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Ch.15 - Chemical Equilibrium
Brown - Chemistry: The Central Science 15th Edition
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Not the one you use?Change textbook
All textbooksBrown 15th EditionCh.15 - Chemical EquilibriumProblem 89a
Chapter 15, Problem 89a

At 700 K, the equilibrium constant for the reaction CCl4(𝑔) β‡Œ C(𝑠) + 2 Cl2(𝑔) is 𝐾𝑝 = 0.76. A flask is charged with 2.00 atm of CCl4, which then reaches equilibrium at 700 K. (a) What fraction of the CCl4 is converted into C and Cl2?

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1
Start by writing the balanced chemical equation for the reaction: CCl4(g) β‡Œ C(s) + 2 Cl2(g). Note that C(s) is a solid and does not appear in the expression for the equilibrium constant Kp.
Express the equilibrium constant Kp in terms of the partial pressures of the gases involved. Since Kp = 0.76, we have: Kp = (P_Cl2)^2 / P_CCl4.
Define the change in pressure for the reaction. Let x be the change in pressure of CCl4 that is converted to products. At equilibrium, the pressure of CCl4 will be (2.00 - x) atm, and the pressure of Cl2 will be 2x atm.
Substitute the equilibrium pressures into the expression for Kp: 0.76 = (2x)^2 / (2.00 - x).
Solve the equation for x to find the change in pressure. The fraction of CCl4 converted is x / 2.00. This will give you the fraction of the initial CCl4 that is converted to C and Cl2 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 (Kp)

The equilibrium constant (Kp) is a dimensionless value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. For gaseous reactions, Kp is calculated using partial pressures. In this case, Kp = 0.76 indicates the relationship between the partial pressures of CCl4, C, and Cl2 when the system reaches equilibrium.
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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, which is essential for understanding how the reaction will shift when CCl4 is introduced into the flask.
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Stoichiometry of the Reaction

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. For the reaction CCl4(g) β‡Œ C(s) + 2 Cl2(g), the stoichiometry indicates that one mole of CCl4 produces one mole of solid carbon and two moles of chlorine gas. Understanding stoichiometry is crucial for calculating the fraction of CCl4 converted to products at equilibrium based on the initial concentration and the equilibrium constant.
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Related Practice
Textbook Question

Nitric oxide (NO) reacts readily with chlorine gas as follows: 2 NO(𝑔) + Cl2(𝑔) β‡Œ 2 NOCl(𝑔) At 700 K, the equilibrium constant 𝐾𝑝 for this reaction is 0.26. For each of the following mixtures at this temperature, indicate whether the mixture is at equilibrium, or, if not, whether it needs to produce more products or reactants to reach equilibrium. (b) 𝑃NO = 0.12atm, 𝑃Cl2 = 0.10atm, π‘ƒNOCl = 0.050atm

Textbook Question

The equilibrium constant constant 𝐾𝑐 for C(𝑠) + CO2(𝑔) β‡Œ 2 CO(𝑔) is 1.9 at 1000 K and 0.133 at 298 K. (a) If excess C is allowed to react with 25.0 g of CO2 in a 3.00-L vessel at 1000 K, how many grams of CO are produced? (b) If excess C is allowed to react with 25.0 g of CO2 in a 3.00-L vessel at 1000 K, how many grams of C are consumed?

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Textbook Question

Consider the hypothetical reaction A(𝑔) + 2 B(𝑔) β‡Œ 2 C(𝑔), for which 𝐾𝑐 = 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol/L of compound A present at equilibrium.

(d) The equation from part (c) is a cubic equation (one that has the form ax3 + bx2 + cx + d = 0). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of x that you specified in part (b). The point at which the cubic equation crosses the x-axis is the solution.

(e) From the plot in part (d), estimate the equilibrium concentrations of A, B, and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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Textbook Question

At 700 K, the equilibrium constant for the reaction CCl4(𝑔) β‡Œ C(𝑠) + 2 Cl2(𝑔) is 𝐾𝑝 = 0.76. A flask is charged with 2.00 atm of CCl4, which then reaches equilibrium at 700 K. (b) What are the partial pressures of CCl4 and Cl2 at equilibrium?

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