Asked by gaby

The following reaction is first order, C2H6==> 2CH3. If the rate constant is equal to 5.5 x 10-4 s^-1 at 1000 K, how long will it take for 0.35 mol of C2H6 in a 1.00 L container to decrease to 0.20 mol in the same container?

Thank you very much
Answered by DrBob222
ln(No/N) = kt
ln(0.35/0.20) = kt
Substitute for k and calculate t. The calculated time will be in seconds.
Answered by gaby
ln(No/N) = kt
ln(0.35/0.20) = kt
ln(0.35/0.20)=1000*t
ln(0.35/0.20)/(1000)=5.59

but that answer is not on my worksheet im i doing something wrong?
Answered by DrBob222
I'm not surprised. Why did you substitute 1000 for k? The problem states k = 5.5 x 10^-4 s^-1. You substituted 1000 Kelvin which is the temperature.
Answered by gaby
oh haha i didn't notice i feel dumb now
Answered by gaby
ok i got i now thank you very much
Answered by DrBob222
Chalk it up to "not focused" and work on focusing on the work at hand.
Answered by Alibaba
i am trying to find tis answer but I m getting different result
Answered by Anonymous
ln(No/N)
=ln(0.35/0.20)
= 0.559
0.559/5.5*10-4
=1017.48
=17mins