Problem 149 - A Complete Resource Book in Chemistry for JEE Main

Question

Assume that the decomposition of \(\mathrm{HNO}_{3}\) can be represented by the following equation \(4 \mathrm{HNO}_{3}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) and the reaction approaches equilibrium at \(400 \mathrm{~K}\) temperature and 30 atm pressure. At equilibrium, partial pressure of \(\mathrm{HNO}_{3}\) is \(2 \mathrm{~atm} .\) Calculate \(\mathrm{K}_{\mathrm{c}}\) in \((\mathrm{mol} / \mathrm{L})^{3}\) at \(400 \mathrm{~K}\) : (Use : \(\mathrm{R}=0.08 \mathrm{~atm}-\mathrm{L} / \mathrm{mol}-\mathrm{K})\) (a) 4 (b) 8 (c) 16 (d) 32

Step-by-Step Solution

Answer

The correct answer is (c) 16.

Step 1: Write the Equilibrium Expression

For the given reaction, the equilibrium expression for the equilibrium constant in terms of pressure \(K_p\) is \[ K_p = \frac{(P_{NO_2})^4 (P_{H_2O})^2 (P_{O_2})}{(P_{HNO_3})^4} \] where \(P_{NO_2}\), \(P_{H_2O}\), and \(P_{O_2}\) are the partial pressures of \(NO_2\), \(H_2O\), and \(O_2\) at equilibrium, respectively, and \(P_{HNO_3}\) is given as 2 atm.

Step 2: Calculate Total Pressure at Equilibrium

The total pressure at equilibrium is 30 atm. The partial pressure of \(HNO_3\) is given as 2 atm. Therefore, the partial pressure of the products combined is \(30 - 2 = 28 \text{ atm}\).

Step 3: Determine Partial Pressures of Products

Using stoichiometry from the balanced equation, let \(x\) be the change in pressure corresponding to \(H_2O\) and \(NO_2\) (since they share similar coefficients), and \(\frac{x}{2}\) be the change for \(O_2\). Thus at equilibrium: \[4x + 2x + x = 28 \Rightarrow 7x = 28 \Rightarrow x = 4 \text{ atm}.\] Then, \[ P_{NO_2} = 4x = 16 \text{ atm},\ \ P_{H_2O} = 2x = 8 \text{ atm},\ \ P_{O_2} = \frac{4}{2} = 2 \text{ atm}. \]

Step 4: Calculate K_p

Substitute the partial pressures back into the equilibrium expression for \(K_p\): \[ K_p = \frac{(16)^4 (8)^2 (2)}{2^4} = \frac{(65536)(64)(2)}{16} = 524288. \]

Step 5: Convert K_p to K_c

Use the equation \[ K_p = K_c (RT)^{\Delta n}, \] where \(\Delta n = 4 + 2 + 1 - 4 = 3\) (change in moles of gas), \(R = 0.08 \text{ atm⋅L/mol⋅K}\), and \(T = 400 \text{ K}\). Thus: \[ K_c = \frac{524288}{(0.08 \times 400)^3} = \frac{524288}{(32)^3} = \frac{524288}{32768} = 16. \]

Step 6: Match with Given Answers

The value of \(K_c\) is 16 \((\mathrm{mol/L})^3\), which matches option (c).

Key Concepts

Equilibrium Constant (Kc)

Chemical reactions that reach a state where the rate of forward reactions equals the rate of reverse reactions are said to be at equilibrium. An important aspect of equilibrium is the equilibrium constant, often denoted as \(K_c\) when concentrations are involved, rather than pressures.
This constant reflects the ratio of the concentrations of products to reactants for a reaction at equilibrium. It is specific to every reaction and depends on temperature, not pressure or concentration changes.
The equation for calculating \(K_c\) for a given reaction is derived from the balanced chemical equation, considering only gas and aqueous states:

• Products are on the numerator (raised to the power of their coefficients in the balanced equation).
• Reactants are on the denominator (also raised to their coefficients).

Understanding \(K_c\) helps predict the direction of the reaction and its position at equilibrium. For instance, if \(K_c > 1\), products are favored. If \(K_c < 1\), reactants are favored.

Partial Pressure

In the realm of chemical equilibrium involving gases, the concept of partial pressure is crucial. Each gas in a mixture exerts a pressure as if it occupied the entire volume alone, and this is known as partial pressure.
To understand a reaction’s dynamics at equilibrium, it’s important to calculate each component’s partial pressure.
The total pressure of the system is the sum of these partial pressures, which is based on the ideal gas law principles:
The partial pressure allows chemists to determine how much of a gas is present in a reaction vessel and how it influences equilibrium. For the decomposition reaction of \(HNO_3\), calculating partial pressures assists in finding the equilibrium constant \(K_p\), which is then converted to \(K_c\) using the ideal gas equation.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. This process is quite common in chemical processes and can be spontaneous or require external conditions like heat or pressure.
Each decomposition reaction has its unique characteristics and equation. For instance, the decomposition of \(HNO_3\) is expressed as:

• \(4 \mathrm{HNO}_{3}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\)

In this balanced equation, 4 molecules of \(HNO_3\) decompose into 4 molecules of \(NO_2\), 2 molecules of \(H_2O\), and 1 molecule of \(O_2\). Decomposition involves breaking chemical bonds, and it plays an important role in studying equilibrium to assess how reactants and products reach balance under a set of conditions.

Equilibrium Expression

The equilibrium expression for a chemical reaction is a mathematical representation that relates the concentrations or partial pressures of reactants and products at equilibrium. For gas-phase reactions, partial pressures are used in place of concentrations, leading to \(K_p\) expressions.
For example, for the decomposition of \(HNO_3\), the equilibrium expression is:

• \[ K_p = \frac{(P_{NO_2})^4 (P_{H_2O})^2 (P_{O_2})}{(P_{HNO_3})^4} \]

Here, each term’s exponent reflects the substance's coefficient in the balanced equation. The equilibrium expression is vital for determining how changes in concentration, pressure, or temperature could shift the equilibrium position.
Understanding this can help predict concentrations and pressures of reactants and products once the reaction reaches equilibrium.