First-Order Reaction
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Theorem
Let a substance decompose spontaneously in a first-order reaction.
Let $x_0$ be a measure of the quantity of that substance at time $t = 0$.
Let the quantity of the substance that remains after time $t$ be $x$.
Then:
- $x = x_0 e^{-k t}$
where $k$ is the rate constant.
Proof 1
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
- $-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
\(\ds \int \dfrac {\d x} x\) | \(=\) | \(\ds \int - k \rd t\) | Solution to Separable Differential Equation | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ln x\) | \(=\) | \(\ds -k t + c\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds e^{- k t + c}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds e^c e^{- k t}\) |
At time $t = 0$ we have that $x = x_0$.
So:
- $x_0 = e^c e^0 = e^c$
and hence the result
- $x = x_0 e^{-k t}$
$\blacksquare$
Proof 2
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
- $-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
This is an instance of the Decay Equation, and has the solution:
- $x = x_0 e^{-k t}$
$\blacksquare$