Problem 54 - 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 concentration of \(\mathrm{H}_{2}, \mathrm{I}_{2}\) and \(\mathrm{HI}\) are \(8.0,3.0\) and \(28.0\) mole/litre, respectively, the equilibrium constant is (a) \(28.34\) (b) \(32.66\) (c) \(34.78\) (d) \(38.88\)

Step-by-Step Solution

Answer

The equilibrium constant is 32.66, which corresponds to option (b).

Step 1: Write the Expression for the Equilibrium Constant

For the equilibrium reaction \( \mathrm{H}_2 + \mathrm{I}_2 \rightleftharpoons 2\mathrm{HI} \), the equilibrium constant \( K_c \) is given by the formula:\[K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\]This expression represents the relationship between the concentrations of the products and the reactants at equilibrium.

Step 2: Substitute Given Concentrations into the Expression

Substitute the given equilibrium concentrations into the equilibrium constant expression:\[K_c = \frac{[28.0]^2}{[8.0][3.0]}\]This step involves placing the numerical concentrations into the formula to calculate \( K_c \).

Step 3: Perform the Calculations

Calculate the numerator: \([\mathrm{HI}]^2 = 28.0^2 = 784.0\).Calculate the denominator: \([\mathrm{H}_2][\mathrm{I}_2] = 8.0 \times 3.0 = 24.0\).Now, divide the numerator by the denominator:\[K_c = \frac{784.0}{24.0} = 32.66\]

Step 4: Determine the Correct Answer Choice

The calculated equilibrium constant \( K_c = 32.66 \). Compare this result to the given options to select the correct choice: (a) 28.34, (b) 32.66, (c) 34.78, (d) 38.88.The correct answer is (b) 32.66.

Key Concepts

Understanding Chemical Equilibrium

Chemical equilibrium occurs when a reversible chemical reaction reaches a state where the concentrations of reactants and products remain constant over time. This doesn’t mean the reactions stop; rather, they continue to occur at the same rate in both directions. In the provided exercise, the equilibrium reaction \( \mathrm{H}_2 + \mathrm{I}_2 \rightleftharpoons 2\mathrm{HI} \) involves hydrogen and iodine reacting to form hydrogen iodide. At equilibrium:

• Neither reactants nor products dominate.
• The reaction continues but no net change in concentration happens.

Being able to determine equilibrium is crucial for predicting the concentration of various species in a reactions’ system. When a reaction is at equilibrium, conditions such as temperature and pressure remain stable, which allows the precise calculation of the equilibrium constant \( K_c \). By examining the equilibrium concentrations of \( \mathrm{H}_2 \), \( \mathrm{I}_2 \), and \( \mathrm{HI} \), students can better understand the proportions in which these components exist at equilibrium.

Role of the Reaction Quotient

The reaction quotient \( Q_c \) is a valuable tool for assessing the progress of a reaction and predicting the direction in which it will proceed. It’s similar in form to the equilibrium constant \( K_c \), but \( Q_c \) is calculated using the current concentrations rather than equilibrium concentrations. The formula mirrors that of \( K_c \):\[ Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \]However, \( Q_c \) helps to assess if a reaction is at equilibrium.

• If \( Q_c = K_c \), the reaction is at equilibrium.
• If \( Q_c < K_c \), the reaction will proceed in the forward direction to reach equilibrium.
• If \( Q_c > K_c \), the reaction will shift backward.

In the provided exercise, since the equilibrium constant \( K_c \) was calculated using equilibrium concentrations, \( Q_c \) isn’t separately calculated but is inherently equal to \( K_c \) at equilibrium. Understanding \( Q_c \) allows students to predict reaction adjustments needed when the system is disturbed.

Performing Concentration Calculations

Concentration calculations involve applying given concentrations to formulas to calculate key quantities like the equilibrium constant. For the reaction \( \mathrm{H}_2 + \mathrm{I}_2 \rightleftharpoons 2\mathrm{HI} \), the equilibrium constant \( K_c \) is determined using:

• Equilibrium concentrations of products and reactants.
• The expression \( K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \).

In the example problem, substituting the concentrations:

• \([\mathrm{HI}] = 28.0 \text{ mol/l}\)
• \([\mathrm{H}_2] = 8.0 \text{ mol/l}\)
• \([\mathrm{I}_2] = 3.0 \text{ mol/l}\)

Students perform calculations:

• Find \([\mathrm{HI}]^2 = 28.0^2 = 784.0\)
• Find \([\mathrm{H}_2] \times [\mathrm{I}_2] = 8.0 \times 3.0 = 24.0\)
• Calculate \( K_c = \frac{784.0}{24.0} = 32.66 \).

Accurate concentration calculations lead to determining the correct value of \( K_c \), letting students match it against provided options. Mastery of these calculations is essential for solving equilibrium-related problems effectively.