L'équilibre HgO(s) ⇌Hg(l)+1 / 2 O2 s'étant établi dans un réacteur dont on peut faire varier le

L'équilibre HgO(s) ⇌Hg(l)+1 / 2 O2 s'étant établi dans un...

L'équilibre $\mathrm{HgO}(\mathrm{s}) \rightleftharpoons \mathrm{Hg}(\mathrm{l})+1 / 2 \mathrm{O}_2$ s'étant établi dans un réacteur dont on peut faire varier le volume, quelles seraient les conséquences des interventions suivantes (toutes faites à température constante)? a) On ajoute une petite quantité d'oxyde de mercure $\mathrm{HgO}$. b) On retire une partie du mercure $\mathrm{Hg}$. c) On ajoute du dioxygène à volume constant. d) On retire une partie du dioxygène à pression constante (en faisant varier le volume pour que la pression reste constante).

L'équilibre $\mathrm{HgO}(\mathrm{s}) \rightleftharpoons \mathrm{Hg}(\mathrm{l})+1 / 2 \mathrm{O}_2$ s'étant établi dans un réacteur dont on peut faire varier le volume, quelles seraient les conséquences des interventions suivantes (toutes faites à température constante)?
a) On ajoute une petite quantité d'oxyde de mercure $\mathrm{HgO}$.
b) On retire une partie du mercure $\mathrm{Hg}$.
c) On ajoute du dioxygène à volume constant.
d) On retire une partie du dioxygène à pression constante (en faisant varier le volume pour que la pression reste constante).

Key Concepts

Effect of Volume and Pressure Changes on Gaseous Equilibria

In reactions involving gases, changes in volume or pressure affect the equilibrium position based on the number of gaseous moles on each side of the equation. A decrease in volume (or an increase in pressure) tends to shift the equilibrium toward the side with fewer moles of gas, while an increase in volume (or a decrease in pressure) favors the side with more moles of gas. This response ensures that the system can partially counteract the imposed change in pressure.

Effect of Concentration Changes

Altering the concentration of a reactant or product in an equilibrium reaction, such as by adding more of a species or removing some of it, can disrupt the equilibrium state. The system will shift its equilibrium position according to Le Chatelier's Principle in order to partially offset the change, leading to either enhanced production of products or increased formation of reactants.

Role of Pure Solids in Equilibrium

In heterogeneous equilibria, pure solids (and pure liquids) do not appear in the equilibrium constant expression because their activities are constant and defined as one. This means that changes in the amount of a pure solid do not affect the equilibrium position directly, as their concentration is not a variable in the equilibrium calculation.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals that of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, even though both reactions continue to occur at the molecular level. This balance is characterized by the equilibrium constant, which is determined by the activities or concentrations of the species involved, excluding pure solids and liquids.

Le Chatelier's Principle

Le Chatelier's Principle is a guideline used to predict how a chemical equilibrium will respond to a change in conditions. When a system at equilibrium is disturbed by changes in concentration, pressure, volume, or temperature, the system adjusts by shifting the reaction in the direction that tends to counteract the disturbance. This principle aids in determining whether a reaction will favor the formation of reactants or products under modified conditions.

Transcript

00:01 We are given this reaction, and we need to use lechatlier's principle in order to predict how each one of the disturbances in parts a through g of this problem affect the equilibrium.

00:12 In part a, we want to know what will happen if the volume is increased.

00:17 Well, if we increase the volume, then that means that we have decreased the pressure.

00:24 According to lechatlier's principle, the system will respond by trying to increase the pressure to reestablish equilibrium.

00:32 And when we consider pressure, we have to look at the total number of moles of gas on each side of the reaction.

00:40 On the reactant side, we have zero moles of gas.

00:42 On the product side, we have this single mole of gas.

00:46 And remember that when we increase the volume, the system will respond by trying to increase the pressure.

00:56 And the way that that will occur is by forming more moles of gas.

01:02 Since one is more than zero, the system will shift to the right to form more products when we increase a volume.

01:09 So that is why for part a, the reaction will shift to the right to reestablish equilibrium.

01:18 And in part b, we want to know what happens when cao is added to the system.

01:26 If we look, we see that cao is a solid.

01:29 We can write out the equilibrium constant expression to be a concentration of co2 gas, since that is the only species that is not a solid or liquid that we can write in terms of concentrations.

01:45 And when that is at its equilibrium concentration, its value is equal to equilibrium constant.

01:52 If we change that concentration of that gaseous species in any way, then it becomes a reaction quotient corresponding.

01:59 To that change in concentration.

02:03 But the solid species do not participate anywhere in that expression.

02:08 So whether we add or remove the solid species, it will have no effect on the equilibrium of the system for that reason...

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