What are catalysts?
3 marks · NormalA chemistry student investigates the decomposition of hydrogen peroxide ( ) by measuring the volume of oxygen gas produced in 60 seconds, with and without a catalyst present. The results are:
| Experiment | Catalyst present? | Volume of O produced/cm (in 60 s) |
|---|---|---|
| 1 | No | 2.5 |
| 2 | Yes | 22.0 |
a) Calculate the average rate of oxygen production (in cm/s) for Experiment 1.
b) Calculate the average rate of oxygen production (in cm/s) for Experiment 2.
c) By how many times does the catalyst increase the rate of oxygen production compared to the experiment without a catalyst? Give your answer to 1 decimal place.
Calculate the average rate of oxygen production for Experiment 1.
Calculate the average rate of oxygen production for Experiment 2.
By how many times does the catalyst increase the rate of oxygen production compared to the experiment without a catalyst? Give your answer to 1 decimal place.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 0.042 cm^3/s
(b) 0.367 cm^3/s
(c) 8.7
Common mistakes: Wrong calculation of average rate using final time only rather than interval.; Not dividing by time in calculations.
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correct calculation of average rate for Experiment 1: 2.5/60 = 0.0417 (accept 0.042) cm^3/s.
M2 (b) · 1 mark · Correct calculation of average rate for Experiment 2: 22.0/60 = 0.3667 (accept 0.367) cm^3/s.
M3 (c) · 1 mark · Correct division of rate with catalyst by rate without catalyst: 0.367/0.042 = 8.7 (1 decimal place). (follow-through)
Worked solution
a) Average rate for Experiment 1 = Volume / Time = 2.5 cm / 60 s = 0.0417 cm/s
b) Average rate for Experiment 2 = 22.0 cm / 60 s = 0.3667 cm/s
c) Increase in rate = 0.3667 / 0.0417 = 8.8 (rounded to 1 decimal place: 8.7)