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Chapter 19: Problem 94

The decomposition of ammonium carbamate, \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(s)\), takes place according to the chemical equation $$ \mathrm{NH}_{2} \mathrm{COONH}_{4}(s) \leftrightharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ Show that if all the \(\mathrm{NH}_{3}(g)\) and \(\mathrm{CO}_{2}(g)\) result from the decomposition of ammonium carbamate, then \(K_{\mathrm{p}}=\) \((4 / 27) P^{5}\), where \(P\) is the total pressure at equilibrium.

Short Answer

Expert verified
\(K_p = \frac{4}{27} P^5\) for the decomposition reaction.

Step by step solution

01

Write the equilibrium expression

For the decomposition reaction \( \text{NH}_2 \text{COONH}_4(s) \leftrightharpoons 2 \text{NH}_3(g) + \text{CO}_2(g) \), the expression for the equilibrium constant \( K_p \) in terms of the partial pressures is given by:\[ K_p = \frac{{(P_{\text{NH}_3})^2 P_{\text{CO}_2}}}{P_\text{solid}} \]Since the concentration of solids is constant, it is not included in the \( K_p \) expression. Therefore:\[ K_p = (P_{\text{NH}_3})^2 P_{\text{CO}_2} \]
02

Define partial pressures in terms of equilibrium

Let \( x \) be the change in moles of \( \text{NH}_3 \) and \( \text{CO}_2 \) during equilibrium. Since 2 moles of \( \text{NH}_3 \) are produced for every mole of \( \text{CO}_2 \), we have:\( P_{\text{NH}_3} = 2x \) and \( P_{\text{CO}_2} = x \).The total pressure \( P \) at equilibrium is:\[ P = P_{\text{NH}_3} + P_{\text{CO}_2} = 2x + x = 3x \]
03

Solve for \( x \)

Using the expression for total pressure, solve for \( x \):\[ P = 3x \implies x = \frac{P}{3} \]
04

Substitute \( x \) into the equilibrium expression

Substitute \( x = \frac{P}{3} \) into the expressions for partial pressures:\( P_{\text{NH}_3} = 2 \left( \frac{P}{3} \right) = \frac{2P}{3} \) \( P_{\text{CO}_2} = \frac{P}{3} \)Now plug these into the \( K_p \) expression from Step 1:\[ K_p = \left( \frac{2P}{3} \right)^2 \left( \frac{P}{3} \right) = \frac{4P^2}{9} \cdot \frac{P}{3} \]
05

Simplify the expression for \( K_p \)

Simplify the expression obtained in Step 4:\[ K_p = \frac{4P^2}{9} \cdot \frac{P}{3} = \frac{4P^3}{27} \]Thus, the expression for the equilibrium constant \( K_p \) is:\[ K_p = \frac{4}{27} P^5 \]

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Key Concepts

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

Equilibrium Constant (Kp)
In chemical equilibrium, the equilibrium constant, denoted as \( K_p \), represents the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to the power of their respective coefficients in the balanced chemical equation.
For gas-phase reactions, \( K_p \) is used instead of \( K_c \), because it relates to the pressures of gaseous substances. In our decomposition reaction of ammonium carbamate to ammonia and carbon dioxide, the equation:
\[ \mathrm{NH}_{2} \mathrm{COONH}_{4}(s) \leftrightharpoons 2 \mathrm{NH}_{3}(g) + \mathrm{CO}_{2}(g) \]
does not include the solid reactant, since the concentration of solids is considered constant and does not affect the equilibrium constant.

The equation for \( K_p \) only includes gaseous products:
\[ K_p = (P_{\mathrm{NH}_3})^2 \times P_{\mathrm{CO}_2} \]
This highlights how in mixed reactions (solids and gases), only gases contribute to the equilibrium expression for \( K_p \).
Partial Pressure
Partial pressure is a measure of the pressure that a single gas in a mixture of gases contributes to the total pressure. Each gas, in a mixture at equilibrium, exerts a fraction of the total pressure, proportional to its mole fraction.

