Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

HI was heated in sealed tube at \(400^{\circ} \mathrm{C}\) till the equilibrium was reached. HI was found to be \(22 \%\) decomposed. The equilibrium constant for dissociation is (a) \(1.99\) (b) \(0.0199\) (c) \(0.0796\) (d) \(0.282\)

Step-by-Step Solution

Answer

The equilibrium constant is 0.0199, option (b).

Step 1: Write the Chemical Equation

The reaction for the dissociation of hydrogen iodide is \( 2 \text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \). Since 22% of the initial HI decomposes, we can use this information to find equilibrium concentrations.

Step 2: Define Variables for Initial and Change

Let the initial concentration of \( \text{HI} \) be \( [\text{HI}]_0 \). If 22% decomposes, the amount of \( \text{HI} \) remaining is \( 0.78[\text{HI}]_0 \). Let \( x \) be the moles of \( \text{HI} \) that decomposed; therefore, \( x = 0.22[\text{HI}]_0 \).

Step 3: Express Equilibrium Concentrations

At equilibrium, the concentration of \( \text{HI} \) is \( [\text{HI}]_e = 0.78[\text{HI}]_0 \). For each \( 2 \text{HI} \) decomposed, 1 mole of \( \text{H}_2 \) and 1 mole of \( \text{I}_2 \) forms, so \( [\text{H}_2] = [\text{I}_2] = 0.11[\text{HI}]_0 \) at equilibrium.

Step 4: Write the Expression for Equilibrium Constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by the formula: \[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]. Substituting the equilibrium concentrations, we have: \[ K_c = \frac{(0.11[\text{HI}]_0)(0.11[\text{HI}]_0)}{(0.78[\text{HI}]_0)^2} \].

Step 5: Simplify and Solve for \( K_c \)

Simplify the expression: \[ K_c = \frac{(0.11)^2}{(0.78)^2} \]. Calculate the values: \[ K_c = \frac{0.0121}{0.6084} \approx 0.0199 \].

Step 6: Identify the Correct Answer

The calculated value \( K_c \approx 0.0199 \) matches option (b).

Key Concepts

Dissociation Reaction

A dissociation reaction is a process where a compound breaks down into two or more simpler substances. In the context of the exercise, we are looking at the dissociation of hydrogen iodide (HI) into hydrogen gas \((\text{H}_2)\) and iodine gas \((\text{I}_2)\). This can be represented by the equation:

• \( 2 \text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \).

This type of reaction is crucial in understanding how different substances interact under certain conditions, such as temperature, to reach a balanced state.The term 'dissociation' indicates that HI is not being completely converted, but rather a portion of it is breaking down into other substances. In our problem, 22% of HI dissociates, meaning only a small fraction of the original compound actually changes form.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products present when a chemical reaction reaches equilibrium. Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in concentrations over time.
In the dissociation of hydrogen iodide, we start with an initial concentration of HI. Upon reaching equilibrium, the concentration changes due to dissociation. Therefore, calculating these concentrations is crucial for determining the equilibrium constant \( K_c \).

• If 22% of HI decomposes, then 78% remains, meaning \( [\text{HI}]_e = 0.78[\text{HI}]_0 \).
• For every 2 moles of HI decomposing, 1 mole of \( \text{H}_2 \) and 1 mole of \( \text{I}_2 \) forms, giving us \( [\text{H}_2] = [\text{I}_2] = 0.11[\text{HI}]_0 \) at equilibrium.

This balancing helps us apply the equilibrium concentrations in the expression for \( K_c \).

Chemical Equilibrium

Chemical equilibrium is the state in a reversible reaction where the forward and reverse reactions occur at the same rate. This balance means that the concentrations of reactants and products remain constant over time. However, it is important to note that it does not imply equal concentrations.
In the context of our exercise, when hydrogen iodide is heated, the system will eventually reach a point where the formation rate of \( \text{H}_2 \) and \( \text{I}_2 \) from HI equals the rate at which \( \text{H}_2 \) and \( \text{I}_2 \) recombine to form HI. The state of equilibrium is dynamic, which means that although the concentrations don't change, the reactions are still occurring.
The equilibrium constant \( K_c \) is a way to quantify this state and is specific to the particular reaction conditions, including temperature, as seen in step 6 of the solution.

Hydrogen Iodide Decomposition

Hydrogen iodide decomposition is crucial because it exemplifies how dynamic equilibrium works in chemical reactions. When HI is heated, it decomposes into \( \text{H}_2 \) and \( \text{I}_2 \). This reaction occurs until equilibrium is established.

• The initial concentration of HI starts to decrease as it breaks down.
• Simultaneously, \( \text{H}_2 \) and \( \text{I}_2 \) start to form until their formation rate equals the rate at which they recombine to form HI.

The equilibrium constant \( K_c \) gives us insight into how far the reaction proceeds towards products at a given temperature. A small \( K_c \) value, like in this exercise, indicates that at equilibrium, there are more reactants than products, meaning the reaction favors the formation of the initial compound, HI.
Understanding this helps students grasp why the dissociation is only partial, which is typical for endothermic reactions like the decomposition of HI at high temperatures.