La constante d'équilibre de la réaction H2( g)+I2( g) ⇄2 HI(g) vaut 13,4 à 250^∘ C a) À cette te

step 1 This question has multiple parts. Each part is quite challenging. So let's get started. We have

La constante d'équilibre de la réaction H2( g)+I2( g) ⇄2...

La constante d'équilibre de la réaction $$ \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftarrows 2 \mathrm{HI}(\mathrm{g}) $$ vaut 13,4 à $250^{\circ} \mathrm{C}$ a) À cette température, quelle est la proportion molaire (\%) d'iodure d'hydrogène $\mathrm{HI}$ dans le mélange gazeux en équilibre, pour une réaction effectuée à partir d'un mélange équimolaire de dihydrogène et de diiode ? b) Dans les mêmes conditions stœchiométriques initiales, à quelle température la proportion de $\mathrm{HI}$ dans le mélange à l'équilibre serait-elle $80 \%$ ? À cette température, quelle proportion (\%) du diiode initial a-t-elle été consommée? c) Le diiode est de loin le plus coûteux des deux réactifs. Pour augmenter son taux de transformation en $\mathrm{HI}$ on peut utiliser un excès de dihydrogène. $\mathrm{Si}$, à la température calculée en (b), on utilisait un mélange initial contenant une quantité cinq fois plus grande de dihydrogène que de diiode, quelle serait la proportion de diiode transformée?

La constante d'équilibre de la réaction
$$
\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftarrows 2 \mathrm{HI}(\mathrm{g})
$$
vaut 13,4 à $250^{\circ} \mathrm{C}$
a) À cette température, quelle est la proportion molaire (\%) d'iodure d'hydrogène $\mathrm{HI}$ dans le mélange gazeux en équilibre, pour une réaction effectuée à partir d'un mélange équimolaire de dihydrogène et de diiode ?
b) Dans les mêmes conditions stœchiométriques initiales, à quelle température la proportion de $\mathrm{HI}$ dans le mélange à l'équilibre serait-elle $80 \%$ ? À cette température, quelle proportion (\%) du diiode initial a-t-elle été consommée?
c) Le diiode est de loin le plus coûteux des deux réactifs. Pour augmenter son taux de transformation en $\mathrm{HI}$ on peut utiliser un excès de dihydrogène. $\mathrm{Si}$, à la température calculée en (b), on utilisait un mélange initial contenant une quantité cinq fois plus grande de dihydrogène que de diiode, quelle serait la proportion de diiode transformée?

Key Concepts

Mole Fraction and Percentage Composition

Mole fraction is a way of expressing the relative abundances of different species in a mixture by dividing the number of moles of a component by the total number of moles present. When these fractions are multiplied by 100, they provide percentage compositions, which are often more intuitive for describing the makeup of a system at equilibrium.

Equilibrium Constant (K)

The equilibrium constant is a numerical value that expresses the ratio of the concentrations or pressures of products to those of reactants at equilibrium, each raised to the power of their stoichiometric coefficients. It provides a snapshot of the reaction’s extent under a given set of conditions and is essential for predicting how far a reaction will proceed.

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative relationships between reactants and products dictated by the balanced chemical equation. This concept is crucial for setting up equilibrium calculations as it determines how the concentrations of different species change in relation to one another as the system approaches equilibrium.

ICE Table Method

The ICE (Initial, Change, Equilibrium) table is a systematic framework used to track the concentrations or pressures of reactants and products from the starting point (initial), through the change during the reaction, to the final equilibrium state. It facilitates the formulation and solution of equations based on the stoichiometry of the reaction and the equilibrium constant.

Le Chatelier's Principle

Le Chatelier's Principle states that if an equilibrium system is disturbed by a change in concentration, pressure, or temperature, the system will adjust to partially counteract that change and re-establish equilibrium. This principle is fundamental for understanding how alterations in reaction conditions, such as using an excess of one reactant, can shift the equilibrium position and affect the yield of the desired product.

Transcript

00:01 This question has multiple parts.

00:03 Each part is quite challenging.

00:05 So let's get started.

00:06 We have an equilibrium.

00:10 One mole h2 plus one mole i2 gas goes to two moles h5 gas.

00:16 It's mentioned that it has an equilibrium constant equal to 50 at 745 kelvin.

00:31 For part a, we add one mole of i2 and three moles of h2 to a 10 -liter flask.

00:38 That means that we're going to have an initial concentration of 0 .1 molar i2 and 0 .3 molar h2.

00:48 We're simply taking the 1 moles and the 3 moles and dividing them by 10 to get our concentration.

00:56 And we have an initial amount of hi of nothing, zero.

01:00 So the reaction needs to shift to the right so that we have some hi at equilibrium.

01:06 So we'll decrease our h2 concentration by x and our i2 concentration by.

01:11 By x, but increase our h .i.

01:14 Concentration by 2x.

01:17 So these will then be the concentrations of each of them at equilibrium.

01:22 To solve for the concentration of everything at equilibrium, which is what we need to do first before we can determine the amount in moles of h .i.

01:31 That are produced, we set up our equilibrium expression.

01:36 So at equilibrium we'll have 2x of h .i.

01:39 In the equilibrium constant expression, we need to square it because of the two.

01:44 We'll then divide by the equilibrium concentrations of h2 and i2.

01:52 Now we have a moderately difficult algebraic expression that we need to solve.

01:57 We'll need to use the quadratic formula or you may choose to use some online solver or a solver in excel or your calculator or whatever your instructor has encouraged you to use.

02:10 And we get x equal to 9 .6 times 10 of the negative 2.

02:15 To determine the moles of h .i.

02:17 X is just the concentration.

02:20 I'm sorry, x is not the concentration of h .i.

02:23 But 2 times x is.

02:24 So what we have to do is we have to figure out the concentration of h .i...

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