Question

An endothermic reaction could be non-spontaneous if the change in enthalpy is less positive than the product of the -TdeltaS is negative? Is this true or false?

Answer 1

consider \[\Delta G = \Delta H - T \Delta S\]]

for a reaction to be spontaneous, delta G has to be negative,

so if the enthalpy is less positive (i think this means closer to zero if you are talking about a number line) and -TdeltaS is more negative (meaning further from zero in the negative side)

so for example, if H=2 and TdeltaS= 3

delta G = 2 - 3 = -1 which is negative, which means the reaction is spontaneous

Answer 2

If deltaH is less positive and -TdeltaS is less negative. For some reason, I cannot wrap my mind around an example of this. Please help.

Answer 3

so, deltaH is not necessarily negative, just less positive; and -TdeltaS is not necessarily positive just "less negative". Would the resulting reaction be spontaneous, or could it indeed be non-spontaneous??

Answer 4

it's saying that if the absolute value of H is less than the absolute value of -TdeltaS, the reaction could be non-spontaenous..which is not the case as i showed you, because the outcome is negative

Answer 5

"so, deltaH is not necessarily negative, just less positive; and -TdeltaS is not necessarily positive just "less negative". Would the resulting reaction be spontaneous, or could it indeed be non-spontaneous??"


it's saying that -TdeltaS is MORE negative, not less negative

Answer 6

Okay, I see your point. So, deltaH is still positive, and -TdeltaS is still negative?

Answer 7

yeah, its just saying in terms of distance regarding a number line, -TdeltaS is further from zero, meaning a greater value.