# Are enthalpy and entropy affected by temperature?

Jan 18, 2016

Yes. In fact, they are state functions of temperature.

$\textcolor{b l u e}{\Delta H \left(T\right) = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} \mathrm{dT}} = {C}_{P} \left({T}_{2} - {T}_{1}\right)$

$\textcolor{b l u e}{\Delta S \left(T\right) = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} / T \mathrm{dT}} = {C}_{P} \ln | {T}_{2} / {T}_{1} |$

If we assume that the constant-pressure heat capacity ${C}_{P}$ is constant (which is a pretty decent assumption for small temperature ranges), then we can pull it out of the integral and focus on integrating the temperature.

(Normally though, ${C}_{P} = {C}_{P} \left(T\right)$.)

From this we can say that enthalpy and entropy change proportionally to a change in temperature.

This is derived below.

The change in enthalpy with respect to temperature at a constant pressure is the definition of the constant-pressure heat capacity.

$\setminus m a t h b f \left({\left(\frac{\partial H}{\partial T}\right)}_{P} = \frac{\partial {q}_{p}}{\partial T} = {C}_{P}\right)$

where ${q}_{p}$ is heat flow at a constant pressure.

From this, we can write at a constant pressure:

$\mathrm{dH} = \partial {q}_{p} = {C}_{P} \mathrm{dT}$

Now integrate everything to get:

$\textcolor{g r e e n}{\Delta H = {q}_{p} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} \mathrm{dT}} = {C}_{P} \left({T}_{2} - {T}_{1}\right)$

Because of that relationship between ${q}_{p}$ and $\Delta H$, notice how we can relate two similar equations, provided the heat flow is reversible:

$\Delta H = {q}_{p , r e v}$

$\textcolor{g r e e n}{\Delta S = {q}_{p , r e v} / T}$

Using the above green relationships, we get:

$\textcolor{b l u e}{\Delta S = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} / T \mathrm{dT}}$

Finally, since both entropy and enthalpy are functions of temperature, we emphasize it by writing:

$\Delta H = \Delta H \left(T\right)$

$\Delta S = \Delta S \left(T\right)$