For the process 2so2(g) + o2(g) mc016-1.jpg 2so3(g), mc016-2.jpgs = –187.9 j/k and mc016-3.jpgh = –198.4 kj at 297.0 k are known. what is the entropy of this reaction? use mc016-4.jpgg = mc016-5.jpgh – tmc016-6.jpgs.

You are given a process of 2SO₂(g) + O₂(g) → 2SO₃(g) with ΔS = –187.9 J/K and ΔH = –198.4 kJ at 297.0 K. You are asked to find the entropy of this reaction. The entropy of this reaction is ΔS = –187.9 J/K. ΔS stands for change in entropy and ΔH stands for change in enthalpy. It follows the second law of thermodynamics, the entropy in which all things go into the chaotic state from a higher form into lower form. A chemical reaction has an increase in entropy and a decrease in enthalpy. A chemical reaction is spontaneous if it has an entropy greater than zero and an enthalpy less than 0. The chemical reaction given is spontaneous.

The correct answer should be "electrons orbit the nucleus in circular motions exactly as shown on the classic bohr model."