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Write a chemical equation for an equilibrium system that would lead to each expression \((\mathrm{a}-\mathrm{c})\) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2} \mathrm{~S}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{2}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{\left(P_{\mathrm{IF}}\right)}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{Cl}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\)

Short Answer

Expert verified
(a) \(2\mathrm{H}_2\mathrm{~S(g)} + 3\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_2(g) + 2\mathrm{H}_2\mathrm{O(g)}\); (b) \(\mathrm{F}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2\mathrm{IF(g)}\); (c) \(\mathrm{Cl}_2(g) + 2 \mathrm{Br}^-(aq) \rightleftharpoons 2 \mathrm{Cl}^-(aq) + \mathrm{Br}_2(g)\).

Step by step solution

01

Understanding Equilibrium Expression (a)

The expression for \( K \) tells us the equilibrium expression based on products over reactants. For expression \( a \), \( K=\frac{(P_{\mathrm{H}_2 \mathrm{~S}})^2 (P_{\mathrm{O}_2})^3}{(P_{\mathrm{SO}_2})^2 (P_{\mathrm{H}_2 \mathrm{O}})^2} \), the \ substances in the numerator \((P_{\mathrm{H}_2 \mathrm{~S}})^2 (P_{\mathrm{O}_2})^3 \) are products and those in the denominator \((P_{\mathrm{SO}_2})^2 (P_{\mathrm{H}_2 \mathrm{O}})^2 \) are reactants.
02

Writing Equation for Expression (a)

Using the products \(2\mathrm{H}_2\mathrm{~S} + 3\mathrm{O}_2\) and reactants \(2\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O}\) from the equilibrium expression, the equation that fits this expression is: \[ \text{2 } \mathrm{H}_2\mathrm{S(g)} + \text{3 } \mathrm{O}_2(g) \rightleftharpoons \text{2 } \mathrm{SO}_2(g) + \text{2 } \mathrm{H}_2\mathrm{O(g)} \]
03

Understanding Equilibrium Expression (b)

For the expression \( b \), \( K=\frac{(P_{\mathrm{F}_2})^{1/2} (P_{\mathrm{I}_2})^{1/2}}{(P_{\mathrm{IF}})} \), the substances in the numerator \((P_{\mathrm{F}_2})^{1/2} (P_{\mathrm{I}_2})^{1/2} \) are products and those in the denominator \((P_{\mathrm{IF}})\) are reactants.
04

Writing Equation for Expression (b)

Assuming what matches simplest proportions, the equation based on the square roots is: \[ \frac{1}{2} \mathrm{F}_2(g) + \frac{1}{2} \mathrm{I}_2(g) \rightleftharpoons \mathrm{IF(g)} \], or simplified as: \[ \mathrm{F}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{IF(g)} \]
05

Understanding Equilibrium Expression (c)

For expression \( c \), \( K=\frac{[\mathrm{Cl}^-]^2}{(P_{\mathrm{Cl}_2}) [\mathrm{Br}^-]^2} \), chloride ions \([\mathrm{Cl}^-]^2\) are products, while \( \mathrm{Cl}_2 \) and bromide ions \([\mathrm{Br}^-]^2\) are reactants.
06

Writing Equation for Expression (c)

Matching the given equilibrium expression, the equation is: \[ \mathrm{Cl}_2(g) + 2 \mathrm{Br}^-(aq) \rightleftharpoons 2 \mathrm{Cl}^-(aq) + \mathrm{Br}_2(g) \]

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Key Concepts

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

Equilibrium Expressions
Equilibrium expressions provide a relationship between the concentrations of products and reactants in a balanced chemical reaction at equilibrium. These expressions are crucial for understanding how the equilibrium state is achieved and maintained. In these expressions, typically written as a ratio, the product concentrations or partial pressures are placed in the numerator, while the reactant concentrations or partial pressures are placed in the denominator. This setup reflects the equilibrium law stating that the ratio remains constant at a given temperature.

When dealing with gases, partial pressures are used, denoted by the letter \( P \). For aqueous solutions, concentrations, denoted by square brackets \( [ ] \), are used. Here are some characteristics:
  • The coefficients from the balanced equation appear as exponents in the expression.
  • Pure solids and liquids are not included in the expression.
  • The expression is dimensionless, meaning it provides a ratio without units.
Understanding equilibrium expressions is fundamental to predicting how changes in conditions can affect the system's equilibrium.
Chemical Reactions
Chemical reactions describe the process where substances, called reactants, are transformed into different substances, known as products. These processes can be reversible, leading to a state of equilibrium where reactants and products coexist. For example, combustion is an irreversible reaction, but many reactions in biological and chemical systems can reach equilibrium.

