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Why does increasing T, speed the ENDO direction *more* than the EXO direction?

On February 13, 2014

In an imminent update to the Equilibrium notes, I have included an ‘explanation’ of why reactions with larger activation energies (i.e., the endothermic process in any equilibrium system – see the diagram below), have their rates of reaction increased to a greater degree than reactions with smaller activation energies (i.e., the exothermic process in any equilibrium system – also see the diagram below), thus ‘proving’ why the endothermic direction is always favored when an equilibrium system has its temperature raised.

ENDO always has higher Ea

I use quotes around the words, explanation and proving, since to be honest, the phrase that I included simply said, ‘which can be proven by computations carried out on the Arrhenius equation‘, which hardly qualifies as either an explanation or a proof! In this post I wanted to do some math to show that statement is indeed true. I have a pretty strong feeling that all of this is beyond any, likely required answer on an AP exam, at least in part because the Arrhenius equation does not appear on the new Equations & Constants sheet, but it makes me feel better to at least attempt to justify my ramblings!

OK, here we go. Using the Arrhenius equation in this form,

$\large&space;k&space;=&space;A&space;e^{\frac{-Ea}{RT}}$

We first take a reaction with an activation energy (Ea) of 100 kJ run at two different temperatures, 200 K and 400 K. Plugging and chugging for each temperature separately, we get two rate constants, thus;

$\large&space;k&space;=&space;A&space;e^{\frac{-100}{(8.314/1000)(200)}}&space;=&space;A(7.62\textrm{&space;x&space;}10^{^{-27}})$

AND

$\large&space;k&space;=&space;A&space;e^{\frac{-100}{(8.314/1000)(400)}}&space;=&space;A(8.73\textrm{&space;x&space;}10^{^{-14}})$

As expected, at the higher temperature of 400 K, the rate constant is larger, but we should focus on the change in k at the two temperatures, which is of an order of magnitude of 1013.

Now, moving on to a reaction with a larger activation energy (Ea) of 150 kJ, we repeat the process of examining k values at the same, two, different temperatures, thus;

$\large&space;k&space;=&space;A&space;e^{\frac{-150}{(8.314/1000)(200)}}&space;=&space;A(6.65\textrm{&space;x&space;}10^{^{-40}})$

AND

$\large&space;k&space;=&space;A&space;e^{\frac{-150}{(8.314/1000)(400)}}&space;=&space;A(2.58\textrm{&space;x&space;}10^{^{-20}})$

Again, as expected, at the higher temperature of 400 K, the rate constant is larger, and again (also as expected), a comparison of each reaction at 200 K, and then each reaction at 400 K, we find that the reaction with the lower Ea has the higher rate constant.

However, we should once again focus on the change in k that occurs, which in the case of the reaction with the higher activation energy is of an order of magnitude of 1020 as opposed to that of approximately 1013 in the reaction with the lower activation energy.

The math shows that reactions with higher activation energies have rates that are more profoundly influenced by increases in temperature than those with lower activation energies. Since the endothermic direction in an equilibrium system will always have the higher Ea, it benefits from the temperature increase to a greater extent than the exothermic reaction, and the system always shifts in the endothermic direction at a higher temperature.