Biology rate of reaction question
The action of the enzyme catalase is shown below.
A student investigated the effect of hydrogen peroxide concentration on the rate of
this reaction. He used catalase from potato tissue.
The student:
• put five potato chips in a flask
• added 20 cm3 of 0.5 mol dm–3 hydrogen peroxide solution to the flask
• measured the time in seconds for production of 10 cm3 of oxygen gas
• repeated this procedure with four different concentrations of hydrogen
peroxide solution.
table is in q7.1 in this link:
https://pmt.physicsandmathstutor.com/download/Biology/A-level/Past-Papers/AQA/AS-Paper-1/June%202019%20QP%20(v1)%20%20Paper%201%20AQA%20Biology%20AS-level.PDF
7.3) The student gave the maximum rate of reaction a value of 1.0 arbitrary units.
Complete Table 5 by calculating the rate of reaction in arbitrary units at each
hydrogen peroxide concentration. Record the rates using an appropriate number of
significant figures.
Here is the mark scheme:
https://pmt.physicsandmathstutor.com/download/Biology/A-level/Past-Papers/AQA/AS-Paper-1/June%202019%20MS%20(v1)%20%20Paper%201%20AQA%20Biology%20AS-level.PDF
Why did they divide 6/time?
A student investigated the effect of hydrogen peroxide concentration on the rate of
this reaction. He used catalase from potato tissue.
The student:
• put five potato chips in a flask
• added 20 cm3 of 0.5 mol dm–3 hydrogen peroxide solution to the flask
• measured the time in seconds for production of 10 cm3 of oxygen gas
• repeated this procedure with four different concentrations of hydrogen
peroxide solution.
table is in q7.1 in this link:
https://pmt.physicsandmathstutor.com/download/Biology/A-level/Past-Papers/AQA/AS-Paper-1/June%202019%20QP%20(v1)%20%20Paper%201%20AQA%20Biology%20AS-level.PDF
7.3) The student gave the maximum rate of reaction a value of 1.0 arbitrary units.
Complete Table 5 by calculating the rate of reaction in arbitrary units at each
hydrogen peroxide concentration. Record the rates using an appropriate number of
significant figures.
Here is the mark scheme:
https://pmt.physicsandmathstutor.com/download/Biology/A-level/Past-Papers/AQA/AS-Paper-1/June%202019%20MS%20(v1)%20%20Paper%201%20AQA%20Biology%20AS-level.PDF
Why did they divide 6/time?
Reply 1
15
because in the previous table you had to solve, to get the Rate of reaction
/ arbitrary units and the and in 7.3 the maximum rate of reaction is the value of 1.0 arbitrary units. If u refer back to the table , 1.0 arbitrary units has the 'Time for production of 10 cm3 of oxygen gas / seconds' containing 1.0, by 6. So 6 will be mainly used throughout.
/ arbitrary units and the and in 7.3 the maximum rate of reaction is the value of 1.0 arbitrary units. If u refer back to the table , 1.0 arbitrary units has the 'Time for production of 10 cm3 of oxygen gas / seconds' containing 1.0, by 6. So 6 will be mainly used throughout.
Reply 2
5
Original post
by scrk009because in the previous table you had to solve, to get the Rate of reaction
/ arbitrary units and the and in 7.3 the maximum rate of reaction is the value of 1.0 arbitrary units. If u refer back to the table , 1.0 arbitrary units has the 'Time for production of 10 cm3 of oxygen gas / seconds' containing 1.0, by 6. So 6 will be mainly used throughout.
/ arbitrary units and the and in 7.3 the maximum rate of reaction is the value of 1.0 arbitrary units. If u refer back to the table , 1.0 arbitrary units has the 'Time for production of 10 cm3 of oxygen gas / seconds' containing 1.0, by 6. So 6 will be mainly used throughout.
Ah I see, I get it now thanks
Reply 3
Hi I still don't understand why 6 would be used
Reply 4
Same here I still cannot see why that would be the reason to divide the time by 6. Can someone break it down step by step please.
Reply 5
Original post
by befairhumanSame here I still cannot see why that would be the reason to divide the time by 6. Can someone break it down step by step please.
From the experimental data, the time to produce 10cm3 of oxygen gas plateaus at 6, i.e. last two rows in table; increasing hydrogen peroxide concentration does not decrease the time taken, therefore, the maximum reaction rate has been reached. The question says that the student gives the maximum reaction rate 1.0 arbitrary unit. So, there are now two ways to look at it. 1) you know that the rate in the last two rows must be 1.0, as this is where the maximum rate is. So, how do you get 1.0? 6/6 or 2) you now know the time for 10 cm3 of oxygen produced is 6s, this is when maximum rate of reaction is reached, so to get the rates for the others you need to see what fraction of 6 (time for maximum rate of reaction) they are.
It's just scaling to give an arbitrary unit.
Reply 6
Hey i just wanted to know why like you do 6/time instead of time/6, like how do you know you have to 6/time and not the other way instead I'm quite confused in that part?
Reply 7
It takes 6s to reach the maximum rate of reaction. You know this because when the concentration is increased, it still takes 6s to produce 10cm3 of oxygen (last two rows). Therefore, to find the rate in arbitrary units at other times, you need to know what fraction of the maximum rate they are, i.e. what fraction of 6 they are.
I also think of it as a rate is something occurring per unit time, i.e. 1/s, therefore the time would have to be on the denominator.
If you did time/6, you're asking how many times longer does that reaction take compared with the maximum rate of reaction. This is not what the question is asking (and not what is conventionally plotted on a graph when looking at reaction rates). Of course, this would just give you the reciprocal of what the question asked.
If you think about it, 1.0 is the maximum rate of reaction, so this can't increase (under these experimental conditions). So, for the slower reactions, the rate of reaction will have to be less than 1.0; else, they would have a rate of reaction greater than the maximum, which does not make sense....
Not sure if that helps.
I also think of it as a rate is something occurring per unit time, i.e. 1/s, therefore the time would have to be on the denominator.
If you did time/6, you're asking how many times longer does that reaction take compared with the maximum rate of reaction. This is not what the question is asking (and not what is conventionally plotted on a graph when looking at reaction rates). Of course, this would just give you the reciprocal of what the question asked.
If you think about it, 1.0 is the maximum rate of reaction, so this can't increase (under these experimental conditions). So, for the slower reactions, the rate of reaction will have to be less than 1.0; else, they would have a rate of reaction greater than the maximum, which does not make sense....
Not sure if that helps.
Reply 8
could you explain a bit clearer?
Reply 9
Original post
by Celeste23could you explain a bit clearer?
We know that the equation for rate of reaction is a fixed number(n) / time. So if we know the time is 6 seconds, the equation must be n/6. Then we are told that the maximum rate of reaction is 1.0 so n/6=1. Therefore we know that n must equal 6 which remains constant i throughout as 6/6=1
Reply 10
btw this question is absolutely terribly written, if anyone wants help, just reply
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