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Chapter 15: Problem 54

At \(218^{\circ} \mathrm{C}, K_{c}=1.2 \times 10^{-4}\) for the equilibrium $$ \mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ Calculate the equilibrium concentrations of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) if a sample of solid \(\mathrm{NH}_{4} \mathrm{SH}\) is placed in a closed vessel at \(218^{\circ} \mathrm{C}\) and decomposes until equilibrium is reached.

Short Answer

Expert verified
The equilibrium concentrations of \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \) are both 0.011 M.

Step by step solution

01

Write the Expression for the Equilibrium Constant

For the reaction \( \mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) \), the equilibrium constant \( K_c \) is based only on the concentrations of gaseous products, because solids do not appear in the equilibrium expression. Therefore, the expression is: \[ K_c = [\mathrm{NH}_3][\mathrm{H}_2 \mathrm{S}] \]
02

Assume Initial Conditions for Reaction

Since we start with only the solid \( \mathrm{NH}_{4} \mathrm{SH} \), we assume the initial concentrations of \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \) are both zero. As the reaction proceeds to equilibrium, let \( x \) be the change in concentration of \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \) because for every mole of \( \mathrm{NH}_3 \) produced, there is a corresponding mole of \( \mathrm{H}_2 \mathrm{S} \) produced.
03

Set Up the Equilibrium Expression

At equilibrium, the concentration of \( \mathrm{NH}_{3} \) is \( x \) and the concentration of \( \mathrm{H}_{2} \mathrm{S} \) is also \( x \). Substitute these into the equilibrium expression: \[ K_c = x \cdot x = x^2 \] Thus, the equation to solve is \( x^2 = 1.2 \times 10^{-4} \).
04

Solve for \( x \)

To find \( x \), take the square root of both sides: \[ x = \sqrt{1.2 \times 10^{-4}} \] Calculate this to find \( x = 0.011 \).
05

State the Equilibrium Concentrations

Since \( x \) represents the equilibrium concentration of both \( \mathrm{NH}_{3} \) and \( \mathrm{H}_2 \mathrm{S} \), the equilibrium concentrations are \{ [\mathrm{NH}_3] = 0.011 \text{ M}, [\mathrm{H}_2 \mathrm{S}] = 0.011 \text{ M} \}.

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Key Concepts

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

Equilibrium Constant
In chemical reactions, the equilibrium constant (\( K_c \)) is a vital tool. It quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. Consider the reaction: \[\mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{H}_{2} \mathrm{~S}(g) \]. At a given temperature, \( K_c \) remains constant. Importantly, for reactions involving gases and solids, \( K_c \) only involves the concentrations of gaseous components.For the decomposition of ammonium hydrosulfide, the equilibrium constant expression (\( K_c \)) is: - \( K_c = [\mathrm{NH}_3][\mathrm{H}_2 \mathrm{S}] \)This formula highlights that only ammonia and hydrogen sulfide gases appear. Solids are not included!
This simplifies calculations because solids maintain constant density and do not change in concentration, making them irrelevant to the expression of the constant.
Concentration Calculations
To determine concentrations of substances at equilibrium, we assume initial conditions. The process involves tracking changes in concentration as the system approaches equilibrium. When solid \( \mathrm{NH}_{4} \mathrm{SH} \) decomposes, it starts with zero concentration of \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \).Changes in concentration are represented by a variable, usually \( x \), which represents the amount of product formed.
Given that one \( \mathrm{NH}_{3} \) and one \( \mathrm{H}_{2} \mathrm{S} \) form per decomposition of \( \mathrm{NH}_{4} \mathrm{SH} \), we have:- \( [\mathrm{NH}_3] = x \)- \( [\mathrm{H}_2 \mathrm{S}] = x \)At equilibrium, both concentrations are equal.
These equal changes reflect the stoichiometry of the reaction, showcasing that for each mole of \( \mathrm{NH}_{4} \mathrm{SH} \) decomposed, one mole each of \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \) is generated.
Equilibrium Expressions
Equilibrium expressions are central to understanding chemical reactions at balance. They involve writing a formula expressing the equilibrium constant (\( K_c \)) based on the concentrations of reactants and products. For gas-solid mixtures like \( \mathrm{NH}_{4} \mathrm{SH} \rightleftharpoons \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \), only gaseous components are included.The equilibrium expression for the given reaction becomes:- \( K_c = x^2 \)Where \( x \) is the change in concentration from initial zero values of gaseous products.
This quadratic equation indicates the delicate balance between products and the position of equilibrium. Consequently, solving \( x \) requires basic algebra: 1. Take the square root of both sides: \( x = \sqrt{1.2 \times 10^{-4}} \)2. Calculate to find the exact concentration at equilibrium.Ultimately, the equilibrium expression offers precise insights into how concentrations evolve until equilibrium is reached.

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Most popular questions from this chapter

Consider the equilibrium $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{array}{lr} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c}=2.0 \\ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c}=2.1 \times 10^{30} \end{array} $$

A flask is charged with \(152.0 \mathrm{kPa}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(101.3 \mathrm{kPa}\) \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C}\), and the following equilibrium is achieved: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(51.9 \mathrm{kPa}\). (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?(\mathbf{b})\) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\). The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .(\mathbf{a})\) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g)\) ?

For the reaction \(\mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g), K_{c}=310\) at \(140^{\circ} \mathrm{C}\). Suppose that \(1.00 \mathrm{~mol}\) IBr in a 5.00-L flask is allowed to reach equilibrium at \(140^{\circ} \mathrm{C}\). What are the equilibrium concentrations of \(\mathrm{IBr}, \mathrm{I}_{2},\) and \(\mathrm{Br}_{2}\) ?

Consider the following exothermic equilibrium (Boudouard reaction) $$ 2 \mathrm{CO}(g) \rightleftharpoons \mathrm{C}(s)+\mathrm{CO}_{2}(g) $$ How will each of the following changes affect an equilibrium mixture of the three gases: \((\mathbf{a})\) a catalyst is added to the mixture; \((\mathbf{b}) \mathrm{CO}_{2}(g)\) is added to the system; (c) \(\mathrm{CO}(g)\) is added from the system; \((\mathbf{d})\) the reaction mixture is heated; (e) the volume of the reaction vessel is doubled; (f) the total pressure of the system is increased by adding a noble gas?
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