Select the correct answer. How does adding oxygen [tex]$\left( O _2\right)$[/tex] to this reaction change the equilibrium? [tex]$2 SO_2(g)+O_2(g) \rightleftharpoons 2 SO_3(g)$[/tex] A. The equilibrium shifts right to produce more [tex]$SO _3$[/tex] molecules. B. The equilibrium shifts left to produce more [tex]$O _2$[/tex] molecules. C. The equilibrium shifts right because of decreased collisions between [tex]$SO _2$[/tex] and [tex]$O _2$[/tex] molecules. D. The equilibrium shifts left with an increase in [tex]$SO _2$[/tex] and [tex]$O _2$[/tex] molecules. E. The equilibrium shifts left because of increased collisions between [tex]$SO _2$[/tex] and [tex]$O _2$[/tex] molecules.
Answer (2)
Adding oxygen to the reaction shifts the equilibrium to relieve the stress.
According to Le Chatelier's principle, the equilibrium shifts to the right.
This shift favors the production of more S O 3 molecules.
The correct answer is A: The equilibrium shifts right to produce more S O 3 molecules. A
Explanation
Understanding the Problem The problem describes a chemical reaction at equilibrium and asks how adding oxygen ( O 2 ) will affect the equilibrium. This is a classic application of Le Chatelier's principle, which states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. In this case, the 'stress' is the addition of a reactant ( O 2 ).
Applying Le Chatelier's Principle Le Chatelier's principle tells us that adding a reactant to a system at equilibrium will cause the equilibrium to shift in the direction that consumes the added reactant. In the given reaction, 2 S O 2 ( g ) + O 2 ( g ) ⇌ 2 S O 3 ( g ) , adding O 2 will cause the equilibrium to shift to the right, favoring the production of more S O 3 . This is because the forward reaction consumes O 2 .
Selecting the Correct Answer Now, let's examine the answer choices to find the one that matches our conclusion: A. The equilibrium shifts right to produce more S O 3 molecules. - This matches our conclusion. B. The equilibrium shifts left to produce more O 2 molecules. - This is the opposite of what we expect. C. The equilibrium shifts right because of decreased collisions between S O 2 and O 2 molecules. - The shift is to the right, but the reasoning about decreased collisions is incorrect; adding O 2 would increase collisions. D. The equilibrium shifts left with an increase in S O 2 and O 2 molecules. - This is incorrect as the equilibrium shifts to the right. E. The equilibrium shifts left because of increased collisions between S O 2 and O 2 molecules. - This is incorrect as the equilibrium shifts to the right.
Final Answer Therefore, the correct answer is A. The equilibrium shifts right to produce more S O 3 molecules.
Examples
Le Chatelier's principle is widely used in industrial chemistry to optimize reaction conditions. For example, in the Haber-Bosch process for synthesizing ammonia ( N 2 + 3 H 2 ⇌ 2 N H 3 ), increasing the pressure shifts the equilibrium towards the side with fewer gas molecules (the product side), resulting in a higher yield of ammonia. Similarly, controlling temperature and reactant concentrations can maximize the efficiency of various chemical processes, making them economically viable and environmentally sustainable. Understanding equilibrium shifts helps in designing efficient chemical reactors and optimizing production processes.
Adding oxygen to the reaction shifts the equilibrium to the right, which leads to the production of more SO_3 molecules, following Le Chatelier's principle. The correct answer is A. The equilibrium shifts right to produce more SO_3 molecules.
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