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In which direction will the position of the equilibrium $$2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)$$ be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\operatorname{HI}(g)\) is removed. d. In a rigid reaction container, some Ar(g) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).

Short Answer

Expert verified
(a) The equilibrium shifts to the right. (b) The equilibrium shifts to the right. (c) The equilibrium shifts to the left. (d) The equilibrium position remains unchanged. (e) The equilibrium shifts to the left. (f) The equilibrium shifts to the right.

Step by step solution

01

(a) Analyze when H2(g) is added

When H2(g) is added to the reaction, it increases the concentration of H2 gas, causing a stress on the equilibrium system. According to Le Chatelier's principle, the system will respond by shifting the equilibrium position to counteract the stress, meaning it will shift in the direction where the added substance is consumed. In this case, the equilibrium will shift to the right, consuming the added H2 gas and producing more HI gas and I2 gas.
02

(b) Analyze when I2(g) is removed

When I2(g) is removed from the reaction, it decreases the concentration of I2 gas, causing a stress on the equilibrium. According to Le Chatelier's principle, the system will respond by shifting the equilibrium position in the direction where the removed substance is produced. In this case, the equilibrium will shift to the right, producing more I2 gas and consuming H2(g) and HI(g).
03

(c) Analyze when HI(g) is removed

When HI(g) is removed from the reaction, it decreases the concentration of HI gas, causing a stress on the equilibrium. According to Le Chatelier's principle, the system will respond by shifting the equilibrium position in the direction where the removed substance is produced. In this case, the equilibrium will shift to the left, producing more HI gas and consuming H2(g) and I2(g).
04

(d) Analyze when Ar(g) is added in a rigid container

When Ar(g) is added to a rigid reaction container, it increases the overall pressure in the system without affecting the concentration of the reactants and products. Since it is an inert gas and does not participate in the reaction, the equilibrium position remains unchanged, as no stress has been applied to the system.
05

(e) Analyze when the volume of the container is doubled

When the volume of the container is doubled, the concentrations of all the gases (H2, I2, and HI) in the system are halved. As the change affects both the reactants and products, the system will shift the equilibrium position to counteract the stress. Since the reaction has a decrease in moles from the reactants to the products (2 moles of HI becomes 1 mole of H2 and 1 mole of I2), the equilibrium will shift to the side where there are more moles of gas, in this case to the left, producing more HI gas and consuming H2(g) and I2(g).
06

(f) Analyze when the temperature is decreased for an exothermic reaction

Since the given reaction is exothermic, it releases heat energy when the reaction occurs. When the temperature is decreased, it introduces a stress on the system by reducing the heat energy available. According to Le Chatelier's principle, the system will respond by shifting the equilibrium position in the direction where heat energy is produced, i.e., in the direction of the exothermic reaction. In this case, the equilibrium will shift to the right, producing more HI gas and consuming H2(g) and I2(g).

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Key Concepts

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

Chemical Equilibrium
In chemical reactions, equilibrium is the state where the concentrations of reactants and products remain constant over time. This does not mean that the reactions stop; rather the forward and backward reactions occur at the same rate. If any changes, such as concentration, pressure, or temperature are applied, the position of the equilibrium may shift to reestablish balance. This principle is known as Le Chatelier's Principle.

Le Chatelier's Principle helps predict how a system at equilibrium responds to changes.
  • If a reactant or product is added, the system shifts to consume the added entities.
  • If a reactant or product is removed, it shifts to produce more of the removed ingredient.
  • If the reaction is exothermic or endothermic, temperature changes can also affect the direction of the shift.
Le Chatelier's Principle is powerful because it helps us understand how equilibrium systems naturally strive to "cancel out" changes.
Exothermic Reactions
Exothermic reactions release heat as they proceed. This means that the energy of the products is lower than that of the reactants, and the difference in energy is released as heat. When the temperature of a system containing an exothermic reaction is decreased, the equilibrium shifts in the direction that produces heat, compensating for the loss of thermal energy.

In the context of our reaction, which is exothermic, reducing the temperature introduces a stress that reduces the system's heat energy. The equilibrium will shift to the right to produce more heat, increasing the concentration of the products, \( ext{H}_2\) and \( ext{I}_2\), by producing more \( ext{HI}\).

This behavior of exothermic reactions counterparts endothermic reactions, where increasing temperature causes an equilibrium shift towards heat absorption
Gas Laws
Gas laws provide valuable insights into understanding equilibrium, particularly when equilibria involve gases. Key concepts include pressure, volume, and the number of moles of gas. When the volume of a container is changed, the pressure and concentration of gaseous substances are affected; a core concept derived from the ideal gas law \(PV = nRT\).

As per Le Chatelier's Principle, when the volume of a reaction container is increased by doubling, the concentrations of all reactants and products drop due to increased space. For the reaction \(\text{2 HI} \rightleftharpoons \text{H}_2 + \text{I}_2\), the equilibrium will shift towards the side with more gas moles. Here, that means a shift to the left, recreating more \(\text{HI}\) to balance the decrease in concentration. Understanding these shifts is crucial for predicting how changes in physical conditions alter chemical equilibria in gaseous reactions.

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Most popular questions from this chapter

Write the equilibrium expression (K) for each of the following gas-phase reactions. a. \(N_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) b. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. \(\operatorname{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g)\) d. \(2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)\)

A 1.00-L flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K,\) for this reaction.

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm} \end{aligned}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$2 \mathrm{A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g)$$ what is the value of \(K\) at the same temperature for the reaction $$\mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{B}(g)$$ What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with 1.50 atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D},\) what is the mole fraction of \(\mathrm{B}\) once equilibrium is reached?

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\) . b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of \(0.30 \mathrm{atm},\) were allowed to come to equilibrium at this temperature?

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