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) RT (c) \(\frac{1}{\sqrt{\mathrm{RT}}}\) (d) 1

Step-by-Step Solution

Answer

The correct answer is (c) \( \frac{1}{\sqrt{RT}} \).

Step 1: Recognize the Relationship Between Kp and Kc

The relationship between the equilibrium constants \( K_p \) and \( K_c \) for a gaseous reaction is given by the formula:\[ K_p = K_c(RT)^{\Delta n} \]where \( \Delta n \) is the change in the number of moles of gas (products - reactants).

Step 2: Determine Δn for the Reaction

For the reaction \( \mathrm{CO} (g) + \frac{1}{2} \mathrm{O}_2 (g) \rightleftharpoons \mathrm{CO}_2 (g) \), calculate \( \Delta n \):- Moles of gases on the product side: 1 (from CO\(_2\))- Moles of gases on the reactant side: \(1 + \frac{1}{2} = 1.5\)- Thus, \( \Delta n = 1 - 1.5 = -0.5 \).

Step 3: Apply Δn to the Kp and Kc Relationship

Using the formula \( K_p = K_c (RT)^{\Delta n} \) and substitute \( \Delta n = -0.5 \):\[ \frac{K_p}{K_c} = (RT)^{-0.5} \]This simplifies to:\[ \frac{K_p}{K_c} = \frac{1}{\sqrt{RT}} \].

Step 4: Match the Result to the Given Options

The expression we derived for \( \frac{K_p}{K_c} \) is \( \frac{1}{\sqrt{RT}} \), which corresponds to option (c).

Key Concepts

Kp and Kc relationship

In the study of chemical equilibria, especially concerning gaseous reactions, two important constants are often discussed: the equilibrium constant in terms of pressure, \(K_p\), and the equilibrium constant in terms of concentration, \(K_c\). These constants help describe the state of a chemical reaction at equilibrium. Their relationship is pivotal for understanding how changes in conditions can affect the reaction's equilibrium position.

The relationship between \(K_p\) and \(K_c\) is defined mathematically as: \[K_p = K_c(RT)^{\Delta n}\] where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) represents the change in the number of moles of gas. This relationship is crucial because it allows us to convert between \(K_p\) and \(K_c\) depending on what information is available about the system.

Understanding this formula helps us adjust our calculations for different conditions—be it temperature or pressure changes—and aids in predicting how these changes shift a reaction's equilibrium.

Delta n calculation

The concept of \(\Delta n\) is essential when dealing with gaseous equilibrium reactions. \(\Delta n\) refers to the difference between the number of moles of gaseous products and the number of moles of gaseous reactants in a balanced chemical equation.

Calculating \(\Delta n\) accurately is vital because it directly affects the relationship between \(K_p\) and \(K_c\). Let’s break down the process of calculating \(\Delta n\) through an example reaction:

• Consider the gaseous reaction: \(\mathrm{CO} (g) + \frac{1}{2} \mathrm{O}_2 (g) \rightleftharpoons \mathrm{CO}_2 (g)\).
• The number of moles on the product side: 1 (from \(\mathrm{CO}_2\)).
• The number of moles on the reactant side: 1 (from \(\mathrm{CO}\)) + 0.5 (from \(\mathrm{O}_2\)) = 1.5 moles.
• Thus, \(\Delta n = 1 - 1.5 = -0.5\).

Recognizing how \(\Delta n\) stems from these mole differences ensures we can correctly apply it in calculations involving \(K_p\) and \(K_c\), making it a foundational part of understanding gaseous reactions.

Chemical equilibrium

Chemical equilibrium occurs when the rate of a forward reaction equals the rate of the backward reaction, resulting in no net change in the concentration of the reactants and products over time. The reaction is still occurring, just that rates forward and backward are balanced, making it a dynamic state.

Equilibrium is a key part of understanding reactions in the real world because it helps chemists predict how changes in conditions—like pressure, temperature, or concentration—affect the outcome of reactions. In the context of gaseous reactions, calculating equilibrium constants like \(K_c\) and \(K_p\) allows us to quantify the position of equilibrium.

By knowing the equilibrium constants, we can determine in which direction the reaction will proceed under different conditions, and thus control product formation. It's crucial for industries relying on chemical synthesis and even for environmental modeling where atmospheric gas equilibria need to be understood.

Gaseous reactions

Gaseous reactions involve substances in their gaseous states, and they often occur in closed systems. In these reactions, changes in pressure, volume, and temperature can significantly affect the outcomes and positions of equilibrium, largely due to the malleability of gases compared to liquids or solids.

Understanding how gaseous reactions behave is necessary for applying the principles of equilibrium constants. Here are key points about gaseous reactions:

• They are affected by changes in pressure and temperature due to the ideal gas law, \(PV = nRT\).
• The equilibrium state depends on the balance between reactant and product gases, often requiring adjustments in reaction conditions to achieve desired outcomes.
• The relationship between \(K_p\) and \(K_c\) is particularly relevant for gaseous reactions, as changes in pressure or temperature alter the moles of gas participating in the system.

Mastering these concepts allows a deeper understanding of not only laboratory experiments but also natural processes, like atmospheric interactions and industrial applications involving gas-phase reactions.