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Chemistry 12 Online
OpenStudy (ashley1nonly):

This graph shows the concentration of the reactant A in the reaction A→B. Determine the average rate of the reaction between 0 and 10 seconds.

OpenStudy (ashley1nonly):

OpenStudy (ashley1nonly):

0.007 M/s 0.014 M/s 0.86 M/s 0.07 M/s

OpenStudy (ashley1nonly):

i know that this one is first order

OpenStudy (ashley1nonly):

rate=k[A]

OpenStudy (aaronq):

The average rate is simple, just add the individual rates and divide by the time

OpenStudy (ashley1nonly):

where do i get the rates

OpenStudy (ashley1nonly):

would it be (.86-1)/(10-0) which would give me .014

OpenStudy (aaronq):

these are the rates 0.007 M/s 0.014 M/s 0.86 M/s 0.07 M/s is there time associated with them? (i havent opened the file, i dont download word documents off the internet)

OpenStudy (ashley1nonly):

its just the graph from react concentration versus time for the first order reaction

OpenStudy (ashley1nonly):

yes there are time. so i add all of them up and divide by 10

OpenStudy (aaronq):

so, 0.07 M/s was the rate for how long?

OpenStudy (aaronq):

i mean, if you have 10 measurements (1 every second), then you divide by 10 seconds. But it doesn't seem like you have 10 measurements

OpenStudy (ashley1nonly):

.007 looks like for about 1 second

OpenStudy (aaronq):

hm what you can do in this scenario is take the end points, then just find the slope.|dw:1413133539485:dw|

OpenStudy (ashley1nonly):

so we could use .77-.86 /20-10

OpenStudy (aaronq):

well do you need the rate between 0 and 10 seconds or 10 to 20 seconds?

OpenStudy (ashley1nonly):

i think the answer is .86

OpenStudy (ashley1nonly):

( .86-1) /10-0 .14/10 =.014

OpenStudy (aaronq):

yep that looks right.

OpenStudy (ashley1nonly):

thank you once more and that was my last question

OpenStudy (ashley1nonly):

i got an 98 thank you so much and for the knowledge i can use on the test.

OpenStudy (aaronq):

great! no problem, glad i could help

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