For the process O2(g)→2O(g), ΔH° = 498 kJ·mol–1. What would be the predicted sign of ΔS°rxn and the conditions under which this reaction would be spontaneous? Question 13 options: 1) ΔS°rxn is positive, and the process is spontaneous at low temperatures only. 2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only. 3) ΔS°rxn is negative, and the process is spontaneous at high temperatures only. 4) ΔS°rxn is negative, and the process is spontaneous at low temperatures only.

2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only.

Explanation:

ΔS = ΔH / T , ΔS is change in entropy , ΔH is change in enthalpy

Since ΔH is positive , ΔS is positive .

ΔG = ΔH - TΔS

For spontaneous reaction . ΔG should be negative .

As ΔS is positive , at high temperature the value of TΔS will be more and hence the value of TΔS will be higher than Δ H . Hence ΔG will be negative.

Hence at higher temperature , the reaction will be spontaneous.

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.