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Ch.15 - Chemical Equilibrium
All textbooksMcMurry 8th EditionCh.15 - Chemical EquilibriumProblem 153c
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Chapter 15, Problem 153c

The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 * 105 at 500 K:
(c) Can you alter the ratio of the two concentrations at equilibrium by adding cyclopropane or by decreasing the volume of the container? Explain.

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Understand the reaction: The decomposition of cyclopropane (C3H6) to propene (C3H6) can be represented as C3H6 (cyclopropane) ightarrow C3H6 (propene).
Recognize the nature of the equilibrium constant (Kc): A large Kc value (1.0 * 10^5) indicates that at equilibrium, the concentration of products (propene) is much higher compared to the concentration of reactants (cyclopropane).
Consider the effect of adding more cyclopropane: Adding more cyclopropane increases the concentration of the reactant. According to Le Chatelier's Principle, the system will adjust to partially counteract this change by producing more propene, thus shifting the equilibrium to the right.
Consider the effect of decreasing the volume: Decreasing the volume of the container increases the pressure of the system. Since the number of moles of gas does not change in this reaction (1 mole of cyclopropane produces 1 mole of propene), the change in pressure does not favor either the forward or reverse reaction significantly.
Conclude the effects: Adding cyclopropane will shift the equilibrium towards the products (propene), increasing their ratio at equilibrium. However, decreasing the volume will not significantly alter the ratio of cyclopropane to propene at equilibrium.

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Key Concepts

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

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. For the reaction involving cyclopropane and propene, Kc indicates the extent to which the reaction favors products or reactants. A high Kc value, like 1.0 * 10^5, suggests that at equilibrium, the concentration of products (propene) is much greater than that of reactants (cyclopropane).
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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 itself to counteract the change and restore a new equilibrium. This principle can be applied to predict how the addition of cyclopropane or a change in volume will affect the concentrations of reactants and products at equilibrium. For instance, adding more cyclopropane will shift the equilibrium towards the products to reduce the concentration of the added reactant.
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Effect of Volume on Gas Equilibrium

In gas-phase reactions, changing the volume of the container affects the concentrations of the gases involved. Decreasing the volume increases the pressure, which can shift the equilibrium position depending on the number of moles of gas on each side of the reaction. For the decomposition of cyclopropane, if the total number of moles of gas increases upon decomposition, reducing the volume will shift the equilibrium towards the reactants, thereby altering the ratio of concentrations at equilibrium.
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Related Practice
Textbook Question

The F-F bond in F2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; Kp = 7.83 at 1500 K for the reaction F2(g) ⇌ 2 F(g). (c) Why is the F-F bond in F2 weaker than the Cl-Cl bond in Cl2?

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Open Question
When 0.500 mol of N2O4 is placed in a 4.00-L reaction vessel and heated at 400 K, 79.3% of the N2O4 decomposes to NO2. (a) Calculate Kc and Kp at 400 K for the reaction N2O4(g) ↔ 2 NO2(g).
Textbook Question

The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 ⨉105 at 500 K:

(a) What is the value of Kp at 500 K?

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Textbook Question

Acetic acid tends to form dimers, (CH3CO2H2), because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51⨉102 in benzene solution but only 3.7⨉10-2 in water solution. (a) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in benzene.

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Textbook Question

Acetic acid tends to form dimers, (CH3CO2H2), because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51⨉102 in benzene solution but only 3.7⨉10-2 in water solution. (b) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in water.

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Open Question
A 125.4 g quantity of water and an equal molar amount of carbon monoxide were placed in an empty 10.0-L vessel, and the mixture was heated to 700 K. At equilibrium, the partial pressure of CO was 9.80 atm. The reaction is CO(g) + H2O(g) ⇌ CO2(g) + H2(g). (a) What is the value of Kp at 700 K? (b) An additional 31.4 g of water was added to the reaction vessel, and a new state of equilibrium was achieved. What are the equilibrium partial pressures of each gas in the mixture? What is the concentration of H2 in molecules/cm³?