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The K. for CO (g) + 3 H2 (g) = CH4 (g) + H2O (g) is 9.17 * 10-2. If a 3.5 L vessel is set up with 0.0155 mol CO, 3.50 * 10-3 mol H2 and 0.0100 mol CH4 and 7.60 * 10-4 mol H20, in which direction will this reaction proceed to reach equilibrium? What is the value of Q.? O left O right O at equilibrium Qc = number or fraction only

The K. for CO (g) + 3 H2 (g) = CH4 (g) + H2O (g) is 9.17 * 10-2. If a 3.5 L vessel is set up with 0.0155 mol CO, 3.50 * 10-3 mol H2 and 0.0100 mol CH4 and 7.60 * 10-4 mol H20, in which direction will this reaction proceed to reach equilibrium? What is the value of Q.? O left O right O at equilibrium Qc = number or fraction only

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 7QRT
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The Ke for CO (g) + 3 H2 (g) = CH4 (g) + H,O (g) is 9.17 * 10-2. If a 3.5 L vessel is set up with 0.0155 mol CO, 3.50 * 10-3 mol H2 and 0.0100 mol CH4 and 7.60 * 10-4 mol H20, in which direction will this reaction proceed to reach equilibrium? What is the value of Q.? left O right O at equilibrium Qc = number or fraction only
Transcribed Image Text:The Ke for CO (g) + 3 H2 (g) = CH4 (g) + H,O (g) is 9.17 * 10-2. If a 3.5 L vessel is set up with 0.0155 mol CO, 3.50 * 10-3 mol H2 and 0.0100 mol CH4 and 7.60 * 10-4 mol H20, in which direction will this reaction proceed to reach equilibrium? What is the value of Q.? left O right O at equilibrium Qc = number or fraction only
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