Separation of Variables and Integrating over an Interval

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Discussion Overview

The discussion revolves around the application of the van't Hoff equation and the method of separation of variables in the context of integrating over an interval. Participants explore the interpretation of the left-hand side of the equation and the implications of the equilibrium constant as a function of temperature.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents the van't Hoff equation and expresses confusion about interpreting the left-hand side of the integral involving the natural logarithm of the equilibrium constant, ##\ln K##, as a function of temperature.
  • Another participant explains the integration process, suggesting that the left-hand side can be interpreted as an integral of the derivative of ##\ln K## with respect to temperature, leading to a substitution that clarifies the relationship between ##K## and ##T##.
  • A third participant reiterates the explanation and emphasizes the conceptual understanding of ##K_T## as the equilibrium constant at a specific temperature, indicating its significance in the context of the discussion.
  • A later reply confirms the interpretation of ##K_T##, suggesting agreement on the conceptual framework presented.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of the equilibrium constant as a function of temperature, but there is some uncertainty regarding the integration process and the implications of the left-hand side of the equation.

Contextual Notes

The discussion highlights the need for clarity in understanding the relationship between variables in the context of the van't Hoff equation, particularly in how the integration is approached and the assumptions involved in interpreting the results.

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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