Problem 155 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant \(\left(\mathrm{K}_{\mathrm{p}}\right)\) for the decomposition of water \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) Is given by (a) \(\mathrm{K}_{\mathrm{p}}=\frac{\alpha^{3} \mathrm{P}^{\frac{1}{2}}}{\sqrt{2}}\) (b) \(\mathrm{K}_{\mathrm{p}}=\frac{\alpha^{3} \mathrm{P}^{\frac{3}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}}\) (c) \(\mathrm{K}_{\mathrm{P}}=\frac{\alpha^{3} \mathrm{P}^{\frac{1}{2}}}{(1+\alpha)(2-\alpha)^{\frac{1}{2}}}\) (d) \(\mathrm{K}_{\mathrm{p}}=\frac{\alpha^{3 / 2} \mathrm{P}^{1 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}\) Where \(\alpha\) is the degree of dissociation and \(\mathrm{p}\) is the pressure of the reaction mixture at equilibrium.

Step-by-Step Solution

Answer

Option (d) is correct: \( \mathrm{K}_{\mathrm{p}}=\frac{\alpha^{3/2} \mathrm{P}^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}} \).

Step 1: Understand the Reaction

The reaction given is the decomposition of water, represented by the equation \[ \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}). \] We need to analyze how the equilibrium constant for this reaction, \( \mathrm{K}_{\mathrm{p}} \), depends on the pressure \( \mathrm{P} \) and the degree of dissociation \( \alpha \).

Step 2: Set up Initial Conditions

Assume the initial moles of \( \mathrm{H}_2\mathrm{O} \) are 1 mole and initially, no \( \mathrm{H}_2 \) or \( \mathrm{O}_2 \) is present. At equilibrium, the moles of \( \mathrm{H}_2\mathrm{O} \) will be \( 1 - \alpha \), \( \mathrm{H}_2 \) will be \( \alpha \), and \( \mathrm{O}_2 \) will be \( \frac{\alpha}{2} \), where \( \alpha \) is the degree of dissociation.

Step 3: Determine Partial Pressures

Using the total pressure \( \mathrm{P} \), the partial pressures at equilibrium are: \( P_{\mathrm{H}_2\mathrm{O}} = \mathrm{P} (1 - \alpha) \), \( P_{\mathrm{H}_2} = \mathrm{P} \alpha \), \( P_{\mathrm{O}_2} = \mathrm{P} \frac{\alpha}{2} \).

Step 4: Write Expression for Equilibrium Constant

The equilibrium constant expression for the decomposition of water is given by \[ \mathrm{K}_{\mathrm{p}} = \frac{(P_{\mathrm{H}_2})(P_{\mathrm{O}_2})^{1/2}}{P_{\mathrm{H}_2\mathrm{O}}}. \] Substituting the expression from partial pressures, we get: \[ \mathrm{K}_{\mathrm{p}} = \frac{(P \alpha)(P \frac{\alpha}{2})^{1/2}}{P (1-\alpha)}. \]

Step 5: Simplify the Expression

Simplify the expression to find \( \mathrm{K}_{\mathrm{p}} \): \[ \mathrm{K}_{\mathrm{p}} = \frac{P^{1/2} \alpha^{3/2}}{\sqrt{2}(1-\alpha)}. \]

Step 6: Compare with Given Options

Given the derived equation \[ \mathrm{K}_{\mathrm{p}} = \frac{\alpha^{3/2} \mathrm{P}^{1/2}}{(1-\alpha)\sqrt{2}}, \] compare this with the options provided: (a) doesn't match, (b) doesn't match, (c) set wrong format, (d) exactly matches our expression.

Key Concepts

Decomposition of Water

Water decomposition refers to the chemical process where a molecule of water (\( \mathrm{H}_2\mathrm{O} \)) breaks down into hydrogen (\( \mathrm{H}_2 \)) and oxygen gases (\( \mathrm{O}_2 \)). This can be represented by the reaction:\[ \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}). \] When this reaction occurs in the gas phase, it becomes important to understand the role of pressure and the amounts of products formed at equilibrium.
The decomposition of water is an endothermic process, meaning it requires energy input to proceed. This type of reaction is vital in many scientific and industrial processes. The decomposition can be influenced by factors such as temperature, pressure, and the use of catalysts.
Understanding this basic idea is essential when analyzing the equilibrium state of any reaction involving water decomposition.

Degree of Dissociation

The degree of dissociation (\( \alpha \)) is a measure of the fraction of the original reactant molecules that have dissociated to form products at equilibrium. It's a crucial concept for understanding reaction dynamics. For the decomposition of water, the degree of dissociation helps to show how much \( \mathrm{H}_2\mathrm{O} \) has broken down into \( \mathrm{H}_2 \) and \( \mathrm{O}_2 \).
At equilibrium, one can calculate \( \alpha \) by observing the initial and equilibrium moles of the substances involved. If you start with 1 mole of \( \mathrm{H}_2\mathrm{O} \) initially, at equilibrium:

• \( \mathrm{H}_2\mathrm{O} \) is \( 1 - \alpha \) moles.
• \( \mathrm{H}_2 \) is \( \alpha \) moles.
• \( \mathrm{O}_2 \) is \( \frac{\alpha}{2} \) moles.

Therefore, \( \alpha \) not only tells us about the reaction progress but also affects the calculation of the equilibrium constant \( \mathrm{K}_p \).

Partial Pressures

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. In the context of chemical equilibrium, calculating partial pressures is essential for understanding how reactants and products are distributed in a reaction like the decomposition of water.
For example, if the total pressure of all gases is \( \mathrm{P} \), the partial pressures at equilibrium are given by:

• \( P_{\mathrm{H}_2\mathrm{O}} = \mathrm{P} (1 - \alpha) \)
• \( P_{\mathrm{H}_2} = \mathrm{P} \alpha \)
• \( P_{\mathrm{O}_2} = \mathrm{P} \frac{\alpha}{2} \)

These expressions allow us to write the equilibrium constant \( \mathrm{K}_p \) in terms of partial pressures, showing the relationship between pressure, degree of dissociation, and the overall equilibrium.

Chemical Equilibrium

Chemical equilibrium is the state in a chemical reaction where the concentrations of reactants and products remain constant over time, meaning that the forward and reverse reactions occur at equal rates. For the decomposition of water, achieving equilibrium involves understanding not only the reactants and products but also the conditions under which they stabilize.

• Equilibrium Constant \( \mathrm{K}_{\mathrm{p}} \) - This is a metric that tells us the relative quantities of products and reactants at equilibrium in terms of partial pressures for gaseous reactions.
• In our reaction, it's expressed as \[ \mathrm{K}_{\mathrm{p}} = \frac{(P_{\mathrm{H}_2})(P_{\mathrm{O}_2})^{1/2}}{P_{\mathrm{H}_2\mathrm{O}}} \].
• Due to its dependence on conditions like temperature and pressure, chemical equilibrium is a dynamic state, not static.

Understanding chemical equilibrium is key to predicting how a system will respond to changes in conditions, crucial for controlling chemical processes in practical applications.