First-Order Reaction

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:

$x = x_0 e^{-k t}$

$\blacksquare$

Also see

• Half-Life of Radioactive Substance