Considere el siguiente equilibrio: N2 O4(g) ⇌2 NO2(g) Los datos termodinámicos para estos gases

step 1 are asked they can consider the following equilibrium. Thermodynamic data through these gases ar

Considere el siguiente equilibrio: N2 O4(g) ⇌2 NO2(g) Los...

Considere el siguiente equilibrio: $$ \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) $$ Los datos termodinámicos para estos gases están dados en el apéndice C. Puede suponer que $\Delta H^{\circ}$ y $\Delta S^{\circ}$ no se modifican con la temperatura. $a$ ) ¿̇A qué temperatura una mezcla de equilibrio contendrá cantidades iguales de los dos gases? $b$ ) ¡¿A qué temperatura una mezcla de equilibrio con una presión total de 1 atm contendrá dos veces más de $\mathrm{NO}_2$ que de $\mathrm{N}_2 \mathrm{O}_4$ ? c) ¡̇A qué temperatura una mezcla de equilibrio con una presión total de 10 atm contendrá dos veces más de $\mathrm{NO}_2$ que de $\mathrm{N}_2 \mathrm{O}_4$ ? d) Explique los resultados de $\operatorname{los}$ incisos b) y c) utilizando el principio de Le Châtelier. [Sección 15.7]

Considere el siguiente equilibrio:
$$
\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)
$$
Los datos termodinámicos para estos gases están dados en el apéndice C. Puede suponer que $\Delta H^{\circ}$ y $\Delta S^{\circ}$ no se modifican con la temperatura. $a$ ) ¿̇A qué temperatura una mezcla de equilibrio contendrá cantidades iguales de los dos gases? $b$ ) ¡¿A qué temperatura una mezcla de equilibrio con una presión total de 1 atm contendrá dos veces más de $\mathrm{NO}_2$ que de $\mathrm{N}_2 \mathrm{O}_4$ ? c) ¡̇A qué temperatura una mezcla de equilibrio con una presión total de 10 atm contendrá dos veces más de $\mathrm{NO}_2$ que de $\mathrm{N}_2 \mathrm{O}_4$ ? d) Explique los resultados de $\operatorname{los}$ incisos b) y c) utilizando el principio de Le Châtelier. [Sección 15.7]

Key Concepts

Partial Pressures in Equilibrium Systems

For gas-phase reactions, the equilibrium state is often described in terms of partial pressures rather than concentrations. The partial pressures of the reactants and products are used to compute the equilibrium constant (Kp). Changes in the total pressure or the distribution of partial pressures can shift the equilibrium composition, making this concept essential when analyzing equilibria under different pressure conditions.

Chemical Equilibrium

This concept refers to the state in a reversible chemical reaction when the rate of the forward reaction equals that of the reverse reaction. At equilibrium, the concentration or partial pressures of the reactants and products remain constant over time, even though both reactions continue to occur. It is a fundamental aspect of reaction dynamics in chemical systems and serves as the foundation for predicting how systems respond to various conditions.

Equilibrium Constant (K)

The equilibrium constant, often denoted as K, quantitatively expresses the ratio of the concentrations or partial pressures of products to reactants in a reaction at equilibrium, each raised to the power of their stoichiometric coefficients. For gas-phase reactions, the equilibrium constant is typically defined in terms of partial pressures (Kp), and its value is temperature-dependent based on the thermodynamic properties of the reaction.

Gibbs Free Energy and Thermodynamic Fundamentals

Gibbs free energy (?G) is a thermodynamic function that indicates the spontaneity of a reaction. The standard change in Gibbs free energy (?G°) is related to the equilibrium constant by the relationship ?G° = -RT ln K. Additionally, the relationship ?G° = ?H° - T?S° connects the enthalpy change (?H°) and the entropy change (?S°) of the reaction, enabling the analysis of how temperature affects the position of the equilibrium.

Van't Hoff Equation

The Van't Hoff equation describes the temperature dependence of the equilibrium constant. By relating the change in the logarithm of the equilibrium constant to the reciprocal of the temperature, it provides insight into how shifts in temperature affect the balance between products and reactants. This is crucial for predicting the conditions under which the concentrations of species in equilibrium will be equal or in a specific ratio.

Le Châtelier's Principle

Le Châtelier's Principle is a qualitative tool used to predict how a system at equilibrium will respond to an external change, such as alterations in pressure, temperature, or concentration. It states that the system will adjust its equilibrium position to counteract the change imposed. This principle is particularly useful for understanding and predicting the behavior of a gaseous equilibrium when subjected to variations in total pressure or temperature.

Transcript

00:02 Are asked they can consider the following equilibrium.

00:05 Thermodynamic data through these gases are given an appendix c.

00:10 We can assume that delta h and delta s, the enthalpy and entropy and entropy, do not vary with temperature.

00:34 At what temperature will an equilibrium mixture contain equal amounts of the two gases? let's figure it out.

01:13 First we're going to calculate delta h, then we're going to calculate calculate delta s and then we're going to find a 10.

01:27 Delta h for this reaction will equal two times the heat the heat formation of no2 minus and that will equal two times times 33 .84 minus 9 .66 equals 58 .02 kilojoules.

02:14 Second, we're going to do the same calculation, except we're going to replace our delta s, or we're going to use delta s to replace delta h.

02:25 So we're going to take 2 times 240 .45 minus 1 times 304 .3.

02:44 And that will equal 0 .176 kilojoules per k.

02:55 And finally, my temperature will equal delta h.

03:03 Or delta s, which will be 58 .02 divided by 0 .176 kilojoules per k.

03:22 And that will equal 328 .5 kelvin.

03:31 And i might, i'm just looking at my values here, i like we can keep that.

03:40 I probably need to go to one fig -fig.

03:42 So i'd go to 329 kelvin.

03:47 I'm not even 100 % sure i'd have to stick that.

03:50 Could be 328.

03:52 I don't have my calculator in front of him anymore because i already rounded it to get the 328 .5.

03:57 So you'd have to do that again.

04:00 B, at what temp would an equilibrium mixture contain one atmosphere total pressure.

04:21 Let me read again.

04:21 What temperature? will an equilibrium mixture of one atmosphere? what temp? equilibrium of one atmosphere total pressure twice as much n02 as n204.

05:02 Is that right? so b, this is going to be similar.

05:13 Calculations here.

05:17 We already have the following information from our last problem.

05:20 This is what we already have.

05:32 And we've got our delta s is, i'm going to change this one to kilojoules, 176.

05:49 .6 joules per chemo.

05:52 We already have those two values from the last question.

05:55 First we're going to calculate this equilibrium constant of the reaction, and the equilibrium constant will be equal to what i'm writing down right here.

06:22 When you're going to use the ratios of the gases, we're going to express the partial pressures as the multiplication of the mole fractions.

08:17 Okay, so then we can plug some values in here.

09:17 Since there are two parts of n02 for one part of n204, then the mole fraction of n2 is equal to x times n204 equals one third.

10:12 So, thus the equivalent constant is 4 over 3.

10:25 Let's go to the next page.

10:27 Now we can use this.

10:37 Our gibbs free energy is equal to negative rt of k.

10:47 And this will equalize these two expressions.

11:21 And then we're going to start playing some values.

11:23 We're going to rearrange salt for t.

11:24 T will be equal to.

12:44 And we'll solve for this.