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September 2, 2014

September 2, 2014

Posted by **biggie** on Saturday, April 10, 2010 at 12:43pm.

My professor did this problem in class:

k = .693 / 10.5 mins

= .066 min ^-1

where did .693 come from?

Thanks

- Chemistry -
**biggie**, Saturday, April 10, 2010 at 12:46pmit's from the half-life formula for first order process rate t1/2 = .693 / k

- Chemistry -
**DBob222**, Saturday, April 10, 2010 at 12:50pm0.693 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.

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