Question

The equilibrium constant for the reaction \(\mathrm{H}_{2}+\) \(\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) at 1 atm and \(1500^{\circ} \mathrm{C}\) is given to be \(K .\) Of the reactions given below, all at \(1500^{\circ} \mathrm{C}\), the reaction that has a different equilibrium constant is \((a) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) at 5 atm \((b) 2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}\) at 1 atm \((c) \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}\) at 2 atm \((d) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}+3 \mathrm{N}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+3 \mathrm{N}_{2} \) at \(5 \mathrm{atm}\) \((e) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}+3 \mathrm{N}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+3 \mathrm{N}_{2} \quad\) at 1 atm

Short answer

Answer: (b) \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{H}_{2}\mathrm{O}\) at 1 atm.

Step-by-step solution

Recall the relationship between the reaction and the equilibrium constant

The equilibrium constant (K) is independent of the pressure or concentration of the reacting substances. However, it depends on the balanced chemical equation and the temperature. Here, the temperature is the same for all reactions (\(1500^{\circ} \mathrm{C}\)), so we need to focus on how the chemical equations differ.

Analyze reaction (a)

Reaction (a) is given in a different pressure (5 atm) compared to the given reaction (1 atm). However, since the equilibrium constant is not dependent on the pressure, the value of K will not change and hence be the same. So, (a) is not the correct answer.

Analyze reaction (b)

Reaction (b) has a different balanced chemical equation: \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{H}_{2}\mathrm{O}\). It is twice the reaction for which the equilibrium constant K is given, so the equilibrium constant will be the square of the initial value: \(K^2\). Therefore, (b) has a different equilibrium constant.

Analyze reaction (c)

Reaction (c) gives: \(\mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2}\mathrm{O}+\frac{1}{2} \mathrm{O}_{2}\). Subtracting \(\frac{1}{2} \mathrm{O}_{2}\) from both the reactants and products, we get the given reaction with equilibrium constant K.

Analyze reactions (d) and (e)

Both reaction (d) and (e) have the same reactants and products as the given reaction, except for the presence of inert \(\mathrm{N}_{2}\) gas. Since the equilibrium constant depends only on the reactants and products that are involved in the reaction, the equilibrium constant will remain the same for these reactions as well.

Identify the correct answer

From our analysis, the reaction that has a different equilibrium constant is \((b) 2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{H}_{2}\mathrm{O}\) at 1 atm.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time, but not necessarily equal. It is critical to understand that chemical equilibrium is dynamic; reactions continue to occur, but with no net change in concentration. Imagine a busy street where the number of people entering a shop equals the number leaving; the shop is not empty or crowded, but in a state of balance. Similarly, at equilibrium, the reactants turn into products at the same rate that products revert to reactants.

Understanding Equilibrium in Terms of Energy

At equilibrium, the system's energy is minimized, and there is no net change in the free energy of the system. While this concept might sound abstract, a practical analogy would be a ball settling at the lowest point in a bowl. The ball moves until it finds a place where it can remain stationary - that's the lowest energy state, analogous to chemical equilibrium.

Role of Catalyst

It's also important to note that a catalyst, while it accelerates both the forward and reverse reactions, does not affect the position of the equilibrium; it simply helps the system reach equilibrium more quickly.

Balanced Chemical Equation

A balanced chemical equation is one where the number of atoms for each element is the same on both the reactant and product sides of the equation. This reflects the law of conservation of mass, ensuring that atoms are neither created nor destroyed - they are simply rearranged in the reaction.

Significance of Balancing Equations

The process of balancing a chemical equation is like a recipe that specifies the exact quantity of each ingredient needed to make a certain dish. When a chemical equation is balanced, it provides the stoichiometric coefficients, which act as the 'recipe' to show the proportions in which reactants combine to form products. These coefficients are essential when calculating the amount of products formed from a given amount of reactants and vice versa.

Impact on Equilibrium Constant

The coefficients in the balanced equation directly affect the equilibrium constant (K). For example, doubling all coefficients in a balanced chemical equation will square the value of K. This is crucial for students to understand when predicting how changes in the equation will influence the value of K.

Temperature Dependence of Equilibrium

The value of the equilibrium constant (K) changes with temperature, reflecting one of the fundamental principles of thermodynamics. In essence, if a reaction releases heat (exothermic), an increase in temperature will generally shift the equilibrium towards the reactants. Conversely, for an endothermic reaction, where heat is absorbed, raising the temperature tends to shift the equilibrium towards the products.

Le Chatelier's Principle

Le Chatelier's Principle can help students understand this concept better. It states that if a system at equilibrium is disturbed, it will adjust to diminish the change. When the temperature increases, the equilibrium shifts to consume the excess heat energy, changing the concentrations of reactants and products - and therefore the value of K.

Understanding how equilibrium responds to temperature changes is critical, especially in industrial processes where reactions are optimized for the most efficient production of desired products. For students, grasping this dependency can help explain why certain reactions must be conducted at specific temperatures to yield the best results.