Posted by Help on Monday, March 15, 2010 at 6:17pm.
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

Chemistry  DrBob222, Monday, March 15, 2010 at 7:18pm
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 halflife 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}
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