Question

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

Answer 1

it's from the half-life formula for first order process rate t1/2 = .693 / k

Answer 2

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</sub>
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.