Question

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

Answer 1

The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide

Step-by-step explanation:

Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen using Lead (IV) oxide as a catalyst.

1. The catalyst surface area is directly proportional to the reaction rate

• So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.

2. Also, Removing lead (IV) oxide from the reaction mixture the reaction rate decreased because as the catalyst is removed.

3. Using 50 cm³ of hydrogen peroxide doesn't affect the rate because the concentration of the reactant doesn't change.

4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased

So, The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide