Problem 95 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reversible reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons-2 \mathrm{NH}_{3}(\mathrm{~g})\) At \(500{ }^{\circ} \mathrm{C}\), the value of \(\mathrm{K}_{\mathrm{p}}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmospheres. The corresponding value of \(\mathrm{K}_{c}\), with concentration in mole \(\mathrm{L}^{-1}\), is (a) \(1.44 \times 10^{-5} /(0.082 \times 500)^{-2}\) (b) \(1.44 \times 10^{-5} /(8.314 \times 773)^{-2}\) (c) \(1.44 \times 10^{-5}(0.082 \times 773)^{2}\) (d) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\)

Step-by-Step Solution

Answer

(c) \(1.44 \times 10^{-5}(0.082 \times 773)^{2}\)

Step 1: Understand the Problem

We need to convert the equilibrium constant from pressure (\( K_p \)) to concentration (\( K_c \)). The relation between these constants is given by the equation: \[ K_c = K_p \left( \frac{1}{RT} \right)^\Delta n \] where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas.

Step 2: Calculate the Change in Moles

Determine \( \Delta n \) for the reaction \( \mathrm{N}_{2}( ext{g}) + 3\mathrm{H}_{2}( ext{g}) \rightleftharpoons 2\mathrm{NH}_{3}( ext{g}) \). Calculate the difference between the moles of products and reactants:\[ \Delta n = (2) - (1 + 3) = -2 \]

Step 3: Determine the Temperature in Kelvin

Convert the given temperature from Celsius to Kelvin:\[ T = 500 ^\circ \text{C} + 273 = 773 \text{ K} \]

Step 4: Choose the Correct Formula

Substitute \( K_p = 1.44 \times 10^{-5} \), \( R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1} \), \( T = 773 \text{ K} \), and \( \Delta n = -2 \) into the formula.\[ K_c = K_p (0.082 \times 773)^{2} \]

Step 5: Find the Correct Answer

Based on the formula and calculation from Step 4, the correct choice matches with option (c):\[ K_c = K_p (0.082 \times 773)^{2} \]

Key Concepts

Conversion of Kp to Kc

When dealing with reversible reactions involving gases, the equilibrium constants are often expressed in terms of either pressure (\( K_p \)) or concentration (\( K_c \)). The conversion between these two forms is crucial for chemists, as different situations might call for different constants.

The relationship between \( K_p \) and \( K_c \) is given by the equation: \[ K_c = K_p \left( \frac{1}{RT} \right)^\Delta n \], where:

• \( R \) is the ideal gas constant, which often takes the value 0.082 L atm K\(^{-1}\) mol\(^{-1}\).
• \( T \) is the temperature in Kelvin.
• \( \Delta n \) is the difference in moles of gaseous products and reactants.

For the conversion, it's imperative to calculate \( \Delta n \) correctly. In the given reaction \( \mathrm{N}_{2}(\mathrm{g}) + 3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_{3}(\mathrm{g}) \), we find \( \Delta n = 2 - (1 + 3) = -2 \). Hence, the formula becomes a straightforward application by inserting known values for targeted results.

Reversible reactions

Reversible reactions are fundamental to understanding chemical equilibrium. In such reactions, the reactants convert to products, but importantly, the products can also convert back to reactants. This bidirectional feature is denoted by a double arrow in the chemical equation and is key to reaching a state known as equilibrium.

At equilibrium, the rates of the forward and reverse reactions are equal. This doesn't mean the amounts of reactants and products are the same, but their concentrations remain constant over time. It’s a dynamic condition where continuous transformation occurs at a molecular level.

Reversible reactions are notable in many natural processes and industrial applications. By adjusting conditions like pressure and temperature, chemists can manipulate the position of equilibrium to favor either the forward or reverse action, thus increasing yield of desirable products. Understanding this concept underpins calculations related to equilibrium constants and their conversion between different forms.

Equilibrium constants

Equilibrium constants provide a way to express the extent of a reaction at equilibrium. They are symbolic of the balance reached between reactants and products in a reversible reaction. In the context of gases, these constants can be either \( K_p \) for pressure-based equations or \( K_c \) for concentration-based equations.

The constant \( K \) informs about the position of equilibrium:

• If \( K \) is large, the reaction favors the formation of products.
• If \( K \) is small, reactants are prevalent at equilibrium.

The mathematical expression for an equilibrium constant relates directly to the balanced chemical equation for the reaction. For example, for a general reversible reaction:
\[ aA + bB \rightleftharpoons cC + dD \] The equilibrium constant \( K_c \) would be expressed as: \[ \frac{[C]^c [D]^d}{[A]^a [B]^b} \].

It's essential for students to understand these constants not only as numbers but as indicators of reaction behavior. Therefore, conversions between \( K_p \) and \( K_c \), as seen in the original exercise, highlight the interactions of temperature and moles on a chemical system at equilibrium.