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

Are you certain of 1.8kPa? I would expect about 108kPa

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)x 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.

To calculate the rate in kPa per second, we need to convert the rate from kPa per minute to kPa per second.

First, we need to find the rate in kPa per minute for 5 drops. Let's assume that the rate is proportional to the enzyme concentration.

From the data given, we can see that as the enzyme concentration increases, the rate also increases. Since the rate roughly doubles when the enzyme concentration doubles, we can assume that the rate is proportional to the enzyme concentration.

So, to find the rate for 5 drops, we can set up a proportion using the rate for 4 drops and the rate for 5 drops:

(4 drops rate) / (4 drops enzyme concentration) = (5 drops rate) / (5 drops enzyme concentration)

Simplifying this proportion, we have:

98.26 / 4 = (5 drops rate) / 5

Cross-multiplying, we get:

(5 drops rate) = (98.26 / 4) * 5

(5 drops rate) = 122.825

So, the rate for 5 drops is approximately 122.825 kPa per minute.

To convert this rate from kPa per minute to kPa per second, we divide by 60 (since there are 60 seconds in a minute):

(5 drops rate) / 60 = 122.825 / 60

(5 drops rate) ≈ 2.047 kPa per second

Therefore, the approximate rate for 5 drops is 2.047 kPa per second.

To calculate the rate in kPa/s, we need to convert the rate from kPa/minute to kPa/second. We have the following data:

Enzyme concentration (drops) Rate (kPa/minute)
1 10.23
2 44.98
3 59.36
4 98.26

To convert from minutes to seconds, we divide the rate by 60, since there are 60 seconds in a minute. Let's calculate the rates in kPa/s for each concentration:

For 1 drop:
Rate = 10.23 kPa/minute / 60 = 0.1705 kPa/s

For 2 drops:
Rate = 44.98 kPa/minute / 60 = 0.7497 kPa/s

For 3 drops:
Rate = 59.36 kPa/minute / 60 = 0.9893 kPa/s

For 4 drops:
Rate = 98.26 kPa/minute / 60 = 1.6377 kPa/s

So, we can see that as the concentration of the enzyme increases, the rate also increases. Now, let's predict the rate for 5 drops:

We observed that the rate roughly doubles when the concentration of the enzyme doubles. Therefore, if we increase the concentration from 4 drops to 5 drops, we would expect the rate to be approximately double the rate for 4 drops, which is 1.6377 kPa/s.

Doubling the rate, we get:
2 * 1.6377 kPa/s ≈ 3.2754 kPa/s

Therefore, we predict that the rate for 5 drops of enzyme concentration would be approximately 3.2754 kPa/s.

Note: These calculations are based on the assumption that the relationship between the enzyme concentration and the rate is linear. It is important to note that actual experimental data should be collected and analyzed to verify any predictions.