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A vessel at equilibrium, contains \(\mathrm{SO}_{3}, \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), Now some helium gas is added, so that total pressure increases while temperature and volume remain constant. According to Le Chatelier's Principle, the dissociation of \(\mathrm{SO}_{3}\) : (a) Decreases (b) Remains unaltered (c) Increases (d) Change unpredictably

Short Answer

Expert verified
The dissociation of \(\mathrm{SO}_3\) remains unaltered.

Step by step solution

01

Analyzing the Reaction

The reaction of interest is the dissociation of sulfur trioxide: \(\mathrm{2SO_3(g) \leftrightarrow 2SO_2(g) + O_2(g)}\). This reaction is at equilibrium.
02

considering Reaction Conditions

Initially, the system is at equilibrium, and the volume is constant. Thus, the addition of helium will only affect the total pressure and not the partial pressures of the reacting gases.
03

Understanding Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium experiences a change in pressure, temperature, or concentration, the system will adjust to counteract that change and restore equilibrium.
04

Effect of Helium Adding to the System

Adding helium, an inert gas, increases the total pressure of the system, but does not affect the partial pressures of \(\mathrm{SO}_3\), \(\mathrm{SO}_2\), or \(\mathrm{O}_2\), since the volume and temperature remain constant.
05

Evaluating the Reaction's Responsiveness

The equilibrium position depends on the partial pressures of the reactants and products. Since the partial pressures remain unaltered by the addition of helium, the reaction does not shift.

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Key Concepts

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

Equilibrium
Equilibrium in chemical reactions occurs when the rates of the forward and reverse reactions become equal. At this point, the concentrations of reactants and products remain constant over time, even though the reactions continue to occur.
One crucial point to understand is that equilibrium is dynamic. This means molecules are constantly reacting with each other, but there's no net change in concentration. Imagine it like a see-saw that balances perfectly: both sides are active, but they're in harmony.
Equilibrium requires a closed system where no substances are added or removed, and the same number of molecules switch back and forth between reactants and products.
In the case of the given reaction, the equilibrium is maintained by the continuous conversion between sulfur trioxide (\(\mathrm{SO}_3\)) and the products sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen (\(\mathrm{O_2}\)).
  • If you make a change to the system, such as altering the concentration or pressure, you disturb this balance.
  • The system will then try to find a new state of balance, highlighting the essence of Le Chatelier's Principle.
Partial Pressure
Partial pressure refers to the pressure exerted by a single component of a gas mixture. When gases mix, each exerts its own pressure independent of the others.
This concept is vital in understanding how gases behave in reactions and at equilibrium.
In the dissociation reaction of sulfur trioxide, the partial pressures of \(\mathrm{SO}_3\), \(\mathrm{SO}_2\), and \(\mathrm{O}_2\) matter because they help determine the position of equilibrium.
The total pressure of the system is the sum of the partial pressures of these gases.
  • However, if you add an inert gas like helium at constant volume and temperature, it increases the total pressure but doesn't change the partial pressures of the reacting gases.
  • This unique behavior occurs because inert gases do not react with the system, so they do not affect the equilibrium position.
The result? The balance within the reaction remains stable, illustrating how partial pressures are crucial for maintaining equilibrium.
Inert Gas Effect
The inert gas effect is an interesting phenomenon observed in chemical equilibria. An inert gas, such as helium, does not react with the chemical substances in a reaction.
Adding an inert gas to a system at equilibrium increases the total pressure, but has no effect on the equilibrium position if the reaction volume remains constant.
This might seem counterintuitive at first, especially when we think of scenarios involving pressure changes.
  • Adding an inert gas does not change the concentration of the gases involved in the reaction, because it neither forms products nor consumes reactants.
  • As a result, the partial pressures of the reacting species remain unchanged, meaning the equilibrium position is not altered.
So, in the textbook exercise example, when helium is added to the system of \(\mathrm{SO}_3\), \(\mathrm{SO}_2\), and \(\mathrm{O_2}\), the reaction's equilibrium remains the same.
This principle highlights the importance of distinguishing between changes in total pressure versus changes in concentration or partial pressures for a given substance.

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Most popular questions from this chapter

In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, relationship between \(\mathrm{K}_{\mathrm{P}}\) and \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{2}\) (c) \(K_{p}=K_{c}(R T)^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, 1.10 molesofNOClremained. Calculatetheequilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for the reaction, \(2 \mathrm{NOC} 1(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}):\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)

The law of chemical equilibrium was first given by: (a) Guldberg (b) Boyle (c) Waage (d) Both (a) and (c)

For the \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), the initial mole ratio of \(\mathrm{N}_{2}: \mathrm{H}_{2}\) is \(1: 3 .\) If at equilibrium only \(50 \%\) has reacted and equilibrium pressure is \(\mathrm{P}\). Find the value of \(\mathrm{P}_{\mathrm{NH}_{3}}\) at equilibrium. (a) \(\frac{\mathrm{P}}{3}\) (b) \(\frac{\mathrm{P}}{5}\) (c) \(\frac{\mathrm{P}}{9}\) (d) \(\frac{\mathrm{P}}{6}\)

In the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are \(0.4\) and \(0.2\) mole/litre respectively. If the value of \(\mathrm{K}_{\mathrm{c}}\) is \(0.5\), what is the concentration of \(\mathrm{Cl}_{2}\) in mole/litre: (a) \(2.0\) (b) \(1.5\) (c) \(1.0\) (d) \(0.5\)

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