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Chapter 7: Problem 31

The cquilibrium \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) shift forward if (1) a catalyst is used (2) an adsorbent is used to remove \(\mathrm{SO}_{3}\) as soon as it is formed (3) small amounts of reactants are used (4) none

Short Answer

Expert verified
Using an adsorbent to remove \(\text{SO}_3\) as soon as it is formed (option 2) shifts the equilibrium forward.

Step by step solution

01

- Understand the Concept of Equilibrium

Chemical equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. Any change that shifts the equilibrium position to favor the formation of products or reactants is influenced by Le Chatelier's Principle.
02

- Analyze the Effect of a Catalyst

Understand that a catalyst speeds up both the forward and reverse reactions equally without altering the equilibrium position. Therefore, using a catalyst does not shift the equilibrium.
03

- Consider the Removal of \(\text{SO}_3\)

According to Le Chatelier's Principle, removing a product will shift the equilibrium towards the product side to compensate for the loss. Thus, using an adsorbent to remove \(\text{SO}_3\) as soon as it is formed will shift the equilibrium forward.
04

- Examine the Effect of Using Small Amounts of Reactants

Using small amounts of reactants does not necessarily shift the equilibrium forward. It may slow down the reaction but does not change the position of equilibrium.
05

- Draw a Conclusion

From the analysis of the effects in the previous steps, it is evident which change will shift the equilibrium towards the formation of \(\text{SO}_3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's Principle is a cornerstone concept in chemical equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the equilibrium will shift to counteract the disturbance. This means that the system will try to restore balance by favoring either the forward or reverse reaction.
For example:
  • Adding more reactants will shift the equilibrium towards the products.
  • Removing products will also shift the equilibrium towards the products to replace what's been taken away.
  • Increasing pressure by decreasing volume in a gas reaction shifts the equilibrium to the side with fewer gas molecules.
Understanding this principle helps us predict how different changes can affect the equilibrium position.
Catalyst Effect
A catalyst is a substance that speeds up the rate of a reaction without being consumed in the process. In the context of chemical equilibrium, a catalyst affects both the forward and reverse reactions equally.
This means that while a catalyst can help a system reach equilibrium faster, it does not alter the position of the equilibrium. For the reaction: \2 \text{SO}_{2}(\text{g})+\text{O}_{2}(\text{g}) \rightleftharpoons 2 \text{SO}_{3}(\text{g}), adding a catalyst like Vanadium(V) oxide will speed up both the production and decomposition of SO\(_3\), but the equilibrium position remains unchanged.
Thus, using a catalyst does not cause the equilibrium to shift to the forward or reverse side.
Equilibrium Shift
An equilibrium shift occurs when conditions are altered, causing the equilibrium to move in the direction that will counteract the change.
This can happen due to:
  • Changes in concentration of reactants or products
  • Changes in temperature
  • Changes in pressure (for gases)
In the given reaction, \2 \text{SO}_{2}(\text{g})+\text{O}_{2}(\text{g}) \rightleftharpoons 2 \text{SO}_{3}(\text{g}), if we remove SO\(_3\) as soon as it is formed, the system will shift to produce more SO\(_3\) to compensate for the loss. This is an application of Le Chatelier's Principle. The equilibrium shift will be towards the product side (SO\(_3\) formation).
Adsorbent Use
Adsorbents are materials that attract and hold particles (like gases or liquids) on their surface. In chemical processes, adsorbents can be used to selectively remove certain products or reactants.
When applied to equilibrium reactions, using an adsorbent to remove a product can cause a shift towards the product side. In the reaction: \2 \text{SO}_{2}(\text{g})+\text{O}_{2}(\text{g}) \rightleftharpoons 2 \text{SO}_{3}(\text{g}), if an adsorbent is used to continuously remove SO\(_3\) as it's formed, it forces the reaction to produce more SO\(_3\) to replace what's removed.
This creates a forward shift in the equilibrium, resulting in higher production of SO\(_3\).

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Most popular questions from this chapter

For the reaction \(2 \mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Z}(\mathrm{g}) ; \Delta H=\) 80 kcal. The highest yicld of \(Z\) at cquilibrium occurs at (1) \(1000 \mathrm{~atm}\) and \(500^{\circ} \mathrm{C}\) (2) \(500 \mathrm{~atm}\) and \(500^{\circ} \mathrm{C}\) (3) \(1000 \mathrm{~atm}\) and \(100^{\circ} \mathrm{C}\) (4) \(500 \mathrm{~atm}\) and \(100^{\circ} \mathrm{C}\)

The aqueous solution of a salt is alkalinc. This shows that salt is made from (1) \(\Lambda\) strong acid and strong base (2) \(\Lambda\) strong acid and weak base (3) \(\Lambda\) wcak acid and weak base (4) \(\Lambda\) weak acid and strong base

\(1.1\) mole of \(\Lambda\) is mixed with \(2.2\) molc of \(B\) and the mixture is then kept in \(1-\mathrm{L}\) flask till the cquilibrium is attained \(\Lambda+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}+\mathrm{D} \cdot \Lambda \mathrm{t}\) the equilibrium \(0.2\) mole of \(C\) are formed. The equilibrium constant of the reaction is (1) \(0.001\) (2) \(0.222\) (3) \(0.003\) (4) \(0.004\)

\(\Lambda\) saturated solution of \(\mathrm{II}_{2} \mathrm{~S}\) in \(0.1 \mathrm{M} \mathrm{IICl}\) at \(25^{\circ} \mathrm{C}\) contains a \(\mathrm{S}^{2-}\) ion concentration of \(10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}\). The solubility products of some sulphidcs are: \(\mathrm{CuS}=10^{-44}\), \(\mathrm{FeS}=10^{-14} ; \mathrm{MnS}=10^{-15}\) and \(\mathrm{CdS}=10^{-25} .\) If \(0.01 \mathrm{M}\) solutions of these salts in \(1 \mathrm{M}\) IICl are saturated with \(\mathrm{H}_{2} \mathrm{~S}\), which of these will be precipitated? (1) All (2) All except MnS (3) All except MnS and FeS (4) Only CuS

For an acid \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} ; K_{1}\) and for a base \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{OH} ; K_{2}\) Then (1) \(K_{1} \cdot K_{2}=K_{\mathrm{w}}\) (2) \(\log K_{1}+\log K_{2}=\log K_{\mathrm{w}}\) (3) \(\mathrm{p} K_{1}+\mathrm{p} K_{2}=\mathrm{p} K_{\mathrm{w}}\) (4) all are correct
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