Question

Problem:

In a first order decomposition reaction, 50% of a compound decomposes in 10.5 minutes. What is the rate constant of the reaction?

The solution my professor gave says:

k = .693 / 10.5min

Where did the .693 come from?

Thanks

Answer 1

That is the natural logarithm of 2. You can do it this way.
Let's start with 100 atoms of something.
No = 100. Since it is 50% decomposed in 10.5 min, then 10.5 is the half-life and N at the end of that time will be 50.
So ln(No/N) = kt
ln(100/50) = k*t_1/2
ln(2) = k*t_1/2
0.693 = k*t_1/2 and
k = 0.693/t_1/2