Hydrogen peroxide solution slowly decomposes to form water and oxygen. The graphs show the volume of oxygen collected against time for the reaction with and...

Assessment: Chemistry 0620 | Paper 3 Mock 01 | Theory (Core) Subject: Chemistry - 0620

Question 1 Report

Hydrogen peroxide solution slowly decomposes to form water and oxygen. The graphs show the volume of oxygen collected against time for the reaction with and without a catalyst.

diagram

(a) Write a word equation for the decomposition of hydrogen peroxide. [2]
(b) Name a catalyst commonly used for this decomposition. [1]
(c) State which curve, P or Q, is the catalysed reaction and give a reason. [2]
(d) Both curves reach the same final volume of oxygen. Explain why. [1]
(e) Give the meaning of the term catalyst. [2]
(f) Describe a test to show that the gas collected is oxygen. [2]

Answer Details

This question is about catalysis: the same decomposition run with and without a catalyst, followed by reading the rate graphs and identifying oxygen.

(a) Hydrogen peroxide breaks down into water and oxygen:

hydrogen peroxide \(\rightarrow\) water + oxygen

One mark for the reactant, one for both products. [2]

(b) A common catalyst for this decomposition is manganese(IV) oxide, MnO2. [1]

(c) Curve P is the catalysed reaction, because P is steeper and reaches its final volume in a shorter time; a catalyst speeds up the reaction. One mark for P, one for the reason. [2]

(d) Both curves reach the same final volume because the same amount of hydrogen peroxide decomposes in each, so the same volume of oxygen is made; the catalyst changes only the speed, not the amount. [1]

(e) A catalyst is a substance that increases the rate of a reaction but is not used up (is chemically unchanged) at the end. Both parts of the definition are needed for the two marks. [2]

(f) To confirm oxygen, place a glowing splint in the gas; if the gas is oxygen the splint relights. One mark for the test, one for the correct result. [2]

Exam takeaway: a catalyst never appears in the overall equation and never changes the yield; it only lowers the activation energy so the same products form faster.

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