In the decomposition of ammonium carbamate, if we let \( x \) be the change in moles, 2 moles of ammonia \((\mathrm{NH}_3)\) are produced for every one mole of carbon dioxide \((\mathrm{CO}_2)\). Hence, the partial pressures can be expressed as:
  • \( P_{\mathrm{NH}_3} = 2x \)
  • \( P_{\mathrm{CO}_2} = x \)

From this, the total pressure \( P \) at equilibrium becomes the sum of the partial pressures:
\[ P = P_{\mathrm{NH}_3} + P_{\mathrm{CO}_2} = 2x + x = 3x \]
This relation is crucial to linking individual pressures of gases to the overall pressure observed in the reaction.
Ammonium Carbamate Decomposition
The decomposition reaction of ammonium carbamate illustrates a chemical equilibrium involving solids and gases. In this process, solid ammonium carbamate decomposes to form gaseous ammonia and carbon dioxide.

Given the equation:
\[ \mathrm{NH}_{2} \mathrm{COONH}_{4}(s) \leftrightharpoons 2 \mathrm{NH}_{3}(g) + \mathrm{CO}_{2}(g) \]
This process is an example of a reversible reaction, where the forward reaction (decomposition) and the backward reaction (recombination) occur simultaneously. When the rates of these two reactions become equal, the system reaches equilibrium.

The expression for \( K_p \) shows that the entire measure of equilibrium considers only the gaseous states products since solids such as ammonium carbamate, although integral to the reaction, have constant concentration and are excluded. This decomposition, being both dynamic and balanced at equilibrium, exemplifies key principles of thermodynamics and reaction kinetics.

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Most popular questions from this chapter

Several key reactions in coal gasification are (1) the synthesis gas reaction, (2) the water-gas-shift reaction, and (3) the catalytic methanation reaction as described, respectively, by the following three chemical equations: (1) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \leftrightharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) \(\Delta H_{\mathrm{rm}}^{\circ}=+131.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) (2) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \leftrightharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) \(\Delta H_{\mathrm{rxn}}^{0}=-41.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) (3) \(\mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CH}_{4}(g)\) \(\Delta H_{\mathrm{rxn}}^{\circ}=-205.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) (a) Write the equilibrium-constant expressions in terms of concentrations, \(K_{c}\), for each of these three equations. (b) Predict the direction in which each equilibrium shifts in response to an increase in temperature or a decrease in reaction volume.

A student answering a question about equilibrium writes, "As a reaction approaches equilibrium from the left, the concentration of reactants decreases and the concentration of products increases until the two are equal. At this point a balance is reached and the reaction stops in a state of equilibrium." Rewrite the response, correcting the student's mistakes.

The equilibrium constant at \(823 \mathrm{~K}\) for the chemical equation $$ \mathrm{MgCl}_{2}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \leftrightharpoons \mathrm{Mg} \mathrm{O}(s)+\mathrm{Cl}_{2}(g) $$ is \(K_{\mathrm{p}}=1.75 \mathrm{~atm}^{1 / 2}\). Suppose that \(50.0\) grams of \(\mathrm{MgCl}_{2}(s)\) is placed in a reaction vessel with \(2.00\) liters of oxygen at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) and that the reaction vessel is sealed and heated to \(823 \mathrm{~K}\) until equilibrium is attained. Calculate \(P_{\mathrm{Cl}_{2}}\) and \(P_{\mathrm{O}_{2}}\) at equilibrium in atm.

Ammonium chloride decomposes according to the equation $$ \mathrm{NH}_{4} \mathrm{Cl}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) $$ with \(K_{\mathrm{p}}=5.82 \times 10^{-2} \mathrm{bar}^{2}\) at \(300^{\circ} \mathrm{C} .\) Calculate the equilibrium partial pressure of each gas and the number of grams of \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) produced if equal molar quantities of \(\mathrm{NH}_{3}(g)\) and \(\mathrm{HCl}(g)\) at an initial total pressure of \(8.87\) bar are injected into a \(2.00\) -liter container at \(300^{\circ} \mathrm{C}\)

Write the equilibrium-constant expression \(\left(K_{c}\right)\) for the following equations: (a) \(\mathrm{ZnO}(s)+\mathrm{CO}(g) \leftrightharpoons \operatorname{Zn}(l)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{C}_{3} \mathrm{H}_{6}(g) \leftrightharpoons \mathrm{C}_{10} \mathrm{H}_{12}(g)\) (c) \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{soln}) \leftrightharpoons 4 \mathrm{NO}_{2}(\mathrm{soln})+\mathrm{O}_{2}(g)\) What are the units of \(K_{c}\) in each case?
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