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Initial rate of reaction ratio- AQA as 2014 specimen paper 1

Hi,
I did 13.5/8=1.7 for 37*C and 10/1=10 for 60*C to get the ratio 10:1.7 to equal 0.17:1 as my final answer.

Now when I check the mark scheme the answer is 'between 4:1 and 5:1'. Can you help?

Thanks 🤓

filestore.aqa.org.uk › AQA-74021-SMS
Original post by MaddieRose99
Hi,
I did 13.5/8=1.7 for 37*C and 10/1=10 for 60*C to get the ratio 10:1.7 to equal 0.17:1 as my final answer.

Now when I check the mark scheme the answer is 'between 4:1 and 5:1'. Can you help?

Thanks 🤓

filestore.aqa.org.uk › AQA-74021-SMS


Hiya! What subject is this for? I'll move it into the specific forum and you'll be more likely to get some good discussion there :smile:
Original post by MaddieRose99
Hi,
I did 13.5/8=1.7 for 37*C and 10/1=10 for 60*C to get the ratio 10:1.7 to equal 0.17:1 as my final answer.

Now when I check the mark scheme the answer is 'between 4:1 and 5:1'. Can you help?

Thanks 🤓

filestore.aqa.org.uk › AQA-74021-SMS


Your accuracy is just a bit off for 37ºC: the tangent goes through (approximately) 5 on both axes, which is rate of reaction of 1.

For 60ºC the tangent goes through about 9-10 on the y axis and 2 on the x axis, which is 4.5-5 per second. So overall its 4.5-5:1. You probably put 1 second instead of 2.
I now understand the 60*C, but how did you get 5 for the Y axis for 37*C? Do you draw the tangent on the line or do you subtract 10 from 15 to find the difference?
Original post by MaddieRose99
I now understand the 60*C, but how did you get 5 for the Y axis for 37*C? Do you draw the tangent on the line or do you subtract 10 from 15 to find the difference?


Just draw a tangent on the curve and assuming you've done it right, it should go through (or close to) 5 on the Y axis. Then just find the gradient of the tangent (change in y / change in x). Because both axes start at 0, its just the Y value for a point on the line divided by the X value, and the obvious point to pick is (5, 5) (or close to that) because its easy to read off the graph
Original post by neon_reaper
Just draw a tangent on the curve and assuming you've done it right, it should go through (or close to) 5 on the Y axis. Then just find the gradient of the tangent (change in y / change in x). Because both axes start at 0, its just the Y value for a point on the line divided by the X value, and the obvious point to pick is (5, 5) (or close to that) because its easy to read off the graph


Ah, so you find the gradient, I understand now

Thank you for your help 😊

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