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For the reaction, \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), \Delta_{r} \mathrm{G}^{0}=3.40 \mathrm{kJ} \mathrm{mor}^{-1}\) at \(298.15 \mathrm{K}\) (a) Calculate the equilibrium constant. (b) Does the reaction favour the products or reactants? (c) If additional \(\mathrm{H}_{2}(\mathrm{g})\) was added to the equilibrium mixture at the same temperature, predict what would happen to the position of equilibrium.

Short Answer

Expert verified
(a) \( K \approx 0.253 \); (b) Reaction favors reactants; (c) Equilibrium shifts towards products.

Step by step solution

01

Understand the Relationship between Gibbs Free Energy and Equilibrium Constant

We know that the standard Gibbs free energy change \( \Delta_r G^0 \) is related to the equilibrium constant \( K \) by the formula:\[ \Delta_r G^0 = -RT \ln K \]where:- \( R \) is the gas constant \(8.314 \text{ J/mol K} \)- \( T \) is the temperature in Kelvin.
02

Calculate the Equilibrium Constant

Rearrange the formula to solve for \( K \):\[ K = e^{-\Delta_r G^0 / RT} \]Substitute the values:- \( \Delta_r G^0 = 3400 \text{ J/mol} \) (converted from kJ)- \( R = 8.314 \text{ J/mol K} \)- \( T = 298.15 \text{ K} \)\[ K = e^{-3400 / (8.314 \times 298.15)} \approx e^{-1.372} \approx 0.253 \]
03

Determine the Favorability of the Reaction

Since \( K < 1 \), the reaction at equilibrium favors the reactants over the products. Recall that if \( K > 1 \), the reaction favors products, and if \( K < 1 \), it favors reactants.
04

Apply Le Chatelier's Principle to Added Reactants

Le Chatelier's Principle predicts that if more \( \text{H}_2 \) is added, the system will shift to counteract the change. Hence, the equilibrium will shift towards the products to reduce the concentration of the added \( \text{H}_2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs Free Energy is a crucial concept in the study of chemical reactions and their spontaneity. It is symbolized as \( \Delta G \) and is a measure that predicts whether a process will occur spontaneously at constant pressure and temperature.
  • If \( \Delta G < 0 \), the reaction proceeds spontaneously in the forward direction.
  • If \( \Delta G > 0 \), the reaction is non-spontaneous in the forward direction and may proceed in reverse.
  • If \( \Delta G = 0 \), the system is at equilibrium, meaning there's no net change in the concentrations of reactants and products.
In the context of equilibrium constant calculations, the standard change in Gibbs free energy (\( \Delta_r G^0 \)) establishes the link between thermodynamics and equilibrium via the formula:\[ \Delta_r G^0 = -RT \ln K \]where \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( K \) is the equilibrium constant. This relationship helps chemists understand how temperature and energy changes influence equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle offers a simple yet powerful way to predict the response of a system in chemical equilibrium to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift in a direction that counteracts the change. This principle can be applied as follows:
  • Concentration: Adding more of a reactant or product will shift the equilibrium to favor the opposite side to balance the increased concentration.
  • Temperature: For exothermic reactions, increasing temperature shifts equilibrium towards the reactants, while for endothermic reactions, it shifts towards the products.
  • Pressure: Increasing pressure favors the side with fewer gas molecules, while decreasing pressure favors the side with more moles of gas.
In the original exercise, when \( \text{H}_2 \) gas was added, Le Chatelier's Principle predicts that the system will shift towards the products to relieve the disturbance, thus increasing the formation of \( \text{HI} \) molecules.
Reaction Favorability
The favorability of a chemical reaction is linked to both the equilibrium constant and the Gibbs free energy change. The equilibrium constant \( K \) offers insight into the extent of the reaction at equilibrium:
  • If \( K > 1 \): The equilibrium position is biased towards products, meaning the forward reaction is favored.
  • If \( K < 1 \): The equilibrium position is biased towards reactants, meaning the reverse reaction is favored.
  • If \( K = 1 \): Both reactants and products are equally favored.
In the exercise, the calculated \( K \approx 0.253 \) suggests that reactants are favored, indicating the reaction does not freely progress towards products. Hence, for the reaction \( \text{H}_2(\text{g}) + \text{I}_2(\text{s}) \rightleftharpoons 2\text{HI}(\text{g}) \), the reactants are more stable under the given conditions.
Chemical Equilibrium
Chemical equilibrium represents a state in which the rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of both reactants and products. This doesn't mean reactants and products are in equal concentrations; rather, it indicates a stable ratio dictated by the equilibrium constant \( K \).
  • Dynamic Nature: At equilibrium, reactions continue to occur, but because the rates are balanced, the macroscopic properties remain unchanged.
  • Macroscopic vs Microscopic: While macroscopically no changes seem to occur, at the molecular level, reactants and products continually interconvert.
  • Equilibrium Shifts: Changes in conditions, such as concentration, temperature, or pressure, can shift this balance, leading to a new state of equilibrium.
Understanding chemical equilibrium helps in managing reactions in industrial processes, ensuring optimal yields by controlling reaction conditions effectively.

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Most popular questions from this chapter

Calculate the maximum quantity (in mol) of \(\mathrm{KIO}_{3}\) that can be added to \(250 \mathrm{cm}^{3}\) of a solution containing \(1.00 \times 10^{-3} \mathrm{mol} \mathrm{dm}^{-3}\) of \(\mathrm{Cu}^{2}\) ' (aq) without precipitating \(\mathrm{Cu}(\mathrm{IO} 3)_{2}(\mathrm{s}) . K_{\mathrm{sp}}=1.4 \times 10^{-7}\) for \(\mathrm{Cu}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s})\).

If the following reaction was at equilibrium in a closed vessel at a controlled temperature, what would be the effect of adding more \(\mathrm{H}_{2}\) to the reaction vessel and permitting the reaction to approach equilibrium again? (Section 15.5 ) \\[ \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \\]

2.0 mol of carbon disulfide and \(4.0 \mathrm{mol}\) of chlorine react at constant temperature according to this equation \\[ \mathrm{CS}_{2}(\mathrm{g})+3 \mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{S}_{2} \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{CCl}_{4}(\mathrm{g}) \\] At equilibrium, \(0.30 \mathrm{mol}\) of tetrachloromethane are formed. How much of each of the other components is present in this equilibrium mixture? (Section 15.4 )

The equilibrium constants for two gas phase reactions at \(1000^{\circ} \mathrm{C}\) are shown. \\[ \begin{array}{l} \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad \mathrm{K}_{1}=9.1 \times 10^{-12} \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{2}=7.1 \times 10^{-12} \end{array} \\] Use these data to find the equilibrium constant at the same temperature for the reaction: \\[ \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\] (Section \(15.1)\)

An important reaction in the formation of smog is: \\[ \mathrm{O}_{3}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{O}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \\] Under certain conditions, the equilibrium constant for this reaction is \(K=6.0 \times 10^{34}\). If the partial pressures of each gas in the air over your home town were \(1.0 \times 10^{-6} \mathrm{bar} \mathrm{O}_{3}\) \(1.0 \times 10^{-5} \mathrm{bar} \mathrm{NO}, 2.5 \times 10^{-4} \mathrm{bar} \mathrm{NO}_{2},\) and \(8.2 \times 10^{-3} \mathrm{bar} \mathrm{O}_{2}\) what could you say about the course of the reaction as it moves to equilibrium? (Section 15.2 )

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