Problem 7
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.
Step-by-Step Solution
Verified 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 \(K = 2\)."
1Step 1: 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:
\[K = \frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}\]
2Step 2: 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:
\[K_1 = \frac{4}{(2)(1)} = 2\]
3Step 3: 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:
\[K_2 = \frac{6}{(1)(3)} = 2\]
4Step 4: 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 EquilibriumReaction QuotientLe Chatelier's PrincipleConcentration Changes in Reactions
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.
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.
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:
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.
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.
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.
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.
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.
Other exercises in this chapter
Problem 5
Consider the reaction \(\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\) in a 1.0-L rigid flask. Answer the following questions fo
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Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told th
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Le Châtelier's principle is stated (Section \(12-7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shif
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The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reacta
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