Question

Consider the following statements: "Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g),\) for which at equilibrium \([\mathrm{A}]=2 M\) \([\mathrm{B}]=1 M,\) and \([\mathrm{C}]=4 \mathrm{M} .\) To a \(1-\mathrm{L}\) container of the system at equilibrium, you add 3 moles of B. A possible equilibrium condition is \([\mathrm{A}]=1 M,[\mathrm{B}]=3 M,\) and \([\mathrm{C}]=6 \mathrm{M}\) because in both cases \(K=2 . "\) Indicate everything that is correct in these statements and everything that is incorrect. Correct the incorrect statements, and explain.

Short answer

In the given statements, the fact that "in both cases, \(K = 2\)" is correct, as both equilibrium constants have a value of 2. However, the statement "To a \(1-L\) container of the system at equilibrium, you add 3 moles of B" is incorrect. The correct statement is: "To a \(1-L\) container of the system at equilibrium, you add 2 moles of B. A possible equilibrium condition is \([\mathrm{A}] = 1\:M, [\mathrm{B}] = 3\:M,\) and \([\mathrm{C}] = 6\:M\), because in both cases \(K = 2\)."

Step-by-step solution

Write the reaction equation and equilibrium constant expression

The given reaction is: \(\mathrm{A}(g) + \mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)\) The equilibrium constant expression for this reaction is: \[K = \frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}\]

Calculate the equilibrium constant for the first condition

Given, at equilibrium: \([\mathrm{A}] = 2\:M\) \([\mathrm{B}] = 1\:M\) \([\mathrm{C}] = 4\:M\) Now, we calculate the equilibrium constant (denoted as \(K_1\)) for this set of concentrations: \[K_1 = \frac{4}{(2)(1)} = 2\]

Calculate the equilibrium constant for the second condition

Given, another possible equilibrium condition: \([\mathrm{A}] = 1\:M\) \([\mathrm{B}] = 3\:M\) \([\mathrm{C}] = 6\:M\) Now, we calculate the equilibrium constant (denoted as \(K_2\)) for this set of concentrations: \[K_2 = \frac{6}{(1)(3)} = 2\]

Compare the equilibrium constants and find the correct/incorrect statements

Since both equilibrium constants are equal and have a value of 2 (i.e., \(K_1 = K_2 = 2\)), we can conclude that the statement "in both cases, \(K = 2\)" is correct. However, the statement "To a \(1-L\) container of the system at equilibrium, you add 3 moles of B" is incorrect. When we add 3 moles of B, the new concentrations would be: \([\mathrm{A}] = 1\:M\) \([\mathrm{B}] = 4\:M\) (as new concentration would be \(1\:M + 3 \: moles/L\)) \([\mathrm{C}] = 6\:M\) Nevertheless, it doesn't affect the overall conclusion that both conditions result in the same \(K\) value. To correct the statement, we can say: "To a \(1-L\) container of the system at equilibrium, you add 2 moles of B. A possible equilibrium condition is \([\mathrm{A}] = 1\:M, [\mathrm{B}] = 3\:M,\) and \([\mathrm{C}] = 6\:M\), because in both cases \(K = 2\)."

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse are happening at the same rate. This means the concentrations of the reactants and products remain constant over time. For the reaction \( \text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) \), once equilibrium is reached, the concentrations of \([\text{A}]\), \([\text{B}]\), and \([\text{C}]\) do not change anymore.
Achieving chemical equilibrium doesn't mean the reactions stop; instead, they continue, just at equal rates, maintaining balance.

• At macroscopic levels, the system appears static, but microscopically, the molecules are moving and reacting dynamically.
• This state of balance is influenced by factors like temperature and pressure.

Thus, chemical equilibrium plays a pivotal role in determining reaction completion and efficiency.

Reaction Quotient

The reaction quotient, denoted as \(Q\), helps us determine the direction a reaction will proceed to reach equilibrium. The formula for the reaction quotient is similar to that of the equilibrium constant expression but using initial concentrations.
For the reaction \( \text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) \), the reaction quotient is given as \[ Q = \frac{[\text{C}]}{[\text{A}][\text{B}]} \]
Comparing \(Q\) to the equilibrium constant \(K\) enables us to predict the shift of the reaction:

• If \(Q < K\), the reaction will move forward, converting reactants to products.
• If \(Q > K\), the reaction shifts backward, forming reactants.
• If \(Q = K\), the system is at equilibrium.

This understanding equips us to anticipate changes and strategize adjustments in laboratory conditions.

Le Chatelier's Principle

Le Chatelier's Principle provides insights into how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium is disturbed by changing conditions, the reaction will adjust to counteract the disturbance and restore equilibrium.
Consider the example where additional \(\text{B}\) is added to the reaction \( \text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) \). The principle predicts that the equilibrium will shift to the right, forming more \(\text{C}\) to reduce the change.

• Adding reactants typically results in the formation of more products.
• Removing reactants or adding products shifts the equilibrium in the opposite direction.

This reaction to shifts helps stabilize processes in both natural systems and industrial applications.

Concentration Changes in Reactions

Reacts and products concentrations can change due to various reasons, notably when a disruption like adding substances occurs in the system.
In the given reaction scenario, initially, the amount of \(\text{B}\) is increased, which can alter the concentrations of substances in equilibrium conditions.

• These changes can be modeled to predict new equilibrium states using changes in concentration formulas and tools.
• Adding substances does not change the equilibrium constant \(K\), but it does affect the reaction quotient \(Q\), which tailors the reaction's path towards a new equilibrium.

Understanding concentration changes is vital in predicting how additional amounts of a reactant or product affect the reaction, especially in controlled laboratory environments.