Consider the reaction $\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) . \mathrm{A}$ friend asks the following: “I know we have been told that if a mixture of A, B, C, and D is at equilibrium and more of A is added, more C and D will form. But how can more C and D form if we do not add more B?” What do you tell your friend?

Short Answer

Expert verified
When more A is added to the system at equilibrium, the concentration of A increases. According to Le Chatelier's Principle, the system will shift towards the right to form more C and D, counteracting the change. The reaction can proceed without adding more B because the initial B concentration was high enough to react with the added A. As the reaction proceeds, the concentrations of A and B decrease, while more C and D are formed, until a new equilibrium is established. This new equilibrium will have a higher concentration of C and D and lower concentrations of A and B compared to the original equilibrium.

Step by step solution

01

State the problem and Le Chatelier's principle

Your friend has a valid question about the reaction at equilibrium. To understand how the reaction responds when more A is added, it is crucial to recall Le Chatelier's Principle. Le Chatelier's Principle states that if a chemical system at equilibrium is perturbed, the system will adjust to minimize the effect of the change.
02

Describe the initial equilibrium state

Initially, the system is at equilibrium, which means that the forward and reverse reaction rates are equal. As a result, the concentrations of A, B, C, and D remain constant over time.
03

Describe the effect of adding more A

When more A is added to the system, the concentration of A increases. According to Le Chatelier's Principle, the system will respond by shifting in the direction that minimizes the effect of this change. In this case, the system will shift towards the right to form more C and D to decrease the concentration of A.
04

Explain how the reaction proceeds without adding more B

The reaction can proceed without adding more B because the overall reaction is still at equilibrium, and the initial B concentration was high enough to react with the added A. As the reaction proceeds, more C and D are formed, while the concentrations of A and B decrease. This change in concentrations continues until a new equilibrium is established.
05

Describe the new equilibrium state

After the reaction has proceeded, a new equilibrium state is reached where the forward and reverse reaction rates are again equal, and the concentrations of A, B, C, and D are constant. This new equilibrium will have a higher concentration of C and D than the original equilibrium because A was added. The concentrations of A and B will end up lower as some of their moles have been converted to form C and D. In conclusion, by applying Le Chatelier's Principle, we can understand that the system can still shift toward forming more products (C and D) when more reactant A is added, even without adding more of reactant B, because the reaction will proceed in a way that counteracts the disturbance in the equilibrium and establish a new equilibrium state.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{p}}\) for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ at \(700^{\circ} \mathrm{C}\) is \(0.76 .\) Determine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 \(\mathrm{atm}\) at \(700^{\circ} \mathrm{C} .\)

For a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

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Hydrogen for use in ammonia production is produced by the reaction $$\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \frac{\text { Nicatalyst }}{750^{\circ} \mathrm{C}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$$ 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?

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