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 \(\mathrm{K}_{i}\), 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

Option c is correct: \(1.44 \times 10^{-5} (0.082 \times 773)^2\).

Step 1: Understand the Relationship Between Kp and Kc

For a gas-phase reaction like \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons=2 \mathrm{NH}_{3}(\mathrm{~g})\), the relationship between the equilibrium constant in terms of partial pressures \(K_p\) and in terms of concentrations \(K_c\) is given by the formula: \[ K_p = K_c (RT)^{\Delta n} \] where \(\Delta n\) is the change in moles of gas (product moles minus reactant moles) and \(R\) is the ideal gas constant. For this reaction, \(\Delta n = 2 - (1 + 3) = -2\).

Step 2: Use the Given Values

Given: \(K_p = 1.44 \times 10^{-5}\), \(R = 0.0821\, \text{L atm K}^{-1} \text{mol}^{-1}\), and temperature \(T = 500^{\circ} \mathrm{C} = 773\, \text{K}\). Using \(R = 0.082 \text{as '\.L atm K}^{-1} \text{mol}^{-1}'\), the expression for \(K_c\) becomes: \[ K_c = \frac{K_p}{(RT)^{\Delta n}} \] Substituting \(\Delta n = -2\), we have \(K_c = \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\).

Step 3: Solve for Kc

Let's evaluate the expression: \[ K_c = 1.44 \times 10^{-5} \times (0.082 \times 773)^2 \] First, calculate \(0.082 \times 773 = 63.386\). Then square this value: \(63.386^2\), and multiply the result by \(1.44 \times 10^{-5}\). Thus, the correct expression is option (c): \[ 1.44 \times 10^{-5} (0.082 \times 773)^2 \].

Step 4: Determine the Correct Option

The expression derived corresponds to option (c), where the conversion from \(K_p\) to \(K_c\) involves multiplying by \((RT)^{-\Delta n}\), and substituting \(RT\) to achieve the necessary conversion.

Key Concepts

Reversible reactions

In chemistry, a reversible reaction is a type of chemical reaction where the reactants form products, which in turn can react together to give back the reactants. These reactions can proceed in both forward and backward directions. Unlike irreversible reactions, reversible reactions do not go to completion and achieve a state of equilibrium.

Equilibrium in a reversible reaction is reached when the rate of the forward reaction equals the rate of the backward reaction. This means the concentrations of reactants and products remain constant over time, though not necessarily equal.

• Characteristics of reversible reactions include dynamic equilibrium and sensitivity to external conditions like temperature and pressure.
• The directionality of a reversible reaction can be altered by changing conditions, such as the concentration of reactants or products.

The reversible reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \,\mathrm{H}_{2}(\mathrm{~g})\rightleftharpoons=2 \,\mathrm{NH}_{3}(\mathrm{~g})\)demonstrates these principles, operating within an equilibrium state at which ammonia concentration remains stable.

Ideal gas constant

The ideal gas constant, often represented by \( R \), is a fundamental constant in chemistry that appears in various equations, including the ideal gas law and equilibrium expressions. It relates pressure, volume, and temperature in the state equation of an ideal gas.

• It plays a crucial role in converting between pressure-based equilibrium constants \( K_p \) and concentration-based constants \( K_c \).
• Its value varies depending on the units used and is often given as \( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \). This version is particularly useful in calculations involving gaseous reactions where pressure is measured in atmospheres.

In the context of converting \( K_p \) to \( K_c \) as in the exercise, \( R \) serves to account for the mole change\( \Delta n \) in the reaction, using the formula:\[K_c = \frac{K_p}{(RT)^{\Delta n}}\]Understanding \( R \) and its applications is pivotal for problem-solving in gas-phase reaction equilibria.

Equilibrium in gas-phase reactions

Equilibrium in gas-phase reactions is a dynamic state where the rates of forward and reverse reactions are equal, leading to no overall change in the system's composition. It involves balancing multiple factors, including pressure and temperature, that can influence the equilibrium position.

• The equilibrium constant \( K_p \) is used for gaseous reactions to describe the ratio of product pressures to reactant pressures at equilibrium.
• \( K_c \) uses concentrations instead and is connected to \( K_p \) by the relationship \( K_p = K_c (RT)^{\Delta n} \), where \( \Delta n \) is the difference in moles between products and reactants.

Calculating \( K_c \) from \( K_p \) involves understanding the changes in moles and accounting for them with the ideal gas constant, ensuring conditions like temperature are properly managed using absolute scales. These conversions are essential to accurately predicting how a system will respond to changes and are key concepts in reaction equilibrium analyses.