Problem 200 - A Complete Resource Book in Chemistry for JEE Main

Question

For the following three reactions \(\mathrm{A}\), B and \(\mathrm{C}\), equilibrium constants are given: (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{1}\) (b) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{2}\) (c) \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{3}\) Which of the following relation is correct? (a) \(\mathrm{K}_{1} \sqrt{\mathrm{K}}_{2}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{2} \mathrm{~K}_{3}=\mathrm{K}_{1}\) (c) \(\mathrm{K}_{3}=\mathrm{K}_{1} \mathrm{~K}_{2}\) (d) \(\mathrm{K}_{3} \cdot \mathrm{K}_{2}^{3}=\mathrm{K}_{1}^{2}\)

Step-by-Step Solution

Answer

Option (c) \( \text{K}_3=\text{K}_1 \cdot \text{K}_2 \) is correct.

Step 1: Write Down Given Reactions

The given reactions are as follows:(a) \( \text{CO(g) + H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g) + H}_2\text{(g)}; \text{K}_1 \)(b) \( \text{CH}_4\text{(g) + H}_2\text{O(g)} \rightleftharpoons \text{CO(g) + 3 H}_2\text{(g)}; \text{K}_2 \)(c) \( \text{CH}_4\text{(g) + 2 H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g) + 4 H}_2\text{(g)}; \text{K}_3 \)

Step 2: Analyze Reaction C

The reaction for \( \text{K}_3 \) is: \( \text{CH}_4\text{(g) + 2 H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g) + 4 H}_2\text{(g)} \). This reaction is essentially a combination of reactions (a) and (b).

Step 3: Combine Reactions A and B

By adding reaction (a) and (b), we try to form reaction (c). Adding reactions: \( \text{CO(g) + H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g) + H}_2\text{(g)} \) \( \text{CH}_4\text{(g) + H}_2\text{O(g)} \rightleftharpoons \text{CO(g) + 3 H}_2\text{(g)} \)Leads to:\( \text{CH}_4\text{(g) + 2 H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g) + 4 H}_2\text{(g)} \)

Step 4: Establish Relationship for Equilibrium Constants

Since we combined reactions (a) and (b) to get reaction (c), according to the properties of equilibrium constants:\[ \text{K}_1 \times \text{K}_2 = \text{K}_3 \]

Step 5: Select Correct Relation

Analyzing the given options:(a) \( \text{K}_1 \sqrt{\text{K}_2}=\text{K}_2 \), incorrect,(b) \( \text{K}_2 \cdot \text{K}_3=\text{K}_1 \), incorrect,(c) \( \text{K}_3=\text{K}_1 \cdot \text{K}_2 \), correct,(d) \( \text{K}_3 \cdot \text{K}_2^{3}=\text{K}_1^{2} \), incorrect.

Key Concepts

Equilibrium Constants

Equilibrium constants are a vital part of understanding chemical equilibrium in reactions. When a chemical reaction reaches equilibrium, the rate at which the reactants are consumed equals the rate at which products are formed. Thus, the concentrations of reactants and products remain constant over time.

The equilibrium constant expression, denoted as \(K\), relates the concentrations of products and reactants at equilibrium. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \(K\) is expressed as:

• \( K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)

Each concentration is raised to the power of its respective stoichiometric coefficient from the balanced equation. The value of \(K\) indicates the extent to which a reaction proceeds:

• If \( K \) is much greater than 1, products are favored at equilibrium.
• If \( K \) is much less than 1, reactants are favored.

It's important to note that changing the temperature affects \(K\), while concentrations, pressures, or volume changes do not permanently alter \(K\). This is because \(K\) is only dependent on temperature, linked to the reaction's enthalpy change.

Reaction Mechanism

A reaction mechanism is a step-by-step sequence of elementary reactions by which an overall chemical change occurs. Understanding the mechanism is crucial for explaining how reactions proceed and for determining the rate laws which dictate the speed of reactions.

In the exercise, each given chemical equation shows a potential step in a complex mechanism. In multi-step reactions, individual steps might have their own equilibrium constants. For example, reactions (a) and (b) in the exercise can be considered steps toward forming reaction (c). The comprehensive relation between equilibrium constants \(K_1\), \(K_2\), and \(K_3\) reflects this connection:

• Combining reactions that happen in sequence results in a combined equilibrium constant: \(K_3 = K_1 \times K_2\).
• This reflects the additive nature of mechanisms where products of one step become reactants of another.

Overall, the combination of elementary steps forms a complete picture of how substances transform from reactants to products.

Chemical Reactions

Chemical reactions involve the transformation of substances by breaking and forming chemical bonds. Each balanced chemical equation provides a map of how reactants turn into products, emphasizing the conservation of mass and rearrangement of atoms.

In this context, reactions (a), (b), and (c) each represent different stages in the possible transformation of methane and water vapor to carbon dioxide and hydrogen. These reactions show:

• Simple to complex transformations, as seen from individual molecules (e.g., CO and H₂) to more complex products (e.g., CH₄ and CO₂).
• The involvement of gaseous reactants and products which emphasizes the influence of pressure and volume changes, though not affecting the equilibrium constant.

The importance of balancing chemical equations cannot be overstated as it ensures the law of conservation of mass and governs the stoichiometry needed for calculating equilibrium constants and reaction yields. This balance is fundamental in studying chemical reactions and predicting reaction outcomes under varying conditions.