Problem 147 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{c}}\) for the reaction \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) will be (a) \(\sqrt{\mathrm{RT}}\) (b) \(\mathrm{RT}\) (c) \(\frac{1}{\sqrt{\mathrm{RT}}}\) (d) 1

Step-by-Step Solution

Answer

The ratio \( \frac{K_P}{K_c} \) is \( \frac{1}{\sqrt{RT}} \), so the answer is (c).

Step 1: Understand the Reaction Equation

The given chemical equilibrium reaction is \( \mathrm{CO}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{g}) \). We need to find the ratio of \( K_P \) to \( K_c \).

Step 2: Use the Relationship between Kₚ and Kc

The equation relating \( K_P \) and \( K_c \) is \( K_P = K_c (RT)^{\Delta n} \), where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas.

Step 3: Calculate Δn for the Reaction

For the reaction \( \mathrm{CO}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{g}) \), \( \Delta n = (1) - (1 + \frac{1}{2}) = -\frac{1}{2} \). This is the change in moles of gas from reactants to products.

Step 4: Substitute Δn into the Equation

Substitute \( \Delta n = -\frac{1}{2} \) into the equation \( K_P = K_c (RT)^{\Delta n} \). This gives \( K_P = K_c (RT)^{-\frac{1}{2}} \).

Step 5: Solve for the Ratio \( \frac{K_P}{K_c} \)

By rearranging the equation from Step 4, we can find \( \frac{K_P}{K_c} = (RT)^{-\frac{1}{2}} \).

Step 6: Identify the Correct Answer Choice

From the expression \( (RT)^{-\frac{1}{2}} = \frac{1}{\sqrt{RT}} \), we find that the correct answer is \( (c) \frac{1}{\sqrt{RT}} \).

Key Concepts

Kp and Kc relationship

In the study of chemical equilibrium, two important constants are often discussed: \( K_P \) and \( K_c \). These constants help us understand how reactions behave in different states. \( K_c \) is the equilibrium constant for reactions in solution, related to concentrations, while \( K_P \) is for reactions involving gases, connected to partial pressures.

There's a formula that links these two concepts:

• \( K_P = K_c (RT)^{\Delta n} \)

Here, \( R \) is the ideal gas constant and \( T \) is the temperature in Kelvin. \( \Delta n \) represents the change in moles of gases.

This equation helps us convert from concentration-based constants to pressure-based ones, revealing how temperature and molecular changes in a reaction affect the equilibrium.

Ideal Gas Constant

The ideal gas constant, represented as \( R \), is a critical value in chemistry, especially when dealing with gases. It links various properties of a gas under ideal conditions, meaning it assumes perfect behavior of gas particles. The value of \( R \) depends on the units involved. Commonly, for calculations involving the atmosphere and liters per mole per Kelvin, \( R = 0.0821 \; \text{L atm} / \text{mol K} \).

\( R \) plays a key role in the equation \( PV = nRT \), where it connects pressure \( P \), volume \( V \), number of moles \( n \), and temperature \( T \). In the context of equilibrium constants like \( K_P \) and \( K_c \), \( R \) helps bridge the gap between concentration and pressure, emphasizing its versatility and importance in different areas of chemistry.

Reaction Stoichiometry

Stoichiometry involves calculating the relationships between reactants and products in chemical reactions. It’s about keeping track of where atoms go during a reaction and ensuring they obey the law of conservation of mass.

In our exercise, the reaction \( \mathrm{CO(g)} + \frac{1}{2} \mathrm{O}_2\mathrm{(g)} \rightleftharpoons \mathrm{CO}_2\mathrm{(g)} \) involves combining carbon monoxide and oxygen to form carbon dioxide. Stoichiometry helps ensure the balanced equation correctly indicates one molecule of carbon monoxide reacts with half a molecule of oxygen to produce one molecule of carbon dioxide.

Understanding stoichiometry is crucial when calculating \( \Delta n \), the change in moles, as it affects how we apply the \( K_P \) and \( K_c \) relationship.

Change in Moles (Δn)

The change in moles, represented as \( \Delta n \), is a key factor in converting between \( K_P \) and \( K_c \). It is calculated by taking the difference in moles of gaseous products and reactants.

• For the reaction \( \mathrm{CO(g)} + \frac{1}{2} \mathrm{O}_2\mathrm{(g)} \rightleftharpoons \mathrm{CO}_2\mathrm{(g)} \), \( \Delta n = 1 - (1 + \frac{1}{2}) = -\frac{1}{2} \).

This value can change how \( RT \) is raised in the \( K_P \) and \( K_c \) relationship. Here, since \( \Delta n \) is negative, \( (RT)^{\Delta n} \) equates to \( (RT)^{-1/2} \), simplifying to \( \frac{1}{\sqrt{RT}} \).

Understanding \( \Delta n \) thus helps predict how equilibrium positions change under varying conditions, offering insight into the behavior and tendencies of chemical reactions.