Problem 5 The standard Gibbs free energy change is 2.60 for the reaction at 25.0° mol C: kJ H2(g) + I2(9) 2 2HI(g); AGn = + 2.60 rxn mol The reaction starts with equal molar amounts of H,(g) and I,(g) in a previously evacuated, constant volume vessel. What is the equilibrium mole fraction of HI? Enter your answers in the box provided with correct units and sig. figs.: The equilibrium mole fraction of HI is Answer:

Problem 5 The standard Gibbs free energy change is 2.60 for the reaction at 25.0° mol C: kJ H2(g) + I2(9) 2 2HI(g); AGn = + 2.60 rxn mol The reaction starts with equal molar amounts of H,(g) and I,(g) in a previously evacuated, constant volume vessel. What is the equilibrium mole fraction of HI? Enter your answers in the box provided with correct units and sig. figs.: The equilibrium mole fraction of HI is Answer:

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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◄ Search 5GE 5:20 PM Question 4 of 6 Kc 1000 11% Consider the equilibrium system described by the chemical reaction below. At equilibrium, a 5.0 L reaction vessel contained a mixture of 1.22 mol CH4, 2.79 mol H₂S, 0.630 mol CS₂, and 1.27 × 10-³ mol H₂ at 420.0 °C. What are the values of Kc and Kp for the reaction at this temperature? Complete Parts 1-2 before submitting your answer. CH₂(g) + 2 H₂S(g) = CS₂(l) + 4 H₂(g) 1 2 Based on the given data, set up the expression for Kc. Each reaction participant must be represented by one tile. Do not combine terms. Once the expression is constructed, solve for Kc. = Submit [1.22] [2.79] [0.630] [0.244] [0.558] [0.126] 2[2.79] [2.79]² 2[0.558] 4[2.54 x 10-41 [2.54 x 10-4]4 4[1.27 x 10-³]| Tap here or pull up for additional resources RESET [1.27 x 10-³] [2.54 x 10-4] [0.558]² [1.27 x 10-³]4