## anonymous 4 years ago Just curious...

1. anonymous

Does this equation have a name? If so, what is it? If not, is it just considered a derivation of the Arrhenius equation?$\ln (\frac{k_2}{k_1})=-\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$If it is a derivation of the Arrhenius equation, could you show me how it's done? Because this is the furthest I've gotten...$\ln(k)=\frac{-E_a}{RT}+\ln(A)$

2. JFraser

this is a manipulation of the arrhenius equation, and when you graph ln(k) vs. 1/T, you get a straight line with slope = -Ea/R and y-intercept of ln(A). $y = mx + b$$\ln(k) = (\frac{-E{_a}}{R})*(\frac{1}{T}) + \ln(A)$

3. JFraser

as far as I know, it doesn't have its own name, just makes turns the equation into a form more convenient for graphing

4. anonymous

One would assume you are calculating k at two different temperatures. Take your furthest point, evaluate it at T1 and T2, then subtract: ln k2 = -Ea/RT2 + lnA ln k1 = -Ea/RT1 + ln A subtract: ln k2 - ln k1 = -Ea/RT2 + lnA - (-Ea/RT1 + lnA) Simplify: ln k2 - ln k1 = -Ea/R (1/T2 - 1/T1) Use the properties of logs: ln (k2/k1) = -Ea/R (1/T2 - 1/T1)

5. anonymous

Brilliant! That makes perfect sense. Thanks to both of you.

6. anonymous

I remember this equation... We got like 5 hours of homework on it on Utexas Quest last year... Its called the Clausius-Clapeyron equation.

7. anonymous

It actually just takes a very similar form to the Clausius-Clapeyron equation. I've seen several equations in chemistry that have the exact same form, just with different parameters. This is the Clausius-Clapeyron equation relates vapor pressures, heat of vaporization, and temperature:$\ln(\frac{P_2}{P_1})=\frac{- \Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$ I find it odd that so many equations take this form. Is there anything particularly special about it? Is it's form derived from some other common general form?

8. anonymous

Well, I haven't taken physical chemistry or differential equations, so I can't say for sure. However, rates of change for many natural phenomena follow similar proportionalities, so they have similar differential equations, which in turn has similar solutions. Both the Clausius Clapeyron & the equation that relates k & t are solutions to first order DE's.

9. anonymous

A good guess, but as it turns out neither equation comes from an underlying differential equation. In fact they are both reflective of ideas from equilibrium statistical mechanics, in particular the Boltzmann distribution, which tells you that the relative probability of observing a system in state s2 versus state s1 P2/P1 = exp(-(E2-E1)/kT) where E1 and E2 are the energies, T is the temperature, and k is the Boltzmann constant, whcih is just R divided by the Avogadro number. You can probably see immediately how this applies to the vapor pressure case: the two states are the molecule in the gas phase and the molecule in the condensed phase, and the difference in energy between the two states is just the enthalpy of vaporization. The Arrhenius equation is trickier, because you'd think it *would* be derived from a differential equation. But it's not. It's derived by assuming a very simple model of chemical kinetics: that the rate of a chemical reaction depends on the probability that on any given collision the collision energy exceeds the activation energy Ea. Clearly that probability, at equilibrium, is again given by the Boltzmann distribution, only this time the energy difference is the activation energy. There is a (very complicated) prefactor in the Arrhenius equation that takes into account genuine kinetic factors, such as the rate of collisions. And if you want to get into genuine chemical dynamics factors, which include the possibility that the reaction may NOT proceed to completion, just because the energy is above the barrier, then you are no longer working in the simple Arrhenius model. There are many more examples of equilibrium and pseudo-equilibrium laws in chemistry that have their roots in the Boltzmann distribution as well. One of the most obvious is the fact that the pressure of the atmosphere falls off exponentially with altitude.