Question

the second order reaction 2C2H4--->C4H8 has half life of 1.51 min when C2H4= .25M. How lomg will it take the concentration of C2H4 to drop from .25M to .08M?

Answer 1

4.53 mins: It would take about 3 half lives for it to go from 0.25 to 0.08 and you would multiply 3 x 1.51 to get 4.53. Hope this was helpful :)

Answer 2

Is there a formula I can use

Answer 3

(Amount given/2) and you keep doing that until you get down to the amount that they ask for. It helps to write the amount left over each time you divide so you can keep track of the amount. When you get the desired amount, you count how many times you divided (# of half lives) and multiply it by the half life of the element.