Question

Determine the change in entropy for the following chemical reaction occurring at standard pressure and the stated temperature: 2H2(g)+O2(g)→2H2O (l) S°[H2 (g)]=130.7 J/mol·K, S°[O2 (g)]=205.1 J/mol·K, and S°[H2O (l)]=69.91 J/mol·K

Answer 1

We have the equation - 2 H₂ (q) + O2(g) 2 H₂O (1) We know, at stams and pressure and at constant tempenature- Change in Entropy, 45° = En sº (products) – En sº Creactants) for this Beaction - 45° = [{2x S Hocus } ] - [{2x 512}+{ $0,}] Given - S of H₂O (1) = 69.91 J/mol.k so of Hy (2) = 130 •7 J/mof.k s of O₂ (8) = 205.1 J/molik

Accondingly, 48° = [2 * Sv.ocus ] - [{2x suscep}+{50,c7}] = [2 x 69.99 g/molk] [{2x 130.7 5/mol.x}+{2015-1 J/mof.x37 = 139.82 J/mo$.k - [263.4 3/60$.K + 205.3 5/mo$. k] = 139.82 J/mol. K - 466.5 J/mol. K =-326.68 J/mol.k. i. The value of change in Entropy is - 326.68 J/mol. K. An.