Question

Consider the reaction 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.84 moles of HCl(g) react at standard conditions.

S°surroundings = J/K

Answer 1

Reaction: 4HC1(8) + O2(g) →→2012() + 2H,0(); AH = -130.8 kJ Standard enthalpy change of the reaction , AHCM = mx AHF podrts) – nxAH i tractares = [(2x AH? cn.ca) +2xAH?w.am) – (4 x AH? fax) +1x 8H10.»)] [2 mol xo hodi +)+(amox-2418 m) - 4mol x-92,30 hode) +(1 mol xo ) = (-483.6 kJ)-(-369.2 kJ) =-114.4 kJ Temperature, T = 298 K les promotes -- CHEMISES - 114.4 kJ =- 298 K = +0.3839 kJ 1000 J 1kJ mol.K = 383,9 J/mol.K For 1.84 moles of HCI, Amorlines = 1.84 mol HCI- 4 mol = +176.6 J/K