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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?


  • Chemistry -

    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*t1/2
    ln(2) = k*t1/2
    0.693 = k*t1/2 and
    k = 0.693/t1/2

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