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Chapter 14: Problem 11

If you added \(\mathrm{SO}_{3}\) to a vessel in which the reaction \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftarrows 2 \mathrm{SO}_{3}\) is at equilibrium, which way would the reaction shift?

Short Answer

Expert verified
If additional SO3 is added to the vessel in which the reaction \(2 SO_{2} + O_{2} \rightleftarrows 2 SO_{3}\) is at equilibrium, the reaction will shift to the left, converting some of the increased SO3 back into SO2 and O2. This occurs according to Le Chatelier's Principle, which states that the equilibrium position will counteract the change made to the system.

Step by step solution

01

Identify the reaction and its equilibrium

The given reaction is: \( 2 SO_{2} + O_{2} \rightleftarrows 2 SO_{3} \) Initially, the reaction is at equilibrium. According to the problem, we are adding more SO3 to the system.
02

Apply Le Chatelier's Principle

According to Le Chatelier's Principle, since we are increasing the concentration of SO3 (a product of the reaction), the reaction will shift to counteract the change and restore the equilibrium. In our case, this means that the reaction will shift to the left, converting some of the increased SO3 into SO2 and O2, which are the reactants.
03

Conclusion

If SO3 is added to the vessel in which the given reaction is at equilibrium, the reaction will shift to the left, converting some of the increased SO3 back into SO2 and O2. This happens in order to counteract the change and restore the equilibrium, according to Le Chatelier's Principle.

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Most popular questions from this chapter

Consider the gas-phase reaction: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)\) Suppose it is at equilibrium inside of a closed vessel. Next, the volume of the closed vessel is suddenly decreased, increasing the overall pressure inside the flask. According to Le Châtelier's principle, which way will the reaction shift, and why? Then explain why the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) would not shift when exposed to the same type of disturbance.

Write the equilibrium constant expression for (a) \(2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) \(+6 \mathrm{HCl}(g)\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftarrows 2 \mathrm{Fe}(l)+3 \mathrm{CO}_{2}(g)\) (c) \(\mathrm{PbSO}_{4}(s) \rightleftarrows \mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) (d) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{CO}_{3}(a q)\)

The process of photosynthesis in plants converts carbon dioxide and water to glucose and oxygen: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \quad \Delta E_{\mathrm{rxn}}=2801 \mathrm{~kJ}\) (a) Write the equilibrium constant expression for this conversion. (b) How would the equilibrium be affected if \(\mathrm{CO}_{2}(g)\) were added? (c) How would the equilibrium be affected if \(\mathrm{H}_{2} \mathrm{O}(l)\) were added? (d) How would the equilibrium be affected if the reaction vessel were warmed? (e) How would the equilibrium be affected if a catalyst were added?

Suppose a reaction has a \(K_{\text {eq }}\) value of \(2.05\). When we write the reaction, can we use a single arrow to the right instead of a double set of equilibrium arrows? Explain your answer.

Which of the following reactions is described by the equilibrium constant expression \(K_{\mathrm{eq}}=\frac{[\mathrm{A}]^{2} \times[\mathrm{B}]^{3}}{[\mathrm{C}]^{3} \times[\mathrm{D}]^{2}}\) (a) \(\mathrm{A}_{2}+\mathrm{B}_{3} \rightleftarrows \mathrm{C}_{3}+\mathrm{D}_{2}\) (b) \(2 \mathrm{~A}+3 \mathrm{~B} \rightleftarrows 3 \mathrm{C}+2 \mathrm{D}\) (c) \(3 \mathrm{C}+2 \mathrm{D} \rightleftarrows 2 \mathrm{~A}+3 \mathrm{~B}\) (d) \(A^{2}+B^{3} \rightleftarrows C^{3}+D^{2}\) (e) \(2 \mathrm{C}+3 \mathrm{D} \rightleftarrows 3 \mathrm{~A}+2 \mathrm{~B}\)
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