Problem 47 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\), the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{l}_{2}\) and \(\mathrm{HI}\) are 8,3 and \(28 \mathrm{~mol} \mathrm{~L}^{-1}\) respectively. Equilibrium constant of the reaction is (a) \(32.67\) (b) \(31.67\) (c) \(34.67\) (d) \(36.67\)

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) is \(32.67\), which is option (a).

Step 1: Write the Expression for Equilibrium Constant

The equilibrium constant, \(K_c\), for the reaction \(\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}\) is given by:\[K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\]This expression is derived from the stoichiometry of the balanced chemical equation.

Step 2: Substitute the Equilibrium Concentrations

Substitute the given equilibrium concentrations into the equilibrium constant expression:\[K_c = \frac{(28)^2}{(8)(3)}\]This step involves placing the provided concentrations of \(\text{H}_2\), \(\text{I}_2\), and \(\text{HI}\) into the equation.

Step 3: Calculate the Equilibrium Constant

Perform the calculation using the substituted values:\[K_c = \frac{784}{24} = 32.67\]You calculate \((28)^2\), which is 784, and then divide by 24, the product of 8 and 3.

Step 4: Verify and Select the Correct Option

The calculated \(K_c\) is \(32.67\). Thus, the correct option from the given choices is option (a) \(32.67\).

Key Concepts

Equilibrium Constant

Chemical reactions often want to reach a state of balance. In chemistry, this balance is quantified by something called the "equilibrium constant," abbreviated as \(K_c\). For a given chemical reaction like \(\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}\), the equilibrium constant \(K_c\) is a number that helps us understand how far the reaction goes toward products or reactants at equilibrium.

This number is obtained through an equation that relates the concentrations of the substances involved. Specifically, for our reaction, the expression for \(K_c\) is \[K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\]. Here, the concentration of hydrogen iodide is squared because two moles are produced for every mole of hydrogen and iodine reacting, demonstrating its dependency on reaction stoichiometry, which we'll discuss next. This means:

• High \(K_c\) values suggest the reaction favors product formation.
• Low \(K_c\) values indicate the reaction favors the reactants.

Understand how to calculate and interpret \(K_c\) to have a complete picture of how reactions behave at equilibrium.

Reaction Stoichiometry

Chemical equations must be balanced, and balancing them is a job for stoichiometry. But what is stoichiometry? It's essentially the part of chemistry focused on the quantitative relationships between reactants and products in a chemical reaction.

For our example reaction \(\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}\), stoichiometry tells us that one molecule of \(\text{H}_2\) reacts with one molecule of \(\text{I}_2\) to create two molecules of \(\text{HI}\). This stoichiometric information is crucial when writing the expression for the equilibrium constant, as it informs us why we square the \([\text{HI}]\) term.

Stoichiometry also helps with:

• Understanding that coefficients in the balanced equation become exponents in the equilibrium constant expression.
• Helping predict the amounts of products formed during the reaction.

Mastering stoichiometry not only helps in equilibrium calculations but is a fundamental skill across all chemical calculations.

Concentration in Equilibrium

When reactions reach equilibrium, the concentrations of reactants and products remain constant. This doesn't mean the amounts are equal; rather, both the forward and backward reaction rates are balanced.

In our example, the concentrations at equilibrium are \([\text{H}_2] = 8 \ \text{mol} \ \text{L}^{-1}\), \([\text{I}_2] = 3 \ \text{mol} \ \text{L}^{-1}\), and \([\text{HI}] = 28 \ \text{mol} \ \text{L}^{-1}\). These values are essential inputs for computing the equilibrium constant.

Managing concentration involves:

• Quantifying how much of each substance is present at equilibrium.
• Using these concentrations to calculate other reaction parameters like \(K_c\).

Understanding how the concentration of reactants and products in equilibrium affects the chemical behavior is crucial for predicting how a reaction progresses over time.

Le Chatelier's Principle

Le Chatelier's Principle is a guiding rule in chemistry that predicts how a system at equilibrium will respond to changes in concentration, pressure, or temperature to restore a new equilibrium state.

If a change occurs (such as adding more \(\text{H}_2\) or increasing temperature), Le Chatelier's Principle can help us guess what will happen next. The reaction will shift to oppose the change:

• Increase in \(\text{H}_2\) will push the equilibrium towards forming more \(\text{HI}\).
• Increasing the temperature usually favors the endothermic direction of the reaction (depending on reaction enthalpy).

By allowing us to predict these shifts, Le Chatelier's Principle becomes an invaluable tool in controlling and understanding chemical reactions, particularly in industrial processes where yield manipulation is critical.