Question

Consider the reaction: 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.84 moles of HCl(g) react at standard conditions. S°system = J/K

Answer 1

4HCl(g) + O2         2H2O(g) + 2Cl2(g)

The entropy change of the system can be calculated by following formula;

∆S° = ∆H°_rex/T

Therefore we first have to calculate ∆H°_rex by using standard heat of formation of reactants and products.

	Moles	species	∆H°f (Kj/mol)
Reactants	4	HCl	-92.3
	1	O2	0
Product	2	H2O	-285.8
	2	2Cl2	0

∆H°_rex = 2 * mol (-285.8 kJ/mol) + 2* mol (0.0 kJ/mol) - [4*mol (-92.3 kJ/mol) + 1*mol (0.0 kJ/mol)] = -200.8 KJ

∆H°_rex = -200.8* 1000= -200,800 J

Now,

∆S°= ∆H°_rex/T

-200,800 J/ 298 = -673.82 J/K

If 1.84 HCl is reacted

1.84 mol HCl(g) x (-673.82 JK^-1 / 4 mol HCl(g)) = -309.9 JK^-1