La pentachlorure de phosphore PCl5, se dissocie en donnant PCl3 et Cl2 : PCl5( g)=PCl3( g)+Cl2(

La pentachlorure de phosphore PCl5, se dissocie en donnant...

La pentachlorure de phosphore $\mathrm{PCl}_5$, se dissocie en donnant $\mathrm{PCl}_3$ et $\mathrm{Cl}_2$ : $$ \mathrm{PCl}_5(\mathrm{~g})=\mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) $$ et la constante d'équilibre de cette réaction vaut 1,8 à $250^{\circ} \mathrm{C}$. a) On porte $1,5 \mathrm{~mol}$ de $\mathrm{PCl}_5$ à $250^{\circ} \mathrm{C}$, sous une pression constante de 1 bar. Quelles sont les pressions partielles des trois gaz à l'équilibre? b) On chauffe à $250^{\circ} \mathrm{C}, 0,7 \mathrm{~mol}$ de $\mathrm{PCl}_5$ dans une enceinte de 5 litres initialement vide. Quelle est la valeur de son coefficient de dissociation? c) On porte à $250^{\circ} \mathrm{C}$ un mélange de $1,5 \mathrm{~mol}$ de $\mathrm{PCl}_5$ et $1 \mathrm{~mol} \mathrm{de} \mathrm{Cl}_2$, sous la pression constante de 1 bar. Quelles sont les pressions partielles des trois gaz à l'équilibre? Quel volume occupent-ils?

La pentachlorure de phosphore $\mathrm{PCl}_5$, se dissocie en donnant $\mathrm{PCl}_3$ et $\mathrm{Cl}_2$ :
$$
\mathrm{PCl}_5(\mathrm{~g})=\mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})
$$
et la constante d'équilibre de cette réaction vaut 1,8 à $250^{\circ} \mathrm{C}$.
a) On porte $1,5 \mathrm{~mol}$ de $\mathrm{PCl}_5$ à $250^{\circ} \mathrm{C}$, sous une pression constante de 1 bar. Quelles sont les pressions partielles des trois gaz à l'équilibre?
b) On chauffe à $250^{\circ} \mathrm{C}, 0,7 \mathrm{~mol}$ de $\mathrm{PCl}_5$ dans une enceinte de 5 litres initialement vide. Quelle est la valeur de son coefficient de dissociation?
c) On porte à $250^{\circ} \mathrm{C}$ un mélange de $1,5 \mathrm{~mol}$ de $\mathrm{PCl}_5$ et $1 \mathrm{~mol} \mathrm{de} \mathrm{Cl}_2$, sous la pression constante de 1 bar. Quelles sont les pressions partielles des trois gaz à l'équilibre? Quel volume occupent-ils?

Key Concepts

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, connects pressure (P), volume (V), the number of moles (n), the universal gas constant (R), and temperature (T). This law is fundamental in calculating the relationships among these variables for gaseous systems, especially when analyzing changes in reaction conditions under constant pressure or constant volume settings.

Degree of Dissociation

The degree of dissociation is defined as the fraction of the initial amount of a reactant that has decomposed into products at equilibrium. It provides a measure of how far a reaction proceeds and is used to quantify the conversion of reactants into products, often expressed as a percentage or fraction.

Partial Pressures

Partial pressures represent the pressure that each gas in a mixture would exert if it occupied the entire volume independently at the same temperature. In equilibrium calculations for gas-phase reactions, understanding partial pressures is essential as they directly relate to the equilibrium constant expression and allow the application of Dalton’s law of partial pressures.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations or pressures of the reactants and products over time. At equilibrium, the system's macroscopic properties remain constant even though the individual molecules continue to interact dynamically.

Equilibrium Constant (Kp)

The equilibrium constant for a gaseous reaction, denoted as Kp, is defined as the ratio of the products’ partial pressures, each raised to the power of its stoichiometric coefficient, to those of the reactants. It quantitatively describes the position of equilibrium and is temperature dependent, providing insight into the relative amounts of reactants and products present at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products as expressed by the coefficients in the balanced chemical equation. This concept is critical for setting up ICE (Initial, Change, Equilibrium) tables and for relating changes in the amounts or pressures of substances during the reaction progress.

Transcript

00:01 This question wouldn't be so challenging if they had given you molarities rather than masses.

00:06 So it is going to take a little bit more in order to use the masses to solve this problem, both parts of this problem.

00:13 The equilibrium is pcl5 as a gas, going to pcl3 as a gas and cl2 as a gas.

00:26 The first thing that we need to do before we can solve part b, where we need to determine the concentrations of everything, is calculate the equilibrium constant.

00:39 They tell you that at equilibrium, there's 3 .120 grams of pcl5, 3 .843 or 45 grams of pcl3, and 1 .78 grams of cl2, all in a one -liter flask.

00:57 So to solve for kc, we need to take these grams and the volume and convert them into concentrations.

01:04 The equilibrium constant is going to be the concentration of pcl2.

01:08 Multiplied by the concentration of cl2, divided by the concentration of pcl5.

01:15 So we'll take the 3 .845 grams of pcl3, divide by the molar mass of pcl3 to get moles pcl3, and then divide by 1, which gives us the same value to get the concentration of pcl3.

01:31 Then we'll calculate the concentration of cl2, which will be the mass of cl2, 1 .77 grams, divided by the molar mass of cl2 divided by 1 liter and then we'll calculate the concentration of pcl5 which will be the mass of pcl5 3 .120 divided by the molar mass of pcl5 208 .24 divided by 1 which will then give us a concentration and we get 0 .071 as our equilibrium constant so now it wants to know if when it's at equilibrium after adding more cl2 in which direction or how will the equilibrium be affected well according to les chattelier's principle if we were to add more cl2 keeping the volume the same would increase the concentration of cl2 increasing the concentration of cl2 which shift the equilibrium to the left in order to partially offset the increase in concentration of cl2 so now we know the direction in which the equilibrium will shift in order to re -establish equilibrium.

02:48 So what i'll do is i'll take the 1 .787 gram cl2, add it to the 1 .418 gram cl2, and then divide by the molar mass of cl2 in order to get the moles of cl2.

03:07 Divide that by 1.

03:08 Now i have my molarity.

03:10 So it's 0452 molarity.

03:11 So it's 0 .0452 molar initially...

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