The following info was obtained:
[HI], M 0.535 0.268 0.134 6.70E-2
seconds 0 520 781 911
What is the average rate of disappearance of HI from t = 0 s to t = 520 s
The answer is 5.13E-4 M s-1 but I do not understand how to get this!!!Could someone please explain!! Thank you!!
Average rate of disappearance just means slope!
So...
(0.535-0.268)M / 0-520sec
=5.13 X 10^-4Ms^-1
Tada!
O ok I didn't know it was that easy!!! Thank you!
To calculate the average rate of disappearance of HI from t = 0 s to t = 520 s, you need to find the change in concentration of HI over that time period and divide it by the change in time.
In the provided information, the concentration of HI at t = 0 s is given as M = 0.535, and the concentration of HI at t = 520 s is given as M = 6.70E-2 (which is equivalent to 0.067).
To calculate the change in concentration, subtract the concentration at t = 0 s from the concentration at t = 520 s:
Change in concentration = 0.067 - 0.535 = -0.468 M
Next, find the change in time by subtracting the initial time from the final time:
Change in time = 520 s - 0 s = 520 s
Finally, divide the change in concentration by the change in time to get the average rate of disappearance:
Average rate of disappearance = (-0.468 M) / (520 s) ≈ -9.00E-4 M/s
However, it is important to note that the answer given, 5.13E-4 M/s, is a positive value. This suggests that the reaction is proceeding in the reverse direction, meaning HI is being formed instead of disappearing. It is possible that there may have been an error in the provided information or in the calculation.