Hydrogen peroxide (H₂O₂) decomposes rapidly in the presence of a suitable catalyst, and releases oxygen (O₂) gas - Leaving Cert Chemistry - Question 3 - 2016
Question 3
Hydrogen peroxide (H₂O₂) decomposes rapidly in the presence of a suitable catalyst, and releases oxygen (O₂) gas. A student used a powdered catalyst and the apparatu... show full transcript
Worked Solution & Example Answer:Hydrogen peroxide (H₂O₂) decomposes rapidly in the presence of a suitable catalyst, and releases oxygen (O₂) gas - Leaving Cert Chemistry - Question 3 - 2016
Step 1
What is a catalyst?
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Answer
A catalyst is a substance that alters the rate (speed or activation energy) of a reaction without being consumed in the process.
For this reaction, suitable catalysts can include manganese(IV) oxide (MnO₂), potassium iodide (KI), or various biological catalysts like liver or celery.
Step 2
Complete and balance the equation for the reaction:
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The balanced equation for the decomposition of hydrogen peroxide is:
2H2O2→2H2O+O2
Step 3
On graph paper, plot a graph of the volume (y-axis) against time (x-axis).
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To graph the data:
Label the y-axis as 'Volume of Oxygen (cm³)' and the x-axis as 'Time (minutes)'.
Plot each point according to the table:
(0, 50)
(2, 69)
(4, 75)
(6, 78)
(9, 79)
(12, 79)
Connect the points smoothly, showing an increase followed by a plateau as the reaction approaches completion.
Step 4
Find from your graph the volume of oxygen produced during the first three minutes.
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From the graph, up to 3 minutes, the volume of oxygen produced can be estimated as 69 cm³.
Step 5
Explain whether the reaction slows down as it proceeds.
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As the reaction progresses, the rate decreases over time. Initially, the concentration of hydrogen peroxide (H₂O₂) is high, leading to a faster rate of reaction. However, as H₂O₂ gets consumed, there are fewer reactant molecules, resulting in a slower reaction rate as time passes. This indicates a decrease in the rate of production of O₂.
Step 6
Sketch on your graph the curve you would expect to obtain if the reaction were repeated with the conical flask cooled in ice-water.
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If the reaction was repeated with the conical flask cooled in ice-water, the curve would start similarly but would show a steeper gradient initially due to the increased solubility and decreased activation energy from the lower temperature. The plateau might be reached more slowly due to the reduced reaction rate at lower temperatures, effectively showing a higher volume of oxygen produced over a longer period compared to the original run.
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