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reaction vessel mol ¹12, mol 12, and 3.91 mol e at 493°C for the reaction of hydrogen and iodine to give hydrogen iodide. The equation is H₂(g) + I2(g) → 2HI(g) c = Assuming that the substances are at equilibrium, find the value

reaction vessel mol ¹12, mol 12, and 3.91 mol e at 493°C for the reaction of hydrogen and iodine to give hydrogen iodide. The equation is H₂(g) + I2(g) → 2HI(g) c = Assuming that the substances are at equilibrium, find the value

Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.27PAE: The following data were collected for a system at equilibrium at 140°C. Calculate the equilibrium...
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A 13.9 L reaction vessel at 493°C contained 0.873 mol H₂, 0.357 mol I2, and 3.91 mol HI. Assuming that the substances are at equilibrium, find the value of Ke at 493°C for the reaction of hydrogen and iodine to give hydrogen iodide. The equation is H₂(g) + I2(g) → 2HI(g) Kc =
Transcribed Image Text:A 13.9 L reaction vessel at 493°C contained 0.873 mol H₂, 0.357 mol I2, and 3.91 mol HI. Assuming that the substances are at equilibrium, find the value of Ke at 493°C for the reaction of hydrogen and iodine to give hydrogen iodide. The equation is H₂(g) + I2(g) → 2HI(g) Kc =
Expert Solution
Step 1

Given ->

Volume=V = 13.9 L 

T = 493°C 

Mole of H2 = 0.873 mole 

Mole of I2 = 0.357 mole 

Mole of HI = 3.91 mole 

 

 

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