IMAT 2014 Q47 [Hydrogen Peroxide]

A student carried out an experiment to find the rate of decomposition of hydrogen peroxide into water and oxygen gas. The student used 100 cm^3 of a 1M solution of hydrogen peroxide at 25°C and 1 atm pressure.

1 g of powdered MnO_2 as a catalyst was added and the solution was constantly stirred. The student measured the total volume of oxygen produced.

The procedure was repeated, but this time using 100 cm^3 of 2M hydrogen peroxide, under identical conditions.

Which option below shows the effect on (R) the rate of reaction, (V) the total volume of oxygen collected, by using the 2M solution compared to the 1M solution?

A. R: Doubled; V: Halved
B. R: No effect; V: Doubled
C. R: No effect; V: No effect
D. R: Doubled; V: Doubled
E. R: Doubled; V: No effect


By using the 2M solution, we must note that we have doubled the concentration of the reactant, hydrogen peroxide. We use the same volume, at a higher concentration. This should mean that our rate of reaction should also increase; since there will be more particles and therefore more successful collisions. (R) is doubled.

If we have doubled the concentration, but used the same volume of reactant, this will double the amount of particles which react. Therefore, (V) of oxygen produced is also doubled.

This makes our answer D.


If we double the concentration and double the volume at the same time, what will happen to R ? Will it still be increased ?

what matters is the ratio so in this case i believe R would stay the same. and increasing the concentration means keeping the volume same while adding more solute. so increasing both the concentration and the volume would change nothing.