Consider the reaction below. 2 POCl3(g) 2 PCl3(g) + O2(g) (a) Calculate ?G Solut
ID: 751665 • Letter: C
Question
Consider the reaction below. 2 POCl3(g) 2 PCl3(g) + O2(g) (a) Calculate ?GExplanation / Answer
2 POCl3(g) ===> 2 PCl3(g) + O2(g) (a) Calculate ?G° for this reaction. The ?Gf° values for POCl3(g) and PCl3(g) are -502 kJ/mol and -270. kJ/mol, dG = dGf's products - reactants dG = [(2mol PCl3 @ -270.kJ/mol) & zero for oxygen] - (2mol POCl3 @ -502kJ/mol) dG = -540.0 - (-1004) dG = + 464kJ ======================================… to use the formula: dG = dH - TdS we need to calculate the dH of this reaction: dH = dHf's products - reactants dH = [(2mol PCl3 @ -288.07kJ/mol) & zero for oxygen] - (2mol POCl3 @ -542.2 kJ/mol) dH = -576.14 - (-1084.4) dH = + 508.26 kJ =============== now we use the formula: dG = dH - TdS but first where dG = 0 when @ equilibrium: 0 = dH - TdS TdS = dH T = dH / dS T = + 508.26 kJ / 0.179 kJ T = 2839 Kelvin