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
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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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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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