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Chapter 10: Problem 19

Calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for each of the following reactions, using data in Appendix \(2 \mathrm{~A}\) : (a) the combustion of hydrogen: \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) the oxidation of carbon monoxide: \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{CO}_{2}(\mathrm{~g})\) (c) the decomposition of limestone: \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

Short Answer

Expert verified
Without having the actual Appendix 2A, it is impossible to calculate the equilibrium constants. The exercise would require the standard Gibbs Free Energy changes for the reactions at 25°C, which could be used to calculate the equilibrium constants using the relation K = exp(-\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\textătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextŭextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătext (RT/\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\textătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextĭextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextătextazătextătextătextătextânătextătextătextătextătextătextătextătextătextătext text ) / (8.314472 J/(mol·K)) and ΔG° in J/mol. Actual values from Appendix 2A are required to calculate K.

Step by step solution

01

Understand the Concept of Equilibrium Constant

The equilibrium constant, denoted as K, for a chemical reaction is the ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients at equilibrium. The general expression for a reaction aA + bB ⇌ cC + dD is K = ([C]^c [D]^d) / ([A]^a [B]^b).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry that occurs when the rates of the forward and reverse reactions are equal, leading to a constant concentration of reactants and products. The point at which this balance occurs is crucial because it tells us the composition of a reaction mixture at any given time.

Chemical reactions can reach equilibrium from either direction of the reaction, whether they start with reactants or products. In the context of a classroom exercise, understanding equilibrium involves being able to visualize and comprehend that dynamic balance is not a static situation, but a continuous process where reactants form products at the same rate that products revert to reactants.

Imagine a seesaw that moves back and forth, eventually settling into a stable position. That's similar to how chemical equilibrium works - the rates are balanced, but the participants, meaning the reactants and products, can still move.
Reaction Stoichiometry
Reaction stoichiometry is all about the quantitative relationship between reactants and products in a chemical reaction. It's the 'recipe' for the reaction, telling you how much of each reactant you need to produce a certain amount of product. In an equilibrium problem, like the one involving the combustion of hydrogen, understanding the stoichiometry is vital because it determines the coefficients that you use in your calculation of the equilibrium constant.

For example, in the combustion of hydrogen, the stoichiometry tells us that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water vapor. This ratio is essential when calculating the equilibrium constant because it directly affects the concentration terms used in the equilibrium expression. It's like following a cooking recipe - if you change the proportions of ingredients, the result won't be what you expect. So in chemical reactions, stoichiometry ensures that the ‘dishes’ or products turn out as planned.
Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the realm of chemistry and particularly in chemical equilibrium, thermodynamics helps to predict the spontaneity of reactions and the position of equilibrium.

It's crucial to understand that the equilibrium constant is deeply rooted in the thermodynamics of a reaction. It’s tied to the Gibbs free energy change, symbolized as \( \Delta G \), which tells us whether a process can occur spontaneously. A negative \( \Delta G \) means a reaction is spontaneous in the forward direction, while a positive value indicates a spontaneous reverse reaction.

When we calculate an equilibrium constant, we're essentially gauging the extent to which a reaction will proceed under a given set of conditions. It's like predicting whether pushing a boulder will make it roll down the hill - thermodynamics informs us about the potential outcome. Thus, by studying the thermodynamics of a chemical reaction, we can anticipate the behavior of reactants and products and know how favorable the reaction is at a given temperature.

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Most popular questions from this chapter

State whether reactants or products will be favored by an increase in the total pressure (resulting from compression) on each of the following equilibria. If there is no change, explain why that is so. (a) \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons \mathrm{H}_{2}\) (g) \(+\mathrm{CO}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(2 \mathrm{HD}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2}\) (g) (e) \(\mathrm{Cl}_{2}\) (g) \(\rightleftharpoons 2 \mathrm{Cl}(\mathrm{g})\)

If \(Q=1.0 \times 10^{40}\) for the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants or will it be at equilibrium?

At \(25^{\circ} \mathrm{C}, K=3.2 \times 10^{-34}\) for the reaction \(2 \mathrm{HCl}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\). If a \(1.0\) - \(\mathrm{L}\) reaction vessel is filled with \(\mathrm{HCl}\) at \(0.22\) bar, what are the equilibrium partial pressures of \(\mathrm{HCl}, \mathrm{H}_{2}\), and \(\mathrm{Cl}_{2}\) ?

In a gas-phase equilibrium mixture of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3}\), and \(\mathrm{Cl}_{2}\) at \(500 . \mathrm{K}, P_{\mathrm{SbCl}_{5}}=0.072\) bar and \(P_{\mathrm{SbCl}_{3}}=5.02 \mathrm{mbar}\). Calculate the equilibrium partial pressure of \(\mathrm{Cl}_{2}\), given that \(K=3.5 \times 10^{-4}\) for the reaction \(\mathrm{SbCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\).

if \(Q=1.0\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants or will it be at equilibrium?
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