# How does calculus relate to chemistry?

May 12, 2015

This is a pretty broad question, but I can give you two examples. Generally, calculus may relate to chemistry when you work with thermodynamics and kinetics.

One simple thermodynamics example is the idea of entropy, which is a measure of disorder in a system. A Physical Chemistry definition of entropy is:

$\mathrm{dS} = \frac{\delta {q}_{r e v}}{T}$ (Eq. A)
where S is entropy, q is heat flow, and T is temperature.

At a constant pressure and temperature (such as during a phase transition), it can be written like this:
$\mathrm{dS} = \frac{\delta {q}_{r e v , p}}{T} = \frac{\mathrm{dH}}{T}$ (Eq. B)
where H is enthalpy. Remember that we're at constant pressure. Now, if you take the partial derivative of entropy with respect to temperature at a constant pressure for Eq. B, you've got the General Chemistry definition of heat capacity---capacity for heat flow at a constant pressure! This is written as ${C}_{P} \left(T\right)$. You get:
${\left(\frac{\delta S}{\delta T}\right)}_{P} = {\left(\frac{\delta H}{\delta T}\right)}_{P} \frac{1}{T} = \frac{{C}_{P} \left(T\right)}{T}$ (Eq. C-1)

If you rewrite this further and take the integral, you get:
$\mathrm{dS} = \frac{{C}_{P} \left(T\right)}{T} \mathrm{dT}$ (Eq. C-2)
or
$\Delta S = S \left({T}_{2}\right) - S \left({T}_{1}\right) = {\int}_{{T}_{1}}^{{T}_{2}} \frac{{C}_{P} \left(T\right)}{T} \mathrm{dT}$ (Eq. D)

So if you know the specific heat capacity, then for example, you can figure out its entropy at $500 K$ if you know its entropy at room temperature ($298 K$) and you assume the heat capacity varies negligibly within the temperature range. The standard state entropy of reaction ($\Delta {S}_{R}^{o}$) can be used for this, for instance.

One more slightly involved example is when you derive half lives from rate laws in kinetics (you may see this in AP Chemistry, or in Physical Chemistry). So maybe you start with the following "classic" decomposition reaction:

${N}_{2} {O}_{4} \to 2 N {O}_{2}$

Since there is only one reactant, you know it is first order. You can write the rate law for this as:
$r \left(t\right) = \frac{1}{2} \frac{d \left[N {O}_{2}\right]}{\mathrm{dt}} = - \cancel{\frac{1}{1}} \frac{d \left[{N}_{2} {O}_{4}\right]}{\mathrm{dt}} = k \left[{N}_{2} {O}_{4}\right]$

If you rearrange this and separate variables, you get:
$- \frac{1}{\left[{N}_{2} {O}_{4}\right]} d \left[{N}_{2} {O}_{4}\right] = k \mathrm{dt}$

then if you integrate this:
$- {\int}_{{\left[{N}_{2} {O}_{4}\right]}_{0}}^{\left[{N}_{2} {O}_{4}\right]} \frac{1}{\left[{N}_{2} {O}_{4}\right]} d \left[{N}_{2} {O}_{4}\right] = {\int}_{{t}_{0}}^{t} k \mathrm{dt}$

you get this:
$- \ln \left(\frac{\left[{N}_{2} {O}_{4}\right]}{{\left[{N}_{2} {O}_{4}\right]}_{0}}\right) = k t - k \left(0\right) = k t$
where ${t}_{0}$ is 0 seconds.

So the half life is then:
$- \ln \left(\frac{{\left[{N}_{2} {O}_{4}\right]}_{\frac{1}{2}}}{{\left[{N}_{2} {O}_{4}\right]}_{0}}\right) = - \ln \left(\frac{\frac{1}{2}}{1}\right) = \ln 2 = k {t}_{\frac{1}{2}}$

${t}_{\frac{1}{2}} = \ln \frac{2}{k}$

So from this you would know that the half life for a substance that is decomposing by itself (such as radioactive decay, which is first order) is independent of its initial concentration.