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If the rate of appearance of o2 is 0.0014 m/s, what is the rate of disappearance of h2o2

User Khoga
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Final answer:

In the decomposition of Hydrogen Peroxide into Water and Oxygen, the rate of disappearance of Hydrogen Peroxide is twice the rate of appearance of Oxygen. Thus, if Oxygen appears at a rate of 0.0014 m/s, Hydrogen Peroxide will disappear at a rate of 0.0028 m/s.

Step-by-step explanation:

The subject in question involves the reaction rate expressions for the decomposition of H₂O₂ (Hydrogen Peroxide) in a chemical reaction to form H₂O (Water) and O₂ (Oxygen). In this reaction, 2 molecules of Hydrogen Peroxide decompose to form 2 molecules of Water and 1 molecule of Oxygen, represented as: 2H₂O₂ -> 2H₂O + O₂.

Since the formation of 1 molecule of O₂ is linked with the decomposition of 2 molecules of H₂O₂, the rate of disappearance of H₂O₂ is twice the rate of appearance of O₂. Hence, if the rate of appearance of O₂ is 0.0014 m/s, the rate of disappearance of H₂O₂ will be 0.0014 * 2 = 0.0028 m/s.

To put this in perspective, for every 0.0028 moles of H₂O₂ that disappear per second, 0.0014 moles of O₂ appear. This demonstrates the stoichiometric relationships inherent in chemical reactions.

Learn more about Chemical Reaction Rates

User Savvas Kleanthous
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