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

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?

Verified step by step guidance
1
Determine the balanced chemical equation for the decomposition of cyclopropane to propene: C_3H_6 (g) \rightarrow C_3H_6 (g).
Use the relationship between K_c and K_p: K_p = K_c(RT)^{\Delta n}, where \Delta n is the change in moles of gas, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
Calculate \Delta n, the change in moles of gas, by subtracting the moles of gaseous reactants from the moles of gaseous products.
Substitute the values of K_c, R, T, and \Delta n into the equation K_p = K_c(RT)^{\Delta n} to find K_p.
Simplify the expression to solve for K_p.

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Key Concepts

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

Equilibrium Constant (Kc and Kp)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction. Kc refers to concentrations in molarity, while Kp refers to partial pressures for gaseous reactions. The relationship between Kc and Kp is given by the equation Kp = Kc(RT)^(Δn), where Δn is the change in moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
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Ideal Gas Law

The ideal gas law (PV = nRT) describes the behavior of ideal gases, relating pressure (P), volume (V), number of moles (n), the ideal gas constant (R), and temperature (T). This law is fundamental in calculating the partial pressures of gases in a reaction, which is essential for determining Kp from Kc. Understanding this law helps in converting between concentration and pressure terms in equilibrium expressions.
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Thermodynamic Relationships

Thermodynamic relationships connect various properties of a system, including temperature, pressure, and concentration. In the context of equilibrium constants, these relationships allow us to derive Kp from Kc using the ideal gas law and the concept of moles of gas. Recognizing how changes in temperature and pressure affect equilibrium can provide insights into the behavior of chemical systems.
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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). (b) What fraction of the F2 molecules dissociate at 1500 K?

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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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Textbook Question

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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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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