Question

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(400 . \mathrm{K}\), \(K=41\). Find the value of \(K\) for each of the following reactions at the same temperature: (a) \(2 \mathrm{NH}_{3}\) (g) \(\rightleftharpoons \mathrm{N}_{2}\) (g) \(+3 \mathrm{H}_{2}\) (g) (b) \(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_{3}(\mathrm{~g})\)

Short answer

\(K' = 1/41\), \(K'' = 41^{1/2}\), \(K''' = 41^2\)

Step-by-step solution

Identify the Original Reaction and Equilibrium Constant

Write down the balanced chemical equation of the original reaction and its corresponding equilibrium constant. In this case, the original reaction is \(\text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g)\) with an equilibrium constant of \(K = 41\).

Relate Equilibrium Constants for Reaction A

For the reaction A, which is the reverse of the original reaction, the equilibrium constant \(K'\) for the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. Therefore, \(K' = 1/K\).

Calculate the Equilibrium Constant for Reaction A

The equilibrium constant for reaction A is the reciprocal of the original constant. So, \(K' = 1/41\).

Analyze the Balanced Equation for Reaction B

For reaction B, the balanced equation is the original reaction divided by 2. As K is raised to the power of the change in coefficients (multiplier), here the power will be 1/2. Thus, the new equilibrium constant \(K''\) will be \(K^{1/2}\).

Calculate the Equilibrium Constant for Reaction B

To find the equilibrium constant for reaction B, take the square root of K. So, \(K'' = K^{1/2} = 41^{1/2}\).

Analyze the Balanced Equation for Reaction C

For reaction C, the balanced equation is the original reaction multiplied by 2. Thus, the new equilibrium constant \(K'''\) will be \(K^2\), where 2 is the factor by which the original reaction is multiplied.

Calculate the Equilibrium Constant for Reaction C

To find the equilibrium constant for reaction C, square the original K. So, \(K''' = K^2 = 41^2\).

Key concepts

Chemical Equilibrium

When we talk about chemical equilibrium, we're referring to a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. It's essential to understand that this doesn't mean the reactants and products are in equal concentrations, but that their amounts are stable. Imagine a busy shopping mall with the same number of people entering as exiting over time—the number of shoppers (reactants and products) inside stays consistent.

Understanding equilibrium is crucial for predicting how a reaction will respond to changes in conditions, such as temperature, pressure, or concentration. These predictions are commonly based on equations that represent the reaction, like the one provided in the exercise \( \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \) and understanding the related math and concepts, which helps in grasping the core idea of balanced chemical processes.

Le Chatelier's Principle

Moving on to Le Chatelier's principle, it's like a natural law for chemistry that predicts how a system at equilibrium will respond to external changes. If you impose a change on the system, such as increasing the pressure or changing the concentration, the equilibrium shifts to counteract the change, exhibiting a 'push back' to maintain balance. Think of it as a seesaw in the playground that self-levels no matter where the weight is added or removed.

For instance, if more reactants are added to a system in equilibrium, Le Chatelier's principle suggests that the reaction will 'shift to the right', producing more products, in an attempt to re-establish equilibrium. This concept helps us not only to predict the direction of the shift but also to understand the reasons behind the adjustments in reactions, making it a cornerstone in the study of chemical reactions and their behavior under different conditions.

Reaction Quotient

The reaction quotient, denoted as \( Q \), helps us determine the direction in which a reaction will proceed to reach equilibrium. It's a calculation that expresses the ratio of the current concentrations (or partial pressures) of the products to the reactants, raised to the power of their stoichiometric coefficients in the balanced equation. Essentially, it’s like a snapshot of where the reaction is at that moment.

If you're taking a picture to compare your position on a hike to the trail map, you're using a 'location quotient' of sorts! In the context of the exercise, we can calculate a reaction quotient for any stage of the reaction, then compare it to the equilibrium constant \( K \), to predict which way the reaction will shift to achieve equilibrium. It's a powerful tool in chemistry for checking the status of reactions and planning the next step.

Equilibrium Constant Calculation

The process of equilibrium constant calculation involves finding the value that expresses the ratio of the concentrations of products to reactants at equilibrium. Each species' concentration is raised to the power of its coefficient in the balanced chemical equation. When calculating for a given reaction, like the synthesis of ammonia in the exercise, \( \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \) with an equilibrium constant \( K \), we’re essentially capturing the essence of the reaction's tendency to form products.

As seen in the step by step solution provided, when the reaction is reversed or the coefficients are altered, the equilibrium constant also changes—highlighting the importance of stoichiometry in these calculations. These constants are invaluable, as they enable chemists to predict the extent of a reaction, understand kinetic data, and even design industrial processes for optimal production yield.