Question

When 1 g of a cube of manganese (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Manganese oxide was not used up in the reaction. Based on this information, which of the following is likely to increase the rate formation of the products?

Using 2 grams of manganese oxide
Using 50 cm3 of hydrogen peroxide
Replacing the cube of manganese (IV) oxide with its powder
Removing manganese (IV) oxide from the reaction mixture

Answer 1

`\boxed{\text{Replacing the cube of manganese(IV) oxide with its powder}}`

Step-by-step explanation:

The MnO₂ is acting as a catalyst: it takes part in the reaction but can be recovered at the end.

A. Using a 2 g cube of MnO₂

The surface area of the catalyst would double, so the rate would probably double. Technically, this option is correct, but I don't think it is the one the questioner intended.

B. Using 50 cm³ of H₂O₂

No effect. The rate depends on how fast the H₂O₂ molecules can reach the surface of the catalyst, and the surface area hasn't changed.

C. Using powdered MnO₂

This would increase the rate enormously, because the powdered catalyst has a much greater surface.

D. Removing MnO₂

The rate would decrease to near-zero, because there is no catalyst.