Problem 63 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of HI was heated in a sealed tube at \(440^{\circ} \mathrm{C}\) till the equilibrium was reached. HI was found to be \(22 \%\) decomposed. The equilibrium constant for dissociation reaction, \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) is (a) \(1.99\) (b) \(0.282\) (c) \(0.01988\) (d) \(0.0796\)

Step-by-Step Solution

Answer

The equilibrium constant is approximately 0.01988.

Step 1: Set Up the Initial Condition

The problem states that you have one mole of HI that decomposes into \(H_2\) and \(I_2\) according to the reaction \(2 \text{ HI} \rightleftharpoons \text{ H}_2 + \text{ I}_2\). Initially, there is 1 mole of HI and 0 moles of \(H_2\) and \(I_2\).

Step 2: Determine the Change in Moles

Since HI is 22% decomposed, 22% of the initial 1 mole decomposes into \(H_2\) and \(I_2\). Therefore, 0.22 moles of HI decompose. According to the stoichiometry of the reaction, this forms 0.11 moles of \(H_2\) and 0.11 moles of \(I_2\).

Step 3: Calculate the Equilibrium Moles

At equilibrium, the moles of HI are \(1 - 0.22 = 0.78\) moles, and both \(H_2\) and \(I_2\) are \(0.11\) moles each.

Step 4: Write the Expression for Equilibrium Constant

The equilibrium constant \(K_c\) for the reaction is given by the expression \[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \] where the concentrations are expressed in terms of moles per unit volume. In this case, if we assume the volume \(V\) is constant, we can replace concentration with moles.

Step 5: Substitute the Values

Substituting the equilibrium moles into the expression we get: \[ K_c = \frac{(0.11)(0.11)}{(0.78)^2} \]

Step 6: Perform the Calculation

Calculate the values: \[ K_c = \frac{0.0121}{0.6084} \approx 0.01988 \]

Step 7: Conclusion

The calculated equilibrium constant \(K_c\) is approximately \(0.01988\), which corresponds to option (c).

Key Concepts

Chemical Equilibrium

In chemistry, equilibrium is when a reaction has reached a point where the concentrations of reactants and products remain constant. This doesn't mean that the reaction has stopped; instead, both the forward and reverse reactions occur at the same rate.
This balance or steady state is called dynamic equilibrium. In the example of the HI dissociation reaction, equilibrium is reached when the conversion of HI into H\(_2\) and I\(_2\) is balanced by their conversion back into HI.

• The position of equilibrium can be affected by external factors like temperature and pressure.
• However, at a given temperature, the equilibrium constant \(K_c\) remains the same.

Understanding equilibrium is key to predicting how a reaction behaves under certain conditions. It's a fundamental concept that governs the behavior of chemical reactions in closed systems.

Dissociation Reaction

A dissociation reaction is a specific type of chemical reaction where a compound breaks down into two or more components. In the exercise, Hydrogen iodide \( (\text{HI}) \) dissociates into Hydrogen \((\text{H}_2)\) and Iodine \((\text{I}_2)\).

• The dissociation is expressed by a balanced chemical equation: \(2 \text{ HI} \rightleftharpoons \text{ H}_2 + \text{ I}_2\).
• The double arrow indicates that the process is reversible, providing a pathway for both the forward and reverse reactions.

The degree of dissociation can be influenced by external conditions like temperature and pressure, shifting the equilibrium toward the production of more products or reactants. Calculating the extent of dissociation is crucial to understanding how much of the original reactant is converted into products.

Stoichiometry

Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the conservation of mass where the total mass of the reactants equals the total mass of the products.

• Writing balanced equations is critical as they provide the precise stoichiometric relationships between different species.
• In the HI dissociation example, stoichiometry dictates that for every 2 moles of HI that dissociate, 1 mole of \(\text{H}_2\) and 1 mole of \(\text{I}_2\) are formed.

These ratios allow you to predict the amount of product formed or the amount of reactant needed. Stoichiometry also helps in calculating the equilibrium constant by providing the necessary mole or concentration values needed in the expression.

Mole Concept

The mole concept is a fundamental unit in chemistry used to quantify substances. One mole corresponds to Avogadro's number \(6.022 \times 10^{23}\) of particles, be they atoms, molecules, ions, or electrons.
In the exercise, one mole of HI is initially present in the reaction system. As the reaction progresses, a percentage of these moles dissociate.
Understanding the mole concept helps you manage:

• How many particles are reacting or produced.
• The conversion between moles and grams using molar mass, which is essential for stoichiometric calculations.

The mole is a crucial concept for quantifying different substances in any chemical reaction, making it easier to perform conversions from microscopic to macroscopic scales.