is this right?
For first order kinetics:
ln[A) = -kt + ln[Ao]
ln[A) = - (6.29 x 10^-4 1/s)(1000. s) + ln (0.00100)
ln[A) = -0.629 - 6.91 = -7.54
[A) = e^-7.54 = 5.33 x 10^-4
The reaction
CH3- N≡C → CH3- C≡N
is a first-order reaction. At 230.3°C, k = 6.29 x 10^-4 s^- 1. If [CH3 -N≡ is 1.00 x 10^-3 initially,
C]
[CH3-N ≡ C] is __________ after 1.000 x 10^3 s.
A) 5.33 x 10^-4
B) 1.00 x 10^-6
C) 4.27 x 10^-3
D) 2.34 x 10^-4
E) 1.88 x 10^-3
I asked this question before and tried to solve it on my own but im confused on how to set it up. i know the answer is A. but i just don't know how to get there.
2 answers
Jes, I worked this problem for you yesterday. You don't go back and look at your posts? Here is how I did it (and Jack's looks ok, too).
ln(No/N) = kt
ln(1 x 10^-3/N) = 6.29 x 10^-4*1 x 10^3
Solve for N. I get 5.33 x 10^-4
ln(No/N) = kt
ln(1 x 10^-3/N) = 6.29 x 10^-4*1 x 10^3
Solve for N. I get 5.33 x 10^-4