In the realm of equilibrium chemical reactions, these are particularly important:
  • The reaction progresses until dynamic equilibrium is achieved, where the forward and reverse reaction rates are equal.
  • At equilibrium, the concentrations of the reactants and products remain constant over time, although their concentrations do not have to be equal.
  • The system remains sensitive to external changes in concentration, pressure, and temperature, which can shift the equilibrium position according to Le Chatelier's Principle.
A balanced chemical equation is essential as it summarizes the stoichiometry of the reaction, providing a blueprint for calculating equilibrium expressions.
Equilibrium Constants
The equilibrium constant, denoted as \( K \), is a vital parameter in chemical reactions that quantify the ratio of product concentrations to reactant concentrations at equilibrium. The specific form of the equilibrium constant depends on the particular reaction being studied and the state of matter involved in the reaction:
  • For gaseous reactions, \( K_p \) uses partial pressures.
  • For reactions in solutions, \( K_c \) refers to molarity (concentration).
Equilibrium constants offer insight into the extent to which a reaction will proceed.

A large \( K \) value indicates that at equilibrium, products predominate over reactants, suggesting a forward-favoring reaction. Conversely, a small \( K \) suggests a reaction favoring reactants, with equilibrium lying far to the left. It is important to remember:
  • \( K \) is constant at a given temperature, underpinning its value in predicting reaction behavior.
  • Temperature changes can alter the equilibrium constant value, thus affecting the position of equilibrium.
Equilibrium constants serve as a quantitative measure for predicting how a mixture of reactants and products will evolve over time to reach equilibrium.

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Most popular questions from this chapter

For the equilibrium $$ \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{CoCl}_{4}^{2-}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) $$ pink blue \(K_{\mathrm{c}}\) is somewhat greater than 1 . If water is added to a blue solution of \(\mathrm{CoCl}_{4}^{2-}(\mathrm{aq}),\) the color changes from blue to pink. (a) Does water appear in the equilibrium constant expression for this reaction? (b) How can adding water shift the equilibrium to the left? (c) Is this shift in the equilibrium in accord with Le Chatelier's principle? Why or why not?

Discuss this statement: "No true chemical equilibrium can exist unless reactant molecules are constantly changing into product molecules, and vice versa."

Write equilibrium constant expressions, in terms of reactant and product concentrations, for each of these reactions. \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{O}_{2}(\mathrm{~g})\) $$ K_{c}=7 \times 10^{56} $$ \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) $$ K_{\mathrm{c}}=1.7 \times 10^{2} $$ $$ \begin{array}{ll} \mathrm{HCOO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{HCOOH}(\text { aq }) & K_{\mathrm{c}}=5.6 \times 10^{3} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{AgI}(\mathrm{s}) & K_{\mathrm{c}}=6.7 \times 10^{15} \end{array} $$ Assume that all gases and solutes have initial concentrations of \(1.0 \mathrm{~mol} / \mathrm{L}\). Then, let the first reactant in each reaction change its concentration by \(-x\). (a) Using the reaction table (ICE table) approach, write equilibrium constant expressions in terms of the unknown variable \(x\) for each reaction. (b) Which of these expressions yield quadratic equations? (c) How would you go about solving the others for \(x\) ?

At high temperature, hydrogen and carbon dioxide react to give water and carbon monoxide. $$ \mathrm{H}_{2 (\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) $$ Laboratory measurements at \(986^{\circ} \mathrm{C}\) show that there is \(0.11 \mathrm{~mol}\) each of \(\mathrm{CO}\) and water vapor and \(0.087 \mathrm{~mol}\) each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) at equilibrium in a sealed 1.0 - \(\mathrm{L}\) container. Calculate the equilibrium constant \(K_{\mathrm{p}}\) for the reac- $$ \text { tion at } 986^{\circ} \mathrm{C} \text { . } $$

Consider the system $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \\ \Delta_{\mathrm{r}} H^{\circ} &=-1530.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ (a) How will the amount of ammonia at equilibrium be affected by (i) removing \(\mathrm{O}_{2}(\mathrm{~g})\) without changing the total gas volume? (ii) adding \(\mathrm{N}_{2}(\mathrm{~g})\) without changing the total gas volume? (iii) adding water without changing the total gas volume? (iv) expanding the container? (v) increasing the temperature? (b) Which of these changes (i to v) increases the value of \(K ?\) Which decreases it?

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