half life for a first order reaction
t(1/2) = 0.693/k
can someone please tell me what the number 0.693 means?
thank u!! (:
2007-04-19 15:18:59 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
0.693 is essentially ln 2, the natural log of 2. This equation says that the time it takes for a first order reaction to be "halfway done" is equal to 0.693 divided by k, the rate constant of the reaction.
2007-04-19 15:22:37 · answer #1 · answered by ihatedecaf 3 · 0⤊ 0⤋
t(1/2) = Ln2/k
Ln2 is the natural logarithm of 2, which means to the base e. Natural logarithms come out of calculus, where *integral* (1/x)dx= Ln x. So if 1/x= 1/2, you get Ln2. People usually use logarithms (log) to the base 10, where Log2 = 0.30103. (You just have to look that up and accept it.) The conversion factor is Ln2 = 2.303Log2 = 0.693
2007-04-19 22:32:58 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋
Check out reference.
Simply stated, -0.693 is the natural log of 0.5.
It appears that your equation has rearranged it to be positive.
2007-04-19 22:39:46 · answer #3 · answered by G_U_C 4 · 0⤊ 0⤋
It is the natural logarithm of 2.
2007-04-19 22:25:59 · answer #4 · answered by Joe 5 · 0⤊ 0⤋
it's just a constant. It is approximately ln(2).
2007-04-19 22:22:26 · answer #5 · answered by egregg14 2 · 0⤊ 0⤋