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Chapter 15: Problem 30

Explain why we normally exclude solvents from liquid-phase reactions in equilibrium-constant expressions.

Short Answer

Expert verified
We normally exclude solvents from liquid-phase reactions in equilibrium constant expressions to simplify the expressions and minimize the impact of the solvent's concentration. This is because solvents are not consumed or produced in the reaction, and their concentration remains almost constant and much higher than that of the reactants and products, making their inclusion less concerning and more complex without providing additional information.

Step by step solution

01

Understanding Equilibrium Constant Expressions

Equilibrium constants are dimensionless numbers that describe the ratio of the concentration of products to the concentration of reactants at equilibrium. For a given balanced chemical equation, the equilibrium constant expression K is given by: \[K = \frac{[Product_1]^{n_1} \times [Product_2]^{n_2} \times ...}{[Reactant_1]^{m_1} \times [Reactant_2]^{m_2} \times ...}\] where [Product_i] and [Reactant_j] represent the equilibrium concentrations of products and reactants, respectively, and n_i and m_j are their corresponding stoichiometric coefficients in the balanced chemical equation.
02

Role of Solvents in Liquid-Phase Reactions

In liquid-phase reactions, the solvent is the substance that dissolves the reactants and products and provides the medium for the reaction to occur. The solvent plays a crucial role in facilitating the reaction but is not consumed in the reaction itself. In many cases, the solvent is present in large excess compared to the reactants and products, which means its concentration remains almost constant during the reaction.
03

Reason to Exclude Solvents from Equilibrium Constant Expressions

There are two main reasons to exclude solvents from equilibrium constant expressions: 1. Since the solvent is not consumed or produced in the reaction, its inclusion in the equilibrium constant expression can often lead to a more complex and less intuitive expression that doesn't provide any additional information. By excluding the solvent, we simplify the formulation of the equilibrium constant expression and make it easier to comprehend and analyze. 2. The concentration of the solvent in a liquid-phase reaction is typically much higher than that of the reactants and products and remains almost constant during the reaction. As a result, including the solvent in the equilibrium constant expression would have an insignificant effect on the overall value of K, as the change in the solvent's concentration would be negligible compared to that of the reactants and products. In conclusion, we exclude solvents from liquid-phase reactions in equilibrium constant expressions to simplify the expressions and minimize the impact of the solvent's concentration, which remains relatively constant and has a negligible effect on the equilibrium constant value.

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Key Concepts

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

Liquid-Phase Reactions
When dealing with liquid-phase reactions, it's important to understand how they differ from reactions in other phases. In liquid-phase reactions, substances are dissolved in a liquid medium, which allows molecules to interact and react with one another efficiently. The solvent, which is the liquid in which the reactants are dissolved, is not only the medium but also plays a significant role in the reaction dynamics.

One common example of such a reaction is the dissolution of salt in water, where water acts as the solvent and facilitates the dispersal and interaction of the sodium and chloride ions. However, despite the importance of solvents, they are commonly excluded from the equilibrium constant expressions for reasons including the vast excess in which they are present, leading to a relatively constant concentration, and the intent to keep formulas less complex and more focused on the changing concentrations of the reactants and products.
Solvent Role in Reactions
The role of a solvent in liquid-phase reactions extends far beyond just being the reaction's setting. Solvents can affect reaction rates, equilibria, and product distributions. They can either participate directly in the reaction or influence reactions through solvent-solute interactions.

For instance, polarity of the solvent can greatly influence a reaction's pathway — polar solvents tend to stabilize charged intermediates, thus altering the reaction mechanism and speed. Solvents can also impact the solubility of reactants, which is critical for the reaction to occur. Nevertheless, due to their generally high and stable concentration, solvents are omitted from equilibrium constant expressions, which primarily reflect the fluctuating concentration of reactants and products involved in the equilibrium.
Stoichiometry in Chemical Equations
Stoichiometry in chemical equations expresses the proportional relationship between reactants and products in a chemical reaction. It is intrinsically tied to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. Thus, stoichiometry ensures that the amount of each element is the same on both sides of the equation.

It's important to grasp that the stoichiometric coefficients represent the mole ratio in which substances react and form products. For equilibrium constant expressions, these coefficients transform into exponents, giving weight to the concentration terms based on their role in the balanced equation. Stoichiometry is fundamental not only for predicting reaction yields but also for understanding the conceptual framework that underpins the formulation of equilibrium constants, an essential component of chemical thermodynamics.

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Most popular questions from this chapter

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24{ }^{\circ} \mathrm{C}\). The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C}\) ?

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15\(), K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure 1 atm \()\) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction?

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) .\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

Write the equilibrium-constant expression for the equilibrium $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ The table that follows shows the relative mole percentages of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) at a total pressure of 1 atm for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic? Explain. $$\begin{array}{rll}\hline{\text { Temperature }\left({ }^{\circ} \mathrm{C}\right)} & \mathrm{CO}_{2}(\mathrm{~mol} \%) & \mathrm{CO}(\mathrm{mol} \%) \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\\1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\\\\hline\end{array}$$
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