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Chapter 16: Problem 42

Ammonium hydrogen sulfide decomposes on heating. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftarrows \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ If \(K_{\mathrm{p}}\) for this reaction is 0.11 at \(25^{\circ} \mathrm{C}\) (when the partial pressures are measured in atmospheres), what is the total pressure in the flask at equilibrium?

Short Answer

Expert verified
The total pressure at equilibrium is approximately 0.66 atm.

Step by step solution

01

Write the expression for equilibrium constant

For the decomposition of ammonium hydrogen sulfide, the equilibrium expression for pressure is \( K_p = P_{NH_3} \cdot P_{H_2S} \). This is because \( K_p \) is defined for gases in terms of partial pressures.
02

Assume initial and change in pressures

Assume the initial partial pressures of \( \mathrm{NH}_3 \) and \( \mathrm{H}_2S \) are zero, increasing by \( x \) at equilibrium. Thus, the equilibrium partial pressures can be written as \( P_{NH_3} = x \) and \( P_{H_2S} = x \).
03

Substitute equilibrium pressures into expression for Kp

Substitute the equilibrium pressures into the equilibrium constant expression: \( K_p = x \cdot x = x^2 \). This simplifies to \( x^2 = 0.11 \).
04

Solve for x

Solve for \( x \) by taking the square root of both sides: \( x = \sqrt{0.11} \). Calculating this gives \( x \approx 0.33 \).
05

Calculate total pressure

The total pressure \( P_{total} \) is the sum of the partial pressures of the gases present at equilibrium: \( P_{total} = P_{NH_3} + P_{H_2S} = x + x = 2x \). Use \( x \approx 0.33 \) to get \( P_{total} = 2 \times 0.33 = 0.66 \) atm.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as either \( K_c \) for concentrations or \( K_p \) for pressures, is a crucial factor in determining the state of a reaction at equilibrium. For reactions involving gases, we use \( K_p \) which is expressed using the partial pressures of the products and reactants. This constant is a reflection of the reaction's tendency to proceed to equilibrium from either side of the equation.

In our specific example with ammonium hydrogen sulfide decomposition, \( K_p \) is defined by the product of the pressures of the gases produced, \( P_{NH_3} \) and \( P_{H_2S} \). Hence, the formula is \( K_p = P_{NH_3} \cdot P_{H_2S} \).
  • The value of \( K_p \) tells us the position of equilibrium and how much product is formed relative to reactants.
  • A large \( K_p \) indicates a reaction that strongly favors products, whereas a small \( K_p \) suggests the reactants are favored.
  • In this exercise, the \( K_p \) value of 0.11 shows a weak preference towards product formation under the given conditions.
Decomposition Reaction
Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. These reactions are essential to various chemical processes in both natural and industrial settings.

For the reaction involving ammonium hydrogen sulfide (\( NH_4HS(s) \)), it decomposes upon heating to form ammonia (\( NH_3(g) \)) and hydrogen sulfide (\( H_2S(g) \)). This is a classic example of a decomposition reaction where a solid turns into two gaseous products.
  • These reactions often require energy, such as heat, to break the bonds in the original compound.
  • They are frequently carried out in closed systems to measure changes such as pressure more effectively.
  • In laboratory terms, understanding decomposition reactions is key as it gives an insight into how compounds can be broken down and how they contribute to the equilibrium.
Partial Pressures
Partial pressure is the pressure that a gas would exert if it were alone in the container. It's particularly important when dealing with mixtures of gases, where each gas contributes to the total pressure independently.

In this example, the decomposition of ammonium hydrogen sulfide produces two gases, ammonia and hydrogen sulfide, both starting from zero pressure. As these gases are produced, their partial pressures, \( P_{NH_3} \) and \( P_{H_2S} \), increase until they reach equilibrium.
  • Partial pressures are used in the calculation of \( K_p \) as it only applies to gaseous components.
  • The summation of partial pressures equates to the total pressure in the system at equilibrium.
  • Understanding partial pressures is crucial for predicting and explaining the behavior of gases in chemical reactions.
Le Chatelier's Principle
Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. This principle is a cornerstone in understanding how systems at equilibrium respond to external changes.

In the decomposition of ammonium hydrogen sulfide, one can consider how changes in pressure or temperature might affect the equilibrium. For instance, increasing the temperature may cause the equilibrium to shift towards the decomposition products (\( NH_3(g) \) and \( H_2S(g) \)), because the reaction is endothermic.
  • This principle helps predict how changes will affect the equilibrium state.
  • It is especially valuable in industrial chemical processes where control over the reaction balance is crucial.
  • By understanding Le Chatelier's Principle, one can manipulate conditions to favor product or reactant formation, depending on the desired outcome.

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Most popular questions from this chapter

Dinitrogen trioxide decomposes to \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in an endothermic process \(\left(\Delta_{\mathrm{r}} H^{\circ}=40.5 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\right)\) $$ \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) $$ Predict the effect of the following changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each of the following changes is made. (a) adding more \(\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g})\) (b) adding more \(\mathrm{NO}_{2}(\mathrm{g})\) (c) increasing the volume of the reaction flask (d) lowering the temperature

The reaction $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) $$ occurs at high temperatures. At \(700^{\circ} \mathrm{C},\) a \(2.0-\mathrm{L}\) flask contains 0.10 mol of \(\mathrm{CO}, 0.20 \mathrm{mol}\) of \(\mathrm{CO}_{2}\) and 0.40 mol of \(\mathrm{C}\) at equilibrium. (a) Calculate \(K_{c}\) for the reaction at \(700^{\circ} \mathrm{C}\) (b) Calculate \(K_{c}\) for the reaction, also at \(700^{\circ} \mathrm{C},\) if the amounts at equilibrium in the 2.0 -I. flask are 0.10 mol of \(\mathrm{CO}, 0.20 \mathrm{mol}\) of \(\mathrm{CO}_{2},\) and \(0.80 \mathrm{mol}\) of C. (c) Compare the results of (a) and (b). Does the quantity of carbon affect the value of \(K_{c} ?\) Explain.

The equilibrium constant, \(K_{c},\) for the following reaction is 1.05 at \(350 \mathrm{K}\) $$ 2 \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CCl}_{4}(\mathrm{g}) $$ If an equilibrium mixture of the three gases at \(350 \mathrm{K}\) contains \(0.0206 \mathrm{M} \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) and \(0.0163 \mathrm{M} \mathrm{CH}_{4},\) what is the equilibrium concentration of \(\mathrm{CCl}_{4} ?\)

Neither \(\mathrm{PbCl}_{2}\) nor \(\mathrm{PbF}_{2}\) is appreciably soluble in water. If solid \(\mathrm{PbCl}_{2}\) and solid \(\mathrm{PbF}_{2}\) are placed in equal amounts of water in separate beakers, in which beaker is the concentration of \(\mathrm{Pb}^{2+}\) greater? Equilibrium constants for these solids dissolving in water are as follows: \(\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \quad K=1.7 \times 10^{-5}\) \(\mathrm{PbF}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq}) \quad K=3.7 \times 10^{-8}\)

The equilibrium reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) has been thoroughly studied (Figure 16.8 ). (a) If the total pressure in a flask containing \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas at \(25^{\circ} \mathrm{C}\) is 1.50 atm and the value of \(K_{\mathrm{p}}\) at this temperature is \(0.148,\) what fraction of the \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to \(\mathrm{NO}_{2} ?\) (b) What happens to the fraction dissociated if the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm?
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