Problem 78 - Chemical Principles: The Quest for Insight

Question

\( \mathrm{Ar} 500^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g}) .\) Calculate the value of \(K_{\varepsilon}\) at \(500^{\circ} \mathrm{C}\) for the following reactions. (a) \(\frac{1}{6} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{3} \mathrm{NH}_{3}(\mathrm{~g})\) (b) \(\mathrm{NH}_{3}(g)=\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g)\) (c) \(4 \mathrm{~N}_{2}(\mathrm{~g})+12 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 8 \mathrm{NH}_{3}(\mathrm{~g})\)

Step-by-Step Solution

Answer

\(K_{\epsilon\text{(a)}} = 0.061^{\frac{1}{6}}, K_{\epsilon\text{(b)}} = \frac{1}{0.061}, K_{\epsilon\text{(c)}} = 0.061^4\)

Step 1: Understand Relationship Between Reaction Quotients

When a balanced chemical equation is multiplied by a factor, the equilibrium constant changes to the power of that factor. The relationship can be represented by the equation: For the reaction 'aA + bB <=> cC + dD', the equilibrium constant is 'K'. If the reaction is multiplied by a factor 'n', so it becomes 'naA + nbB <=> ncC + ndD', the new equilibrium constant 'K' becomes 'K^n'.

Step 2: Calculate the Equilibrium Constant for Reaction (a)

Given reaction (a) is obtained by dividing all stoichiometric coefficients in the original reaction by 6. To find the new equilibrium constant, raise the original equilibrium constant to the power of 1/6. Hence: \(K_{\epsilon\text{(a)}} = (K_c)^{\frac{1}{6}} = 0.061^{\frac{1}{6}}\).

Step 3: Calculate the Equilibrium Constant for Reaction (b)

Reaction (b) is the reverse reaction of the original equation. The equilibrium constant for a reverse reaction is the inverse of the equilibrium constant for the forward reaction. Therefore: \(K_{\epsilon\text{(b)}} = \frac{1}{K_c} = \frac{1}{0.061}\).

Step 4: Calculate the Equilibrium Constant for Reaction (c)

Given reaction (c) is obtained by multiplying all stoichiometric coefficients in the original reaction by 4. To find the new equilibrium constant, raise the original equilibrium constant to the power of 4. Hence: \(K_{\epsilon\text{(c)}} = (K_c)^4 = 0.061^4\).

Key Concepts

Equilibrium Constant Calculation

Understanding the concept of equilibrium constant calculation is fundamental in grasping how chemical reactions behave at equilibrium. This value, represented by the symbol \( K \), indicates the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients, when the reaction has reached a state of balance and no further concentration changes are observed.

In the provided exercise, the equilibrium constant \( K_c \) for the reaction between nitrogen (\r{N2}) and hydrogen (\r{H2}) to form ammonia (\r{NH3}) is given as 0.061. Calculation of equilibrium constants for related reactions, like (a), can be done by manipulating the original balanced equation. When the stoichiometric coefficients of the entire reaction are divided by a common factor, as in reaction (a), the equilibrium constant of the new reaction is the original constant raised to the power of the reciprocal of that common factor.

This important concept allows us to predict the extent of a reaction, understand reaction yields, and how changes in conditions can affect the reaction equilibrium.

Reaction Quotient

The reaction quotient, \( Q \), is similar to the equilibrium constant but is calculated using the initial concentrations or partial pressures of the reactants and products, not at equilibrium. It serves as a predictive tool to determine the direction in which a reaction will proceed to reach equilibrium. If \( Q < K \), the reaction shifts to the right, forming more products; if \( Q > K \), the reaction shifts to the left, forming more reactants.

In the context of the example, we didn’t calculate the reaction quotient, but understanding this concept would help in predicting how the reaction will shift if it wasn’t at equilibrium. The relationship between \( Q \) and \( K \) is crucial for chemists to manipulate reaction conditions and yields, especially in industrial processes where maximizing production is essential.

Equilibrium Constant for Reverse Reaction

The directionality of a reaction plays a pivotal role in the value of the equilibrium constant. For the reverse reaction, the equilibrium constant, \( K' \), is the inverse of the equilibrium constant for the forward reaction, \( K \). This means that \( K' = \frac{1}{K} \).

In the exercise step 3, reaction (b) represents the reverse process of the original equilibrium. Applying the reverse reaction concept to the given equilibrium constant, \( K_c \), allows us to calculate the equilibrium constant for the production of nitrogen and hydrogen from ammonia. This knowledge is particularly useful when evaluating reversible reactions in chemical industries, such as the Haber process for synthesizing ammonia.

Manipulating Equilibrium Constants

An interesting aspect of equilibrium constants is that they can be manipulated mathematically depending on changes to the balanced equations. When the coefficients in a chemical equation are multiplied by a factor, the equilibrium constant is raised to that power. Likewise, when the reaction is reversed, the equilibrium constant is inverted, and when the stoichiometry is divided, the exponent becomes a fraction.

In our calculation in step 4, reaction (c)'s stoichiometric coefficients were multiplied by a factor of 4. Consequently, we raised the original equilibrium constant, \( K_c \), to the power of 4 to determine the new equilibrium constant. These manipulations are not just theoretical exercises but are practical tools in the synthesis and analysis of chemical reactions, impacting concentrations, pressures, and temperature conditions to optimize particular reactions. Such control is particularly useful in the design of chemical reactors and predicting the impact of process variables.