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A system at equilibrium is described by the equation of fixed temperature \(T\). $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ What effect will an increases in the total pressure caused by a decrease in volume have on the equilibrium?

Short Answer

Expert verified
Increasing the total pressure by decreasing the volume will cause the equilibrium to shift to the left, resulting in an increased concentration of \(\text{SO}_{2}\text{Cl}_{2}(g)\) and decreased concentrations of \(\text{SO}_{2}(g)\) and \(\text{Cl}_{2}(g)\).

Step by step solution

01

Recognize Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In this case, changing the pressure by decreasing the volume will affect the equilibrium position.
02

Assess the Reaction Stoichiometry

Analyze the balanced equation to determine the moles of gas on each side of the equilibrium. In the reaction \(\text{SO}_{2}\text{Cl}_{2}(g) \rightleftharpoons \text{SO}_{2}(g) + \text{Cl}_{2}(g)\), there is 1 mole of gas on the left side and 2 moles of gas on the right side.
03

Predict the Direction of the Shift

When the volume of the system is decreased, the total pressure increases, causing the equilibrium to shift toward the side with fewer moles of gas. Since there are more moles of gas on the right side (2 moles) compared to the left side (1 mole), the equilibrium will shift to the left, towards the formation of \(\text{SO}_{2}\text{Cl}_{2}(g)\).
04

Consider the Effect on Concentrations

The shift to the left will result in an increase in the concentration of \(\text{SO}_{2}\text{Cl}_{2}(g)\) and a decrease in the concentrations of \(\text{SO}_{2}(g)\) and \(\text{Cl}_{2}(g)\).

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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry, representing a state in which the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. It is important to note that this does not mean that reactions stop at equilibrium, but that the reactants are converted to products at the same rate as products revert back to reactants.

The dynamic nature of chemical equilibrium is described by an equilibrium constant, which is a ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. Understanding this concept is critical as it influences how changes in conditions, like pressure or temperature, will affect the system.
Examining Reaction Stoichiometry
Reaction stoichiometry outlines the proportional relationship between reactants and products in a chemical reaction. It is essential for predicting the amounts of substances consumed and produced, and allows chemists to calculate yields and efficiencies.

When discussing chemical equilibrium and Le Chatelier's Principle, stoichiometry helps us to understand how changing physical conditions, such as pressure, affect the equilibrium. By analyzing the stoichiometric coefficients in a balanced chemical equation, we can determine the number of moles of gas on each side and predict the response of the system when subjected to a change in conditions, such as an increase in total pressure due to a decrease in volume.
Predicting Equilibrium Shifts
Le Chatelier's Principle serves as a guide to predict an equilibrium shift when a system at equilibrium is disturbed. According to this principle, a system will react to minimize the impact of a change in conditions, such as temperature, pressure, or concentration.

In the context of the given exercise, when the volume of the gas is decreased, pressure increases. The system seeks to counter this by shifting towards the side with fewer moles of gas, which, in this case, is the side of the reactants. This shift decreases the pressure that was increased by reducing the volume, thus demonstrating the system's attempt to maintain equilibrium. Understanding the direction of the equilibrium shift is crucial for predicting the outcome of changing conditions on a chemical system.

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Most popular questions from this chapter

In a system \(A(s) \rightleftharpoons 2 B(g)+3 C(g)\), if the concentration of \(C\) at equilibrium is increased by a factor of 2 , it will cause the equilibrium concentration of \(B\) to change to : (a) two times the original value (b) one half of its original value (c) \(2 \sqrt{2}\) times to the original value (d) \(\frac{1}{2 \sqrt{2}}\) times the original value

In the presence of excess of anhydrous \(\mathrm{SrCl}_{2}\), the amount of water taken up is governed by \(K_{p}=10^{12} \mathrm{~atm}^{-4}\) for the following reaction at \(273 \mathrm{~K}\) $$ \mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{SrCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) $$ What is equilibrium vapour pressure (in torr) of water in a closed vessel that contains \(\mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) ?\) (a) \(0.001\) torr (b) \(10^{3}\) torr (c) \(0.76\) torr (d) \(1.31\) torr

A nitrogen-hydrogen mixture initially in the molar ratio of \(1: 3\) reached equilibrium to form ammonia when \(25 \%\) of the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) had reacted. If the total pressure of the system was 21 atm, the partial pressure of ammonia at the equilibrium was: (a) \(4.5\) atm (b) \(3.0 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.5 \mathrm{~atm}\)

Two moles of \(\mathrm{NH}_{3}\) when put into a previously evacuated vessel (one litre), partially dissociated into \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\). If at equilibrium one mole of \(\mathrm{NH}_{3}\) is present, the equilibrium constant is : (a) \(3 / 4 \mathrm{~mol}^{2}\) litre \(^{-2}\) (b) \(27 / 64 \mathrm{~mol}^{2}\) litre \(^{-2}\) (c) \(27 / 32 \mathrm{~mol}^{2}\) litre \(^{-2}\) (d) \(27 / 16 \mathrm{~mol}^{2}\) litre \(^{-2}\)

What is the unit of \(K_{p}\) for the reaction? $$ \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) $$ (a) atm (b) \(\mathrm{atm}^{-2}\) (c) \(\mathrm{atm}^{2}\) (d) \(\mathrm{atm}^{-1}\)

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