Question

Consider the reaction 4HCl(g) + O_2(g) rightarrow 2H_2O(g) + 2Cl_2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.75 moles of HCl(g) react at standard conditions.

Answer 1

Enthalpy of HCl ( in gas phase) = -92.30 KJ/mol

Enthalpy of O2( in gas phase) = 0 KJ/mol

Enthalpy of H2O ( in gas phase) = -241.8 KJ/mol

Enthalpy of Cl2( in gas phase) = 0 KJ/mol

Enthalpy of the reaction = (-241.8) + 92.3 = -149.5 KJ/mol

Moles of HCL added = 1.75 mole

Final Enthalpy of the reaction = -149.5 KJ/mol * 1.75 mole = -261.625 KJ

Delta S (surroundings) = -Enthalpy/Temp = 261.625 KJ/298 = 877.93 J/K