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Hydrogen peroxide decomposes rapidly in the presence of a manganese(IV) oxide (MnO₂) catalyst - Leaving Cert Chemistry - Question 3 - 2005
Question 3
Hydrogen peroxide decomposes rapidly in the presence of a manganese(IV) oxide (MnO₂) catalyst. (a) Write a balanced equation for the decomposition of hydrogen perox... show full transcript
Worked Solution & Example Answer:Hydrogen peroxide decomposes rapidly in the presence of a manganese(IV) oxide (MnO₂) catalyst - Leaving Cert Chemistry - Question 3 - 2005
Step 1
Step 2
Draw a labelled diagram of an apparatus a student could assemble to measure the rate of decomposition of hydrogen peroxide in the presence of a manganese(IV) oxide (MnO₂) catalyst.
Answer
- Apparatus Setup: The diagram should include a conical flask with hydrogen peroxide and manganese(IV) oxide inside.
- Delivery Tube: Connect the flask to a gas collection system using a delivery tube.
- Start Timing: Use a stop clock or timer to ensure that the reaction starts at a known time.
- Gas Collection: Collect the oxygen produced in a graduated cylinder or gas syringe, ensuring it is level with the liquid in the flask.
Step 3
Which would you expect to have a greater average rate of reaction over the first minute of the reaction? Give a reason for your answer.
Answer
Using finely powdered manganese(IV) oxide is expected to yield a greater average rate of reaction. This is because the larger surface area of the powdered catalyst allows for more collisions between reactants, accelerating the reaction.
Step 4
Plot a graph to illustrate the volume of oxygen produced versus time.
Answer
- X-axis and Y-axis: Label the X-axis as 'Time (minutes)' and the Y-axis as 'Volume of O₂ (cm³)'.
- Data Points: Plot the points from the provided table:
- (0, 0)
- (1, 13.5)
- (2, 23.0)
- (3, 34.5)
- (4, 38.5)
- (5, 39.5)
- (6, 39.5)
- (7, 39.5)
- Connecting the Points: Draw a smooth curve through the points to illustrate the volume of oxygen produced over time.
Step 5
Determine the volume of oxygen produced during the first 2.5 minutes.
Answer
To find the volume of oxygen produced during the first 2.5 minutes, calculate:
Volume at 2 minutes (23.0 cm³) - Volume at 0 minutes (0 cm³) = 23.0 cm³
Then, find the volume produced in the remaining 0.5 minutes by estimating from the graph, which may be approximately 8.5 cm³. Therefore, total volume = 23.0 cm³ + 8.5 cm³ = 31.5 cm³.
Step 6
Determine the instantaneous rate of the reaction at 2.5 minutes.
Answer
The instantaneous rate of the reaction at 2.5 minutes can be calculated by determining the slope of the tangent line to the curve at that point on the graph. If we estimate from the graph, it might be around 5.5 cm³/min, depending on how steep the curve is at that point.
Step 7
What changes would you expect in the graph if the experiment were repeated using a solution of the same volume but exactly half the concentration?
Answer
If the experiment were repeated with half the concentration of hydrogen peroxide, we would expect the overall volume of oxygen produced to be lower, and the graph would likely show a slower initial rate of reaction. The maximum volume would be lower, and the shape of the curve may also be less steep during the first few minutes.