Consider the formation of ammonia in two experiments. (a) To a 1.00 -L container at \(727^{\circ} \mathrm{C}, 1.30 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) and \(1.65 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) are added. At equilibrium, \(0.100 \mathrm{~mol}\) of \(\mathrm{NH}_{3}\) is present. Calculate the equilibrium concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\), and find \(K_{\mathrm{c}}\) for the reaction $$ 2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) $$ (b) In a different 1.00 -L container at the same temperature, equilibrium is established with \(8.34 \times 10^{-2} \mathrm{~mol}\) of \(\mathrm{NH}_{3}, 1.50 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) and \(1.25 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) $$ (c) What is the relationship between the \(K_{\mathrm{c}}\) values in parts (a) and (b)? Why aren't these values the same?

Step by step solution

01

- Write the Balancing Equations

Write the balanced equations for the given reactions: (a) \[2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g)\] (b) \[\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(g) + \frac{3}{2} \mathrm{H}_{2}(g)\]

02

- Initial Concentrations for Case (a)

Calculate the initial concentrations: Since the volume is 1.00 L, the initial concentrations are \([\mathrm{N}_{2}]_{0} = 1.30 \mathrm{M}\), \([\mathrm{H}_{2}]_{0} = 1.65 \mathrm{M}\) and \([\mathrm{NH}_{3}]_{0} = 0\).

03

- Changes in Concentrations at Equilibrium for Case (a)

Given that \([\mathrm{NH}_{3}]_{e} = 0.100 \mathrm{M}\) at equilibrium. To form 0.100 M of \(\mathrm{NH}_{3}\): \([\mathrm{N}_{2}]\) decreases by \(\Delta \mathrm{N}_{2} = \frac{0.100}{2} = 0.050 \mathrm{M}\), \([\mathrm{H}_{2}]\) decreases by \(\Delta \mathrm{H}_{2} = \frac{3}{2} \times 0.100 = 0.150 \mathrm{M}\).

04

- Calculate Equilibrium Concentrations for Case (a)

Using the changes, the equilibrium concentrations are: \([\mathrm{N}_{2}]_{e} = 1.30 - 0.050 = 1.25 \mathrm{M}\) \([\mathrm{H}_{2}]_{e} = 1.65 - 0.150 = 1.50 \mathrm{M}\) \([\mathrm{NH}_{3}]_{e} = 0.100 \mathrm{M}\)

05

- Calculate the Equilibrium Constant for Case (a)

Using the equilibrium concentrations, the equilibrium constant \(K_{c}\) is: \[K_{c} = \frac{[\mathrm{N}_{2}][\mathrm{H}_{2}]^{3}}{2[\mathrm{NH}_{3}]^{2}}\] \[K_{c} = \frac{(1.25 \mathrm{M})(1.50 \mathrm{M})^{3}}{2(0.100 \mathrm{M})^{2}} = 421.875 \]

06

- Equilibrium Concentrations for Case (b)

The given equilibrium concentrations are: \([\mathrm{NH}_{3}]_{e} = 8.34 \times 10^{-2} \mathrm{M}\), \([\mathrm{N}_{2}]_{e} = 1.50 \mathrm{M}\), \([\mathrm{H}_{2}]_{e} = 1.25 \mathrm{M}\).

07

- Calculate the Equilibrium Constant for Case (b)

Use the equilibrium concentrations to calculate the equilibrium constant \(K_{c}\): \[K_{c} = \frac{[\mathrm{N}_{2}]^{1/2}[\mathrm{H}_{2}]^{3/2}}{[\mathrm{NH}_{3}]}\] \[K_{c} = \frac{(1.50 \mathrm{M})^{1/2}(1.25 \mathrm{M})^{3/2}}{8.34 \times 10^{-2} \mathrm{M}} = 23.59 \]

08

- Relationship Between Both Cases

Observe the relationship between \(K_{c}\) values: \[K_{c(base)} = (K_{c(current)})^{power}\] For Example, \[K_{c(B)} = (421.875)^{1/2} = 421.875^{1/2} = 23.59\]. They are different due to different reaction stoichiometries.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium

Chemical equilibrium occurs in a chemical reaction when the concentrations of the reactants and products remain constant over time. This state is reached when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the reaction appears to stop, but at the molecular level, there is a dynamic balance where reactants are continuously forming products, and products are continuously forming reactants. This balance can be disturbed by changing the conditions such as temperature, pressure, or concentration.

Concentration Changes

Concentration changes play a crucial role in achieving and understanding chemical equilibrium. When the initial concentrations of reactants and products are known, you can determine the changes each undergoes to reach equilibrium. For instance, in the provided exercise, we start with 1.30 mol of \(\text{N}_2\) and 1.65 mol of \(\text{H}_2\) in a 1.00-L container. As the reaction proceeds towards equilibrium, the concentrations of these reactants decrease while the concentration of the product \(\text{NH}_3\) increases. Using stoichiometry, the changes in the concentrations can be mathematically described and used to calculate equilibrium concentrations. This way, we can understand how much reactants are converted into products and vice versa.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It allows us to calculate the amount of reactants needed or products formed based on balanced chemical equations. In the given exercise, stoichiometric coefficients from the equations: \[2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \] and \[\text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \] help determine the changes in the concentrations of \(\text{N}_2\) and \(\text{H}_2\) as \(\text{NH}_3\) is formed. For every mole of \(\text{NH}_3\) produced or consumed, specific amounts of \(\text{N}_2\) and \(\text{H}_2\) will be accordingly produced or consumed due to the stoichiometric ratios.

Reaction Quotient

The reaction quotient, often represented as \(\text{Q}\), is a measure used to determine the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant \(\text{K}_c\), but with the initial concentrations of reactants and products. Comparing \(\text{Q}\) to \(\text{K}_c\) helps predict whether the reaction will proceed forward or backward:

• If \(\text{Q} < \text{K}_c\), the reaction will proceed forward to form more products.
• If \(\text{Q} > \text{K}_c\), the reaction will proceed in reverse to form more reactants.
• If \(\text{Q} = \text{K}_c\), the reaction is at equilibrium.

In the exercise, once the equilibrium concentrations are known, the \(\text{K}_c\) values can be compared and related to understand the reaction conditions in different setups and their influence on the balance point.