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Ch.15 - Chemical Equilibrium
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook
All textbooksMcMurry 8th EditionCh.15 - Chemical EquilibriumProblem 15
Chapter 15, Problem 15

The decomposition of solid ammonium carbamate, (NH4)2NH2CO2, to gaseous ammonia and carbon dioxide is an endothermic reaction: (NH4)2NH2CO2(s) ⇌ 2 NH3(g) + CO2(g). (a) When solid (NH4)2NH2CO2 is introduced into an evacuated flask at 25 °C, the total pressure of gas at equilibrium is 0.116 atm. What is the value of Kp at 25 °C?(b) Given that the decomposition reaction is at equilibrium, how would the following changes affect the total quantity of NH3 in the flask once equilibrium is reestablished? (i) Adding CO2 (ii) Removing CO2 (iii) Adding (NH4)2NH2CO2 (iv) Increasing the total volume (v) Adding neon (vi) Increasing the temperature

Verified step by step guidance
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Step 1: To find the equilibrium constant Kp for the reaction, start by writing the expression for Kp based on the balanced chemical equation: \( K_p = \frac{(P_{NH_3})^2 \cdot P_{CO_2}}{1} \). Since the reaction involves only gases, the pressure of the solid ammonium carbamate is not included in the expression.
Step 2: At equilibrium, the total pressure is given as 0.116 atm. Let the partial pressure of NH3 be \( P_{NH_3} \) and the partial pressure of CO2 be \( P_{CO_2} \). Since the stoichiometry of the reaction is 2:1 for NH3 to CO2, we can express the total pressure as \( P_{total} = 2P_{NH_3} + P_{CO_2} = 0.116 \) atm.
Step 3: Solve the system of equations to find \( P_{NH_3} \) and \( P_{CO_2} \). Use the stoichiometric relationship \( P_{CO_2} = P_{NH_3} \) to substitute into the total pressure equation: \( 2P_{NH_3} + P_{NH_3} = 0.116 \). This simplifies to \( 3P_{NH_3} = 0.116 \).
Step 4: For part (b), consider Le Chatelier's principle to predict the effect of changes on the equilibrium position: (i) Adding CO2 will shift the equilibrium to the left, decreasing NH3. (ii) Removing CO2 will shift the equilibrium to the right, increasing NH3. (iii) Adding more solid ammonium carbamate will not affect the equilibrium position as solids do not appear in the expression for Kp.
Step 5: Continuing with part (b): (iv) Increasing the total volume will decrease the pressure, shifting the equilibrium to the side with more moles of gas, which is the right side, increasing NH3. (v) Adding neon, an inert gas, will not affect the equilibrium position as it does not change the partial pressures of the reactants or products. (vi) Increasing the temperature will favor the endothermic direction, which is the decomposition, increasing NH3.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (Kp)

The equilibrium constant, Kp, is a measure of the ratio of the partial pressures of the products to the reactants at equilibrium for a gaseous reaction. It is calculated using the formula Kp = (P_NH3^2 * P_CO2) / (P_(NH4)2NH2CO2), where P represents the partial pressures of the gases involved. For endothermic reactions, an increase in temperature typically shifts the equilibrium to favor the formation of products, thus affecting Kp.
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Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to counteract the change and restore a new equilibrium. This principle helps predict how changes in concentration, pressure, or temperature will affect the position of equilibrium in a reaction, such as the addition or removal of reactants or products in the given decomposition reaction.
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Endothermic Reactions

Endothermic reactions absorb heat from their surroundings, resulting in a decrease in temperature of the environment. In the context of the given reaction, the decomposition of ammonium carbamate requires heat input, which means that increasing the temperature will favor the production of gaseous products (NH3 and CO2). Understanding the nature of endothermic reactions is crucial for predicting the effects of temperature changes on equilibrium.
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Related Practice
Textbook Question
For the general, single-step reaction A1g2 + B1g2 ∆ AB1g2, Kc = 4.5 * 10-6, which of the following statements is true? (LO 15.16) (a) Ea 1forward2 6 Ea 1reverse2 (b) The equilibrium mixture contains mostly products. (c) kr 7 kf (d) The reaction is exothermic.
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Textbook Question

Calculate the value of the equilibrium constant at 427 °C for the reaction

Na O1s2 + 1>2 O 1g2 ∆ Na O 1s2

given the following equilibrium constants at 427 °C.

Na2O1s2 ∆ 2 Na1l2 + 1>2 O21g2 Kc = 2 * 10-25 Na O 1s2 ∆ 2 Na1l2 + O 1g2 K = 5 * 10-29

Textbook Question

Calculate the value of the equilibrium constant for the reaction 4 NH31g2 + 3 O21g2 ∆ 2 N21g2 + 6 H2O1g2

given the following equilibrium constants at a certain temperature.

2 H21g2 + O21g2 ∆ 2 H2O1g2 Kc = 3.2 * 1081

N21g2 + 3 H21g2 ∆ 2 NH31g2 Kc = 3.5 * 108

Textbook Question

The reaction

2 PH31g2 + As21g2 ∆ 2 AsH31g2 + P21g2

has Kp = 2.9 * 10-5 at 873 K. At the same temperature,

what is Kp for each of the following reactions?

(a) 2 AsH31g2 + P21g2 ∆ 2 PH31g2 + As21g2

Textbook Question

If Kc = 7.5×10−9 at 1000 K for the reaction N2(g) + O2(g) → 2 NO(g), give the value of Kc at 1000 K for the reaction

(a) 2 NO(g) → N2(g) + O2(g)