Separation of Variables and Integrating over an Interval

[Mayhem]

TL;DR Summary

How integrating over an interval works for separation of variables

I was solving the van't Hoff equation over an interval $[T_1 , T_2]$:

The van't Hoff equation

$\frac{\mathrm{d} \ln K}{\mathrm{d} T} = \frac{\Delta_r H^{\circ}}{RT^2}$

which can be solved with separation of variables:

$d \ln K = \frac{\Delta_rH^\circ}{RT^2}dT$
$\Updownarrow$
$\int_{T_1}^{T_2} d \ln K = \int_{T_1}^{T_2} \frac{\Delta_rH^\circ}{RT^2}dT$
$\Updownarrow$
$\int_{T_1}^{T_2} d \ln K = \frac{\Delta_rH^\circ}{R} \int_{T_1}^{T_2}\frac{1}{T^2}dT$

Here is the confusing part. Intuitively, I want to evaluate the LHS over the integral, but I can see that this doesn't exactly work as $\ln K$ isn't a value of $T$, but rather a function of $T$. So we would write it as:

$\ln K_{T_2} - \ln {K_{T_1}} = -\frac{\Delta_rH^\circ}{R}\left (\frac{1}{T_2}+\frac{1}{T_1} \right )$

This is the result that my textbook writes, to the symbol, and I am wondering how I should interpret the LHS. Does this mean that $K_{T_1}$ and $K_{T_2}$ is some value dependent on the $T_1$ and $T_2$ respectively? I understand the RHS perfectly, I'm just a little confused as to how I should interpret the LHS. Bonus question: how do we kind of generalize this for these kinds of problems, where we separate the values and integrate over an interval?

 

[vela]

What you're actually doing is this:
\begin{align*}
\frac{\mathrm{d} \ln K}{\mathrm{d} T} &= \frac{\Delta_r H^{\circ}}{RT^2} \\
\int_{T_1}^{T_2} \frac{\mathrm{d} \ln K}{\mathrm{d} T}\,dT &= \int_{T_1}^{T_2} \frac{\Delta_r H^{\circ}}{RT^2}\,dT
\end{align*} Then you're making the substitution $\ln K = f(T)$ so that $d(\ln K) = f'(T)\,dT$. Then the LHS becomes
$$\int_{T_1}^{T_2} \frac{\mathrm{d} \ln K}{\mathrm{d} T}\,dT = \int_{T_1}^{T_2} f'(T)\,dT =\int_{f(T_1)}^{f(T_2)} d(\ln K).$$

 

[Mayhem]

What you're actually doing is this:
\begin{align*}
\frac{\mathrm{d} \ln K}{\mathrm{d} T} &= \frac{\Delta_r H^{\circ}}{RT^2} \\
\int_{T_1}^{T_2} \frac{\mathrm{d} \ln K}{\mathrm{d} T}\,dT &= \int_{T_1}^{T_2} \frac{\Delta_r H^{\circ}}{RT^2}\,dT
\end{align*} Then you're making the substitution $\ln K = f(T)$ so that $d(\ln K) = f'(T)\,dT$. Then the LHS becomes
$$\int_{T_1}^{T_2} \frac{\mathrm{d} \ln K}{\mathrm{d} T}\,dT = \int_{T_1}^{T_2} f'(T)\,dT =\int_{f(T_1)}^{f(T_2)} d(\ln K).$$

This is chemistry, where #K# is the equilibrium constant. Conceptually I should understand this as #K_T# meaning "the equilibrium constant at a temperature T" then? Seems like a very powerful equation in that case.

 

[vela]

Yes, that’s what it looks like.