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Chapter 12: Problem 7

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

Expert verified
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

01

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: \[K = \frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}\]
02

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: \[K_1 = \frac{4}{(2)(1)} = 2\]
03

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: \[K_2 = \frac{6}{(1)(3)} = 2\]
04

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 \(K = 2\)."

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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 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.

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Most popular questions from this chapter

Lexan is a plastic used to make compact discs, eyeglass lenses, and bulletproof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right),\) an extremely poisonous gas. Phosgene decomposes by the reaction,$$\operatorname{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$,for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

Le Ch芒telier'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 shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease, and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\text { Ni catalyst }}{=\frac{750^{\circ} \mathrm{C}}{7}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8} .\) What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

For the following endothermic reaction at equilibrium: $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T\) ) d. decreasing the volume (constant \(T\) ) e. adding Ne(g) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )
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