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Chapter 15: Problem 54

Consider the following equilibrium system involving \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (sulfuryl dichloride): $$ \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) $$ Predict how the equilibrium position would change if (a) \(\mathrm{Cl}_{2}\) gas were added to the system, (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) were removed from the system, (c) \(\mathrm{SO}_{2}\) were removed from the system. The temperature remains constant in each case.

Short Answer

Expert verified
(a) Shift right; (b) Shift right; (c) Shift left.

Step by step solution

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01

Understanding the Equilibrium System

The equilibrium system is represented by the equation: \( \text{SO}_{2}(g) + \text{Cl}_{2}(g) \rightleftarrows \text{SO}_{2}\text{Cl}_{2}(g) \). According to the principle of chemical equilibrium, when a change (or stress) is applied to a system at equilibrium, the system will adjust in such a way as to counteract that change.
02

Applying Le Chatelier's Principle Part 1

(a) If \( \text{Cl}_{2} \) gas is added, the concentration of a reactant increases. According to Le Chatelier's principle, the system will shift to the right, towards the formation of more \( \text{SO}_{2}\text{Cl}_{2} \), in order to reduce the concentration of \( \text{Cl}_{2} \).
03

Applying Le Chatelier's Principle Part 2

(b) If \( \text{SO}_{2} \text{Cl}_{2} \) is removed from the system, the concentration of the product decreases. The system will shift to the right (towards the formation of \( \text{SO}_{2}\text{Cl}_{2} \)) to oppose the reduction in \( \text{SO}_{2} \text{Cl}_{2} \).
04

Applying Le Chatelier's Principle Part 3

(c) If \( \text{SO}_{2} \) is removed, the concentration of a reactant decreases. The system will shift to the left, promoting the reverse reaction and forming more \( \text{SO}_{2} \) and \( \text{Cl}_{2} \) to oppose the removal of \( \text{SO}_{2} \).

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction reaches a point where the concentrations of reactants and products remain constant over time. This does not mean that the reactions have stopped; instead, they continue to proceed at equal rates in both the forward and reverse directions. In our system, the reaction between sulfur dioxide \((\text{SO}_2)\) and chlorine \((\text{Cl}_2)\) to form sulfuryl dichloride \((\text{SO}_2\text{Cl}_2)\) is in equilibrium when the rate of formation of the product equals the rate of its decomposition back into reactants. Even though there's dynamic activity at the molecular level, no net change in the concentrations occurs.
Equilibrium Shifts
An equilibrium shift is a response to a change in conditions of a reaction at equilibrium. According to Le Chatelier's Principle, the system will adjust itself to counteract any changes imposed upon it. If a disturbance, such as a concentration change, pressure change, or temperature change, is applied to the system, the position of equilibrium will shift to restore balance.
  • Adding a reactant or removing a product generally shifts the equilibrium to the right, favoring the production of more product.
  • Conversely, removing a reactant or adding a product shifts the equilibrium to the left, favoring the production of more reactants.
This principle is a useful tool in predicting how a system will react to various changes.
Concentration Changes
The concentration of substances in an equilibrium reaction plays a crucial role in determining the direction of the equilibrium shift. When the concentration of one reactant or product is altered, the system responds by partially offsetting this change.
  • If the concentration of \(\text{Cl}_2\) is increased, the system will shift to the right to form more \(\text{SO}_2\text{Cl}_2\) and reduce the excess \(\text{Cl}_2\).
  • If \(\text{SO}_2\text{Cl}_2\) is removed, the equilibrium shifts to the right to increase its concentration.
  • If \(\text{SO}_2\) is removed, the equilibrium will shift to the left, increasing the concentration of \(\text{SO}_2\) and \(\text{Cl}_2\).
These shifts help restore the balance in the system effectively.
Equilibrium Reactions
Equilibrium reactions are characterized by their ability to reach a state of balance between reactants and products. In such reactions, unlike irreversible reactions, reactants can form products that can also revert to become reactants again. This back-and-forth process continues until the rates of the forward and reverse reactions are equal.
In our given example with \(\text{SO}_2,\) \(\text{Cl}_2,\) and \(\text{SO}_2\text{Cl}_2,\) the system can dynamically readjust to changes in concentration and maintain equilibrium.
Understanding these principles can help predict the outcomes in an equilibrium system when subjected to external changes.

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Most popular questions from this chapter

What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant, and, in each case, the reactants are in a cylinder fitted with a movable piston. (a) \(\mathrm{A}(s) \rightleftarrows 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftarrows \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftarrows \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftarrows \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g)\)

Consider the following reaction at \(1600^{\circ} \mathrm{C}\) $$\mathrm{Br}_{2}(g) \rightleftarrows 2 \mathrm{Br}(g)$$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 -L flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Consider the following reaction at a certain temperature: $$ \mathrm{A}_{2}+\mathrm{B}_{2} \rightleftarrows 2 \mathrm{AB} $$ The mixing of 1 mole of \(\mathrm{A}_{2}\) with 3 moles of \(\mathrm{B}_{2}\) gives rise to \(x\) mole of \(\mathrm{AB}\) at equilibrium. The addition of 2 more moles of \(\mathrm{A}_{2}\) produces another \(x\) mole of \(\mathrm{AB}\). What is the equilibrium constant for the reaction?

The following equilibrium constants were determined at \(1123 \mathrm{~K}:\) $$ \begin{array}{l} \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) \quad K_{P}^{\prime}=1.3 \times 10^{14} \\ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{COCl}_{2}(g) \quad K_{P}^{\prime \prime}=6.0 \times 10^{-3} \end{array} $$ Write the equilibrium constant expression \(K_{P}\), and calculate the equilibrium constant at \(1123 \mathrm{~K}\) for $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{COCl}_{2}(g) $$

Consider the decomposition of ammonium chloride at a certain temperature: $$\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftarrows \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$$ Calculate the equilibrium constant \(K_{P}\) if the total pressure is \(2.2 \mathrm{~atm}\) at that temperature.
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