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

In a first order decomposition reaction, 50% of a compound decomposes in 10.5 min. What is the rate constant of the reaction? How long does it take for 75% of the compound to decompose?

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 12:46pm
it's from the half-life formula for first order process rate t1/2 = .693 / k

- Chemistry -
**DBob222**, Saturday, April 10, 2010 at 12:50pm
0.693 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.

## Answer This Question

## Related Questions

- Chemistry - What is the value of the rate constant for a reaction in which 75.0 ...
- Chemistry - What is the value of the rate constant for a reaction in which 75.0 ...
- Chemistry - Problem: In a first order decomposition reaction, 50% of a compound ...
- Chemistry - Time, (min): 0 10 20 30 Moles,(trans): 1.00 0.90 0.81 0.73 Given the...
- Chemistry - A reaction is first order and it takes 324 minutes for the reaction ...
- Half-Life - Can someone please show me a detailed solution to this problem? ...
- chemistry kinetics - Decomposition of an organic compound A follows first order ...
- Chemistry - a certain first order decomposition reaction has a half-life of 15.0...
- Chemistry - Dinitrogen pentoxide, N2O5, decomposes by a first-order reaction. ...
- chemistry - If the half-life of a first-order decomposition reaction is 24.2 ...

More Related Questions