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September 30, 2014

September 30, 2014

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

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 7:18pmThat 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}

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