2 HI(g) ----> H2(g) + I2(g)?

Question

100.0 mmoles of HI are initially in a 1.00 L container at 800 K After 900.0 seconds the concentration reduces to 0.0100 M.


What is the average rate of production of H2 in mmole / min ?

Answer 1

First you have to calculate the molarity of the HI initially. Knowing 100mmoles HI = 0.1 mole HI and that the volume of the containter is 1 liter, then the molarity of HI is:

0.1mole HI/ 1 liter = 0.1 M

You then calculate the amount of HI that is converted to H2 and I2 gas:

HI lost = HI original conc'n- HI final conc'n
HI lost = 0.1M - 0.01M
HI lost = 0.09M

From your original equation you know that every two moles HI gas will be converted to one mole H2 gas, so the concentration of H2 gas produced will be:

H2 conc'n = HI conc'n lost/2
H2 conc'n = 0.09M/2
H2 conc'n = 0.045M

You then calculate the moles H2 produced from the above knowing the original volume of the container has not changed. So the moles of H2 produced is:

H2 = 0.045M * 1L
H2 = 0.045 moles
H2 = 45 mmoles

You can then calculate the rate using the mmoles H2 calculated above and the time elapsed from your original question:

900 sec = 15 min

rate H2 produced = 45mmoles/15min = 3mmoles/min

Answer 2

Please stop publishing my locker combination. THANK-YOU!