Undergrad Separation of Variables and Integrating over an Interval

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The discussion centers on solving the van't Hoff equation using separation of variables over an interval [T1, T2]. The left-hand side (LHS) of the equation represents the change in the natural logarithm of the equilibrium constant K between two temperatures, K_T1 and K_T2, indicating that K is temperature-dependent. The right-hand side (RHS) involves integrating the enthalpy change over the inverse of the temperature squared. There is clarification on interpreting the LHS, emphasizing that K_T1 and K_T2 are values dependent on their respective temperatures. Overall, the equation illustrates the relationship between temperature and the equilibrium constant, highlighting its significance in chemical thermodynamics.
Mayhem
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TL;DR
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?
 
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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
$$\int_{T_1}^{T_2} \frac{\mathrm{d} \ln K}{\mathrm{d} T}\,dT = \int_{T_1}^{T_2} f'(T)\,dT =\int_{f(T_1)}^{f(T_2)} d(\ln K).$$
 
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vela said:
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.
 
Yes, that’s what it looks like.
 
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