas Law makes it feasible to reflect real-world conditions into calculable elements, making problem solving in chemistry much more intuitive.

Equilibrium Constants

Equilibrium constants are fundamental in determining how a chemical reaction proceeds at a given temperature. These constants, denoted by

• Kc
• Kp

, represent the ratio of the concentrations, or pressures, of products to reactants when a system is at equilibrium. Importantly, equilibrium constants tell us the inherent tendency of a reaction to proceed towards products or remain as reactants.
A large equilibrium constant implies products are favored at equilibrium, while a small constant indicates reactants are favored. Understanding equilibrium constants is crucial for predicting how changes in conditions—such as temperature or pressure—will affect a system.
Adjusting any variable can shift the position of equilibrium, highlighting the dynamic nature of chemical processes. For scientists and students alike, mastering this concept is crucial in fields ranging from pharmaceuticals to environmental science.