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

Question

The ratio of \(\mathrm{K}_{p} / \mathrm{K}_{c}\) 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) \((\mathrm{RT})^{L / 2}\) (d) \((\mathrm{RT})^{-1 / 2}\)

Step-by-Step Solution

Answer

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

Step 1: Identify Relation between Kp and Kc

For a general reaction: \[ aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) \]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.

Step 2: Determine Δn for the Given Reaction

For the reaction \(\mathrm{CO}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)\), the change in moles of gas, \(\Delta n\), is calculated as follows:- Products: 1 mole \(\mathrm{CO}_{2}(g)\)- Reactants: 1 mole \(\mathrm{CO}(g)\) + 0.5 mole \(\mathrm{O}_{2}(g)\) = 1.5 moles\(\Delta n = 1 - 1.5 = -0.5\)

Step 3: Apply the Formula to Find Kp/Kc

Substitute \(\Delta n = -0.5\) into the \(K_p\) and \(K_c\) relation to find:\[ \frac{K_p}{K_c} = (RT)^{-0.5} \]Thus, the correct answer is option (d), which corresponds to \((\mathrm{RT})^{-1/2}\).

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant is a crucial concept that helps us understand the balance in chemical reactions. It comes in two forms: the concentration-based constant, \( K_c \), and the pressure-based constant, \( K_p \).
The equilibrium constant reflects the relative amounts of products and reactants at equilibrium. For reactions involving gases, we tend to use \( K_p \), while \( K_c \) is applied for reactions expressed in terms of concentrations. Both constants indicate the extent to which a reaction will proceed and are linked through the equation:

• \( K_p = K_c (RT)^{\Delta n} \)

This formula shows how pressure and concentration descriptors relate when considering their equilibrium states. Here, \( R \) is the gas constant and \( T \) represents the temperature in Kelvin. Understanding this equation is vital in chemical thermodynamics, as it helps predict how changes in conditions will influence the positions of equilibria.

Chemical Equilibrium

Chemical equilibrium is reached in a reaction when the rate of the forward reaction equals the rate of the reverse reaction. This stops macroscopic concentration changes of reactants and products, creating a steady state.
At this point, the concentrations of reactants and products are constant, but not necessarily equal. The equilibrium constant provides a measure of how far the reaction runs towards products or stays towards reactants:

• If \( K \gg 1 \), the equilibrium position lies far to the right, favoring products.
• If \( K \ll 1 \), the equilibrium favors the reactants, staying on the left.

Factors such as temperature and pressure can shift equilibria positions, but the principle of equilibrium itself remains unchanged. This principle governs not only chemical systems but also biochemistry, industrial processes, and environmental sciences.

Gas Reactions

Gas reactions are particularly interesting because they can show significant changes in volume and pressure. This makes understanding the relationship between \( K_p \) and \( K_c \) so important.
The reaction \[ \mathrm{CO}(g) + \frac{1}{2}\mathrm{O}_2(g) \rightleftharpoons \mathrm{CO}_2(g) \] serves as a prime example. Here, we calculate the change in moles of gas, \( \Delta n \), which is crucial for finding the precise relation between \( K_p \) and \( K_c \). For this reaction:

• Moles of gas on products: \( 1 \) mole \( \mathrm{CO}_2(g) \)
• Moles of gas on reactants: \( 1.5 \) (\( 1 \) mole \( \mathrm{CO}(g) + 0.5 \) mole \( \mathrm{O}_2(g) \))
• \( \Delta n = 1 - 1.5 = -0.5 \)

Using \( \Delta n \) helps determine how the reaction will behave under different conditions, allowing chemists to control and optimize these types of reactions in industrial applications.