Dans un autoclave (réacteur en acier pouvant résister à des...

Dans un autoclave (réacteur en acier pouvant résister à des pressions internes élevées) dont le volume est 5 litres, se trouve un mélange de méthane $\mathrm{CH}_4$ et d'éthylène $\mathrm{C}_2 \mathrm{H}_4$ sous une pression de 15 bars, à $20^{\circ} \mathrm{C}$. On y introduit en outre du dihydrogène $\mathrm{H}_2$, jusqu'à ce que le pression dans le réacteur soit égale à 30 bars. Le réacteur contient aussi le catalyseur nécessaire (de volume négligeable) pour que la réaction d'hydrogénation de l'éthylène en éthane $$ \left(\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})=\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})\right) $$ ait lieu; on observe que la pression baisse jusqu'à 20 bars puis reste stable. Quelle était la composition (\% en volume et en masse) du mélange gazeux avant l'addition de $\mathrm{H}_2$ et quelle est celle du mélange final après l'achèvement de la réaction?

Dans un autoclave (réacteur en acier pouvant résister à des pressions internes élevées) dont le volume est 5 litres, se trouve un mélange de méthane $\mathrm{CH}_4$ et d'éthylène $\mathrm{C}_2 \mathrm{H}_4$ sous une pression de 15 bars, à $20^{\circ} \mathrm{C}$.
On y introduit en outre du dihydrogène $\mathrm{H}_2$, jusqu'à ce que le pression dans le réacteur soit égale à 30 bars. Le réacteur contient aussi le catalyseur nécessaire (de volume négligeable) pour que la réaction d'hydrogénation de l'éthylène en éthane
$$
\left(\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})=\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})\right)
$$
ait lieu; on observe que la pression baisse jusqu'à 20 bars puis reste stable. Quelle était la composition (\% en volume et en masse) du mélange gazeux avant l'addition de $\mathrm{H}_2$ et quelle est celle du mélange final après l'achèvement de la réaction?

Key Concepts

Ideal Gas Law

This concept relates the pressure, volume, temperature, and number of moles of a gas using the equation PV = nRT. It underpins the calculation of gas amounts in a closed system and is crucial when analyzing changes in pressure resulting from addition or consumption of gases under constant volume and temperature conditions.

Dalton's Law of Partial Pressures

This law states that the total pressure of a gas mixture is the sum of the partial pressures of its individual components. It is essential in determining the relative amounts of each gas in a mixture, both before and after additional gases are introduced or chemical reactions occur.

Reaction Stoichiometry in Gas-Phase Reactions

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. In gas-phase hydrogenation reactions, understanding the mole ratios (such as one mole of C2H4 reacting with one mole of H2 to form one mole of C2H6) is key to predicting how the reaction will affect the total number of moles and, consequently, the pressure in the system.

Conversion between Mole and Mass Percentages

This concept covers the transformation from mole fractions (or volume percentages, when gases behave ideally) to mass percentages by considering the molecular masses of the different gases. It is necessary for reporting the composition of gas mixtures in different formats and for understanding the practical implications of gas mixtures in terms of weight.

Catalysis

Catalysis involves the use of a catalyst to accelerate a chemical reaction without being consumed in the process. In the context of gas-phase hydrogenation, a catalyst facilitates the conversion of more unsaturated molecules to saturated ones, thereby affecting the reaction rate and the equilibrium without altering the overall mass balance directly.

Transcript

00:01 In this problem, let's see here.

00:02 We have, what are we got? in a step in the production of a synthetic liquid fuel from organic waste matter, we use one kilomol of ethylene that's converted from waste at room temperature and 5 ,000, or 5 megapascals, and 2 kilomoles of steam at 300 degrees c and 5 megapaascal.

00:28 And they enter a catalytic reactor.

00:30 An ideal gas mixture of ethanol, ethylene, and water in chemical equilibrium leaves the reactor at 700 kelvin and 500 megapascal, 5 ,000 megapascals.

00:43 Oh, no.

00:44 Five megapacels.

00:46 And determine the composition of the mixture and the heat transfer for the reactor.

00:50 Okay, so we got stuff going in.

00:53 So here's our chemical equilibrium.

00:55 We got stuff going in is the ethylene and the water.

01:02 Temperatures and pressures.

01:05 Coming out, we have ethanol, what are we here? yeah, we have ethanol, ethylene, left over, and water left over.

01:18 So we're getting some of this guy stuff out to you.

01:22 Let's see here, and that is all the temperature and pressure, that stuff.

01:26 So we need to figure out the gibbs function or the changing gibbs function for this process.

01:33 And again, there's a lot of calculations behind this.

01:35 These numbers, but i assume you can do those by looking everything up for 700 calvin.

01:43 And so we get a change in the entropy is, let's see here, is minus 38 .6 megajoules, again from the here to here.

01:57 And the change in the entropy is minus 0 .11 megajoules per calvin.

02:02 So we get a change in gibbs function of 38, 0 .8 megajoules, which means the natural log of k is minus 6 .67, which means k is 0 .00127.

02:19 So if we do our perturbation of our equilibrium equation and then use the equilibrium, the numerical equation here, so our equilibrium reaction, and use their equilibrium equation, with the k we found and then the mole fractions of everything.

02:37 And now we have this pressure is because why.

02:47 Oh, we just have p .0 here, that's right.

02:50 So we're up at 5 micapascals.

02:53 So again, we can, let's see here, this is just a quadratic for x, so we could use quadratic equation and get the appropriate root, and that is 0 .0404.

03:05 And so we have, if we look at our moles, moles, number of moles per mole of, whether it's, no, we have a total number, we have a total number of moles.

03:23 We have one kilomole.

03:24 So, so we have, in the end we have 0 .0404 kilomoles of ethanol, which is 1 .37%, 1 .37%.

03:40 We have a little less, ethylene than we had before.

03:44 We had one before...

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