Problem 23 - Chemistry An Atoms First Approach

Question

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\). b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\). d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).

Step-by-Step Solution

Answer

The equilibrium constants for each reaction at the given temperature are: a. \(K_a = 0.114\) b. \(K_b = 76.9\) c. \(K_c = 8.74\) d. \(K_d = 1.69 \times 10^{-4}\)

Step 1: Relationship between equilibrium constants and stoichiometric coefficients

When multiplying a reaction by a constant, the equilibrium constant must be raised to the power of that same constant. For example, if we multiplied the given reaction by 2: \[ 2(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)) \] \[ (2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g)) \] Then the equilibrium constant would be: \[ K' = K^2 \] Now we can apply this to each of the reactions (a, b, c, d).

Step 2: Calculate \(K\) for reaction a

In reaction a, each stoichiometric coefficient in the original reaction is multiplied by 0.5: \[ 0.5N_2(g) + 1.5H_2(g) \rightleftharpoons NH_3(g) \] The equilibrium constant for this reaction (\(K_a\)) will be: \[ K_a = K^{0.5} \] Here, we substitute the given K value and calculate \(K_a\): \[ K_a = (1.3 \times 10^{-2})^{0.5} \] \[ K_a = 0.114 \]

Step 3: Calculate \(K\) for reaction b

In reaction b, the given reaction is simply reversed: \[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \] The equilibrium constant for this reaction (\(K_b\)) will be the inverse of the original \(K\): \[ K_b = \frac{1}{K} \] Here, we substitute the given K value and calculate \(K_b\): \[ K_b = \frac{1}{1.3 \times 10^{-2}} \] \[ K_b = 76.9 \]

Step 4: Calculate \(K\) for reaction c

In reaction c, the stoichiometric coefficients of the products of the original reaction are multiplied by 0.5, and the given reaction is reversed: \[ NH_3(g) \rightleftharpoons 0.5N_2(g) + 1.5H_2(g) \] This is equivalent to multiplying the original reaction by -0.5, so the equilibrium constant for this reaction (\(K_c\)) will be: \[ K_c = K^{-0.5} \] Here, we substitute the given K value and calculate \(K_c\): \[ K_c = (1.3 \times 10^{-2})^{-0.5} \] \[ K_c = 8.74 \]

Step 5: Calculate \(K\) for reaction d

In reaction d, the given reaction is multiplied by 2: \[ 2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g) \] The equilibrium constant for this reaction (\(K_d\)) will be: \[ K_d = K^{2} \] Here, we substitute the given K value and calculate \(K_d\): \[ K_d = (1.3 \times 10^{-2})^{2} \] \[ K_d = 1.69 \times 10^{-4} \] Now we have the equilibrium constant values for each of the reactions (a, b, c, d) at the given temperature: a. \(K_a = 0.114\) b. \(K_b = 76.9\) c. \(K_c = 8.74\) d. \(K_d = 1.69 \times 10^{-4}\)

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a fundamental concept in chemical equilibrium. It expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. Every chemical reaction at a given temperature has a specific \( K \) value that reflects the relative concentrations of reactants and products at equilibrium.
Key points about the equilibrium constant include:

• It is specific to the reaction and the temperature at which the reaction occurs.
• Its value indicates the extent to which a reaction proceeds to form products.
• A large \( K \) value implies a higher concentration of products, whereas a small \( K \) value suggests a reaction that favors reactants.

When stoichiometric coefficients in a balanced chemical equation change, the equilibrium constant adjusts accordingly. For instance, multiplying or dividing coefficients by a constant changes the power to which the \( K \) value is raised, as demonstrated in the given examples.
In a reversible reaction, the equilibrium constant of the reverse reaction is the reciprocal \( (1/K) \), highlighting the principle that reactions move towards equilibrium from both forward and backward directions.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a balanced chemical equation. It dictates the proportions in which substances react or are produced. Understanding stoichiometry is crucial for calculating the equilibrium constant when the reaction equation is altered.
To adjust the equilibrium constant for a modified reaction:

• If you multiply the entire reaction by a factor \( n \), the new equilibrium constant is raised to the power of \( n \): \( K' = K^n \).
• If the reaction is divided or reversed, appropriate alterations to \( K \) must be made, such as taking reciprocals for reverse reactions.

When the coefficients are changed, a reliable approach is to ensure the equation remains balanced while calculating its new \( K \) value accurately. This helps in determining how shifts in reactant and product ratios impact the chemical equilibrium.

Reversible Reactions

Reversible reactions can proceed in both forward and backward directions. At equilibrium, the forward and reverse reactions occur at the same rate, leading to no net change in the concentration of reactants and products. This balance is a hallmark of reversible reactions.
Characteristics of reversible reactions in equilibrium:

• They never go to completion but reach a state where reactant and product concentrations remain constant over time.
• The equilibrium position depends on initial concentrations and external conditions such as temperature and pressure.

Understanding reversibility is crucial when calculating equilibrium constants, as these constants represent the dynamic balance between forward and reverse reaction rates. Hence, alterations in reaction conditions require accurate adjustments in \( K \) values through stoichiometric considerations, ensuring that the principles governing reversible reactions are upheld.