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.

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b

the answer is d.) potassium