Posted by biggie on .
In a first order decomposition reaction, 50% of a compound decomposes in 10.5 min. What is the rate constant of the reaction? How long does it take for 75% of the compound to decompose?
My professor did this problem in class:
k = .693 / 10.5 mins
= .066 min ^1
where did .693 come from?
Thanks

Chemistry 
biggie,
it's from the halflife formula for first order process rate t1/2 = .693 / k

Chemistry 
DBob222,
0.693 is the natural log of 2.
It comes from using the expression
ln(No/N) = kt
ln(100/50) = k*t_{1/2}
ln 2 = k*t<sub<1/2
k = ln2/t_{1/2} and
k = 9.693/t_{1/2}
The No/N I used above as 100 and 50 can be any number choose as long as N is 1/2 of No.