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Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products. (a) \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g)\) (b) \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) (c) \(\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q)\) (d) \(\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Short Answer

Expert verified
\(K_{a} = \frac{[\mathrm{Cl}_{2}(g)]}{[\mathrm{PCl}_{5}(g)]}\), \(K_{b} = [\mathrm{O}_{2}(g)]^{3}\), \(K_{c} = \frac{[\mathrm{H}_{3}O^{+}][\mathrm{F}^{-}]}{[\mathrm{HF}]}\), \(K_{d} = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{3}]}\).

Step by step solution

01

Equilibrium Expression for (a)

In an equilibrium expression, the products are in the numerator and the reactants in the denominator with their coefficients as exponents. Only species in gaseous (g) and aqueous (aq) states are included. For the reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g)\), the equilibrium expression is \(K = \frac{[\mathrm{Cl}_{2}(g)]}{[\mathrm{PCl}_{5}(g)]}\), because \mathrm{PCl}_{3} is a liquid and is not included.
02

Equilibrium Expression for (b)

For the reaction \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\), solids \(\mathrm{KClO}_{3}\) and \(\mathrm{KCl}\) are not included in the equilibrium expression. Thus, the expression is \(K = [\mathrm{O}_{2}(g)]^{3}\).
03

Equilibrium Expression for (c)

The reaction \(\mathrm{HF}(aq)+\mathrm{H}_{2}O(l) \rightleftharpoons \mathrm{H}_{3}O^{+}(aq)+\mathrm{F}^{-}(aq)\) includes one liquid, \mathrm{H}_{2}O, which is not included in the expression. Therefore, the equilibrium constant is \(K = \frac{[\mathrm{H}_{3}O^{+}][\mathrm{F}^{-}]}{[\mathrm{HF}]}\).
04

Equilibrium Expression for (d)

In the reaction \(\mathrm{NH}_{3}(aq)+\mathrm{H}_{2}O(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(aq)+\mathrm{OH}^{-}(aq)\), the water is a liquid and is omitted from the expression. The equilibrium constant is given by \(K = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{3}]}\).

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is essential to grasp why certain chemical reactions do not proceed to completion and instead reach a state where the concentrations of reactants and products remain constant. At the beginning of a reaction, reactants are converted to products, leading to an increase in product concentration. However, as products accumulate, they can also start to react and form reactants in a reverse reaction.

Over time, the forward and reverse reactions occur at the same rate, meaning the concentrations of reactants and products no longer change. This state is known as chemical equilibrium. It's important to note that reaching equilibrium does not mean reactants and products are present in equal amounts – rather, their ratios do not change over time.

An equilibrium expression is written to quantify the equilibrium state, usually denoted by the equilibrium constant, K. This expression is tailored for each specific reaction and only involves the concentrations of gaseous and aqueous species, as these are the states of matter affected by changes in concentration. Solids and liquids are omitted because their concentrations remain constant under typical conditions.
Reaction Quotient
The reaction quotient, Q, is a concept closely related to the equilibrium constant. It is calculated in the same way as the equilibrium constant, using the same formula, but with the initial concentrations of the reactants and products, not the equilibrium concentrations.

Using the formula for Q can provide insight into which direction a reaction will proceed to reach equilibrium. If Q is less than K, the forward reaction is favored, and the system will produce more products. Conversely, if Q is greater than K, the reverse reaction is favored, and the system will consume products to form more reactants. When Q equals K, the system is at equilibrium.

Understanding the reaction quotient helps to predict how a reaction will respond to changes in conditions, such as concentration, pressure, or temperature. Knowing whether a system is at equilibrium can also help you decide which conditions to alter to move the reaction in a desired direction.
Le Chatelier's Principle
Le Chatelier's principle provides insight into how a chemical system at equilibrium responds to external changes. It states that if an external stress is applied to a system at equilibrium, the system will adjust itself in such a way as to counteract the imposed change and re-establish equilibrium.

For example, if the concentration of a reactant is increased, the system will respond by favoring the reaction that consumes that reactant, producing more products until a new equilibrium is established. Similarly, if the temperature of an exothermic reaction is increased, the system can be thought of as having an increase in product (heat). The reaction will shift toward the reactants to absorb the extra heat.

Le Chatelier's principle is a critical tool in chemical engineering and various industries that rely on chemical reactions because it allows for the prediction and control of reactions to optimize yields and processes. Knowing how to manipulate conditions to favor the production of certain products can make chemical processes more efficient and cost-effective.

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Most popular questions from this chapter

The solubility of nickel(II) carbonate at \(25^{\circ} \mathrm{C}\) is \(0.042 \mathrm{~g} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for nickel(II) carbonate.

Describe three ways a reaction at equilibrium can be changed such that it is no longer in equilibrium. For each case, indicate which way the reaction will shift to return to equilibrium.

This reaction is endothermic. $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Predict the effect (shift right, shift left, or no effect) of these changes. (a) increasing the reaction temperature (b) decreasing the reaction temperature

This reaction is exothermic. $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g) $$ If you were a chemist trying to maximize the amount of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) produced, which of the following might you try? Assume that the reaction mixture reaches equilibrium. (a) increasing the reaction volume (b) removing \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) from the reaction mixture as it forms (c) lowering the reaction temperature (d) adding \(\mathrm{Cl}_{2}\)

This reaction is endothermic. $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{I}_{2}(g) $$ If you were a chemist trying to maximize the amount of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{I}_{2}\) produced, which of the following might you try? Assume that the reaction mixture reaches equilibrium. (a) decreasing the reaction volume (b) removing \(\mathrm{I}_{2}\) from the reaction mixture (c) raising the reaction temperature (d) adding \(\mathrm{C}_{2} \mathrm{H}_{4}\) to the reaction mixture

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