Asked by sammi
Question: what do you think will happen to the rate of reaction if the concentration of enzyme is increased to five drops?Predict what thte rate would be for 5 drops.
Answer:Roughly, the rate doubles when the concentration of enzyme doubles. Since the data are
somewhat linear, the rate is proportional to the concentration. At a concentration of 5 drops,the rate in the above experiment should be about 1.8 kPa/s
But I don't understand how the calculations work to 1.8 kpa PER SECOND. Ca someone show me. Here is the sample data,
Enzyme concentration Rate(kPa/MINUTE)
1 Drops 10.23
2 Drops 44.98
3 Drops 59.36
4 Drops 98.26
Answer:Roughly, the rate doubles when the concentration of enzyme doubles. Since the data are
somewhat linear, the rate is proportional to the concentration. At a concentration of 5 drops,the rate in the above experiment should be about 1.8 kPa/s
But I don't understand how the calculations work to 1.8 kpa PER SECOND. Ca someone show me. Here is the sample data,
Enzyme concentration Rate(kPa/MINUTE)
1 Drops 10.23
2 Drops 44.98
3 Drops 59.36
4 Drops 98.26
Answers
Answered by
bobpursley
Are you certain of 1.8kPa? I would expect about 108kPa
Answered by
DrBob222
Where do you get the "roughly doubles when the concn is doubled."
If I do 1 drop to 2 drops I come up with rate = k(A)<sup>x</sup> with x = VERY close to 2. 1 drop to 3 drops gives me x of 1.6 and 1 drop to 4 drops gives me x = 1.6.
If I do 1 drop to 2 drops I come up with rate = k(A)<sup>x</sup> with x = VERY close to 2. 1 drop to 3 drops gives me x of 1.6 and 1 drop to 4 drops gives me x = 1.6.
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