On introduit dans un réacteur de 10 litres, qu'on porte...

On introduit dans un réacteur de 10 litres, qu'on porte ensuite à $1000 \mathrm{~K}$, 1 mol de monoxyde de carbone $\mathrm{CO}$ et $100 \mathrm{~g}$ d'oxyde de fer $\mathrm{Fe}_3 \mathrm{O}_4$. On obtient l'équilibre : $$ \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons 3 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) $$ dont la constante d'équilibre, à cette température est 1,1 . a) Quelle est la composition de la phase gazeuse lorsque le système n'évolue plus? b) Quelles sont alors les pressions partielles des deux gaz?

On introduit dans un réacteur de 10 litres, qu'on porte ensuite à $1000 \mathrm{~K}$, 1 mol de monoxyde de carbone $\mathrm{CO}$ et $100 \mathrm{~g}$ d'oxyde de fer $\mathrm{Fe}_3 \mathrm{O}_4$. On obtient l'équilibre :
$$
\mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons 3 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})
$$
dont la constante d'équilibre, à cette température est 1,1 .
a) Quelle est la composition de la phase gazeuse lorsque le système n'évolue plus?
b) Quelles sont alors les pressions partielles des deux gaz?

Key Concepts

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative relationships between reactants and products as indicated by the balanced chemical equation. It is used to relate the changes in amounts or concentrations of the reactants to those of the products as the reaction proceeds, ensuring that the law of conservation of mass is upheld. In equilibrium problems, stoichiometry helps in formulating the changes in concentration or pressure used to build the equilibrium expressions.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of the reactants and products. This concept is pivotal because it provides a criterion for when a reaction has reached a steady state, allowing us to define and use equilibrium constants to quantify the relative amounts of species present at that state.

Equilibrium Constant Expression

The equilibrium constant (often denoted as K, Kc, or Kp) is a numerical value that relates the concentrations or partial pressures of the products and reactants at equilibrium for a given reaction. In heterogeneous equilibria where solids are involved, only the concentrations or partial pressures of the gaseous or aqueous species are included in the expression. This allows us to predict the direction of the reaction and calculate the composition of the system once equilibrium is achieved.

Heterogeneous Equilibria

Heterogeneous equilibria involve reactions where more than one phase is present, typically including solids and gases or liquids. In such systems, the concentrations of pure solids and pure liquids are treated as constants and are not included in the equilibrium constant expression. This simplification is essential for correctly formulating the mathematical expression that governs the behavior of the system.

Partial Pressures

Partial pressure is the pressure that an individual gas in a mixture would exert if it occupied the entire volume of the container alone at the same temperature. In the context of gas-phase equilibria, the contributions of each gas are considered separately and are essential for calculating the total pressure of the system. Partial pressures are directly used in the equilibrium constant expressions for gaseous reactions.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, relates the pressure, volume, number of moles, and temperature of an ideal gas. It is a key tool in equilibrium problems involving gaseous species, as it allows the conversion between moles and pressures. This law helps in determining the total pressure and individual partial pressures at equilibrium when the number of moles and the volume of the reactor are known.

Transcript

00:01 In this problem, we are given this reaction and we are told that we have one atmosphere of partial pressure of co gas reacting with excess feo.

00:11 We are also given a value for the equilibrium constant kp.

00:17 We want to determine what the equilibrium partial pressures are for both co and co2 gas.

00:23 When we are writing out our reaction table, we are only concerned with the equilibrium for the the non -solid or liquid species.

00:34 That's why we can ignore these two solid species, since we can't write them in terms of concentrations.

00:41 We are writing out our expression for the equilibrium constant kp, since these are both gases, and the p corresponds to partial pressure concentration units.

00:52 And based on this reaction, that would come out to the partial pressure of co2, divided by the partial pressure of c .o.

01:02 Co.

01:05 Notice that the solid species that are part of this reaction do not appear in this equilibrium expression.

01:12 So we are told that we initially have one atmosphere of co and we have no products.

01:18 So for the change for co, we have minus x and for co2, we have plus x.

01:26 So for the equilibrium for co, that's one minus x and for co2 that corresponds to x.

01:32 And for co2, that corresponds to x.

01:35 And for the equilibrium expression, both of these partial pressures correspond to the equilibrium partial pressures.

01:47 We are given a value of kp.

01:51 These expressions are used to find those equilibrium partial pressures, and we want to determine each one of the equilibrium partial pressures in terms of atmospheres...

Please give Ace some feedback