Problem 30 - Chemistry and Chemical Reactivity

Question

The equilibrium constant for the reaction $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ is \(1.7 \times 10^{-3}\) at \(2300 K\) (a) What is \(K\) for the reaction when written as follows? $$ 1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g}) $$ (b) What is \(K\) for the following reaction? $$ 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

(a) 4.12 x 10^{-2}; (b) 588.24

Step 1: Understanding the Problem

We are given a reaction with an equilibrium constant \( K_c = 1.7 \times 10^{-3} \) at \( 2300 \, K \) for the formation of NO from \( N_2 \) and \( O_2 \). We need to find the equilibrium constants for two different but related reactions.

Step 2: Find K for Reaction (a)

The reaction given is \( \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO(g) \). This is achieved by dividing the original reaction by 2. When a reaction is divided by a number, its equilibrium constant is the square root of the original. Therefore, \( K' = \sqrt{1.7 \times 10^{-3}} \).

Step 3: Calculate Square Root for Reaction (a)

Compute the value: \( \sqrt{1.7 \times 10^{-3}} = 4.12 \times 10^{-2} \). So, \( K' = 4.12 \times 10^{-2} \) for the new reaction.

Step 4: Find K for Reaction (b)

The reaction given is \( 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \). This is simply the reverse of the original reaction. The equilibrium constant for a reverse reaction is the reciprocal of the forward reaction's constant. Thus, \( K'' = \frac{1}{1.7 \times 10^{-3}} \).

Step 5: Calculate Reciprocal for Reaction (b)

Compute the reciprocal: \( \frac{1}{1.7 \times 10^{-3}} = 588.24 \). Thus, \( K'' = 588.24 \) for this reverse reaction.

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant, denoted as \(K\), plays a crucial role in understanding the balance point of a chemical reaction at a given temperature. It is a number that expresses the relationship between the concentrations of the reactants and products at equilibrium for a reversible reaction. When a reaction reaches equilibrium, the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of the products and reactants remain constant. The equilibrium constant can be expressed in terms of concentration as \(K_c\) or in terms of partial pressures as \(K_p\) for gas-phase reactions.

• \(K_c = \frac{[products]}{[reactants]}\)
• If \(K\) is much greater than 1, products are favored at equilibrium.
• If \(K\) is much less than 1, reactants are favored at equilibrium.

Remember, the equilibrium constant changes with temperature and may also change when the reaction is altered (e.g., by changing stoichiometric coefficients). For example, when a reaction equation is multiplied by a coefficient, the equilibrium constant is raised to the power of the same coefficient, as evident in the half reactions used in the exercise.

Reversible Reactions

Reversible reactions are chemical reactions where the conversion of reactants to products and the conversion of products back to reactants occur simultaneously. This back-and-forth process establishes a dynamic equilibrium where the reaction rates of both directions are equal leading to stable concentrations of reactants and products.In notations, reversible reactions are represented by the symbol \(\rightleftharpoons\).

• Example: For the reaction \( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g) \), both nitrogen and oxygen can combine to form nitric oxide, and nitric oxide can decompose back into nitrogen and oxygen.
• Factors affecting equilibrium include concentration, temperature, and pressure, often analyzed through Le Chatelier's Principle.

The thing with reversible reactions is, when a condition changes, like pressure or temperature, the system adjusts to counter the change and establish a new equilibrium point.

Gas Phase Reactions

Gas phase reactions are those where all reactants and products are in the gas state. These reactions frequently occur in both industrial and natural processes, such as combustion or atmospheric reactions.The behavior of gases in these reactions often follows the ideal gas law, \( PV = nRT \), where pressure (\(P\)), volume (\(V\)), amount of gas in moles (\(n\)), ideal gas constant (\(R\)), and temperature (\(T\)) are interrelated.In the context of chemical equilibrium with gases:

• The equilibrium constant based on partial pressures is \( K_p \).
• Reactions often show a change in the number of moles of gases, indicated by \(\Delta n\) (moles of products minus moles of reactants).

For the reactions discussed in the exercise, understanding gas behavior helps in calculating equilibrium constants and predicting the direction of reaction shifts when conditions change.