H2o2+3i+2h -----> I3+2H2o
In the first 10.0 second of the reaction the concentration of I dropped from 1.000M to 0.868M
a) calculate the average rate of reaction for iodine in this time interval.
b) Determine the rate of change in the concentration of H during this time interval.
You really need to watch the caps. Is that I^- and H^+? If so, then
a.
(1.000-0.868)/(3*10)= ?
b. rate H^+ = rate I^- x (2/3) = ?
To calculate the average rate of reaction for iodine (I) in this time interval, we will use the equation:
Average rate of reaction = (change in concentration of I) / (change in time)
a) The change in concentration of I is calculated by subtracting the final concentration from the initial concentration:
change in concentration of I = final concentration - initial concentration = 0.868M - 1.000M = -0.132M
The change in time is given as 10.0 seconds. Now we can calculate the average rate of reaction for iodine:
Average rate of reaction for I = (-0.132M) / (10.0 seconds) = -0.0132M/s
Therefore, the average rate of reaction for iodine in this time interval is -0.0132M/s.
b) To determine the rate of change in the concentration of hydrogen (H), we need to analyze the balanced chemical equation:
H2O2 + 3I + 2H --> I3 + 2H2O
From the equation, we can see that for every 2 moles of H2O2 consumed, 2 moles of H2O are produced. This means that the change in concentration of H2O is equal to half the change in concentration of H2O2.
Since the change in concentration of I is given, we can find the change in concentration of H2O2:
change in concentration of H2O2 = (change in concentration of I) / 3 = -0.132M / 3 = -0.044M
The rate of change in the concentration of H2O2 is equal to the rate of change in the concentration of H2O. Therefore, the rate of change in the concentration of H during this time interval is -0.044M/s.
Note: It is important to consider stoichiometry when calculating the rate of change in the concentration of different species in a chemical reaction.