When 2 grams of powdered lead (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Lead (IV) oxide was not used up in the reaction. Based on the information, which of the following is likely to decrease the rate of formation of the products?

Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 50 cm3 of hydrogen peroxide
Using 1.0 gram of lead (IV) oxide

The large crystals have less surface area exposed to the other reagents than the powdered lead oxide. High surface area leads to a high rate of reaction thus the products are formed faster, while a low surface area leads to a lower rate of reaction since the reagent is less exposed to the other reagents.

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sarajevo sarajevo · Answer 2 · 2019-07-04T19:06:22+02:00

sarajevo sarajevo

04-07-2019

Answer:

Replacing the powdered lead oxide with its large crystals

Explanation:

Powdered lead (IV) oxide has larger surface area than its crystal form. Surface area increases the rate of the reaction. Larger the surface larger is the effective collision thereby faster reaction completion.

Using large of crystals of lead (IV) oxide is likely to decrease the rate of formation of the products.

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