Average Rate of Change problem

[jahaddow]

delete

[LCKurtz]

When a certain reaction was allowed to run for t minutes, the amount of substance A(t) was recorded as:

t (min) 10 15 20 25 30 35 40
A(t) 26.5 36.5 44.8 52.1 ? ? ?

This is also attached to this post, if this is unreadable above


The average rate of change between 20 and 30 minutes was 1.23 mol/m
and between 0 and 40 mins was 0.9 mol/min.


What was the average rate of reaction between 30 and 40 mins?


Ok so I got given this problem, Average rate of change, but all I have managed to do is work out A at 30 mins, like this:

1.23 x 10 = 12.3
20 = 44.8
44.8+12.3= 57.1
30mins = 57.1



Don't write 20 = 44.8 when you mean A(20) = 44.8 and similarly for A(30).
Your calculation for A(30) is ok. Now use the average change from 0 to 40 to similarly calculate A(40). Then once you have A(40) and A(30) you can answer the question.

[Edit] When I read your 0 to 40 I was thinking your first reading was at t = 0, but it is at t = 10. My remarks apply only if you were actually given the average rate from 10 to 40, not 0 to 40.

[HallsofIvy]

You were told that "The average rate of change between 0 and 40 mins was 0.9 mol/min." So in 40 minutes the total change would have been ???

If you can assume the amount of A at 0 was A(0)= 0, then A(40) is just that total change.

[jahaddow]

soo, what do i do with the A(0)=0

[HallsofIvy]

You add the total change to the initial amount to get the final amount! That was why I said that if A(0)= 0, then the change is the final amount.