BE.180:FirstOrderDecay

First Order Decay (of anything)

Givens:

• A pile of some thing, X.
• A first-order chemical process by which X is destroyed (or transformed into something else).

Tasks:

• Compute amount of X remaining as a function of time.
• Compute amount of time until there is half as much X as there is now (this length of time is called the "half-life" of X or [math]\displaystyle{ t_{1/2} }[/math]).

Approach:

• Write differential equation for change in X over time.

[math]\displaystyle{ \frac{dX}{dt} = -k_d * [X] }[/math]

• Solve equation for [X] as a function of time, t.

[math]\displaystyle{ \frac{dX}{[X]} = -k_d * dt }[/math]

• Integrating from [math]\displaystyle{ X_{(t=0)} }[/math] to [math]\displaystyle{ X_{(t=t)} }[/math]

[math]\displaystyle{ [lnX]_{X_{(t=0)}}^{X_{(t=t)}} = [-k_d*t]_{(t=0)}^{(t=t)} }[/math]

• Solving at the limits produces...

[math]\displaystyle{ ln \Bigg( \frac{X_{(t=t)}}{X_{(t=0)}} \Bigg) = -k_d*t }[/math]

• Which provides a general analytical solution for X as a function of time, t

[math]\displaystyle{ X_{t=t} = X_{t=0} * e^{-k_d*t} }[/math]

• Now, note that at [math]\displaystyle{ t_{1/2} }[/math], [math]\displaystyle{ X_{(t=t)}/X_{(t=0)} = 0.5 }[/math] by definition. So we can substitute and get...

[math]\displaystyle{ ln(0.5) = -k_d*t_{1/2} }[/math]

• Which is the same as...

[math]\displaystyle{ 0.69 = k_d*t_{1/2} }[/math]