Question

Consider the reactions \(\alpha\) (i) \(2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2}(g)\); Eqm. Constant \(=K_{1}\) (ii) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) ;\) Eqm. Constant \(=K_{2}\) (iii) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) ;\) Eqm. Constant \(=K_{3}\) Which of the following relation is correct? (a) \(K_{3}=\frac{K_{1}}{K_{2}}\) (b) \(K_{3}=\frac{K_{1}^{2}}{K_{2}^{2}}\) (c) \(K_{3}=K_{1} K_{2}\) (d) \(K_{3}=\sqrt{K_{1}} \cdot K_{2}\)

Short answer

Correct relation: (b) \(K_3=\frac{K_{1}^{2}}{K_{2}^{2}}\

Step-by-step solution

- Analyze the Equilibrium Constants

For reaction (i), the equilibrium constant is given as \( K_1 \) for the conversion of \( CO \) and \( H_2O \) to \( CO_2 \) and \( H_2 \). Similarly, \( K_2 \) for reaction (ii), which converts \( CH_4 \) and \( H_2O \) into \( CO \) and \( H_2 \). The equilibrium constant \( K_3 \) is for reaction (iii), which is the conversion of \( CH_4 \) and \( H_2O \) to \( CO_2 \) and \( H_2 \) directly.

- Combine Reactions (i) and (ii)

To derive reaction (iii) from (i) and (ii), observe that adding reaction (i) to twice the reverse of reaction (ii) yields reaction (iii). Reverse reaction (ii) and multiply it by 2 to prepare for combination: \[ 2\times \left(\mathrm{CO}(g) + 3 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g) + \mathrm{H}_{2}O(g)\right) \]

- Calculate \( K_3 \) Using \( K_1 \) and \( K_2 \)

Since the reverse of a reaction has the reciprocal of the equilibrium constant, reversing reaction (ii) gives us \( \frac{1}{K_2^2} \). Now, add the modified reaction (ii) to reaction (i) to obtain reaction (iii), and apply the rule that when reactions are added, their equilibrium constants are multiplied. Therefore, \( K_3 = K_1 \times \left( \frac{1}{K_2^2} \right) = \frac{K_1}{K_2^2} \)

Key concepts

Chemical Equilibrium

To first understand equilibrium constants you need to grasp the concept of chemical equilibrium. This is the state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction, meaning the concentrations of products and reactants remain unchanged over time. It's dynamic because the reactions are still occurring, but with no net effect on the concentrations.

In the case of our exercise, the reactions provided are all reversible, indicating they can reach equilibrium. The equilibrium constant, represented as K, is crucial because it provides a ratio of product concentrations to reactant concentrations at equilibrium. Each reaction will have its own unique equilibrium constant. For instance, reaction (i) with the equilibrium constant K₁ shows how carbon monoxide and water, when converted to carbon dioxide and hydrogen gas, will balance out at a certain point depending on the temperature when equilibrium is achieved.

Reaction Quotient

Moving forward, the reaction quotient, Q, is a measure that tells us the direction in which a reaction mixture will shift to reach equilibrium. It's calculated in the same way as the equilibrium constant, using the concentrations (or partial pressures for gases) of the products raised to the power of their stoichiometric coefficients divided by the reactants raised to their coefficients. However, Q can be measured at any point during the reaction, not just at equilibrium.

When comparing Q to the equilibrium constant K, if Q < K, the forward reaction is favored; if Q > K, the reverse reaction is favored; and if Q = K, the system is at equilibrium. In the textbook exercise, understanding how to manipulate and compare these values was crucial for finding the relationship between the equilibrium constants of the combined reactions.

Le Chatelier’s Principle

Lastly, Le Chatelier’s Principle predicts how a change in conditions can affect the chemical equilibrium of a reaction. It states that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the position of equilibrium will shift to counteract the change. It's a qualitative tool that helps us understand the behavior of a system at equilibrium when it's subjected to external stresses.

For example, when referring to our problem, if we were to change the concentration of any reactant or product, Le Chatelier’s Principle would lead us to predict whether the system would produce more products or reactants in response. It's important to note, though, that Le Chatelier's Principle doesn't apply to calculating the equilibrium constants. In the step-by-step solution provided, we didn't use Le Chatelier's Principle directly, but it is worth understanding its implications when considering the effects of changes to a system already at equilibrium.