ΔH

[Andrew Yoon 3L]

In lecture #14, Lavelle went over the relationship between ΔS with T(temp). What did he mean when he said that "the ΔH(rxn) plays an important role at low temperatures"? Can someone clarify how it relates back to the relationship between ΔS and T?

[Rich Luong 1D]

If you're referring to the Gibbs free energy, then yes, ΔH would play a big role at low temperatures. If you recall, the equation for Gibbs free energy is
ΔG = ΔH - TΔS. If in this case temperature was a very low value, then the significance of ΔS also goes down alongside it. As a result, all you're basically left with is ΔH and because of that, depending on the sign of ΔH, it determines the sign of ΔG.

[Jeffrey Doeve 2I]

Hi Andrew, I believe that Lavelle was referencing the importance of delta H in comparison to delta S for the Gibbs Free Energy Equation(delta G = delta H - T* delta S). For example, if the T was low, that means that the impact of delta S would be low as well. This means that delta H would dominate and determine whether the reaction is spontaneous(sign of delta G).

[allyssa bradley 1H]

Agreeing with the above replies, in a more abstract approach, I think it would be difficult to store energy with respect to the "order" of the system since those particles aren't moving around very much, so even if there are a lot of "occupy-able" states related to entropy and degeneracy, the system is so SLOW that it sort of relies on heat and any work/pressure to store its energy, which would of course be enthalpy.

[Jaden Joodi 3J]

From the math point of view, the reason that ΔH plays such a large role at low temperatures is because it is being divided by temperature. The difference between being divided by 1, 2, 3 is quite large compared to being divided by 50,51, 52.

[Ethan Laureano 3H]

If you recall the sapling homework, if both delta H and delta S are positive, then the equation would be deltaG = deltaH - TdeltaS. In order to label a reaction spontaneous, deltaG must be negative. At low temperatures, -TdeltaS might not have a high enough magnitude to overcome deltaH and result in a negative deltaG, depending on the magnitude of deltaH.

[Brennan McGurrr 3C]

It's due to the equation to calculate Gibbs Free Energy. When T is very low, it will lower the term in the equation with T and delta S. This makes delta G way more dependent on delta H.

[Ranen_Chang_2G]

You could also look at the relationship S of surroundings and H of reaction have, . At low T, influences the value of more than T does.

[Susanna Givan 2B]

Delta S has a very important relationship with temperature--as the temp. increases, the number of particles that can strike each other increases, and the entropy increases. If temp. is very low, then Delta S is effectively zero or very small. Then, the value of Delta H would effectively control Delta G and change it's value.

[Nandini_Parmar_1I]

The change in enthalpy relates to the change in entropy times the temperature. ie. change in heat minus the temperature difference times the disorder gives the free energy. Gibb's free energy is calculated by constant temperature and pressure.

[Aanjaneyaa]

Due to the equation deltaG=deltaH-TdeltaS, it shows that both delta S and delta T have an effect on delta G.

[Caitlin_Doak_2H]

Does delta H also equal delta U?

[Arshaun Faraji 2H]

it shows that when analyzing the equation deltaG=deltaH-TdeltaS, we should know that deltaH and deltaS are the factors that determine the deltaG

[Josh Richter 2C]

Those values relate to Gibbs free energy, where they can see whether or not a reaction will be spontaneous

[raynebunado]

It relates to the Gibbs Free Energy equation!