Problem 183 - A Complete Resource Book in Chemistry for JEE Main

Question

The degree of dissociation of HI at a particular temperature is \(0.8\). Calculate the volume (in litres) of \(1.6\) M \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) required to neutralize the iodine present in a equilibrium mixture of a reaction when 2 mole each of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) are heated in a closed vessel of \(2 \mathrm{~L}\).

Step-by-Step Solution

Answer

0.5 litres of Na2S2O3 are needed to neutralize the iodine.

Step 1: Write the Dissociation Reaction

The dissociation of \( ext{HI} \) is given by the reaction \[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]. The degree of dissociation is the fraction of the initial amount that dissociates, which is 0.8 in this case.

Step 2: Calculate the Change in Moles Due to Dissociation

Let the initial moles of \( \text{HI} \) at equilibrium be represented as \( x \). Then, at equilibrium, \[ x = 2 + 2 \times 0.8 = 3.6 \] because 0.8 of \( \text{HI} \) has dissociated and each mole forms two moles of \( \text{HI} \).

Step 3: Determine Equilibrium Concentrations

Each species at equilibrium will have a concentration based on the volume of the reaction vessel. Assume initial moles of \( \text{H}_2 \) and \( \text{I}_2 \) were 2 moles each. Then, the concentration in a 2 L vessel becomes:- \( [\text{H}_2] = [\text{I}_2] = 1 - 0.8 = 0.2 \text{ Moles per Litre}\)- \( [\text{HI}] = 3.6/2 = 1.8 \,\text{M} \).

Step 4: Determine Iodine Concentration from Dissociation

For \( \text{I}_2 \), \[ 1 - 0.8 = 0.2 \text{ Moles per Litre} \]; hence, in 2 L, total moles = 0.2 \( \times 2 = 0.4 \,\text{moles of I}_2 \) remain undissociated.

Step 5: Calculate Moles of Iodine to be Neutralized

The reaction that occurs for neutralization is \( \text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-} \). Thus, 0.4 moles of \( \text{I}_2 \) will need \( 0.8 \) moles of \( \text{Na}_2\text{S}_2\text{O}_3 \).

Step 6: Calculate Volume Required to Neutralize

Given: \( c = 1.6 \) M for \( \text{Na}_2\text{S}_2\text{O}_3 \). Use \( n = c \times V \), rearrange to find: \[ V = \frac{n}{c} = \frac{0.8}{1.6} = 0.5 \,\text{Litres} \].

Key Concepts

Degree of Dissociation

When studying chemical equilibrium, the degree of dissociation is a fundamental concept. It represents the fraction of the initial substance that undergoes dissociation during a reaction. In our exercise, the degree of dissociation for hydrogen iodide (HI) is 0.8. This indicates that 80% of the HI molecules dissociate at equilibrium.
Understanding this concept is crucial, because dissociation affects the concentrations of species in the reaction mixture. To calculate how many moles dissociate, we multiply the initial number of moles by the degree of dissociation. Hence, if there are 2 moles of both hydrogen and iodine initially, dissociation forms additional HI, changing the system dynamics. Recognizing the degree of dissociation helps predict these concentrations at equilibrium.

Reaction Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. The given reaction, \[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \], illustrates the stoichiometric coefficients. It tells us that one molecule of hydrogen reacts with one molecule of iodine to form two molecules of hydrogen iodide.
In our example, knowing these relationships is vital for computing equilibrium concentrations and understanding how the reaction progresses. If the degree of dissociation is known, stoichiometry lets us judge the quantity of each substance at equilibrium. It forms the backbone for tidily connecting the initial quantities, dissociation fractions, and quantities at equilibrium, allowing precise calculations and predictions.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products in a reaction mixture when the reaction has reached a state of balance. For the given reaction, the initial moles of hydrogen and iodine were 2 each. Assuming complete mixing in a 2-liter vessel, their concentrations would be 1 mole per liter initially.
Due to the degree of dissociation (0.8 for HI), however, the concentrations at equilibrium shift. The equilibrium concentration of unreacted \(\text{H}_2\) and \(\text{I}_2\) becomes 0.2 moles per liter, and \(\text{HI}\) concentration increases to 1.8 moles per liter. These values help assess the position of equilibrium, crucial for predicting reaction behavior in chemical processes.

Neutralization Reaction

Neutralization reactions are vital in chemistry, especially when dealing with titrations or analyses involving acids and bases. In our exercise, the given neutralization setup involves iodine and sodium thiosulfate:\[ \text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-} \]This reaction highlights the electron transfer from thiosulfate ions to iodine.
To calculate how much \(\text{Na}_2\text{S}_2\text{O}_3\) is needed, we rely on stoichiometry. Each mole of \(\text{I}_2\) needs two moles of sodium thiosulfate. For the task, 0.4 moles of \(\text{I}_2\) require 0.8 moles of thiosulfate, leading to the conditionally calculated 0.5 liters of solution needed. This illustrates using stoichiometry to balance reactions and ensure the complete neutralization of iodine.