posted by Paul .
N2O5 dissolved in CCl4 decomposes to give N2O4 and O2. The reaction is first order with a half-life of 1234 seconds. How long will it take, in seconds, for the concentration of N2O5 to fall to 1.1% of its initial value?
Didn't I do this a couple of days ago for you?
k = 0.693/t1/2
Substitute and solve for k.
ln(No/N) = kt
Set any convenient number, such as 100, for No.
N, then, is 1.1% No. For No = 100, N = 1.10
k = from above.
t = solve for t in seconds.
Please do not answer this post. This question is a part of the MIT edx 3.091 solid state chemistry curriculum, and posting or using this information is in direct violation of the honor code. Thank you :)
Thank you DrBob222,
the fraction left after t seconds is (1/2)^(t/1234)
(1/2)^(t/1234) = 0.011
t = 8028.84