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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 25c
Chapter 15, Problem 25c

At 1000 K, 𝐾𝑝 = 1.85 for the reaction SO2(𝑔) + 12 O2(𝑔) β‡Œ SO3(𝑔) (c) What is the value of 𝐾𝑐 for the reaction in part (b)?

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1
Step 1: Understand the relationship between Kp and Kc. The relationship between Kp and Kc is given by the equation Kp = Kc(RT)^(Ξ”n), where R is the ideal gas constant (0.0821 LΒ·atm/KΒ·mol), T is the temperature in Kelvin, and Ξ”n is the change in moles of gas in the reaction.
Step 2: Calculate Ξ”n for the reaction. Ξ”n is calculated as the difference between the total moles of gaseous products and the total moles of gaseous reactants. In this case, Ξ”n = (moles of SO3) - (moles of SO2 + moles of O2) = 1 - (1 + 1/2) = -1/2.
Step 3: Substitute the known values into the equation. We know that Kp = 1.85, R = 0.0821 LΒ·atm/KΒ·mol, T = 1000 K, and Ξ”n = -1/2. Substituting these values into the equation gives us 1.85 = Kc(0.0821*1000)^(-1/2).
Step 4: Solve the equation for Kc. To isolate Kc on one side of the equation, we need to divide both sides by (0.0821*1000)^(-1/2). This gives us Kc = 1.85 / (0.0821*1000)^(-1/2).
Step 5: Calculate the value of Kc. Now you can calculate the value of Kc using a calculator. Remember to follow the order of operations (PEMDAS/BODMAS) when performing the calculation.

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Key Concepts

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

Equilibrium Constant (Kp and Kc)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium. Kp is used for gas-phase reactions and is expressed in terms of partial pressures, while Kc is used for reactions in solution and is expressed in terms of molar concentrations. The relationship between Kp and Kc is given by the equation Kp = Kc(RT)^(Ξ”n), where Ξ”n is the change in moles of gas.
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Relationship between Kp and Kc

The relationship between Kp and Kc is crucial for converting between the two constants. This relationship depends on the temperature and the change in the number of moles of gas during the reaction. Specifically, Kp can be calculated from Kc using the formula Kp = Kc(RT)^(Ξ”n), where R is the ideal gas constant and T is the temperature in Kelvin.
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Stoichiometry of the Reaction

Understanding the stoichiometry of the reaction is essential for determining Ξ”n, which is the difference in moles of gaseous products and reactants. In the given reaction, the stoichiometry indicates that for every mole of SO2 and 12 moles of O2, 1 mole of SO3 is produced. This information is necessary to calculate the change in moles of gas, which directly affects the conversion between Kp and Kc.
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Related Practice
Textbook Question

Consider the equilibrium N2(𝑔) + O2(𝑔) + Br2(𝑔) β‡Œ 2 NOBr(𝑔) Calculate the equilibrium constant 𝐾𝑝 for this reaction, given the following information at 298 K:

2 NO(𝑔) + Br2(𝑔) β‡Œ 2 NOBr(𝑔) 𝐾𝑐 = 2.02

NO(𝑔) β‡Œ N2(𝑔) + O2(𝑔) 𝐾𝑐 = 2.1Γ—1030

Textbook Question

Consider the following equilibrium: 2 H2(g) + S2(g) β‡Œ 2 H2S(g) Kc = 1.08 Γ— 107 at 700Β°C (c) Calculate the value of 𝐾𝑐 if you rewrote the equation H2(g) + 1/2 S2(g) β‡Œ H2S(g)

Textbook Question

The following equilibria were attained at 823 K:

CoO(s) + H2(g) β†’ Co(s) + H2O(g) Kc = 67

CoO(s) + CO(g) β†’ Co(s) + CO2(g) Kc = 490

Based on these equilibria, calculate the value of 𝐾𝑐 for H2(𝑔)+ CO2(𝑔) β‡Œ CO(𝑔) + H2O(𝑔) at 823 K.

Textbook Question

Consider the following equilibrium: 2 H2(g) + S2(g) β‡Œ 2 H2S(g) Kc = 1.08 Γ— 107 at 700Β°C (b) Does the equilibrium mixture contain mostly H2 and S2 or mostly H2S?

Textbook Question

Consider the following equilibrium, for which at 𝐾𝑝 = 0.0752 at 480Β°C: 2 Cl2(𝑔) + 2 H2O(𝑔) β‡Œ 4 HCl(𝑔) + O2(𝑔) (a) What is the value of 𝐾𝑝 for the reaction 4 HCl(𝑔) + O2(𝑔) β‡Œ 2 Cl2(𝑔) + 2 H2O(𝑔)?